https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=First&feedformat=atom AoPS Wiki - User contributions [en] 2022-08-18T11:21:26Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_5&diff=173018 2018 AIME II Problems/Problem 5 2022-03-28T18:04:35Z <p>First: /* Solution 1 (Euler's formula and Substitution) */</p> <hr /> <div>==Problem==<br /> <br /> Suppose that &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are complex numbers such that &lt;math&gt;xy = -80 - 320i&lt;/math&gt;, &lt;math&gt;yz = 60&lt;/math&gt;, and &lt;math&gt;zx = -96 + 24i&lt;/math&gt;, where &lt;math&gt;i&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\sqrt{-1}&lt;/math&gt;. Then there are real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;x + y + z = a + bi&lt;/math&gt;. Find &lt;math&gt;a^2 + b^2&lt;/math&gt;.<br /> <br /> ==Solution 1 (Euler's formula and Substitution)==<br /> <br /> The First (pun intended) thing to notice is that &lt;math&gt;xy&lt;/math&gt; and &lt;math&gt;zx&lt;/math&gt; have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is &lt;math&gt;e^{\frac{3\pi i}{2}}&lt;/math&gt;. Anyway this results in getting that &lt;math&gt;\left(\frac{-3i}{10}\right)y=z&lt;/math&gt;. Then substitute this into &lt;math&gt;yz&lt;/math&gt; to get, after some calculation, that &lt;math&gt;y=10e^{\frac{5\pi i}{4}}\sqrt{2}&lt;/math&gt; and &lt;math&gt;z=-3e^{\frac{7\pi i}{4}}\sqrt{2}&lt;/math&gt;. Then plug &lt;math&gt;z&lt;/math&gt; into &lt;math&gt;zx&lt;/math&gt;, you could do the same thing with &lt;math&gt;xy&lt;/math&gt; but &lt;math&gt;zx&lt;/math&gt; looks like it's easier due to it being smaller. Anyway you get &lt;math&gt;x=20+12i&lt;/math&gt;. Then add all three up, it turns out easier than it seems because for &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; the &lt;math&gt;\sqrt{2}&lt;/math&gt; disappears after you expand the root of unity (e raised to a specific power). Long story short, you get &lt;math&gt;x=20+12i, y=-3+3i, z=-10-10i \implies x+y+z=7+5i \implies a^2+b^2=\boxed{074}&lt;/math&gt;.<br /> <br /> ~First<br /> <br /> ==Solution 2==<br /> <br /> First we evaluate the magnitudes. &lt;math&gt;|xy|=80\sqrt{17}&lt;/math&gt;, &lt;math&gt;|yz|=60&lt;/math&gt;, and &lt;math&gt;|zx|=24\sqrt{17}&lt;/math&gt;. Therefore, &lt;math&gt;|x^2y^2z^2|=17\cdot80\cdot60\cdot24&lt;/math&gt;, or &lt;math&gt;|xyz|=240\sqrt{34}&lt;/math&gt;. Divide to find that &lt;math&gt;|z|=3\sqrt{2}&lt;/math&gt;, &lt;math&gt;|x|=4\sqrt{34}&lt;/math&gt;, and &lt;math&gt;|y|=10\sqrt{2}&lt;/math&gt;.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((4,0),red);<br /> draw((0,0)--(-4,0));<br /> draw((0,0)--(0,-4));<br /> draw((0,0)--(-4,1));<br /> dot((-4,1),red);<br /> draw((0,0)--(-1,-4));<br /> dot((-1,-4),red);<br /> draw((0,0)--(4,4),red);<br /> draw((0,0)--(4,-4),red);<br /> &lt;/asy&gt;<br /> This allows us to see that the argument of &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;\frac{\pi}{4}&lt;/math&gt;, and the argument of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;-\frac{\pi}{4}&lt;/math&gt;. We need to convert the polar form to a standard form. Simple trig identities show &lt;math&gt;y=10+10i&lt;/math&gt; and &lt;math&gt;z=3-3i&lt;/math&gt;. More division is needed to find what &lt;math&gt;x&lt;/math&gt; is. &lt;cmath&gt;x=-20-12i&lt;/cmath&gt; &lt;cmath&gt;x+y+z=-7-5i&lt;/cmath&gt; &lt;cmath&gt;(-7)^2+(-5)^2=\boxed{74}&lt;/cmath&gt;<br /> &lt;cmath&gt;QED\blacksquare&lt;/cmath&gt;<br /> Written by [[User:A1b2|a1b2]]<br /> <br /> == Solution 3 (Pretty easy, no hard stuff, just watch ur arithmetic) ==<br /> Solve this system the way you would if the RHS of all equations were real. Multiply the first and 3rd equations out and then factor out &lt;math&gt;60&lt;/math&gt; to find &lt;math&gt;x^2&lt;/math&gt;, then use standard techniques that are used to evaluate square roots of irrationals. let &lt;cmath&gt;x = c+di&lt;/cmath&gt;, then you get &lt;cmath&gt;c^2 - d^2 = 256&lt;/cmath&gt; and &lt;cmath&gt;2cd = 480&lt;/cmath&gt; Solve to get &lt;math&gt;x&lt;/math&gt; as &lt;math&gt;20+12i&lt;/math&gt; and &lt;math&gt;-20-12i&lt;/math&gt;. Both will give us the same answer, so use the positive one. Divide &lt;math&gt;-80-320i&lt;/math&gt; by &lt;math&gt;x&lt;/math&gt;, and you get &lt;math&gt;10+10i&lt;/math&gt; as &lt;math&gt;y&lt;/math&gt;. This means that &lt;math&gt;z&lt;/math&gt; is a multiple of &lt;math&gt;1-i&lt;/math&gt; to get a real product, so you find &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3-3i&lt;/math&gt;. Now, add the real and imaginary parts separately to get &lt;math&gt;-7-5i&lt;/math&gt;, and calculate &lt;math&gt;a^2 + b^2&lt;/math&gt; to get &lt;math&gt;\boxed{74}&lt;/math&gt;. <br /> ~minor latex improvements done by jske25 and jdong2006<br /> <br /> ==Solution 4==<br /> Dividing the first equation by the second equation given, we find that &lt;math&gt;\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)&lt;/math&gt;. Substituting this into the third equation, we get &lt;math&gt;z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i&lt;/math&gt;. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of &lt;math&gt;y&lt;/math&gt; is the negative of that of &lt;math&gt;z&lt;/math&gt;, and their magnitudes multiply to &lt;math&gt;60&lt;/math&gt;. Thus, we have &lt;math&gt;z=\sqrt{-18i}=3-3i&lt;/math&gt; and &lt;math&gt;3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i&lt;/math&gt;. To find &lt;math&gt;x&lt;/math&gt;, we can use the previous substitution we made to find that &lt;math&gt;x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i&lt;/math&gt;.<br /> Therefore, &lt;math&gt;x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{74}&lt;/math&gt;<br /> <br /> Solution by ktong<br /> <br /> ==Solution 5 ==<br /> <br /> We are given that &lt;math&gt;xy=-80-320i&lt;/math&gt;. Thus &lt;math&gt;y=\frac{-80-320i}{x}&lt;/math&gt;. We are also given that &lt;math&gt;xz= -96+24i&lt;/math&gt;. Thus &lt;math&gt;z=\frac{-96+24i}{x}&lt;/math&gt;. We are also given that &lt;math&gt;yz&lt;/math&gt; = &lt;math&gt;60&lt;/math&gt;. Substitute &lt;math&gt;y=\frac{-80-320i}{x}&lt;/math&gt; and &lt;math&gt;z=\frac{-96+24i}{x}&lt;/math&gt; into &lt;math&gt;yz&lt;/math&gt; = &lt;math&gt;60&lt;/math&gt;. We have &lt;math&gt; \frac{(-80-320i)(-96+24i)}{x^2}=60&lt;/math&gt;. Multiplying out &lt;math&gt;(-80-320i)(-96+24i)&lt;/math&gt; we get &lt;math&gt;(1920)(8+15i)&lt;/math&gt;. Thus &lt;math&gt;\frac{1920(8+15i)}{x^2} =60&lt;/math&gt;. Simplifying this fraction we get &lt;math&gt;\frac{32(8+15i)}{x^2}=1&lt;/math&gt;. Cross-multiplying the fractions we get &lt;math&gt;x^2=32(8+15i)&lt;/math&gt; or &lt;math&gt;x^2= 256+480i&lt;/math&gt;. Now we can rewrite this as &lt;math&gt;x^2-256=480i&lt;/math&gt;. Let &lt;math&gt;x= (a+bi)&lt;/math&gt;.Thus &lt;math&gt;x^2=(a+bi)^2&lt;/math&gt; or &lt;math&gt;a^2+2abi-b^2&lt;/math&gt;. We can see that &lt;math&gt;a^2+2abi-b^2-256=480i&lt;/math&gt; and thus &lt;math&gt;2abi=480i&lt;/math&gt; or &lt;math&gt;ab=240&lt;/math&gt;.We also can see that &lt;math&gt;a^2-b^2-256=0&lt;/math&gt; because there is no real term in &lt;math&gt;480i&lt;/math&gt;. Thus &lt;math&gt;a^2-b^2=256&lt;/math&gt; or &lt;math&gt;(a+b)(a-b)=256&lt;/math&gt;. Using the two equations &lt;math&gt;ab=240&lt;/math&gt; and &lt;math&gt;(a+b)(a-b)=256&lt;/math&gt; we solve by doing system of equations that &lt;math&gt;a=-20&lt;/math&gt; and &lt;math&gt;b=-12&lt;/math&gt;. And &lt;math&gt;x=a+bi&lt;/math&gt; so &lt;math&gt;x=-20-12i&lt;/math&gt;. Because &lt;math&gt;y=\frac{-80-320i}{x}&lt;/math&gt;, then &lt;math&gt;y=\frac{-80-320i}{-20-12i}&lt;/math&gt;. Simplifying this fraction we get &lt;math&gt;y=\frac{-80(1+4i)}{-4(5+3i)}&lt;/math&gt; or &lt;math&gt;y=\frac{20(1+4i)}{(5+3i)}&lt;/math&gt;. Multiplying by the conjugate of the denominator (&lt;math&gt;5-3i&lt;/math&gt;) in the numerator and the denominator and we get &lt;math&gt;y=\frac{20(17-17i)}{34}&lt;/math&gt;. Simplifying this fraction we get &lt;math&gt;y=10-10i&lt;/math&gt;. Given that &lt;math&gt;yz&lt;/math&gt; = &lt;math&gt;60&lt;/math&gt; we can substitute &lt;math&gt;(10-10i)(z)=60&lt;/math&gt; We can solve for z and get &lt;math&gt;z=3+3i&lt;/math&gt;. Now we know what &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are, so all we have to do is plug and chug. &lt;math&gt;x+y+z= (-20-12i)+(10+10i)+(3-3i)&lt;/math&gt; or &lt;math&gt;x+y+z= -7-5i&lt;/math&gt; Now &lt;math&gt;a^2 +b^2=(-7)^2+(-5)^2&lt;/math&gt; or &lt;math&gt;a^2 +b^2 = 74&lt;/math&gt;. Thus &lt;math&gt;74&lt;/math&gt; is our final answer.(David Camacho)<br /> <br /> ==Solution 6==<br /> <br /> We observe that by multiplying &lt;math&gt;xy,&lt;/math&gt; &lt;math&gt;yz,&lt;/math&gt; and &lt;math&gt;zx,&lt;/math&gt; we get &lt;math&gt;(xyz)^2=(-80-320i)(60)(-96+24i).&lt;/math&gt; Next, we divide &lt;math&gt;(xyz)^2&lt;/math&gt; by &lt;math&gt;(yz)^2&lt;/math&gt; to <br /> <br /> get &lt;math&gt;x^2.&lt;/math&gt; We have &lt;math&gt;x^2=\frac{(-80-320i)(60)(-96+24i)}{3600}=256+480i.&lt;/math&gt; We can write &lt;math&gt;x&lt;/math&gt; in the form of &lt;math&gt;a+bi,&lt;/math&gt; so we get <br /> <br /> &lt;math&gt;(a+bi)^2=256+480i.&lt;/math&gt; Then, &lt;math&gt;a^2-b^2+2abi=256+480i,&lt;/math&gt; &lt;math&gt;a^2-b^2=256,&lt;/math&gt; and &lt;math&gt;2ab=480.&lt;/math&gt; Solving this system of equations is relatively <br /> <br /> simple. We have two cases, &lt;math&gt;a=20, b=12,&lt;/math&gt; and &lt;math&gt;a=-20, b=-12.&lt;/math&gt; <br /> <br /> Case 1: &lt;math&gt;a=20, b=12,&lt;/math&gt; so &lt;math&gt;x=20+12i.&lt;/math&gt; We solve for &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; by plugging in &lt;math&gt;x&lt;/math&gt; to the two equations. We see<br /> <br /> &lt;math&gt;y=\frac{-80-320i}{20+12i}=-10-10i&lt;/math&gt; and &lt;math&gt;z=\frac{-96+24i}{20+12i}=-3+3i.&lt;/math&gt; &lt;math&gt;x+y+z=7+5i,&lt;/math&gt; so &lt;math&gt;a=7&lt;/math&gt; and &lt;math&gt;b=5.&lt;/math&gt; Solving, we end up with <br /> <br /> &lt;math&gt;7^2+5^2=\boxed{074}&lt;/math&gt; as our answer. <br /> <br /> Case 2: &lt;math&gt;a=-20, b=-12,&lt;/math&gt; so &lt;math&gt;x=-20-12i.&lt;/math&gt; Again, we solve for &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z.&lt;/math&gt; We find &lt;math&gt;y=\frac{-80-320i}{-20-12i}=10+10i,&lt;/math&gt; <br /> <br /> &lt;math&gt;z=\frac{-96+24i}{-20-12i}=3-3i,&lt;/math&gt; so &lt;math&gt;x+y+z=-7-5i.&lt;/math&gt; We again have &lt;math&gt;(-7)^2+(-5)^2=\boxed{074}.&lt;/math&gt;<br /> <br /> Solution by Airplane50<br /> <br /> ==Solution 7 (Based on advanced mathematical knowledge)==<br /> According to the Euler's Theory, we can rewrite &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; as &lt;cmath&gt;x=r_{1}e^{i{\theta}_1}&lt;/cmath&gt; &lt;cmath&gt;y=r_{2}e^{i{\theta}_2}&lt;/cmath&gt; &lt;cmath&gt;x=r_{3}e^{i{\theta}_3}&lt;/cmath&gt; As a result, &lt;cmath&gt;|xy|=r_{1}r_{2}=\sqrt{80^2+320^2}=80\sqrt{17}&lt;/cmath&gt; &lt;cmath&gt;|yz|=r_{2}r_{3}=60&lt;/cmath&gt; &lt;cmath&gt;|xz|=r_{1}r_{3}=\sqrt{96^2+24^2}=24\sqrt{17}&lt;/cmath&gt; Also, it is clear that &lt;cmath&gt;yz=r_{2}e^{i{\theta}_2}r_{3}e^{i{\theta}_3}=|yz|e^{i({\theta}_2+{\theta}_3)}=|yz|=60&lt;/cmath&gt; So &lt;math&gt;{\theta}_2+{\theta}_3=0&lt;/math&gt;, or &lt;cmath&gt;{\theta}_2=-{\theta}_3&lt;/cmath&gt; Also, we have &lt;cmath&gt;xy=-80\sqrt{17}e^{i\arctan{4}}&lt;/cmath&gt; &lt;cmath&gt;yz=60&lt;/cmath&gt; &lt;cmath&gt;xz=-24\sqrt{17}e^{i\arctan{-\frac{1}{4}}}&lt;/cmath&gt; So now we have &lt;math&gt;r_{1}r_{2}=80\sqrt{17}&lt;/math&gt;, &lt;math&gt;r_{2}r_{3}=60&lt;/math&gt;, &lt;math&gt;r_{1}r_{3}=24\sqrt{17}&lt;/math&gt;, &lt;math&gt;{\theta}_1+{\theta}_2=\arctan{4}&lt;/math&gt; and &lt;math&gt;{\theta}_1-{\theta}_2=\arctan {-\frac{1}{4}}&lt;/math&gt;. Solve these above, we get &lt;cmath&gt;r_{1}=4\sqrt{34}&lt;/cmath&gt; &lt;cmath&gt;r_{2}=10\sqrt{2}&lt;/cmath&gt; &lt;cmath&gt;r_{3}=3\sqrt{2}&lt;/cmath&gt; &lt;cmath&gt;{\theta}_2=\frac{\arctan{4}-\arctan{-\frac{1}{4}}}{2}=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}&lt;/cmath&gt; So we can get &lt;cmath&gt;y=r_{2}e^{i{\theta}_2}=10\sqrt{2}e^{i\frac{\pi}{4}}=10+10i&lt;/cmath&gt; &lt;cmath&gt;z=r_{3}e^{i{\theta}_3}=r_{3}e^{-i{\theta}_2}=3\sqrt{2}e^{-i\frac{\pi}{4}}=3-3i&lt;/cmath&gt; Use &lt;math&gt;xy=-80-320i&lt;/math&gt; we can find that &lt;cmath&gt;x=-20-12i&lt;/cmath&gt; So &lt;cmath&gt;x+y+z=-20-12i+10+10i+3-3i=-7-5i&lt;/cmath&gt; So we have &lt;math&gt;a=-7&lt;/math&gt; and &lt;math&gt;b=-5&lt;/math&gt;.<br /> <br /> As a result, we finally get &lt;cmath&gt;a^2+b^2=(-7)^2+(-5)^2=\boxed{074}&lt;/cmath&gt;<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Solution 8==<br /> <br /> We can turn the expression &lt;math&gt;x+y+z&lt;/math&gt; into &lt;math&gt;\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}&lt;/math&gt;, and this would allow us to plug in the values after some computations.<br /> <br /> Based off of the given products, we have<br /> <br /> &lt;math&gt;xy^2z=60(-80-320i)&lt;/math&gt;<br /> <br /> &lt;math&gt;xyz^2=60(-96+24i)&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2yz=(-96+24i)(-80-320i)&lt;/math&gt;.<br /> <br /> Dividing by the given products, we have<br /> <br /> &lt;math&gt;y^2=\frac{60(-80-320i)}{-96+24i}&lt;/math&gt;<br /> <br /> &lt;math&gt;z^2=\frac{60(-96+24i)}{-80-320i}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2=\frac{(-96+24i)(-80-320i)}{60}&lt;/math&gt;.<br /> <br /> Simplifying, we get that this expression becomes &lt;math&gt;\sqrt{24+70i}&lt;/math&gt;. This equals &lt;math&gt;\pm{(7+5i)}&lt;/math&gt;, so the answer is &lt;math&gt;7^2+5^2=\boxed{74}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{-RootThreeOverTwo}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=II|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_5&diff=173017 2018 AIME II Problems/Problem 5 2022-03-28T18:03:46Z <p>First: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> Suppose that &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are complex numbers such that &lt;math&gt;xy = -80 - 320i&lt;/math&gt;, &lt;math&gt;yz = 60&lt;/math&gt;, and &lt;math&gt;zx = -96 + 24i&lt;/math&gt;, where &lt;math&gt;i&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\sqrt{-1}&lt;/math&gt;. Then there are real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;x + y + z = a + bi&lt;/math&gt;. Find &lt;math&gt;a^2 + b^2&lt;/math&gt;.<br /> <br /> ==Solution 1 (Euler's formula and Substitution)==<br /> <br /> The [b]First[/b] thing to notice is that &lt;math&gt;xy&lt;/math&gt; and &lt;math&gt;zx&lt;/math&gt; have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is &lt;math&gt;e^{\frac{3\pi i}{2}}&lt;/math&gt;. Anyway this results in getting that &lt;math&gt;\left(\frac{-3i}{10}\right)y=z&lt;/math&gt;. Then substitute this into &lt;math&gt;yz&lt;/math&gt; to get, after some calculation, that &lt;math&gt;y=10e^{\frac{5\pi i}{4}}\sqrt{2}&lt;/math&gt; and &lt;math&gt;z=-3e^{\frac{7\pi i}{4}}\sqrt{2}&lt;/math&gt;. Then plug &lt;math&gt;z&lt;/math&gt; into &lt;math&gt;zx&lt;/math&gt;, you could do the same thing with &lt;math&gt;xy&lt;/math&gt; but &lt;math&gt;zx&lt;/math&gt; looks like it's easier due to it being smaller. Anyway you get &lt;math&gt;x=20+12i&lt;/math&gt;. Then add all three up, it turns out easier than it seems because for &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; the &lt;math&gt;\sqrt{2}&lt;/math&gt; disappears after you expand the root of unity (e raised to a specific power). Long story short, you get &lt;math&gt;x=20+12i, y=-3+3i, z=-10-10i \implies x+y+z=7+5i \implies a^2+b^2=\boxed{074}&lt;/math&gt;.<br /> <br /> ~First<br /> <br /> ==Solution 2==<br /> <br /> First we evaluate the magnitudes. &lt;math&gt;|xy|=80\sqrt{17}&lt;/math&gt;, &lt;math&gt;|yz|=60&lt;/math&gt;, and &lt;math&gt;|zx|=24\sqrt{17}&lt;/math&gt;. Therefore, &lt;math&gt;|x^2y^2z^2|=17\cdot80\cdot60\cdot24&lt;/math&gt;, or &lt;math&gt;|xyz|=240\sqrt{34}&lt;/math&gt;. Divide to find that &lt;math&gt;|z|=3\sqrt{2}&lt;/math&gt;, &lt;math&gt;|x|=4\sqrt{34}&lt;/math&gt;, and &lt;math&gt;|y|=10\sqrt{2}&lt;/math&gt;.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((4,0),red);<br /> draw((0,0)--(-4,0));<br /> draw((0,0)--(0,-4));<br /> draw((0,0)--(-4,1));<br /> dot((-4,1),red);<br /> draw((0,0)--(-1,-4));<br /> dot((-1,-4),red);<br /> draw((0,0)--(4,4),red);<br /> draw((0,0)--(4,-4),red);<br /> &lt;/asy&gt;<br /> This allows us to see that the argument of &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;\frac{\pi}{4}&lt;/math&gt;, and the argument of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;-\frac{\pi}{4}&lt;/math&gt;. We need to convert the polar form to a standard form. Simple trig identities show &lt;math&gt;y=10+10i&lt;/math&gt; and &lt;math&gt;z=3-3i&lt;/math&gt;. More division is needed to find what &lt;math&gt;x&lt;/math&gt; is. &lt;cmath&gt;x=-20-12i&lt;/cmath&gt; &lt;cmath&gt;x+y+z=-7-5i&lt;/cmath&gt; &lt;cmath&gt;(-7)^2+(-5)^2=\boxed{74}&lt;/cmath&gt;<br /> &lt;cmath&gt;QED\blacksquare&lt;/cmath&gt;<br /> Written by [[User:A1b2|a1b2]]<br /> <br /> == Solution 3 (Pretty easy, no hard stuff, just watch ur arithmetic) ==<br /> Solve this system the way you would if the RHS of all equations were real. Multiply the first and 3rd equations out and then factor out &lt;math&gt;60&lt;/math&gt; to find &lt;math&gt;x^2&lt;/math&gt;, then use standard techniques that are used to evaluate square roots of irrationals. let &lt;cmath&gt;x = c+di&lt;/cmath&gt;, then you get &lt;cmath&gt;c^2 - d^2 = 256&lt;/cmath&gt; and &lt;cmath&gt;2cd = 480&lt;/cmath&gt; Solve to get &lt;math&gt;x&lt;/math&gt; as &lt;math&gt;20+12i&lt;/math&gt; and &lt;math&gt;-20-12i&lt;/math&gt;. Both will give us the same answer, so use the positive one. Divide &lt;math&gt;-80-320i&lt;/math&gt; by &lt;math&gt;x&lt;/math&gt;, and you get &lt;math&gt;10+10i&lt;/math&gt; as &lt;math&gt;y&lt;/math&gt;. This means that &lt;math&gt;z&lt;/math&gt; is a multiple of &lt;math&gt;1-i&lt;/math&gt; to get a real product, so you find &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3-3i&lt;/math&gt;. Now, add the real and imaginary parts separately to get &lt;math&gt;-7-5i&lt;/math&gt;, and calculate &lt;math&gt;a^2 + b^2&lt;/math&gt; to get &lt;math&gt;\boxed{74}&lt;/math&gt;. <br /> ~minor latex improvements done by jske25 and jdong2006<br /> <br /> ==Solution 4==<br /> Dividing the first equation by the second equation given, we find that &lt;math&gt;\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)&lt;/math&gt;. Substituting this into the third equation, we get &lt;math&gt;z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i&lt;/math&gt;. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of &lt;math&gt;y&lt;/math&gt; is the negative of that of &lt;math&gt;z&lt;/math&gt;, and their magnitudes multiply to &lt;math&gt;60&lt;/math&gt;. Thus, we have &lt;math&gt;z=\sqrt{-18i}=3-3i&lt;/math&gt; and &lt;math&gt;3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i&lt;/math&gt;. To find &lt;math&gt;x&lt;/math&gt;, we can use the previous substitution we made to find that &lt;math&gt;x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i&lt;/math&gt;.<br /> Therefore, &lt;math&gt;x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{74}&lt;/math&gt;<br /> <br /> Solution by ktong<br /> <br /> ==Solution 5 ==<br /> <br /> We are given that &lt;math&gt;xy=-80-320i&lt;/math&gt;. Thus &lt;math&gt;y=\frac{-80-320i}{x}&lt;/math&gt;. We are also given that &lt;math&gt;xz= -96+24i&lt;/math&gt;. Thus &lt;math&gt;z=\frac{-96+24i}{x}&lt;/math&gt;. We are also given that &lt;math&gt;yz&lt;/math&gt; = &lt;math&gt;60&lt;/math&gt;. Substitute &lt;math&gt;y=\frac{-80-320i}{x}&lt;/math&gt; and &lt;math&gt;z=\frac{-96+24i}{x}&lt;/math&gt; into &lt;math&gt;yz&lt;/math&gt; = &lt;math&gt;60&lt;/math&gt;. We have &lt;math&gt; \frac{(-80-320i)(-96+24i)}{x^2}=60&lt;/math&gt;. Multiplying out &lt;math&gt;(-80-320i)(-96+24i)&lt;/math&gt; we get &lt;math&gt;(1920)(8+15i)&lt;/math&gt;. Thus &lt;math&gt;\frac{1920(8+15i)}{x^2} =60&lt;/math&gt;. Simplifying this fraction we get &lt;math&gt;\frac{32(8+15i)}{x^2}=1&lt;/math&gt;. Cross-multiplying the fractions we get &lt;math&gt;x^2=32(8+15i)&lt;/math&gt; or &lt;math&gt;x^2= 256+480i&lt;/math&gt;. Now we can rewrite this as &lt;math&gt;x^2-256=480i&lt;/math&gt;. Let &lt;math&gt;x= (a+bi)&lt;/math&gt;.Thus &lt;math&gt;x^2=(a+bi)^2&lt;/math&gt; or &lt;math&gt;a^2+2abi-b^2&lt;/math&gt;. We can see that &lt;math&gt;a^2+2abi-b^2-256=480i&lt;/math&gt; and thus &lt;math&gt;2abi=480i&lt;/math&gt; or &lt;math&gt;ab=240&lt;/math&gt;.We also can see that &lt;math&gt;a^2-b^2-256=0&lt;/math&gt; because there is no real term in &lt;math&gt;480i&lt;/math&gt;. Thus &lt;math&gt;a^2-b^2=256&lt;/math&gt; or &lt;math&gt;(a+b)(a-b)=256&lt;/math&gt;. Using the two equations &lt;math&gt;ab=240&lt;/math&gt; and &lt;math&gt;(a+b)(a-b)=256&lt;/math&gt; we solve by doing system of equations that &lt;math&gt;a=-20&lt;/math&gt; and &lt;math&gt;b=-12&lt;/math&gt;. And &lt;math&gt;x=a+bi&lt;/math&gt; so &lt;math&gt;x=-20-12i&lt;/math&gt;. Because &lt;math&gt;y=\frac{-80-320i}{x}&lt;/math&gt;, then &lt;math&gt;y=\frac{-80-320i}{-20-12i}&lt;/math&gt;. Simplifying this fraction we get &lt;math&gt;y=\frac{-80(1+4i)}{-4(5+3i)}&lt;/math&gt; or &lt;math&gt;y=\frac{20(1+4i)}{(5+3i)}&lt;/math&gt;. Multiplying by the conjugate of the denominator (&lt;math&gt;5-3i&lt;/math&gt;) in the numerator and the denominator and we get &lt;math&gt;y=\frac{20(17-17i)}{34}&lt;/math&gt;. Simplifying this fraction we get &lt;math&gt;y=10-10i&lt;/math&gt;. Given that &lt;math&gt;yz&lt;/math&gt; = &lt;math&gt;60&lt;/math&gt; we can substitute &lt;math&gt;(10-10i)(z)=60&lt;/math&gt; We can solve for z and get &lt;math&gt;z=3+3i&lt;/math&gt;. Now we know what &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are, so all we have to do is plug and chug. &lt;math&gt;x+y+z= (-20-12i)+(10+10i)+(3-3i)&lt;/math&gt; or &lt;math&gt;x+y+z= -7-5i&lt;/math&gt; Now &lt;math&gt;a^2 +b^2=(-7)^2+(-5)^2&lt;/math&gt; or &lt;math&gt;a^2 +b^2 = 74&lt;/math&gt;. Thus &lt;math&gt;74&lt;/math&gt; is our final answer.(David Camacho)<br /> <br /> ==Solution 6==<br /> <br /> We observe that by multiplying &lt;math&gt;xy,&lt;/math&gt; &lt;math&gt;yz,&lt;/math&gt; and &lt;math&gt;zx,&lt;/math&gt; we get &lt;math&gt;(xyz)^2=(-80-320i)(60)(-96+24i).&lt;/math&gt; Next, we divide &lt;math&gt;(xyz)^2&lt;/math&gt; by &lt;math&gt;(yz)^2&lt;/math&gt; to <br /> <br /> get &lt;math&gt;x^2.&lt;/math&gt; We have &lt;math&gt;x^2=\frac{(-80-320i)(60)(-96+24i)}{3600}=256+480i.&lt;/math&gt; We can write &lt;math&gt;x&lt;/math&gt; in the form of &lt;math&gt;a+bi,&lt;/math&gt; so we get <br /> <br /> &lt;math&gt;(a+bi)^2=256+480i.&lt;/math&gt; Then, &lt;math&gt;a^2-b^2+2abi=256+480i,&lt;/math&gt; &lt;math&gt;a^2-b^2=256,&lt;/math&gt; and &lt;math&gt;2ab=480.&lt;/math&gt; Solving this system of equations is relatively <br /> <br /> simple. We have two cases, &lt;math&gt;a=20, b=12,&lt;/math&gt; and &lt;math&gt;a=-20, b=-12.&lt;/math&gt; <br /> <br /> Case 1: &lt;math&gt;a=20, b=12,&lt;/math&gt; so &lt;math&gt;x=20+12i.&lt;/math&gt; We solve for &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; by plugging in &lt;math&gt;x&lt;/math&gt; to the two equations. We see<br /> <br /> &lt;math&gt;y=\frac{-80-320i}{20+12i}=-10-10i&lt;/math&gt; and &lt;math&gt;z=\frac{-96+24i}{20+12i}=-3+3i.&lt;/math&gt; &lt;math&gt;x+y+z=7+5i,&lt;/math&gt; so &lt;math&gt;a=7&lt;/math&gt; and &lt;math&gt;b=5.&lt;/math&gt; Solving, we end up with <br /> <br /> &lt;math&gt;7^2+5^2=\boxed{074}&lt;/math&gt; as our answer. <br /> <br /> Case 2: &lt;math&gt;a=-20, b=-12,&lt;/math&gt; so &lt;math&gt;x=-20-12i.&lt;/math&gt; Again, we solve for &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z.&lt;/math&gt; We find &lt;math&gt;y=\frac{-80-320i}{-20-12i}=10+10i,&lt;/math&gt; <br /> <br /> &lt;math&gt;z=\frac{-96+24i}{-20-12i}=3-3i,&lt;/math&gt; so &lt;math&gt;x+y+z=-7-5i.&lt;/math&gt; We again have &lt;math&gt;(-7)^2+(-5)^2=\boxed{074}.&lt;/math&gt;<br /> <br /> Solution by Airplane50<br /> <br /> ==Solution 7 (Based on advanced mathematical knowledge)==<br /> According to the Euler's Theory, we can rewrite &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; as &lt;cmath&gt;x=r_{1}e^{i{\theta}_1}&lt;/cmath&gt; &lt;cmath&gt;y=r_{2}e^{i{\theta}_2}&lt;/cmath&gt; &lt;cmath&gt;x=r_{3}e^{i{\theta}_3}&lt;/cmath&gt; As a result, &lt;cmath&gt;|xy|=r_{1}r_{2}=\sqrt{80^2+320^2}=80\sqrt{17}&lt;/cmath&gt; &lt;cmath&gt;|yz|=r_{2}r_{3}=60&lt;/cmath&gt; &lt;cmath&gt;|xz|=r_{1}r_{3}=\sqrt{96^2+24^2}=24\sqrt{17}&lt;/cmath&gt; Also, it is clear that &lt;cmath&gt;yz=r_{2}e^{i{\theta}_2}r_{3}e^{i{\theta}_3}=|yz|e^{i({\theta}_2+{\theta}_3)}=|yz|=60&lt;/cmath&gt; So &lt;math&gt;{\theta}_2+{\theta}_3=0&lt;/math&gt;, or &lt;cmath&gt;{\theta}_2=-{\theta}_3&lt;/cmath&gt; Also, we have &lt;cmath&gt;xy=-80\sqrt{17}e^{i\arctan{4}}&lt;/cmath&gt; &lt;cmath&gt;yz=60&lt;/cmath&gt; &lt;cmath&gt;xz=-24\sqrt{17}e^{i\arctan{-\frac{1}{4}}}&lt;/cmath&gt; So now we have &lt;math&gt;r_{1}r_{2}=80\sqrt{17}&lt;/math&gt;, &lt;math&gt;r_{2}r_{3}=60&lt;/math&gt;, &lt;math&gt;r_{1}r_{3}=24\sqrt{17}&lt;/math&gt;, &lt;math&gt;{\theta}_1+{\theta}_2=\arctan{4}&lt;/math&gt; and &lt;math&gt;{\theta}_1-{\theta}_2=\arctan {-\frac{1}{4}}&lt;/math&gt;. Solve these above, we get &lt;cmath&gt;r_{1}=4\sqrt{34}&lt;/cmath&gt; &lt;cmath&gt;r_{2}=10\sqrt{2}&lt;/cmath&gt; &lt;cmath&gt;r_{3}=3\sqrt{2}&lt;/cmath&gt; &lt;cmath&gt;{\theta}_2=\frac{\arctan{4}-\arctan{-\frac{1}{4}}}{2}=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}&lt;/cmath&gt; So we can get &lt;cmath&gt;y=r_{2}e^{i{\theta}_2}=10\sqrt{2}e^{i\frac{\pi}{4}}=10+10i&lt;/cmath&gt; &lt;cmath&gt;z=r_{3}e^{i{\theta}_3}=r_{3}e^{-i{\theta}_2}=3\sqrt{2}e^{-i\frac{\pi}{4}}=3-3i&lt;/cmath&gt; Use &lt;math&gt;xy=-80-320i&lt;/math&gt; we can find that &lt;cmath&gt;x=-20-12i&lt;/cmath&gt; So &lt;cmath&gt;x+y+z=-20-12i+10+10i+3-3i=-7-5i&lt;/cmath&gt; So we have &lt;math&gt;a=-7&lt;/math&gt; and &lt;math&gt;b=-5&lt;/math&gt;.<br /> <br /> As a result, we finally get &lt;cmath&gt;a^2+b^2=(-7)^2+(-5)^2=\boxed{074}&lt;/cmath&gt;<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Solution 8==<br /> <br /> We can turn the expression &lt;math&gt;x+y+z&lt;/math&gt; into &lt;math&gt;\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}&lt;/math&gt;, and this would allow us to plug in the values after some computations.<br /> <br /> Based off of the given products, we have<br /> <br /> &lt;math&gt;xy^2z=60(-80-320i)&lt;/math&gt;<br /> <br /> &lt;math&gt;xyz^2=60(-96+24i)&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2yz=(-96+24i)(-80-320i)&lt;/math&gt;.<br /> <br /> Dividing by the given products, we have<br /> <br /> &lt;math&gt;y^2=\frac{60(-80-320i)}{-96+24i}&lt;/math&gt;<br /> <br /> &lt;math&gt;z^2=\frac{60(-96+24i)}{-80-320i}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2=\frac{(-96+24i)(-80-320i)}{60}&lt;/math&gt;.<br /> <br /> Simplifying, we get that this expression becomes &lt;math&gt;\sqrt{24+70i}&lt;/math&gt;. This equals &lt;math&gt;\pm{(7+5i)}&lt;/math&gt;, so the answer is &lt;math&gt;7^2+5^2=\boxed{74}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{-RootThreeOverTwo}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=II|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_5&diff=173016 2018 AIME II Problems/Problem 5 2022-03-28T18:03:35Z <p>First: </p> <hr /> <div>==Problem==<br /> <br /> Suppose that &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are complex numbers such that &lt;math&gt;xy = -80 - 320i&lt;/math&gt;, &lt;math&gt;yz = 60&lt;/math&gt;, and &lt;math&gt;zx = -96 + 24i&lt;/math&gt;, where &lt;math&gt;i&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;\sqrt{-1}&lt;/math&gt;. Then there are real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;x + y + z = a + bi&lt;/math&gt;. Find &lt;math&gt;a^2 + b^2&lt;/math&gt;.<br /> <br /> ==Solution 1 (Euler's formula and Substitution)<br /> <br /> The [b]First[/b] thing to notice is that &lt;math&gt;xy&lt;/math&gt; and &lt;math&gt;zx&lt;/math&gt; have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is &lt;math&gt;e^{\frac{3\pi i}{2}}&lt;/math&gt;. Anyway this results in getting that &lt;math&gt;\left(\frac{-3i}{10}\right)y=z&lt;/math&gt;. Then substitute this into &lt;math&gt;yz&lt;/math&gt; to get, after some calculation, that &lt;math&gt;y=10e^{\frac{5\pi i}{4}}\sqrt{2}&lt;/math&gt; and &lt;math&gt;z=-3e^{\frac{7\pi i}{4}}\sqrt{2}&lt;/math&gt;. Then plug &lt;math&gt;z&lt;/math&gt; into &lt;math&gt;zx&lt;/math&gt;, you could do the same thing with &lt;math&gt;xy&lt;/math&gt; but &lt;math&gt;zx&lt;/math&gt; looks like it's easier due to it being smaller. Anyway you get &lt;math&gt;x=20+12i&lt;/math&gt;. Then add all three up, it turns out easier than it seems because for &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; the &lt;math&gt;\sqrt{2}&lt;/math&gt; disappears after you expand the root of unity (e raised to a specific power). Long story short, you get &lt;math&gt;x=20+12i, y=-3+3i, z=-10-10i \implies x+y+z=7+5i \implies a^2+b^2=\boxed{074}&lt;/math&gt;.<br /> <br /> ~First<br /> <br /> ==Solution 2==<br /> <br /> First we evaluate the magnitudes. &lt;math&gt;|xy|=80\sqrt{17}&lt;/math&gt;, &lt;math&gt;|yz|=60&lt;/math&gt;, and &lt;math&gt;|zx|=24\sqrt{17}&lt;/math&gt;. Therefore, &lt;math&gt;|x^2y^2z^2|=17\cdot80\cdot60\cdot24&lt;/math&gt;, or &lt;math&gt;|xyz|=240\sqrt{34}&lt;/math&gt;. Divide to find that &lt;math&gt;|z|=3\sqrt{2}&lt;/math&gt;, &lt;math&gt;|x|=4\sqrt{34}&lt;/math&gt;, and &lt;math&gt;|y|=10\sqrt{2}&lt;/math&gt;.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((4,0),red);<br /> draw((0,0)--(-4,0));<br /> draw((0,0)--(0,-4));<br /> draw((0,0)--(-4,1));<br /> dot((-4,1),red);<br /> draw((0,0)--(-1,-4));<br /> dot((-1,-4),red);<br /> draw((0,0)--(4,4),red);<br /> draw((0,0)--(4,-4),red);<br /> &lt;/asy&gt;<br /> This allows us to see that the argument of &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;\frac{\pi}{4}&lt;/math&gt;, and the argument of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;-\frac{\pi}{4}&lt;/math&gt;. We need to convert the polar form to a standard form. Simple trig identities show &lt;math&gt;y=10+10i&lt;/math&gt; and &lt;math&gt;z=3-3i&lt;/math&gt;. More division is needed to find what &lt;math&gt;x&lt;/math&gt; is. &lt;cmath&gt;x=-20-12i&lt;/cmath&gt; &lt;cmath&gt;x+y+z=-7-5i&lt;/cmath&gt; &lt;cmath&gt;(-7)^2+(-5)^2=\boxed{74}&lt;/cmath&gt;<br /> &lt;cmath&gt;QED\blacksquare&lt;/cmath&gt;<br /> Written by [[User:A1b2|a1b2]]<br /> <br /> == Solution 3 (Pretty easy, no hard stuff, just watch ur arithmetic) ==<br /> Solve this system the way you would if the RHS of all equations were real. Multiply the first and 3rd equations out and then factor out &lt;math&gt;60&lt;/math&gt; to find &lt;math&gt;x^2&lt;/math&gt;, then use standard techniques that are used to evaluate square roots of irrationals. let &lt;cmath&gt;x = c+di&lt;/cmath&gt;, then you get &lt;cmath&gt;c^2 - d^2 = 256&lt;/cmath&gt; and &lt;cmath&gt;2cd = 480&lt;/cmath&gt; Solve to get &lt;math&gt;x&lt;/math&gt; as &lt;math&gt;20+12i&lt;/math&gt; and &lt;math&gt;-20-12i&lt;/math&gt;. Both will give us the same answer, so use the positive one. Divide &lt;math&gt;-80-320i&lt;/math&gt; by &lt;math&gt;x&lt;/math&gt;, and you get &lt;math&gt;10+10i&lt;/math&gt; as &lt;math&gt;y&lt;/math&gt;. This means that &lt;math&gt;z&lt;/math&gt; is a multiple of &lt;math&gt;1-i&lt;/math&gt; to get a real product, so you find &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3-3i&lt;/math&gt;. Now, add the real and imaginary parts separately to get &lt;math&gt;-7-5i&lt;/math&gt;, and calculate &lt;math&gt;a^2 + b^2&lt;/math&gt; to get &lt;math&gt;\boxed{74}&lt;/math&gt;. <br /> ~minor latex improvements done by jske25 and jdong2006<br /> <br /> ==Solution 4==<br /> Dividing the first equation by the second equation given, we find that &lt;math&gt;\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)&lt;/math&gt;. Substituting this into the third equation, we get &lt;math&gt;z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i&lt;/math&gt;. Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of &lt;math&gt;y&lt;/math&gt; is the negative of that of &lt;math&gt;z&lt;/math&gt;, and their magnitudes multiply to &lt;math&gt;60&lt;/math&gt;. Thus, we have &lt;math&gt;z=\sqrt{-18i}=3-3i&lt;/math&gt; and &lt;math&gt;3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i&lt;/math&gt;. To find &lt;math&gt;x&lt;/math&gt;, we can use the previous substitution we made to find that &lt;math&gt;x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i&lt;/math&gt;.<br /> Therefore, &lt;math&gt;x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{74}&lt;/math&gt;<br /> <br /> Solution by ktong<br /> <br /> ==Solution 5 ==<br /> <br /> We are given that &lt;math&gt;xy=-80-320i&lt;/math&gt;. Thus &lt;math&gt;y=\frac{-80-320i}{x}&lt;/math&gt;. We are also given that &lt;math&gt;xz= -96+24i&lt;/math&gt;. Thus &lt;math&gt;z=\frac{-96+24i}{x}&lt;/math&gt;. We are also given that &lt;math&gt;yz&lt;/math&gt; = &lt;math&gt;60&lt;/math&gt;. Substitute &lt;math&gt;y=\frac{-80-320i}{x}&lt;/math&gt; and &lt;math&gt;z=\frac{-96+24i}{x}&lt;/math&gt; into &lt;math&gt;yz&lt;/math&gt; = &lt;math&gt;60&lt;/math&gt;. We have &lt;math&gt; \frac{(-80-320i)(-96+24i)}{x^2}=60&lt;/math&gt;. Multiplying out &lt;math&gt;(-80-320i)(-96+24i)&lt;/math&gt; we get &lt;math&gt;(1920)(8+15i)&lt;/math&gt;. Thus &lt;math&gt;\frac{1920(8+15i)}{x^2} =60&lt;/math&gt;. Simplifying this fraction we get &lt;math&gt;\frac{32(8+15i)}{x^2}=1&lt;/math&gt;. Cross-multiplying the fractions we get &lt;math&gt;x^2=32(8+15i)&lt;/math&gt; or &lt;math&gt;x^2= 256+480i&lt;/math&gt;. Now we can rewrite this as &lt;math&gt;x^2-256=480i&lt;/math&gt;. Let &lt;math&gt;x= (a+bi)&lt;/math&gt;.Thus &lt;math&gt;x^2=(a+bi)^2&lt;/math&gt; or &lt;math&gt;a^2+2abi-b^2&lt;/math&gt;. We can see that &lt;math&gt;a^2+2abi-b^2-256=480i&lt;/math&gt; and thus &lt;math&gt;2abi=480i&lt;/math&gt; or &lt;math&gt;ab=240&lt;/math&gt;.We also can see that &lt;math&gt;a^2-b^2-256=0&lt;/math&gt; because there is no real term in &lt;math&gt;480i&lt;/math&gt;. Thus &lt;math&gt;a^2-b^2=256&lt;/math&gt; or &lt;math&gt;(a+b)(a-b)=256&lt;/math&gt;. Using the two equations &lt;math&gt;ab=240&lt;/math&gt; and &lt;math&gt;(a+b)(a-b)=256&lt;/math&gt; we solve by doing system of equations that &lt;math&gt;a=-20&lt;/math&gt; and &lt;math&gt;b=-12&lt;/math&gt;. And &lt;math&gt;x=a+bi&lt;/math&gt; so &lt;math&gt;x=-20-12i&lt;/math&gt;. Because &lt;math&gt;y=\frac{-80-320i}{x}&lt;/math&gt;, then &lt;math&gt;y=\frac{-80-320i}{-20-12i}&lt;/math&gt;. Simplifying this fraction we get &lt;math&gt;y=\frac{-80(1+4i)}{-4(5+3i)}&lt;/math&gt; or &lt;math&gt;y=\frac{20(1+4i)}{(5+3i)}&lt;/math&gt;. Multiplying by the conjugate of the denominator (&lt;math&gt;5-3i&lt;/math&gt;) in the numerator and the denominator and we get &lt;math&gt;y=\frac{20(17-17i)}{34}&lt;/math&gt;. Simplifying this fraction we get &lt;math&gt;y=10-10i&lt;/math&gt;. Given that &lt;math&gt;yz&lt;/math&gt; = &lt;math&gt;60&lt;/math&gt; we can substitute &lt;math&gt;(10-10i)(z)=60&lt;/math&gt; We can solve for z and get &lt;math&gt;z=3+3i&lt;/math&gt;. Now we know what &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are, so all we have to do is plug and chug. &lt;math&gt;x+y+z= (-20-12i)+(10+10i)+(3-3i)&lt;/math&gt; or &lt;math&gt;x+y+z= -7-5i&lt;/math&gt; Now &lt;math&gt;a^2 +b^2=(-7)^2+(-5)^2&lt;/math&gt; or &lt;math&gt;a^2 +b^2 = 74&lt;/math&gt;. Thus &lt;math&gt;74&lt;/math&gt; is our final answer.(David Camacho)<br /> <br /> ==Solution 6==<br /> <br /> We observe that by multiplying &lt;math&gt;xy,&lt;/math&gt; &lt;math&gt;yz,&lt;/math&gt; and &lt;math&gt;zx,&lt;/math&gt; we get &lt;math&gt;(xyz)^2=(-80-320i)(60)(-96+24i).&lt;/math&gt; Next, we divide &lt;math&gt;(xyz)^2&lt;/math&gt; by &lt;math&gt;(yz)^2&lt;/math&gt; to <br /> <br /> get &lt;math&gt;x^2.&lt;/math&gt; We have &lt;math&gt;x^2=\frac{(-80-320i)(60)(-96+24i)}{3600}=256+480i.&lt;/math&gt; We can write &lt;math&gt;x&lt;/math&gt; in the form of &lt;math&gt;a+bi,&lt;/math&gt; so we get <br /> <br /> &lt;math&gt;(a+bi)^2=256+480i.&lt;/math&gt; Then, &lt;math&gt;a^2-b^2+2abi=256+480i,&lt;/math&gt; &lt;math&gt;a^2-b^2=256,&lt;/math&gt; and &lt;math&gt;2ab=480.&lt;/math&gt; Solving this system of equations is relatively <br /> <br /> simple. We have two cases, &lt;math&gt;a=20, b=12,&lt;/math&gt; and &lt;math&gt;a=-20, b=-12.&lt;/math&gt; <br /> <br /> Case 1: &lt;math&gt;a=20, b=12,&lt;/math&gt; so &lt;math&gt;x=20+12i.&lt;/math&gt; We solve for &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; by plugging in &lt;math&gt;x&lt;/math&gt; to the two equations. We see<br /> <br /> &lt;math&gt;y=\frac{-80-320i}{20+12i}=-10-10i&lt;/math&gt; and &lt;math&gt;z=\frac{-96+24i}{20+12i}=-3+3i.&lt;/math&gt; &lt;math&gt;x+y+z=7+5i,&lt;/math&gt; so &lt;math&gt;a=7&lt;/math&gt; and &lt;math&gt;b=5.&lt;/math&gt; Solving, we end up with <br /> <br /> &lt;math&gt;7^2+5^2=\boxed{074}&lt;/math&gt; as our answer. <br /> <br /> Case 2: &lt;math&gt;a=-20, b=-12,&lt;/math&gt; so &lt;math&gt;x=-20-12i.&lt;/math&gt; Again, we solve for &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z.&lt;/math&gt; We find &lt;math&gt;y=\frac{-80-320i}{-20-12i}=10+10i,&lt;/math&gt; <br /> <br /> &lt;math&gt;z=\frac{-96+24i}{-20-12i}=3-3i,&lt;/math&gt; so &lt;math&gt;x+y+z=-7-5i.&lt;/math&gt; We again have &lt;math&gt;(-7)^2+(-5)^2=\boxed{074}.&lt;/math&gt;<br /> <br /> Solution by Airplane50<br /> <br /> ==Solution 7 (Based on advanced mathematical knowledge)==<br /> According to the Euler's Theory, we can rewrite &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; as &lt;cmath&gt;x=r_{1}e^{i{\theta}_1}&lt;/cmath&gt; &lt;cmath&gt;y=r_{2}e^{i{\theta}_2}&lt;/cmath&gt; &lt;cmath&gt;x=r_{3}e^{i{\theta}_3}&lt;/cmath&gt; As a result, &lt;cmath&gt;|xy|=r_{1}r_{2}=\sqrt{80^2+320^2}=80\sqrt{17}&lt;/cmath&gt; &lt;cmath&gt;|yz|=r_{2}r_{3}=60&lt;/cmath&gt; &lt;cmath&gt;|xz|=r_{1}r_{3}=\sqrt{96^2+24^2}=24\sqrt{17}&lt;/cmath&gt; Also, it is clear that &lt;cmath&gt;yz=r_{2}e^{i{\theta}_2}r_{3}e^{i{\theta}_3}=|yz|e^{i({\theta}_2+{\theta}_3)}=|yz|=60&lt;/cmath&gt; So &lt;math&gt;{\theta}_2+{\theta}_3=0&lt;/math&gt;, or &lt;cmath&gt;{\theta}_2=-{\theta}_3&lt;/cmath&gt; Also, we have &lt;cmath&gt;xy=-80\sqrt{17}e^{i\arctan{4}}&lt;/cmath&gt; &lt;cmath&gt;yz=60&lt;/cmath&gt; &lt;cmath&gt;xz=-24\sqrt{17}e^{i\arctan{-\frac{1}{4}}}&lt;/cmath&gt; So now we have &lt;math&gt;r_{1}r_{2}=80\sqrt{17}&lt;/math&gt;, &lt;math&gt;r_{2}r_{3}=60&lt;/math&gt;, &lt;math&gt;r_{1}r_{3}=24\sqrt{17}&lt;/math&gt;, &lt;math&gt;{\theta}_1+{\theta}_2=\arctan{4}&lt;/math&gt; and &lt;math&gt;{\theta}_1-{\theta}_2=\arctan {-\frac{1}{4}}&lt;/math&gt;. Solve these above, we get &lt;cmath&gt;r_{1}=4\sqrt{34}&lt;/cmath&gt; &lt;cmath&gt;r_{2}=10\sqrt{2}&lt;/cmath&gt; &lt;cmath&gt;r_{3}=3\sqrt{2}&lt;/cmath&gt; &lt;cmath&gt;{\theta}_2=\frac{\arctan{4}-\arctan{-\frac{1}{4}}}{2}=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}&lt;/cmath&gt; So we can get &lt;cmath&gt;y=r_{2}e^{i{\theta}_2}=10\sqrt{2}e^{i\frac{\pi}{4}}=10+10i&lt;/cmath&gt; &lt;cmath&gt;z=r_{3}e^{i{\theta}_3}=r_{3}e^{-i{\theta}_2}=3\sqrt{2}e^{-i\frac{\pi}{4}}=3-3i&lt;/cmath&gt; Use &lt;math&gt;xy=-80-320i&lt;/math&gt; we can find that &lt;cmath&gt;x=-20-12i&lt;/cmath&gt; So &lt;cmath&gt;x+y+z=-20-12i+10+10i+3-3i=-7-5i&lt;/cmath&gt; So we have &lt;math&gt;a=-7&lt;/math&gt; and &lt;math&gt;b=-5&lt;/math&gt;.<br /> <br /> As a result, we finally get &lt;cmath&gt;a^2+b^2=(-7)^2+(-5)^2=\boxed{074}&lt;/cmath&gt;<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Solution 8==<br /> <br /> We can turn the expression &lt;math&gt;x+y+z&lt;/math&gt; into &lt;math&gt;\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}&lt;/math&gt;, and this would allow us to plug in the values after some computations.<br /> <br /> Based off of the given products, we have<br /> <br /> &lt;math&gt;xy^2z=60(-80-320i)&lt;/math&gt;<br /> <br /> &lt;math&gt;xyz^2=60(-96+24i)&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2yz=(-96+24i)(-80-320i)&lt;/math&gt;.<br /> <br /> Dividing by the given products, we have<br /> <br /> &lt;math&gt;y^2=\frac{60(-80-320i)}{-96+24i}&lt;/math&gt;<br /> <br /> &lt;math&gt;z^2=\frac{60(-96+24i)}{-80-320i}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2=\frac{(-96+24i)(-80-320i)}{60}&lt;/math&gt;.<br /> <br /> Simplifying, we get that this expression becomes &lt;math&gt;\sqrt{24+70i}&lt;/math&gt;. This equals &lt;math&gt;\pm{(7+5i)}&lt;/math&gt;, so the answer is &lt;math&gt;7^2+5^2=\boxed{74}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{-RootThreeOverTwo}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=II|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_10&diff=172995 2018 AIME II Problems/Problem 10 2022-03-27T17:38:46Z <p>First: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Find the number of functions &lt;math&gt;f(x)&lt;/math&gt; from &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; to &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; that satisfy &lt;math&gt;f(f(x)) = f(f(f(x)))&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Just to visualize solution 1. If we list all possible &lt;math&gt;(x,f(x))&lt;/math&gt;, from &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; to &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; in a specific order, we get &lt;math&gt;5 \cdot 5 = 25&lt;/math&gt; different &lt;math&gt;(x,f(x))&lt;/math&gt; 's.<br /> Namely:<br /> <br /> &lt;cmath&gt;(1,1) (1,2) (1,3) (1,4) (1,5)&lt;/cmath&gt;<br /> &lt;cmath&gt;(2,1) (2,2) (2,3) (2,4) (2,5)&lt;/cmath&gt; <br /> &lt;cmath&gt;(3,1) (3,2) (3,3) (3,4) (3,5)&lt;/cmath&gt; <br /> &lt;cmath&gt;(4,1) (4,2) (4,3) (4,4) (4,5)&lt;/cmath&gt;<br /> &lt;cmath&gt;(5,1) (5,2) (5,3) (5,4) (5,5)&lt;/cmath&gt;<br /> <br /> To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of &lt;math&gt;(x,x)&lt;/math&gt; where &lt;math&gt;x\in{1,2,3,4,5}&lt;/math&gt; must exist.In this case I rather &quot;go backwards&quot;. First fixing &lt;math&gt;5&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt;, (the diagonal of our table) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{5!} = 1&lt;/math&gt; way. Then fixing &lt;math&gt;4&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;1&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in <br /> &lt;math&gt;4\cdot\frac{5!}{4!} = 20&lt;/math&gt; ways. Then fixing &lt;math&gt;3&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;2&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\tfrac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \tfrac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150&lt;/math&gt; ways.<br /> Fixing &lt;math&gt;2&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;3&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!3!} + \frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!2!} + \frac{(5\cdot4\cdot6\cdot2\cdot1)}{2!2!} = 380&lt;/math&gt; ways.<br /> Lastly, fixing &lt;math&gt;1&lt;/math&gt; pair &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;4&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\tfrac{5!}{4!} + 4\cdot\tfrac{5!}{3!} + 5! = 205&lt;/math&gt; ways.<br /> <br /> So &lt;math&gt;1 + 20 + 150 + 380 + 205 = \framebox{756}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We perform casework on the number of fixed points (the number of points where &lt;math&gt;f(x) = x&lt;/math&gt;). To better visualize this, use the grid from Solution 1.<br /> <br /> '''Case 1:''' 5 fixed points<br /> <br /> :- Obviously, there must be &lt;math&gt;1&lt;/math&gt; way to do so.<br /> <br /> '''Case 2:''' 4 fixed points<br /> <br /> :- &lt;math&gt;\binom 54&lt;/math&gt; ways to choose the &lt;math&gt;4&lt;/math&gt; fixed points. For the sake of conversation, let them be &lt;math&gt;(1, 1), (2, 2), (3, 3), (4, 4)&lt;/math&gt;.<br /> :- The last point must be &lt;math&gt;(5, 1), (5, 2), (5, 3),&lt;/math&gt; or &lt;math&gt;(5, 4)&lt;/math&gt;.<br /> :- There are &lt;math&gt;\binom 54 \cdot 4 = 20&lt;/math&gt; solutions for this case.<br /> <br /> '''Case 3:''' 3 fixed points<br /> <br /> :- &lt;math&gt;\binom 53&lt;/math&gt; ways to choose the &lt;math&gt;3&lt;/math&gt; fixed points. For the sake of conversation, let them be &lt;math&gt;(1, 1), (2, 2), (3, 3)&lt;/math&gt;.<br /> :- '''Subcase 3.1:''' None of &lt;math&gt;4&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt; map to each other<br /> ::- The points must be &lt;math&gt;(4, 1), (4, 2), (4, 3)&lt;/math&gt; and &lt;math&gt;(5, 1), (5, 2), (5, 3)&lt;/math&gt;, giving &lt;math&gt;3 \cdot 3 = 9&lt;/math&gt; solutions.<br /> :- '''Subcase 3.2:''' &lt;math&gt;4&lt;/math&gt; maps to &lt;math&gt;5&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt; maps to &lt;math&gt;4&lt;/math&gt;<br /> ::- The points must be &lt;math&gt;(4, 5)&lt;/math&gt; and &lt;math&gt;(5, 1), (5, 2), (5, 3)&lt;/math&gt; or &lt;math&gt;(5, 4)&lt;/math&gt; and &lt;math&gt;(4, 1), (4, 2), (4, 3)&lt;/math&gt;, giving &lt;math&gt;3+3 = 6&lt;/math&gt; solutions.<br /> :- There are &lt;math&gt;\binom 53 \cdot (9+6) = 150&lt;/math&gt; solutions for this case.<br /> <br /> '''Case 4:''' 4 fixed points<br /> <br /> :- &lt;math&gt;\binom 52&lt;/math&gt; ways to choose the &lt;math&gt;2&lt;/math&gt; fixed points. For the sake of conversation, let them be &lt;math&gt;(1, 1), (2, 2)&lt;/math&gt;.<br /> :- '''Subcase 4.1:''' None of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, or &lt;math&gt;5&lt;/math&gt; map to each other<br /> ::- There are &lt;math&gt;2 \cdot 2 \cdot 2 = 8&lt;/math&gt; solutions for this case, by similar logic to '''Subcase 3.1'''.<br /> :- '''Subcase 4.2:''' exactly one of &lt;math&gt;3, 4, 5&lt;/math&gt; maps to another.<br /> ::- Example: &lt;math&gt;(3, 4), (4, 1), (5, 2)&lt;/math&gt;<br /> ::- &lt;math&gt;\binom 32 \cdot 2! = 6&lt;/math&gt; ways to choose the 2 which map to each other, and &lt;math&gt;2\cdot 2&lt;/math&gt; ways to choose which ones of &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; they map to, giving &lt;math&gt;24&lt;/math&gt; solutions for this case.<br /> :- '''Subcase 4.3:''' two of &lt;math&gt;3, 4, 5&lt;/math&gt; map to the third<br /> ::- Example: &lt;math&gt;(3, 5), (4, 5), (5, 2)&lt;/math&gt;<br /> ::- &lt;math&gt;\binom 31&lt;/math&gt; ways to choose which point is being mapped to, and &lt;math&gt;2&lt;/math&gt; ways to choose which one of &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; it maps to, giving &lt;math&gt;6&lt;/math&gt; solutions for this case.<br /> :- There are &lt;math&gt;\binom 52 \cdot (8+24+6) = 380&lt;/math&gt; solutions.<br /> <br /> '''Case 5:''' 1 fixed point<br /> :- &lt;math&gt;\binom 51&lt;/math&gt; ways to choose the fixed point. For the sake of conversation, let it be &lt;math&gt;(1, 1)&lt;/math&gt;.<br /> :- '''Subcase 5.1:''' None of &lt;math&gt;2, 3, 4, 5&lt;/math&gt; map to each other<br /> ::- Obviously, there is &lt;math&gt;1^4 = 1&lt;/math&gt; solution to this; all of them map to &lt;math&gt;1&lt;/math&gt;.<br /> :- '''Subcase 5.2:''' One maps to another, and the other two map to &lt;math&gt;1&lt;/math&gt;.<br /> ::- Example: &lt;math&gt;(2, 3), (3, 1), (4, 1), (5, 1)&lt;/math&gt;<br /> ::- There are &lt;math&gt;\binom 42 \cdot 2!&lt;/math&gt; ways to choose which two map to each other, and since each must map to &lt;math&gt;1&lt;/math&gt;, this gives &lt;math&gt;12&lt;/math&gt;.<br /> :- '''Subcase 5.3:''' One maps to another, and of the other two, one maps to the other as well.<br /> ::- Example: &lt;math&gt;(2, 3), (3, 1), (5, 4), (4, 1)&lt;/math&gt;<br /> ::- There are &lt;math&gt;\binom 42 \cdot 2! \cdot 2! / 2!&lt;/math&gt; ways to choose which ones map to another. This also gives &lt;math&gt;12&lt;/math&gt;.<br /> :- '''Subcase 5.4:''' 2 map to a third, and the fourth maps to &lt;math&gt;1&lt;/math&gt;.<br /> ::- Example: &lt;math&gt;(4, 2), (5, 2), (2, 1), (3, 1)&lt;/math&gt;<br /> ::- There are &lt;math&gt;\binom 42 \cdot \binom 21 = 12&lt;/math&gt; ways again.<br /> :- '''Subcase 5.5:''' 3 map to the fourth.<br /> ::- Example: &lt;math&gt;(2, 4), (3, 4), (5, 4), (4, 1)&lt;/math&gt;<br /> ::- There are &lt;math&gt;\binom 41&lt;/math&gt; ways to choose which one is being mapped to, giving &lt;math&gt;4&lt;/math&gt; solutions. <br /> :- There are &lt;math&gt;\binom 51 \cdot (1+12+12+12+4) = 205&lt;/math&gt; solutions.<br /> <br /> Therefore, the answer is &lt;math&gt;1+20+150+380+205 = \boxed{756}&lt;/math&gt;<br /> <br /> ~First<br /> <br /> ==Solution 3==<br /> <br /> We can do some caseworks about the special points of functions &lt;math&gt;f&lt;/math&gt; for &lt;math&gt;x\in\{1,2,3,4,5\}&lt;/math&gt;. Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be three different elements in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt;. There must be elements such like &lt;math&gt;k&lt;/math&gt; in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; satisfies &lt;math&gt;f(k)=k&lt;/math&gt;, and we call the points such like &lt;math&gt;(k,k)&lt;/math&gt; on functions &lt;math&gt;f&lt;/math&gt; are &quot;Good Points&quot; (Actually its academic name is &quot;fixed-points&quot;). The only thing we need to consider is the &quot;steps&quot; to get &quot;Good Points&quot;. Notice that the &quot;steps&quot; must less than &lt;math&gt;3&lt;/math&gt; because the highest iterations of function &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;. Now we can classify &lt;math&gt;3&lt;/math&gt; cases of “Good points” of &lt;math&gt;f&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; One &quot;step&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=x&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(x))=f(x)=x&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; Two &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt; and &lt;math&gt;f(y)=y&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=y&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(y)=y&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 3:}&lt;/math&gt; Three &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt;, &lt;math&gt;f(y)=z&lt;/math&gt; and &lt;math&gt;f(z)=z&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=z&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(z)=z&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> Divide set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; into three parts which satisfy these three cases, respectively. Let the first part has &lt;math&gt;a&lt;/math&gt; elements, the second part has &lt;math&gt;b&lt;/math&gt; elements and the third part has &lt;math&gt;c&lt;/math&gt; elements, it is easy to see that &lt;math&gt;a+b+c=5&lt;/math&gt;. First, there are &lt;math&gt;\binom{5}{a}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 1. Second, we have &lt;math&gt;\binom{5-a}{b}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is &lt;math&gt;a^b&lt;/math&gt;. Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is &lt;math&gt;b^c&lt;/math&gt;. As a result, the number of such functions &lt;math&gt;f&lt;/math&gt; can be represented in an algebraic expression contains &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;\boxed{\binom{5}{a}\cdot \binom{5-a}{b}\cdot a^b\cdot b^c}&lt;/math&gt;<br /> <br /> Now it's time to consider about the different values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; and the total number of functions &lt;math&gt;f&lt;/math&gt; satisfy these values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;:<br /> <br /> For &lt;math&gt;a=5&lt;/math&gt;, &lt;math&gt;b=0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{5}=1&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{4}\cdot \binom{1}{1}\cdot 4^1\cdot 1^0=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{1}\cdot 3^1\cdot 1^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{2}\cdot 3^2\cdot 2^0=90&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{1}\cdot 2^1\cdot 1^2=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{2}\cdot 2^2\cdot 2^1=240&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{3}\cdot 2^3\cdot 3^0=80&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=3&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{1}\cdot 1^1\cdot 1^3=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{2}\cdot 1^2\cdot 2^2=120&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{3}\cdot 1^3\cdot 3^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=4&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt;s is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{4}\cdot 1^4\cdot 4^0=5&lt;/math&gt;<br /> <br /> Finally, we get the total number of function &lt;math&gt;f&lt;/math&gt;, the number is &lt;math&gt;1+20+60+90+60+240+80+20+120+60+5=\boxed{756}&lt;/math&gt;<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Note (fun fact)==<br /> This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test.<br /> This problem also showed up on the 2010 Mock AIME 2 here: https://artofproblemsolving.com/wiki/index.php/Mock_AIME_2_2010_Problems<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Intermediate Combinatorics Problems]]</div> First https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_6&diff=172984 2018 AIME II Problems/Problem 6 2022-03-27T16:20:24Z <p>First: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A real number &lt;math&gt;a&lt;/math&gt; is chosen randomly and uniformly from the interval &lt;math&gt;[-20, 18]&lt;/math&gt;. The probability that the roots of the polynomial<br /> <br /> &lt;math&gt;x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2&lt;/math&gt;<br /> <br /> are all real can be written in the form &lt;math&gt;\dfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> The polynomial we are given is rather complicated, so we could use [[Rational Root Theorem]] to turn the given polynomial into a degree-2 polynomial. With Rational Root Theorem, &lt;math&gt;x = 1, -1, 2, -2&lt;/math&gt; are all possible rational roots. Upon plugging these roots into the polynomial, &lt;math&gt;x = -2&lt;/math&gt; and &lt;math&gt;x = 1&lt;/math&gt; make the polynomial equal 0 and thus, they are roots that we can factor out.<br /> <br /> The polynomial becomes:<br /> <br /> &lt;math&gt;(x - 1)(x + 2)(x^2 + (2a - 1)x + 1)&lt;/math&gt;<br /> <br /> Since we know &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;-2&lt;/math&gt; are real numbers, we only need to focus on the quadratic.<br /> <br /> We should set the discriminant of the quadratic greater than or equal to 0.<br /> <br /> &lt;math&gt;(2a - 1)^2 - 4 \geq 0&lt;/math&gt;.<br /> <br /> This simplifies to:<br /> <br /> &lt;math&gt;a \geq \dfrac{3}{2}&lt;/math&gt;<br /> <br /> or<br /> <br /> &lt;math&gt;a \leq -\dfrac{1}{2}&lt;/math&gt;<br /> <br /> This means that the interval &lt;math&gt;\left(-\dfrac{1}{2}, \dfrac{3}{2}\right)&lt;/math&gt; is the &quot;bad&quot; interval. The length of the interval where &lt;math&gt;a&lt;/math&gt; can be chosen from is 38 units long, while the bad interval is 2 units long. Therefore, the &quot;good&quot; interval is 36 units long.<br /> <br /> &lt;math&gt;\dfrac{36}{38} = \dfrac{18}{19}&lt;/math&gt;<br /> <br /> &lt;math&gt;18 + 19 = \boxed{037}&lt;/math&gt;<br /> <br /> ~First<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=q2oc7n-n6aA<br /> ~Shreyas S<br /> <br /> ==See Also:==<br /> <br /> {{AIME box|year=2018|n=II|num-b=5|num-a=7}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_8&diff=172982 2018 AIME II Problems/Problem 8 2022-03-27T16:16:35Z <p>First: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> A frog is positioned at the origin of the coordinate plane. From the point &lt;math&gt;(x, y)&lt;/math&gt;, the frog can jump to any of the points &lt;math&gt;(x + 1, y)&lt;/math&gt;, &lt;math&gt;(x + 2, y)&lt;/math&gt;, &lt;math&gt;(x, y + 1)&lt;/math&gt;, or &lt;math&gt;(x, y + 2)&lt;/math&gt;. Find the number of distinct sequences of jumps in which the frog begins at &lt;math&gt;(0, 0)&lt;/math&gt; and ends at &lt;math&gt;(4, 4)&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> We solve this problem by working backwards. Notice, the only points the frog can be on to jump to &lt;math&gt;(4,4)&lt;/math&gt; in one move are &lt;math&gt;(2,4),(3,4),(4,2),&lt;/math&gt; and &lt;math&gt;(4,3)&lt;/math&gt;. This applies to any other point, thus we can work our way from &lt;math&gt;(0,0)&lt;/math&gt; to &lt;math&gt;(4,4)&lt;/math&gt;, recording down the number of ways to get to each point recursively. <br /> <br /> &lt;math&gt;(0,0): 1&lt;/math&gt;<br /> <br /> &lt;math&gt;(1,0)=(0,1)=1&lt;/math&gt;<br /> <br /> &lt;math&gt;(2,0)=(0, 2)=2&lt;/math&gt;<br /> <br /> &lt;math&gt;(3,0)=(0, 3)=3&lt;/math&gt;<br /> <br /> &lt;math&gt;(4,0)=(0, 4)=5&lt;/math&gt;<br /> <br /> &lt;math&gt;(1,1)=2&lt;/math&gt;, &lt;math&gt;(1,2)=(2,1)=5&lt;/math&gt;, &lt;math&gt;(1,3)=(3,1)=10&lt;/math&gt;, &lt;math&gt;(1,4)=(4,1)= 20&lt;/math&gt;<br /> <br /> &lt;math&gt;(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71&lt;/math&gt;<br /> <br /> &lt;math&gt;(3,3)=84, (3,4)=(4,3)=207&lt;/math&gt;<br /> <br /> &lt;math&gt;(4,4)=2\cdot \left( (4,2)+(4,3)\right) = 2\cdot \left( 207+71\right)=2\cdot 278=\boxed{556}&lt;/math&gt;<br /> <br /> A diagram of the numbers:<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> add(shift(0,0)*grid(4,4));<br /> label((0,0), &quot;1&quot;, SW);<br /> label((1,0), &quot;1&quot;, SW);<br /> label((2,0), &quot;2&quot;, SW);<br /> label((3,0), &quot;3&quot;, SW);<br /> label((4,0), &quot;5&quot;, SW);<br /> <br /> label((0,1), &quot;1&quot;, SW);<br /> label((1,1), &quot;2&quot;, SW);<br /> label((2,1), &quot;5&quot;, SW);<br /> label((3,1), &quot;10&quot;, SW);<br /> label((4,1), &quot;20&quot;, SW);<br /> <br /> label((0,2), &quot;2&quot;, SW);<br /> label((1,2), &quot;5&quot;, SW);<br /> label((2,2), &quot;14&quot;, SW);<br /> label((3,2), &quot;32&quot;, SW);<br /> label((4,2), &quot;71&quot;, SW);<br /> <br /> label((0,3), &quot;3&quot;, SW);<br /> label((1,3), &quot;10&quot;, SW);<br /> label((2,3), &quot;32&quot;, SW);<br /> label((3,3), &quot;84&quot;, SW);<br /> label((4,3), &quot;207&quot;, SW);<br /> <br /> label((0,4), &quot;5&quot;, SW);<br /> label((1,4), &quot;20&quot;, SW);<br /> label((2,4), &quot;71&quot;, SW);<br /> label((3,4), &quot;207&quot;, SW);<br /> label((4,4), &quot;556&quot;, SW);<br /> &lt;/asy&gt;<br /> <br /> ~First<br /> <br /> ==Solution 2==<br /> We'll refer to the moves &lt;math&gt;(x + 1, y)&lt;/math&gt;, &lt;math&gt;(x + 2, y)&lt;/math&gt;, &lt;math&gt;(x, y + 1)&lt;/math&gt;, and &lt;math&gt;(x, y + 2)&lt;/math&gt; as &lt;math&gt;R_1&lt;/math&gt;, &lt;math&gt;R_2&lt;/math&gt;, &lt;math&gt;U_1&lt;/math&gt;, and &lt;math&gt;U_2&lt;/math&gt;, respectively. Then the possible sequences of moves that will take the frog from &lt;math&gt;(0,0)&lt;/math&gt; to &lt;math&gt;(4,4)&lt;/math&gt; are all the permutations of &lt;math&gt;U_1U_1U_1U_1R_1R_1R_1R_1&lt;/math&gt;, &lt;math&gt;U_2U_1U_1R_1R_1R_1R_1&lt;/math&gt;, &lt;math&gt;U_1U_1U_1U_1R_2R_1R_1&lt;/math&gt;, &lt;math&gt;U_2U_1U_1R_2R_1R_1&lt;/math&gt;, &lt;math&gt;U_2U_2R_1R_1R_1R_1&lt;/math&gt;, &lt;math&gt;U_1U_1U_1U_1R_2R_2&lt;/math&gt;, &lt;math&gt;U_2U_2R_2R_1R_1&lt;/math&gt;, &lt;math&gt;U_2U_1U_1R_2R_2&lt;/math&gt;, and &lt;math&gt;U_2U_2R_2R_2&lt;/math&gt;. We can reduce the number of cases using symmetry.<br /> <br /> Case 1: &lt;math&gt;U_1U_1U_1U_1R_1R_1R_1R_1&lt;/math&gt;<br /> <br /> There are &lt;math&gt;\frac{8!}{4!4!} = 70&lt;/math&gt; possibilities for this case.<br /> <br /> Case 2: &lt;math&gt;U_2U_1U_1R_1R_1R_1R_1&lt;/math&gt; or &lt;math&gt;U_1U_1U_1U_1R_2R_1R_1&lt;/math&gt;<br /> <br /> There are &lt;math&gt;2 \cdot \frac{7!}{4!2!} = 210&lt;/math&gt; possibilities for this case.<br /> <br /> Case 3: &lt;math&gt;U_2U_1U_1R_2R_1R_1&lt;/math&gt;<br /> <br /> There are &lt;math&gt;\frac{6!}{2!2!} = 180&lt;/math&gt; possibilities for this case.<br /> <br /> Case 4: &lt;math&gt;U_2U_2R_1R_1R_1R_1&lt;/math&gt; or &lt;math&gt;U_1U_1U_1U_1R_2R_2&lt;/math&gt;<br /> <br /> There are &lt;math&gt;2 \cdot \frac{6!}{2!4!} = 30&lt;/math&gt; possibilities for this case.<br /> <br /> Case 5: &lt;math&gt;U_2U_2R_2R_1R_1&lt;/math&gt; or &lt;math&gt;U_2U_1U_1R_2R_2&lt;/math&gt;<br /> <br /> There are &lt;math&gt;2 \cdot \frac{5!}{2!2!} = 60&lt;/math&gt; possibilities for this case.<br /> <br /> Case 6: &lt;math&gt;U_2U_2R_2R_2&lt;/math&gt;<br /> <br /> There are &lt;math&gt;\frac{4!}{2!2!} = 6&lt;/math&gt; possibilities for this case.<br /> <br /> Adding up all these cases gives us &lt;math&gt;70+210+180+30+60+6=\boxed{556}&lt;/math&gt; ways.<br /> <br /> ==Solution 3 (General Case)==<br /> <br /> Mark the total number of distinct sequences of jumps for the frog to reach the point &lt;math&gt;(x,y)&lt;/math&gt; as &lt;math&gt;\varphi (x,y)&lt;/math&gt;. Consider for each point &lt;math&gt;(x,y)&lt;/math&gt; in the first quadrant, there are only &lt;math&gt;4&lt;/math&gt; possible points in the first quadrant for frog to reach point &lt;math&gt;(x,y)&lt;/math&gt;, and these &lt;math&gt;4&lt;/math&gt; points are &lt;cmath&gt;(x-1,y); (x-2,y); (x,y-1); (x,y-2)&lt;/cmath&gt;. As a result, the way to count &lt;math&gt;\varphi (x,y)&lt;/math&gt; is &lt;cmath&gt;\varphi (x,y)=\varphi (x-1,y)+\varphi (x-2,y)+\varphi (x,y-1)+\varphi (x,y-2)&lt;/cmath&gt;<br /> <br /> Also, for special cases, &lt;cmath&gt;\varphi (0,y)=\varphi (0,y-1)+\varphi (0,y-2)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\varphi (x,0)=\varphi (x-1,0)+\varphi (x-2,0)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\varphi (x,1)=\varphi (x-1,1)+\varphi (x-2,1)+\varphi (x,0)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\varphi (1,y)=\varphi (1,y-1)+\varphi (1,y-2)+\varphi (0,y)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\varphi (1,1)=\varphi (1,0)+\varphi (0,1)&lt;/cmath&gt;<br /> <br /> Start with &lt;math&gt;\varphi (0,0)=1&lt;/math&gt;, use this method and draw the figure below, we can finally get &lt;cmath&gt;\varphi (4,4)=556&lt;/cmath&gt; (In order to make the LaTeX thing more beautiful to look at, I put &lt;math&gt;0&lt;/math&gt; to make every number &lt;math&gt;3&lt;/math&gt; digits)<br /> <br /> &lt;cmath&gt;005-020-071-207-\boxed{556}&lt;/cmath&gt; &lt;cmath&gt;003-010-032-084-207&lt;/cmath&gt; &lt;cmath&gt;002-005-014-032-071&lt;/cmath&gt; &lt;cmath&gt;001-002-005-010-020&lt;/cmath&gt; &lt;cmath&gt;001-001-002-003-005&lt;/cmath&gt;<br /> <br /> So the total number of distinct sequences of jumps for the frog to reach &lt;math&gt;(4,4)&lt;/math&gt; is &lt;math&gt;\boxed {556}&lt;/math&gt;.<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Solution 4 (Casework)==<br /> <br /> Casework Solution:<br /> x-distribution: 1-1-1-1 (1 way to order)<br /> y-distribution: 1-1-1-1 (1 way to order)<br /> &lt;math&gt;\dbinom{8}{4} = 70&lt;/math&gt; ways total<br /> <br /> x-distribution: 1-1-1-1 (1 way to order)<br /> y-distribution: 1-1-2 (3 ways to order)<br /> &lt;math&gt;\dbinom{7}{3} \times 3= 105&lt;/math&gt; ways total<br /> <br /> x-distribution: 1-1-1-1 (1 way to order)<br /> y-distribution: 2-2 (1 way to order)<br /> &lt;math&gt;\dbinom{6}{4} = 15&lt;/math&gt; ways total<br /> <br /> x-distribution: 1-1-2 (3 ways to order)<br /> y-distribution: 1-1-1-1 (1 way to order)<br /> &lt;math&gt;\dbinom{7}{3} \times 3= 105&lt;/math&gt; ways total<br /> <br /> x-distribution: 1-1-2 (3 ways to order)<br /> y-distribution: 1-1-2 (3 ways to order)<br /> &lt;math&gt;\dbinom{6}{3} \times 9= 180&lt;/math&gt; ways total<br /> <br /> x-distribution: 1-1-2 (3 ways to order)<br /> y-distribution: 2-2 (1 way to order)<br /> &lt;math&gt;\dbinom{5}{3} \times 3 = 30&lt;/math&gt; ways total<br /> <br /> x-distribution: 2-2 (1 way to order)<br /> y-distribution: 1-1-1-1 (1 way to order)<br /> &lt;math&gt;\dbinom{6}{4} = 15&lt;/math&gt; ways total<br /> <br /> x-distribution: 2-2 (1 way to order)<br /> y-distribution: 1-1-2 (3 ways to order)<br /> &lt;math&gt;\dbinom{5}{3} \times 3 = 30&lt;/math&gt; ways total<br /> <br /> x-distribution: 2-2 (1 way to order)<br /> y-distribution: 2-2 (1 way to order)<br /> &lt;math&gt;\dbinom{4}{2} = 6&lt;/math&gt; ways total<br /> <br /> &lt;math&gt;6+30+15+105+180+70+30+15+105=\boxed{556}&lt;/math&gt;<br /> -fidgetboss_4000<br /> <br /> ==Video Solution==<br /> <br /> On The Spot STEM : <br /> https://www.youtube.com/watch?v=v2fo3CaAhmM<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=II|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_13&diff=172981 2018 AIME II Problems/Problem 13 2022-03-27T16:12:31Z <p>First: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Misha rolls a standard, fair six-sided die until she rolls &lt;math&gt;1-2-3&lt;/math&gt; in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is &lt;math&gt;\dfrac{m}{n}&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution 0==<br /> Let &lt;math&gt;P_i&lt;/math&gt; be the probability of getting it in an &lt;math&gt;n&lt;/math&gt; number of times. Let &lt;math&gt;x = P_3+P_5+...=1-(P_4+P_6+..)&lt;/math&gt;. It's easy to see that<br /> &lt;cmath&gt;P_{2n+1}=P_{2n}-\frac{P_{2n-2}}{6^3}&lt;/cmath&gt;<br /> for all positive integers &lt;math&gt;n \ge 2&lt;/math&gt;. Summing for &lt;math&gt;n=2,4,...&lt;/math&gt; gives<br /> &lt;cmath&gt;x-\frac{1}{6^3} = \frac{6^3-1}{6^3}(1-x)&lt;/cmath&gt;<br /> &lt;cmath&gt; \implies x = \frac{216}{431} &lt;/cmath&gt;<br /> ==Solution 1==<br /> Let &lt;math&gt;P_{odd}=\frac{m}{n}&lt;/math&gt;, with the subscript indicating an odd number of rolls. Then &lt;math&gt;P_{even}=1-\frac{m}{n}&lt;/math&gt;.<br /> <br /> The ratio of &lt;math&gt;\frac{P_{odd}}{P_{even}}&lt;/math&gt; is just &lt;math&gt;\frac{P_{odd}}{1-P_{odd}}&lt;/math&gt;.<br /> <br /> We see that &lt;math&gt;P_{odd}&lt;/math&gt; is the sum of &lt;math&gt;P_3&lt;/math&gt;,&lt;math&gt;P_5&lt;/math&gt;,&lt;math&gt;P_7&lt;/math&gt;,... , while &lt;math&gt;P_{even}&lt;/math&gt; is the sum of &lt;math&gt;P_4&lt;/math&gt;, &lt;math&gt;P_6&lt;/math&gt;, &lt;math&gt;P_8&lt;/math&gt;,... .<br /> <br /> &lt;math&gt;P_3&lt;/math&gt;, the probability of getting rolls of 1-2-3 in exactly 3 rolls, is obviously &lt;math&gt;\frac{1}{216}&lt;/math&gt;.<br /> <br /> We set this probability of &lt;math&gt;P_3&lt;/math&gt; aside, meaning we totally remove the chance of getting 1-2-3 in 3 rolls. Now the ratio of &lt;math&gt;P_4+P_6+P_8+...&lt;/math&gt; to &lt;math&gt;P_5+P_7+P_9+...&lt;/math&gt; should be equal to the ratio of &lt;math&gt;\frac{P_{odd}}{P_{even}}&lt;/math&gt;, because in this case the 1st roll no longer matters, so we can disregard the very existence of it in counting how many times of rolls, and thus, 4 rolls, 6 rolls, 8 rolls... would become an odd number of rolls (while 5 rolls, 7 rolls, 9 rolls... would become even number of rolls).<br /> <br /> Notice &lt;math&gt;P_4+P_6+P_8+...=P_{even}&lt;/math&gt;, and also &lt;math&gt;P_5+P_7+P_9+...=P_{odd}-P_3=P_{odd}-\frac{1}{216}&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;\frac{P_{even}}{P_{odd}-\frac{1}{216}}=\frac{P_{odd}}{P_{even}}&lt;/math&gt;.<br /> <br /> Finally, we get &lt;math&gt;P_{odd}=\frac{m}{n}=\frac{216}{431}&lt;/math&gt;. <br /> Therefore, &lt;math&gt;m+n = \boxed{647}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Call the probability you win on a certain toss &lt;math&gt;f_n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is the toss number.<br /> Obviously, since the sequence has length 3, &lt;math&gt;f_1=0&lt;/math&gt; and &lt;math&gt;f_2=0&lt;/math&gt;.<br /> Additionally, &lt;math&gt;f_3=\left(\frac{1}{6}\right)^3&lt;/math&gt;. We can call this value &lt;math&gt;x&lt;/math&gt;, to keep our further equations looking clean.<br /> We can now write our general form for &lt;math&gt;f&lt;/math&gt; as &lt;math&gt;f_n=x(1-\sum_{i=1}^{n-3}f_i)&lt;/math&gt;. This factors the probability of the last 3 rolls being 1-2-3, and the important probability that the sequence has not been rolled in the past (because then the game would already be over).<br /> Note that &lt;math&gt;\sum_{i=1}^{\infty}f_i=1&lt;/math&gt; since you'll win at some point.<br /> An intermediate step here is figuring out &lt;math&gt;f_n-f_{n+1}&lt;/math&gt;. This is equal to &lt;math&gt;x(1-\sum_{i=1}^{n-3}f_i)-x(1-\sum_{i=1}^{n-2}f_i)=x(\sum_{i=1}^{n-2}f_i-\sum_{i=1}^{n-3}f_i)=xf_{n-2}&lt;/math&gt;.<br /> Adding up all the differences, i.e. &lt;math&gt;\sum_{i=2}^{\infty}(f_{2n-1}-f_{2n})&lt;/math&gt; will give us the amount by which the odds probability exceeds the even probability. Since they sum to 1, that means the odds probability will be half of the difference above one-half. Subbing in our earlier result from the intermediate step, the odd probability is equal to &lt;math&gt;\frac{1}{2}+\frac{1}{2}x\sum_{i=2}^{\infty}f_{2n-3}&lt;/math&gt;.<br /> Another way to find the odd probability is simply summing it up, which turns out to be &lt;math&gt;\sum_{i=1}^{\infty}f_{2n-1}&lt;/math&gt;. Note the infinite sums in both expressions are equal; let's call it &lt;math&gt;P&lt;/math&gt;. Equating them gives &lt;math&gt;\frac{1}{2}+\frac{1}{2}xP=P&lt;/math&gt;, or &lt;math&gt;P=\frac{1}{2-x}&lt;/math&gt;.<br /> Finally, substituting &lt;math&gt;x=\frac{1}{216}&lt;/math&gt;, we find that &lt;math&gt;P=\frac{216}{431}&lt;/math&gt;, giving us a final answer of &lt;math&gt;216 + 431 = \boxed{647}&lt;/math&gt;.<br /> <br /> ~First<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; be the number of strings of length &lt;math&gt;n&lt;/math&gt; containing the digits &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;6&lt;/math&gt; that do not contain the string &lt;math&gt;123&lt;/math&gt;. Then we have &lt;math&gt;S(n) = 6 \cdot S(n-1) - S(n-3)&lt;/math&gt; because we can add any digit to end of a string with length &lt;math&gt;n-1&lt;/math&gt; but we have to subtract all the strings that end in &lt;math&gt;123&lt;/math&gt;. We rewrite this as<br /> &lt;cmath&gt;\begin{align*}<br /> S(n) &amp;= 6 \cdot S(n-1) - S(n-3) \\ &amp;= 6 \cdot (6 \cdot S(n-2) - S(n-4)) - (6 \cdot (S(n-4) - S(n-6)) \\ &amp;= 36 \cdot S(n-2) - 12 \cdot S(n-4) + S(n-6)<br /> \end{align*}&lt;/cmath&gt; We wish to compute &lt;math&gt;P=\sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}}&lt;/math&gt; since the last three rolls are &lt;math&gt;123&lt;/math&gt; for the game to end. Summing over the recursion, we obtain<br /> &lt;cmath&gt;\sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}} =36 \cdot \sum_{n=0}^\infty \frac{S(2n-2)}{6^{2n+3}} - 12 \cdot \sum_{n=0}^\infty \frac{S(2n-4)}{6^{2n+3}}+ \sum_{n=0}^\infty \frac{S(2n-6)}{6^{2n+3}} &lt;/cmath&gt;Now shift the indices so that the inside term is the same:<br /> &lt;cmath&gt;\begin{align*}<br /> \sum_{n=3}^\infty \frac{S(2n)}{6^{2n+3}} &amp;= \frac{36}{6^2} \cdot \sum_{n=2}^\infty \frac{S(2n)}{6^{2n+3}} - \frac{12}{6^4} \cdot \sum_{n=1}^\infty \frac{S(2n)}{6^{2n+3}} + \frac{1}{6^6} \cdot \sum_{n=0}^\infty \frac{S(2n)}{6^{2n+3}} \\ \left(P - \frac{S(0)}{6^3} - \frac{S(2)}{6^5} -\frac{S(4)}{6^7} \right) &amp;= \frac{36}{6^2} \cdot \left( P - \frac{S(0)}{6^3} - \frac{S(2)}{6^5}\right) - \frac{12}{6^4} \cdot \left( P - \frac{S(0)}{6^3} \right) + \frac{1}{6^6} \cdot P <br /> \end{align*}&lt;/cmath&gt;Note that &lt;math&gt;S(0) = 1, S(2) = 36&lt;/math&gt; and &lt;math&gt;S(4) = 6^4 - 2 \cdot 6 = 1284&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\begin{align*}<br /> \left(P - \frac{1}{6^3} - \frac{36}{6^5} -\frac{1284}{6^7} \right) = \frac{36}{6^2} \cdot \left( P - \frac{1}{6^3} - \frac{36}{6^5}\right) - \frac{12}{6^4} \cdot \left( P - \frac{1}{6^3} \right) + \frac{1}{6^6} \cdot P <br /> \end{align*}&lt;/cmath&gt;Solving for &lt;math&gt;P&lt;/math&gt;, we obtain &lt;math&gt;P = \frac{216}{431} \implies m+n = \boxed{647}&lt;/math&gt;. <br /> <br /> -Vfire<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;A=\frac{1}{6} \begin{bmatrix}<br /> 5 &amp; 1 &amp; 0 &amp; 0 \\<br /> 4 &amp; 1 &amp; 1 &amp; 0 \\<br /> 4 &amp; 1 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> \end{bmatrix}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is a transition matrix for the prefix of 1-2-3 matched so far. The state corresponding to a complete match has no outgoing probability mass. The probability that we roll the dice exactly &lt;math&gt;k&lt;/math&gt; times is &lt;math&gt;(A^k)_{1,4}&lt;/math&gt;. Thus the probability that we roll the dice an odd number of times is &lt;math&gt;1-\left(\sum_{k=0}^\infty A^{2k}\right)_{1,4} = 1-\left((I - A^2)^{-1}\right)_{1,4} = \frac{216}{431}&lt;/math&gt;. Thus the answer is &lt;math&gt;216+431=\boxed{647}&lt;/math&gt;.<br /> ==Solution 5 quick cheat ==<br /> Consider it as a contest of Odd and Even. Let &lt;math&gt;P_o&lt;/math&gt; and &lt;math&gt;P_e&lt;/math&gt; be probability that Odd and Even wins, respectively. If we consider every 3 rolls as an atomic action, then we can have a simple solution. If the rolls is 1-2-3, Odd wins; otherwise, Odd and Even switch the odds of winning. Therefore, we have<br /> &lt;cmath&gt; P_o = \frac{1}{216} + \frac{215}{216}P_e&lt;/cmath&gt;<br /> Plug in &lt;math&gt;P_e = 1 - P_o&lt;/math&gt; and we can easily solve for &lt;math&gt;Po=\frac{216}{431}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\boxed{647}&lt;/math&gt;.<br /> <br /> Of course this is not a rigorous solution. I think it works because the requirement is a strict sequence of pure random events.<br /> <br /> -Mathdummy<br /> <br /> ==Solution 6 Elementary Probability ==<br /> Consider it a contest for Odd or Even to win. Let &lt;math&gt;P_o&lt;/math&gt;, &lt;math&gt;P_e&lt;/math&gt; be the winning probabilities respectively. We call Odd is &quot;in position&quot; when a new sequence of 1-2-3 starts at odd position, and likewise, call Even is &quot;in position&quot; when a new sequence starts at even position. Now consider the situation when the first roll is 1. The conditional probability for Odd or Even to eventually win out depends on whose is in position. So let's denote by &lt;math&gt;P_o(1), P_e(1)&lt;/math&gt; the probabilities of Odd and Even winning out, respectively, both when Odd is in position. Remember that the probabilities simply switch if Even is in position. Similarly, after 1-2 is rolled, we denote by &lt;math&gt;P_o(2), P_e(2)&lt;/math&gt; the conditional probabilities of Odd and Even winning out, when Odd is in position.<br /> <br /> Consider the first roll. If it's not a 1, the sequence restart, but Even is now in position; if it's a 1, then Odd's winning probability becomes &lt;math&gt;P_o(1)&lt;/math&gt;. So,<br /> &lt;cmath&gt;P_o = \frac{1}{6}P_o(1) + \frac{5}{6}P_e&lt;/cmath&gt;<br /> In the next roll, there are 3 outcomes. If the roll is 2, then Odd's winning probability becomes &lt;math&gt;P_o(2)&lt;/math&gt;; if the roll is 1, then we stay in the sequence, but Even is now in position, so the probability of Odd winning now becomes P_e(1); if the rolls is any other number, then the sequence restarts, and Odd is still in position. So,<br /> &lt;cmath&gt; P_o(1) = \frac{1}{6}P_o(2) + \frac{1}{6}P_e(1) + \frac{4}{6}P_o&lt;/cmath&gt;<br /> In the next roll after a 1-2 sequence, there are 3 outcomes. If the roll is a 3, Odd wins; if it's a 1, we go back to the state when 1 is just rolled, and Odd is in position; if it's any other number, then the sequence restarts, and Even is in position. So,<br /> &lt;cmath&gt; P_o(2) = \frac{1}{6} + \frac{1}{6}P_o(1) + \frac{4}{6}P_e&lt;/cmath&gt;<br /> <br /> Plug in &lt;math&gt;P_e = 1-P_o&lt;/math&gt; and &lt;math&gt;P_e(1) = 1 - P_o(1)&lt;/math&gt;, we have a 3-equation linear system which is not hard to solve. The final answer is &lt;math&gt;Po=\frac{216}{431}&lt;/math&gt;. &lt;math&gt;\boxed{647}&lt;/math&gt;.<br /> <br /> -Mathdummy<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=II|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_15&diff=172951 2015 AIME II Problems/Problem 15 2022-03-27T02:59:36Z <p>First: </p> <hr /> <div>==Problem==<br /> <br /> Circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; have radii &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, respectively, and are externally tangent at point &lt;math&gt;A&lt;/math&gt;. Point &lt;math&gt;B&lt;/math&gt; is on &lt;math&gt;\mathcal{P}&lt;/math&gt; and point &lt;math&gt;C&lt;/math&gt; is on &lt;math&gt;\mathcal{Q}&lt;/math&gt; so that line &lt;math&gt;BC&lt;/math&gt; is a common external tangent of the two circles. A line &lt;math&gt;\ell&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt;\mathcal{P}&lt;/math&gt; again at &lt;math&gt;D&lt;/math&gt; and intersects &lt;math&gt;\mathcal{Q}&lt;/math&gt; again at &lt;math&gt;E&lt;/math&gt;. Points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; lie on the same side of &lt;math&gt;\ell&lt;/math&gt;, and the areas of &lt;math&gt;\triangle DBA&lt;/math&gt; and &lt;math&gt;\triangle ACE&lt;/math&gt; are equal. This common area is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> import cse5;<br /> pathpen=black; pointpen=black;<br /> size(6cm);<br /> <br /> pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689);<br /> <br /> filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7));<br /> filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7));<br /> D(CR((0,1),1)); D(CR((4,4),4,150,390));<br /> D(L(MP(&quot;D&quot;,D(D),N),MP(&quot;A&quot;,D((0.8,1.6)),NE),1,5.5));<br /> D((-1.2,0)--MP(&quot;B&quot;,D((0,0)),S)--MP(&quot;C&quot;,D((4,0)),S)--(8,0));<br /> D(MP(&quot;E&quot;,E,N));<br /> &lt;/asy&gt;<br /> <br /> ==Hint==<br /> &lt;math&gt;[ABC] = \frac{1}{2}ab \text{sin} C&lt;/math&gt; is your friend for a quick solve. If you know about homotheties, go ahead, but you'll still need to do quite a bit of computation. If you're completely lost and you have a lot of time left in your mocking of this AIME, go ahead and use analytic geometry.<br /> <br /> ==Solution 1 (guys trig is fast)==<br /> Let &lt;math&gt;M&lt;/math&gt; be the intersection of &lt;math&gt;\overline{BC}&lt;/math&gt; and the common internal tangent of &lt;math&gt;\mathcal P&lt;/math&gt; and &lt;math&gt;\mathcal Q.&lt;/math&gt; We claim that &lt;math&gt;M&lt;/math&gt; is the circumcenter of right &lt;math&gt;\triangle{ABC}.&lt;/math&gt; Indeed, we have &lt;math&gt;AM = BM&lt;/math&gt; and &lt;math&gt;BM = CM&lt;/math&gt; by equal tangents to circles, and since &lt;math&gt;BM = CM, M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BC},&lt;/math&gt; implying that &lt;math&gt;\angle{BAC} = 90.&lt;/math&gt; Now draw &lt;math&gt;\overline{PA}, \overline{PB}, \overline{PM},&lt;/math&gt; where &lt;math&gt;P&lt;/math&gt; is the center of circle &lt;math&gt;\mathcal P.&lt;/math&gt; Quadrilateral &lt;math&gt;PAMB&lt;/math&gt; is cyclic, and by Pythagorean Theorem &lt;math&gt;PM = \sqrt{5},&lt;/math&gt; so by Ptolemy on &lt;math&gt;PAMB&lt;/math&gt; we have &lt;cmath&gt;AB \sqrt{5} = 2 \cdot 1 + 2 \cdot 1 = 4 \iff AB = \dfrac{4 \sqrt{5}}{5}.&lt;/cmath&gt; Do the same thing on cyclic quadrilateral &lt;math&gt;QAMC&lt;/math&gt; (where &lt;math&gt;Q&lt;/math&gt; is the center of circle &lt;math&gt;\mathcal Q&lt;/math&gt; and get &lt;math&gt;AC = \frac{8 \sqrt{5}}{5}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;\angle A = \angle{DAB}.&lt;/math&gt; By Law of Sines, &lt;math&gt;BD = 2R \sin A = 2 \sin A.&lt;/math&gt; Note that &lt;math&gt;\angle{D} = \angle{ABC}&lt;/math&gt; from inscribed angles, so <br /> &lt;cmath&gt;\begin{align*}<br /> [ABD] &amp;= \dfrac{1}{2} BD \cdot AB \cdot \sin{\angle B} \\<br /> &amp;= \dfrac{1}{2} \cdot \dfrac{4 \sqrt{5}}{5} \cdot 2 \sin A \sin{\left(180 - \angle A - \angle D\right)} \\<br /> &amp;= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \sin{\left(\angle A + \angle D\right)} \\<br /> &amp;= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos D + \cos A \sin D\right) \\<br /> &amp;= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos{\angle{ABC}} + \cos A \sin{\angle{ABC}}\right) \\<br /> &amp;= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\dfrac{\sqrt{5} \sin A}{5} + \dfrac{2 \sqrt{5} \cos A}{5}\right) \\<br /> &amp;= \dfrac{4}{5} \cdot \sin A \left(\sin A + 2 \cos A\right)<br /> \end{align*}&lt;/cmath&gt; after angle addition identity.<br /> <br /> Similarly, &lt;math&gt;\angle{EAC} = 90 - \angle A,&lt;/math&gt; and by Law of Sines &lt;math&gt;CE = 8 \sin{\angle{EAC}} = 8 \cos A.&lt;/math&gt; Note that &lt;math&gt;\angle{E} = \angle{ACB}&lt;/math&gt; from inscribed angles, so<br /> &lt;cmath&gt;\begin{align*}<br /> [ACE] &amp;= \dfrac{1}{2} AC \cdot CE \sin{\angle C} \\<br /> &amp;= \dfrac{1}{2} \cdot \dfrac{8 \sqrt{5}}{5} \cdot 8 \cos A \sin{\left[180 - \left(90 - \angle A\right) - \angle E\right]} \\<br /> &amp;= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \sin{\left[\left(90 - \angle A\right) + \angle{ACB}\right]} \\<br /> &amp;= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \left(\dfrac{2 \sqrt{5} \cos A}{5} + \dfrac{\sqrt{5} \sin A}{5}\right) \\<br /> &amp;= \dfrac{32}{5} \cdot \cos A \left(\sin A + 2 \cos A\right)<br /> \end{align*}&lt;/cmath&gt; after angle addition identity.<br /> Setting the two areas equal, we get &lt;cmath&gt;\tan A = \frac{\sin A}{\cos A} = 8 \iff \sin A = \frac{8}{\sqrt{65}}, \cos A = \frac{1}{\sqrt{65}}&lt;/cmath&gt; after Pythagorean Identity. Now plug back in and the common area is &lt;math&gt;\frac{64}{65} \iff \boxed{129}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> &lt;asy&gt;<br /> unitsize(35);<br /> draw(Circle((-1,0),1));<br /> draw(Circle((4,0),4));<br /> pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;<br /> A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);<br /> label(&quot;$A$&quot;,A,NE);label(&quot;$O_1$&quot;,O_1,NE);label(&quot;$O_2$&quot;,O_2,NE);label(&quot;$B$&quot;,B,SW);label(&quot;$C$&quot;,C,SW);label(&quot;$D$&quot;,D,NE);label(&quot;$E$&quot;,E,NE);label(&quot;$N$&quot;,N,W);label(&quot;$K$&quot;,(-24/15,0.2));label(&quot;$L$&quot;,(24/15,0.2));label(&quot;$n$&quot;,(-0.8,-0.12));label(&quot;$p$&quot;,((29/15,-48/15)));label(&quot;$\mathcal{P}$&quot;,(-1.6,1.1));label(&quot;$\mathcal{Q}$&quot;,(6,4));<br /> draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed);<br /> dot(O_1);dot(O_2);<br /> draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X);<br /> <br /> &lt;/asy&gt;<br /> <br /> Call &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; the centers of circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt;, respectively, and extend &lt;math&gt;CB&lt;/math&gt; and &lt;math&gt;O_2O_1&lt;/math&gt; to meet at point &lt;math&gt;N&lt;/math&gt;. Call &lt;math&gt;K&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt; the feet of the altitudes from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;O_1N&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;O_2N&lt;/math&gt;, respectively. Using the fact that &lt;math&gt;\triangle{O_1BN} \sim \triangle{O_2CN}&lt;/math&gt; and setting &lt;math&gt;NO_1 = k&lt;/math&gt;, we have that &lt;math&gt;\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}&lt;/math&gt;. We can do some more length chasing using triangles similar to &lt;math&gt;O_1BN&lt;/math&gt; to get that &lt;math&gt;AK = AL = \frac{24}{15}&lt;/math&gt;, &lt;math&gt;BK = \frac{12}{15}&lt;/math&gt;, and &lt;math&gt;CL = \frac{48}{15}&lt;/math&gt;. Now, consider the circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; on the coordinate plane, where &lt;math&gt;A&lt;/math&gt; is the origin. If the line &lt;math&gt;\ell&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt;\mathcal{P}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt; then &lt;math&gt;4 \cdot DA = AE&lt;/math&gt;. To verify this, notice that &lt;math&gt;\triangle{AO_1D} \sim \triangle{EO_2A}&lt;/math&gt; from the fact that both triangles are isosceles with &lt;math&gt;\angle{O_1AD} \cong \angle{O_2AE}&lt;/math&gt;, which are corresponding angles. Since &lt;math&gt;O_2A = 4\cdot O_1A&lt;/math&gt;, we can conclude that &lt;math&gt;4 \cdot DA = AE&lt;/math&gt;.<br /> <br /> <br /> Hence, we need to find the slope &lt;math&gt;m&lt;/math&gt; of line &lt;math&gt;\ell&lt;/math&gt; such that the perpendicular distance &lt;math&gt;n&lt;/math&gt; from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; is four times the perpendicular distance &lt;math&gt;p&lt;/math&gt; from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;AE&lt;/math&gt;. This will mean that the product of the bases and heights of triangles &lt;math&gt;ACE&lt;/math&gt; and &lt;math&gt;DBA&lt;/math&gt; will be equal, which in turn means that their areas will be equal. Let the line &lt;math&gt;\ell&lt;/math&gt; have the equation &lt;math&gt;y = -mx \implies mx + y = 0&lt;/math&gt;, and let &lt;math&gt;m&lt;/math&gt; be a positive real number so that the negative slope of &lt;math&gt;\ell&lt;/math&gt; is preserved. Setting &lt;math&gt;A = (0,0)&lt;/math&gt;, the coordinates of &lt;math&gt;B&lt;/math&gt; are &lt;math&gt;(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)&lt;/math&gt;, and the coordinates of &lt;math&gt;C&lt;/math&gt; are &lt;math&gt;(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)&lt;/math&gt;. Using the point-to-line distance formula and the condition &lt;math&gt;n = 4p&lt;/math&gt;, we have &lt;cmath&gt;\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}&lt;/cmath&gt; &lt;cmath&gt;\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.&lt;/cmath&gt; If &lt;math&gt;m &gt; 2&lt;/math&gt;, then clearly &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; would not lie on the same side of &lt;math&gt;\ell&lt;/math&gt;. Thus since &lt;math&gt;m &gt; 0&lt;/math&gt;, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have &lt;cmath&gt;\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.&lt;/cmath&gt; Thus, the equation of &lt;math&gt;\ell&lt;/math&gt; is &lt;math&gt;y = -\frac{3}{2}x&lt;/math&gt;.<br /> <br /> <br /> Then we can find the coordinates of &lt;math&gt;D&lt;/math&gt; by finding the point &lt;math&gt;(x,y)&lt;/math&gt; other than &lt;math&gt;A = (0,0)&lt;/math&gt; where the circle &lt;math&gt;\mathcal{P}&lt;/math&gt; intersects &lt;math&gt;\ell&lt;/math&gt;. &lt;math&gt;\mathcal{P}&lt;/math&gt; can be represented with the equation &lt;math&gt;(x + 1)^2 + y^2 = 1&lt;/math&gt;, and substituting &lt;math&gt;y = -\frac{3}{2}x&lt;/math&gt; into this equation yields &lt;math&gt;x = 0, -\frac{8}{13}&lt;/math&gt; as solutions. Discarding &lt;math&gt;x = 0&lt;/math&gt;, the &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;D&lt;/math&gt; is &lt;math&gt;-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}&lt;/math&gt;. The distance from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; is then &lt;math&gt;\frac{4}{\sqrt{13}}.&lt;/math&gt; The perpendicular distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; or the height of &lt;math&gt;\triangle{DBA}&lt;/math&gt; is &lt;math&gt;\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.&lt;/math&gt; Finally, the common area is &lt;math&gt;\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}&lt;/math&gt;, and &lt;math&gt;m + n = 64 + 65 = \boxed{129}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> By [[homothety]], we deduce that &lt;math&gt;AE = 4 AD&lt;/math&gt;. (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt;.) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt; is four times that from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt;. Let the distance from &lt;math&gt;C&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt; and the distance from &lt;math&gt;B&lt;/math&gt; be &lt;math&gt;4x&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the centers of their respective circles. Then dropping a perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; creates a &lt;math&gt;3-4-5&lt;/math&gt; right triangle, from which &lt;math&gt;BC = 4&lt;/math&gt; and, if &lt;math&gt;\alpha = \angle{AQC}&lt;/math&gt;, that &lt;math&gt;\cos \alpha = \dfrac{3}{5}&lt;/math&gt;. Then &lt;math&gt;\angle{BPA} = 180^\circ - \alpha&lt;/math&gt;, and the Law of Cosines on triangles &lt;math&gt;APB&lt;/math&gt; and &lt;math&gt;AQC&lt;/math&gt; gives &lt;math&gt;AB = \dfrac{4}{\sqrt{5}}&lt;/math&gt; and &lt;math&gt;AC = \dfrac{8}{\sqrt{5}}.&lt;/math&gt;<br /> <br /> Now, using the Pythagorean Theorem to express the length of the projection of &lt;math&gt;BC&lt;/math&gt; onto line &lt;math&gt;l&lt;/math&gt; gives<br /> &lt;cmath&gt;\sqrt{\frac{16}{5} - 16x^2} + \sqrt{\frac{64}{5} - x^2} = \sqrt{16 - 9x^2}.&lt;/cmath&gt;<br /> Squaring and simplifying gives<br /> &lt;cmath&gt;\sqrt{\left(\frac{1}{5} - x^2\right)\left(\frac{64}{5} - x^2\right)} = x^2,&lt;/cmath&gt;<br /> and squaring and solving gives &lt;math&gt;x = \dfrac{8}{5\sqrt{13}}.&lt;/math&gt;<br /> <br /> By the Law of Sines on triangle &lt;math&gt;ABD&lt;/math&gt;, we have<br /> &lt;cmath&gt;\frac{BD}{\sin A} = 2.&lt;/cmath&gt;<br /> But we know &lt;math&gt;\sin A = \dfrac{4x}{AB}&lt;/math&gt;, and so a small computation gives &lt;math&gt;BD = \dfrac{16}{\sqrt{65}}.&lt;/math&gt; The Pythagorean Theorem now gives<br /> &lt;cmath&gt;AD = \sqrt{BD^2 - (4x)^2} + \sqrt{AB^2 - (4x)^2} = \frac{4}{\sqrt{13}},&lt;/cmath&gt;<br /> and so the common area is &lt;math&gt;\dfrac{1}{2} \cdot \frac{4}{\sqrt{13}} \cdot \frac{32}{5\sqrt{13}} = \frac{64}{65}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{129}.&lt;/math&gt;<br /> <br /> ==Alternate Path to x==<br /> Call the intersection of lines &lt;math&gt;l&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; &lt;math&gt;E&lt;/math&gt;.You can use similar triangles to find that the distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt; is four times the distance from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;. Then draw a perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call the point &lt;math&gt;F&lt;/math&gt;. &lt;math&gt;AF = \frac{8}{5}&lt;/math&gt; and &lt;math&gt;FE = FC + CE = \frac{16}{5} + \frac{4}{3} = \frac{68}{15}&lt;/math&gt;, so by the Pythagorean Theorem, &lt;math&gt;AE = \dfrac{20\sqrt{13}}{5}&lt;/math&gt;. You can now use similar triangles to find that &lt;math&gt;x = \dfrac{8}{5\sqrt{13}}&lt;/math&gt; and continue on like in solution 2.<br /> <br /> ==Solution 4==<br /> &lt;math&gt;DE&lt;/math&gt; goes through &lt;math&gt;A&lt;/math&gt;, the point of tangency of both circles. So &lt;math&gt;DE&lt;/math&gt; intercepts equal arcs in circle &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;: [[homothety]]. Hence, &lt;math&gt;AE=4AD&lt;/math&gt;. We will use such similarity later.<br /> <br /> The diagonal distance between the centers of the circles is &lt;math&gt;4+1=5&lt;/math&gt;. The difference in heights is &lt;math&gt;4-1=3&lt;/math&gt;. So &lt;math&gt;BC=\sqrt{5^2-3^2}=4&lt;/math&gt;.<br /> <br /> The triangle connecting the centers with a side parallel to &lt;math&gt;BC&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; right triangle. Since &lt;math&gt;O_PA=1&lt;/math&gt;, the height of &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;1+3/5=8/5&lt;/math&gt;. Drop an altitude from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call it &lt;math&gt;I&lt;/math&gt;: &lt;math&gt;IB=4/5&lt;/math&gt; and &lt;math&gt;IC=4-4/5=32/5&lt;/math&gt;. Since right &lt;math&gt;\triangle AIB\sim\triangle CIB&lt;/math&gt;, &lt;math&gt;ABC&lt;/math&gt; is a right triangle also; &lt;math&gt;IB:IA:IC&lt;/math&gt; form a geometric progression &lt;math&gt;\times 2&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;BA&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;G&lt;/math&gt; on the other side of &lt;math&gt;\circ Q&lt;/math&gt;. By [[homothety]], &lt;math&gt;\triangle DAB\sim\triangle EAG&lt;/math&gt;. By angle chasing &lt;math&gt;\triangle DAB&lt;/math&gt; through right triangle &lt;math&gt;ABC&lt;/math&gt;, we deduce that &lt;math&gt;\angle CEG&lt;/math&gt; is a right angle. Since &lt;math&gt;ACEG&lt;/math&gt; is cyclic, &lt;math&gt;\angle GAC&lt;/math&gt; is also right. So &lt;math&gt;CG&lt;/math&gt; is a diameter of &lt;math&gt;\circ G&lt;/math&gt;. Because of this, &lt;math&gt;CG \perp BC&lt;/math&gt;, the tangent line. &lt;math&gt;\triangle BCG&lt;/math&gt; is right and &lt;math&gt;\triangle BCG\sim\triangle ABC\sim\triangle CAG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;AC=\sqrt{(8/5)^2+(16/5)^2}=8\sqrt{5}/5&lt;/math&gt; so &lt;math&gt;AG=2AC=16\sqrt{5}/5&lt;/math&gt; and &lt;math&gt;[\triangle CAG]=64/5&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;[\triangle DAB]=[\triangle ACE]&lt;/math&gt;, the common area is &lt;math&gt;[ACEG]/17&lt;/math&gt;. &lt;math&gt;16[\triangle DAB]=[\triangle GAE]&lt;/math&gt; because the triangles are similar with a ratio of &lt;math&gt;1:4&lt;/math&gt;. So we only need to find &lt;math&gt;[\triangle CEG]&lt;/math&gt; now.<br /> <br /> <br /> Extend &lt;math&gt;DE&lt;/math&gt; through &lt;math&gt;E&lt;/math&gt; to intersect the tangent at &lt;math&gt;F&lt;/math&gt;. Because &lt;math&gt;4DA=AE&lt;/math&gt;, the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; times the height from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;EA&lt;/math&gt;. So &lt;math&gt;BC=3/4BF&lt;/math&gt; and &lt;math&gt;BF=16/3&lt;/math&gt;. We look at right triangle &lt;math&gt;\triangle AIF&lt;/math&gt;. &lt;math&gt;IF=68/15&lt;/math&gt; and &lt;math&gt;AI=8/5&lt;/math&gt;. &lt;math&gt;\triangle AIF&lt;/math&gt; is a &lt;math&gt;17-6-5\sqrt{13}&lt;/math&gt; right triangle. Hypotenuse &lt;math&gt;AF&lt;/math&gt; intersects &lt;math&gt;CG&lt;/math&gt; at a point, we call it &lt;math&gt;H&lt;/math&gt;. &lt;math&gt;CH=4/3\div 68/15\cdot 8/5=8/17&lt;/math&gt;. So &lt;math&gt;HG=8-8/17=128/17&lt;/math&gt;.<br /> <br /> <br /> By [[Power of a Point Theorem|Power of a Point]], &lt;math&gt;CH\cdot HG=AH\cdot HE&lt;/math&gt;. &lt;math&gt;AH=16/5\cdot 5\sqrt{13}/17=16\sqrt{13}/17.&lt;/math&gt; So &lt;math&gt;HE=1024/289\cdot 17/(16\sqrt{13})=64/(17\sqrt{13})&lt;/math&gt;. The height from &lt;math&gt;E&lt;/math&gt; to &lt;math&gt;CG&lt;/math&gt; is &lt;math&gt;17/(5\sqrt{13})\cdot 64/(17\sqrt{13})=64/65&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;[\triangle CEG]=64/65\cdot 8\div 2=256/65&lt;/math&gt;. The area of the whole cyclic quadrilateral is &lt;math&gt;64/5+256/65=(832+256)/65=1088/65&lt;/math&gt;. Lastly, the common area is &lt;math&gt;1/17&lt;/math&gt; the area of the quadrilateral, or &lt;math&gt;64/65&lt;/math&gt;. So &lt;math&gt;64+65=\boxed{129}&lt;/math&gt;.<br /> <br /> ==Solution 5 (HARD computation)==<br /> &lt;asy&gt;<br /> unitsize(35);<br /> draw(Circle((-1,0),1));<br /> draw(Circle((4,0),4));<br /> pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;<br /> A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);<br /> label(&quot;$A$&quot;,A,NE);label(&quot;$O_1$&quot;,O_1,NE);label(&quot;$O_2$&quot;,O_2,NE);label(&quot;$B$&quot;,B,SW);label(&quot;$C$&quot;,C,SW);label(&quot;$D$&quot;,D,NE);label(&quot;$E$&quot;,E,NE);label(&quot;$N$&quot;,N,W);label(&quot;$K$&quot;,(-24/15,0.2));label(&quot;$L$&quot;,(24/15,0.2));label(&quot;$n$&quot;,(-0.8,-0.12));label(&quot;$p$&quot;,((29/15,-48/15)));label(&quot;$\mathcal{P}$&quot;,(-1.6,1.1));label(&quot;$\mathcal{Q}$&quot;,(6,4));<br /> draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed);<br /> dot(O_1);dot(O_2);<br /> draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));<br /> //draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X);<br /> path circle2 = Circle((4,0),4);<br /> N = (-8/3,0);<br /> pair X =rotate(180,O_2)*E;<br /> pair Y = (8,0);<br /> draw(X--Y,dashed); draw(E--Y,dashed);draw(E--X,dashed); draw(Y--C,dashed); draw(C--X,dashed); draw(O_2--Y);<br /> dot(&quot;$X$&quot;, X, NE);dot(&quot;$Y$&quot;, Y, NE);<br /> &lt;/asy&gt;<br /> <br /> Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because &lt;math&gt;AE=XY&lt;/math&gt; and &lt;math&gt;AE \parallel XY&lt;/math&gt;, &lt;math&gt;XYE&lt;/math&gt; is right angle. <br /> <br /> First, &lt;math&gt;\frac{NO_1}{NO_1+5} = \frac{1}{4}&lt;/math&gt;, so &lt;math&gt;NO_1=\frac{5}{3}&lt;/math&gt;. And,<br /> &lt;cmath&gt;\cos{\angle{AO_2C}}=\cos{2\angle{AYC}} = \frac{O_2C}{NO_1+5} = \frac{3}{5}&lt;/cmath&gt;<br /> &lt;cmath&gt;\sin{\angle{AYC}} = \sqrt{\dfrac{1-\cos{2\angle{AYC}}}{2}}=\frac{1}{\sqrt{5}}&lt;/cmath&gt;<br /> &lt;cmath&gt; \cos{\angle{AYC}} = \frac{2}{\sqrt{5}}&lt;/cmath&gt;<br /> Then,<br /> &lt;cmath&gt;[AEC] = \frac{1}{2}AE*CE*\sin{\angle{ACE}}=\frac{1}{2}AE*8\sin{\angle{CYE}}*\frac{1}{\sqrt{5}}&lt;/cmath&gt;<br /> &lt;cmath&gt;[CXY] = \frac{1}{2}CX*XY*\sin{\angle{CXY}}=\frac{1}{2}*8\sin{\angle{XYC}}*XY*\sin{\angle{CAY}}&lt;/cmath&gt;<br /> Since &lt;math&gt;\angle{CAY} = 90 - \angle{AYC}&lt;/math&gt;, &lt;math&gt;\angle{XYC} = 90 - \angle{CYE}&lt;/math&gt;, &lt;math&gt;XY = AE&lt;/math&gt;, we have<br /> &lt;cmath&gt;[CXY] = \frac{1}{2}*8AE\cos{\angle{CYE}}*\cos{\angle{AYC}}=\frac{1}{2}*8AE\cos{\angle{CYE}}*\frac{2}{\sqrt{5}}&lt;/cmath&gt;<br /> Since &lt;math&gt;\triangle{CXY}&lt;/math&gt; is four times in scale to &lt;math&gt;\triangle{AEC}&lt;/math&gt;, their area ratio is 16. Divide the two equations for the two areas, we have<br /> &lt;cmath&gt;\tan{\angle{CYE}} = \frac{1}{8}&lt;/cmath&gt;<br /> With this angle found, everything else just follows. <br /> &lt;cmath&gt;\sin{\angle{CYE}} = \dfrac{\tan{\angle{CYE}}}{\sqrt{1+\tan^2{\angle{CYE}}}}=\dfrac{1}{\sqrt{65}}&lt;/cmath&gt;<br /> &lt;cmath&gt;\cos{\angle{CYE}} = \dfrac{8}{\sqrt{65}}&lt;/cmath&gt;<br /> &lt;cmath&gt;\sin{\angle{AYE}} = \sin({\angle{AYC}+\angle{CYE}} )= \frac{1}{\sqrt{5}}*\frac{8}{\sqrt{65}} + \frac{2}{\sqrt{5}}*\frac{1}{\sqrt{65}} = \frac{2}{\sqrt{13}}&lt;/cmath&gt;<br /> &lt;cmath&gt;AE = 8\sin{\angle{AYE}} = \frac{16}{\sqrt{13}}&lt;/cmath&gt;<br /> &lt;cmath&gt;[AEC] = \frac{1}{2}*8*\frac{16}{\sqrt{13}}*\dfrac{1}{\sqrt{65}}*\frac{1}{\sqrt{5}}=\frac{64}{65}&lt;/cmath&gt;<br /> <br /> ==Solution 6 (Simple computation)==<br /> <br /> Let &lt;math&gt;K&lt;/math&gt; be the intersection of &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt;. Since the radii of the two circles are 1:4, so we have &lt;math&gt;AD:AE=1:4&lt;/math&gt;, and the distance from &lt;math&gt;B&lt;/math&gt; to line &lt;math&gt;l&lt;/math&gt; and the distance from &lt;math&gt;C&lt;/math&gt; to line &lt;math&gt;l&lt;/math&gt; are in a ratio of 4:1, so &lt;math&gt;BK:CK=4:1&lt;/math&gt;. We can easily calculate the length of &lt;math&gt;BC&lt;/math&gt; to be 4, so &lt;math&gt;CK=\frac{4}{3}&lt;/math&gt;. Let &lt;math&gt;J&lt;/math&gt; be the foot of perpendicular line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt;, we can know that &lt;math&gt;BJ:CJ=1:4&lt;/math&gt;, so &lt;math&gt;BJ = 0.8&lt;/math&gt;, &lt;math&gt;CJ=3.2&lt;/math&gt;, &lt;math&gt;AJ=1.6&lt;/math&gt;, and &lt;math&gt;AK=\sqrt{1.6^2+\left(3.2+\frac{4}{3}\right)^2}=\frac{4}{3}\sqrt{13}&lt;/math&gt;. Since &lt;math&gt;CK^2 = EK\cdot AK&lt;/math&gt;, so &lt;math&gt;EK=\frac{4}{39}\sqrt{13}&lt;/math&gt;, and &lt;math&gt;AE = \frac{4}{3}\sqrt{13} - \frac{4}{39}\sqrt{13}= \frac{16}{13}\sqrt{13}&lt;/math&gt;. &lt;math&gt;\sin\angle AKB=\frac{AJ}{AK} = \frac{1.6}{\frac{4}{3}\sqrt{13}}=\frac{1.2}{\sqrt{13}}&lt;/math&gt;, so the distance from &lt;math&gt;C&lt;/math&gt; to line &lt;math&gt;l&lt;/math&gt; is &lt;math&gt;d=CK\cdot \sin\angle AKB = \frac{4}{3}\cdot \frac{1.2}{\sqrt{13}}=\frac{1.6}{\sqrt{13}}&lt;/math&gt;. so the area is<br /> &lt;cmath&gt;<br /> [ACE] = \frac{1}{2}\cdot AE\cdot d = \frac{1}{2}\cdot\frac{16}{13}\sqrt{13}\frac{1.6}{\sqrt{13}} = \frac{64}{65}<br /> &lt;/cmath&gt;<br /> The final answer is &lt;math&gt;\boxed{129}&lt;/math&gt;.<br /> <br /> --- by Dan Li<br /> <br /> ==Solution 7==<br /> Consider the common tangent from &lt;math&gt;A&lt;/math&gt; to both circles. Let this intersect &lt;math&gt;BC&lt;/math&gt; at point &lt;math&gt;K&lt;/math&gt;. From equal tangents, we have &lt;math&gt;BK=AK=CK&lt;/math&gt;, which implies that &lt;math&gt;\angle BAC = 90^\circ&lt;/math&gt;.<br /> <br /> Let the center of &lt;math&gt;\mathcal{P}&lt;/math&gt; be &lt;math&gt;O_1&lt;/math&gt;, and the center of &lt;math&gt;\mathcal{Q}&lt;/math&gt; be &lt;math&gt;O_2&lt;/math&gt;. Angle chasing, we find that &lt;math&gt;\triangle O_1DA \sim \triangle O_2EA&lt;/math&gt; with a ratio of &lt;math&gt;1:4&lt;/math&gt;. Hence &lt;math&gt;4AD = AE&lt;/math&gt;.<br /> <br /> We can easily deduce that &lt;math&gt;BC=4&lt;/math&gt; by dropping an altitude from &lt;math&gt;O_1&lt;/math&gt; to &lt;math&gt;O_2C&lt;/math&gt;. Let &lt;math&gt;\angle ABC = \theta&lt;/math&gt;. By some simple angle chasing, we obtain that &lt;math&gt;\angle BO_1A = 2\angle BDA = 2\angle ABC = 2\theta,&lt;/math&gt; and similarly &lt;math&gt;\angle CO_2A = 180 - 2\theta&lt;/math&gt;.<br /> <br /> Using LoC, we get that &lt;math&gt;AB = \sqrt{2-2\cos2\theta}&lt;/math&gt; and &lt;math&gt;AC = \sqrt{32+32\cos2\theta}&lt;/math&gt;. From Pythagorean theorem, we have &lt;cmath&gt;AB^2 + AC^2 = BC^2 \implies \cos 2\theta = -\frac{3}{5} \implies \cos \theta = \frac{1}{\sqrt5}, \sin \theta = \frac{2}{\sqrt 5}&lt;/cmath&gt;<br /> In other words, &lt;math&gt;AB = \frac{4}{\sqrt5}, AC = \frac{8}{\sqrt5}&lt;/math&gt;.<br /> <br /> Using the area condition, we have:<br /> &lt;cmath&gt;\begin{align*}<br /> \frac12 AD*AB \sin \angle DAB &amp;= \frac 12 AE*AC \sin(90-\angle DAB) \\<br /> AD*\frac{4}{\sqrt5} \sin \angle DAB &amp;= 4AD*\frac{8}{\sqrt5} \cos \angle DAB \\<br /> \sin \angle DAB &amp;= 8 \cos \angle DAB \\<br /> \implies \sin \angle DAB &amp;= \frac{8}{\sqrt{65}}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Now, for brevity, let &lt;math&gt;\angle D = \angle ADB&lt;/math&gt; and &lt;math&gt;\angle A = \angle DAB&lt;/math&gt;.<br /> <br /> From Law of Sines on &lt;math&gt;\triangle ABD&lt;/math&gt;, we have<br /> &lt;cmath&gt;\begin{align*}<br /> \frac{AB}{\sin \angle D} &amp;= \frac{AD}{\sin (180-\angle A - \angle D)} \\<br /> \frac{\frac{4}{\sqrt5}}{\frac{2}{\sqrt5}} &amp;= \frac{AD}{\sin \angle A \cos\angle D + \sin \angle D\cos\angle A} \\<br /> 2 &amp;= \frac{AD}{\frac{2}{\sqrt{13}}} \\<br /> AD &amp;= \frac{4}{\sqrt{13}}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> It remains to find the area of &lt;math&gt;\triangle ABD&lt;/math&gt;. This is just &lt;cmath&gt;\frac12 AD*AB*\sin \angle A = \frac12 * \frac{4}{\sqrt{13}}*\frac{4}{\sqrt5}*\frac{8}{\sqrt{65}} = \frac{64}{65}&lt;/cmath&gt; for an answer of &lt;math&gt;\boxed{129}.&lt;/math&gt;<br /> <br /> This solution was brought to you by Leonard_my_dude.<br /> <br /> ==Solution 8 (Synthetic-Trigonometry)==<br /> Add in the line &lt;math&gt;k&lt;/math&gt; as the internal tangent between the two circles. Let &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;BC&lt;/math&gt;; It is well-known that &lt;math&gt;M&lt;/math&gt; is on &lt;math&gt;k&lt;/math&gt; and because &lt;math&gt;k&lt;/math&gt; is the radical axis of the two circles, &lt;math&gt;AM=BM=CM&lt;/math&gt;. Therefore because &lt;math&gt;M&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{BAC}&lt;/math&gt;, &lt;math&gt;\angle{BAC}=90^{\circ}&lt;/math&gt;. Let &lt;math&gt;O_P&lt;/math&gt; be the center of circle &lt;math&gt;\mathcal{P}&lt;/math&gt; and likewise let &lt;math&gt;O_Q&lt;/math&gt; be the center of circle &lt;math&gt;\mathcal{Q}&lt;/math&gt;. It is well known that by homothety &lt;math&gt;O_P, A,&lt;/math&gt; and &lt;math&gt;O_Q&lt;/math&gt; are collinear. It is well-known that &lt;math&gt;\angle{ABC}=\angle{ADB}=b&lt;/math&gt;, and likewise &lt;math&gt;\angle{ACB}=\angle{AEC}=c&lt;/math&gt;. By homothety, &lt;math&gt;AD=4AE&lt;/math&gt;, therefore since the two triangles mentioned in the problem, the length of the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; is four times the length of the altitude from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;AE&lt;/math&gt;. Using the Pythagorean Theorem, &lt;math&gt;BC = 4&lt;/math&gt;. By angle-chasing, &lt;math&gt;O_{P}AMB&lt;/math&gt; is cyclic, and likewise &lt;math&gt;O_{Q}CMA&lt;/math&gt; is cyclic. Use the Pythagorean Theorem for &lt;math&gt;\triangle{O_{P}BM}&lt;/math&gt; to get &lt;math&gt;O_{P}M=\sqrt{5}&lt;/math&gt;. Then by Ptolemy's Theorem &lt;math&gt;AB=\frac{4\sqrt{5}}{5} \implies AC=\frac{8\sqrt{5}}{5}&lt;/math&gt;. Now to compute the area, using what we know about the length of the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; is four times the length of the altitude from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;AE&lt;/math&gt;, letting &lt;math&gt;x&lt;/math&gt; be the length of the altitude from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;AE&lt;/math&gt;, &lt;math&gt;x=\frac{8}{5\sqrt{13}}&lt;/math&gt;. From the Law of Sines, &lt;math&gt;\frac{BD}{\sin{A}}=2 \implies \sin{A}=\frac{4x}{AB} \implies BD=\frac{8x}{AB}=\frac{16\sqrt{65}}{65}&lt;/math&gt;. Then use the Pythagorean Theorem twice and add up the lengths to get &lt;math&gt;AD=\frac{4}{\sqrt{13}}&lt;/math&gt;. Use the formula &lt;math&gt;\frac{b \times h}{2}&lt;/math&gt; to get &lt;math&gt;\frac{64}{65} = \boxed{129}&lt;/math&gt; as the answer.<br /> <br /> ~First<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=169902 2019 AIME I Problems/Problem 14 2022-01-14T22:21:36Z <p>First: /* Solution */</p> <hr /> <div>==Problem==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> We know that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we find &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is a positive divisor of &lt;math&gt;16&lt;/math&gt;.<br /> <br /> However, if the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;1, 2, 4,&lt;/math&gt; or &lt;math&gt;8,&lt;/math&gt; then &lt;math&gt;2019^8&lt;/math&gt; will be equivalent to &lt;math&gt;1 \pmod{p},&lt;/math&gt; which contradicts the given requirement that &lt;math&gt;2019^8\equiv -1\pmod{p}&lt;/math&gt;.<br /> <br /> Therefore, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;. Because all orders modulo &lt;math&gt;p&lt;/math&gt; divide &lt;math&gt;\phi(p)&lt;/math&gt;, we see that &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of &lt;math&gt;16&lt;/math&gt;. As &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p\equiv 1 \pmod{16}&lt;/math&gt;. The two smallest primes equivalent to &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. Because &lt;math&gt;16 | p - 1&lt;/math&gt;, and &lt;math&gt; p - 1 \geq 16&lt;/math&gt;, each possible value of &lt;math&gt;p&lt;/math&gt; must be verified by manual calculation to make sure that &lt;math&gt;p | 2019^8+1&lt;/math&gt;. As &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt; and &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, the smallest possible &lt;math&gt;p&lt;/math&gt; is thus &lt;math&gt;\boxed{097}&lt;/math&gt;.<br /> <br /> ===Note to solution===<br /> &lt;math&gt;\phi(k)&lt;/math&gt; is the Euler Totient Function of integer &lt;math&gt;k&lt;/math&gt;. &lt;math&gt;\phi(k)&lt;/math&gt; is the number of positive integers less than &lt;math&gt;k&lt;/math&gt; relatively prime to &lt;math&gt;k&lt;/math&gt;. Define the numbers &lt;math&gt;k_1,k_2,k_3,\cdots,k_n&lt;/math&gt; to be the prime factors of &lt;math&gt;k&lt;/math&gt;. Then, we have&lt;cmath&gt;\phi(k)=k\cdot \prod^n_{i=1}\left(1-\dfrac{1}{k_i}\right).&lt;/cmath&gt;A property of the Totient function is that, for any prime &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;\phi(p)=p-1&lt;/math&gt;.<br /> <br /> Euler's Totient Theorem states that&lt;cmath&gt;a^{\phi(k)} \equiv 1\pmod k&lt;/cmath&gt;if &lt;math&gt;\gcd(a,k)=1&lt;/math&gt;.<br /> <br /> Furthermore, the order &lt;math&gt;a&lt;/math&gt; modulo &lt;math&gt;n&lt;/math&gt; for an integer &lt;math&gt;a&lt;/math&gt; relatively prime to &lt;math&gt;n&lt;/math&gt; is defined as the smallest positive integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;a^{d} \equiv 1\pmod n&lt;/math&gt;. An important property of the order &lt;math&gt;d&lt;/math&gt; is that &lt;math&gt;d|\phi(n)&lt;/math&gt;.<br /> <br /> ==Solution 2 (Basic Modular Arithmetic)==<br /> <br /> In this solution, &lt;math&gt;k&lt;/math&gt; will represent an arbitrary nonnegative integer.<br /> <br /> We will show that any potential prime &lt;math&gt;p&lt;/math&gt; must be of the form &lt;math&gt;16k+1&lt;/math&gt; through a proof by contradiction. Suppose that there exists some prime &lt;math&gt;p&lt;/math&gt; that can not be expressed in the form &lt;math&gt;16k+1&lt;/math&gt; that is a divisor of &lt;math&gt;2019^8+1&lt;/math&gt;. First, note that if the prime &lt;math&gt;p&lt;/math&gt; is a divisor of &lt;math&gt;2019&lt;/math&gt;, then &lt;math&gt;2019^8&lt;/math&gt; is a multiple of &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;2019^8+1&lt;/math&gt; is not. Thus, &lt;math&gt;p&lt;/math&gt; is not a divisor of &lt;math&gt;2019&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;2019^8+1&lt;/math&gt; is a multiple of &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;2019^8+1\equiv0\pmod{p}&lt;/math&gt;. This means that &lt;math&gt;2019^8\equiv-1\pmod{p}&lt;/math&gt;, and by raising both sides to an arbitrary odd positive integer, we have that &lt;math&gt;2019^{16k+8}\equiv-1\pmod{p}&lt;/math&gt;.<br /> <br /> Then, since the problem requires an odd prime, &lt;math&gt;p&lt;/math&gt; can be expressed as &lt;math&gt;16k+m&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; is an odd integer ranging from &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;15&lt;/math&gt;, inclusive. By Fermat's Little Theorem, &lt;math&gt;2019^{p-1}\equiv1\pmod{p}&lt;/math&gt;, and plugging in values, we get &lt;math&gt;2019^{16k+n}\equiv1\pmod{p}&lt;/math&gt;, where &lt;math&gt;n=m-1&lt;/math&gt; and is thus an even integer ranging from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;14&lt;/math&gt;, inclusive.<br /> <br /> If &lt;math&gt;n=8&lt;/math&gt;, then &lt;math&gt;2019^{16k+8}\equiv1\pmod{p}&lt;/math&gt;, which creates a contradiction. If &lt;math&gt;n&lt;/math&gt; is not a multiple of &lt;math&gt;8&lt;/math&gt; but is a multiple of &lt;math&gt;4&lt;/math&gt;, squaring both sides of &lt;math&gt;2019^{16k+n}\equiv1\pmod{p}&lt;/math&gt; also results in the same contradictory equivalence. For all remaining &lt;math&gt;n&lt;/math&gt;, raising both sides of &lt;math&gt;2019^{16k+n}\equiv1\pmod{p}&lt;/math&gt; to the &lt;math&gt;4&lt;/math&gt;th power creates the same contradiction. (Note that &lt;math&gt;32k&lt;/math&gt; and &lt;math&gt;64k&lt;/math&gt; can both be expressed in the form &lt;math&gt;16k&lt;/math&gt;.)<br /> <br /> Since we have proved that no value of &lt;math&gt;n&lt;/math&gt; can work, this means that a prime must be of the form &lt;math&gt;16k+1&lt;/math&gt; in order to be a factor of &lt;math&gt;2019^8+1&lt;/math&gt;. The smallest prime of this form is &lt;math&gt;17&lt;/math&gt;, and testing it, we get<br /> &lt;cmath&gt;2019^8+1\equiv13^8+1\equiv169^4+1\equiv(-1)^4+1\equiv1+1\equiv2\pmod{17},&lt;/cmath&gt;<br /> so it does not work. The next smallest prime of the required form is &lt;math&gt;97&lt;/math&gt;, and testing it, we get<br /> &lt;cmath&gt;2019^8+1\equiv(-18)^8+1\equiv324^4+1\equiv33^4+1\equiv1089^2+1\equiv22^2+1\equiv484+1\equiv-1+1\equiv0\pmod{97},&lt;/cmath&gt;<br /> so it works. Thus, the answer is &lt;math&gt;\boxed{097}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 3 (Official MAA)==<br /> Suppose prime &lt;math&gt;p&gt;2&lt;/math&gt; divides &lt;math&gt;2019^8+1.&lt;/math&gt; Then &lt;math&gt;2019^8\equiv -1\pmod p.&lt;/math&gt; Squaring gives &lt;math&gt;2019^{16}\equiv 1\pmod p.&lt;/math&gt; If &lt;math&gt;2019^m\equiv 1 \pmod p&lt;/math&gt; for some &lt;math&gt;0&lt;m&lt;16,&lt;/math&gt; it follows that &lt;cmath&gt;2019^{\gcd(m,16)}\equiv 1\pmod p.&lt;/cmath&gt; But &lt;math&gt;2019^8\equiv -1\pmod p,&lt;/math&gt; so &lt;math&gt;\gcd(m,16)&lt;/math&gt; cannot divide &lt;math&gt;8,&lt;/math&gt; which is a contradiction. Thus &lt;math&gt;2019^{16}&lt;/math&gt; is the least positive power congruent to &lt;math&gt;1\pmod p.&lt;/math&gt; By Fermat's Little Theorem, &lt;math&gt;2019^{p-1}\equiv 1\pmod p.&lt;/math&gt; It follows that &lt;math&gt;p=16k+1&lt;/math&gt; for some positive integer &lt;math&gt;k.&lt;/math&gt; The least two primes of this form are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97.&lt;/math&gt; The least odd factor of &lt;math&gt;2019^8+1&lt;/math&gt; is not &lt;math&gt;17&lt;/math&gt; because &lt;cmath&gt;2019\equiv 13\pmod{17}\qquad \text{and}\qquad 13^2\equiv 169\equiv -1\pmod{17},&lt;/cmath&gt; which implies &lt;math&gt;2019^8\equiv 1\not\equiv -1\pmod {17}.&lt;/math&gt; But &lt;math&gt;2019\equiv -18\pmod{97},&lt;/math&gt; so &lt;cmath&gt;\begin{align*}<br /> (-18)^2=324&amp;\equiv 33\pmod{97}, \\<br /> 33^2=1089&amp;\equiv 22\pmod{97},\,\text{and} \\<br /> 22^2=484&amp;\equiv -1\pmod{97}.\end{align*}&lt;/cmath&gt; Thus the least odd prime factor is &lt;math&gt;97.&lt;/math&gt;<br /> <br /> In fact, &lt;math&gt;2019^8+1=2\cdot97\cdot p,&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; is the &lt;math&gt;25&lt;/math&gt;-digit prime &lt;cmath&gt;1423275002072658812388593.&lt;/cmath&gt;<br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_11&diff=141432 2019 AIME II Problems/Problem 11 2021-01-03T19:04:46Z <p>First: </p> <hr /> <div>==Problem==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB=7, BC=8, &lt;/math&gt; and &lt;math&gt;CA=9.&lt;/math&gt; Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and is tangent to line &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;C&lt;/math&gt; and is tangent to line &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Let &lt;math&gt;K&lt;/math&gt; be the intersection of circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; not equal to &lt;math&gt;A.&lt;/math&gt; Then &lt;math&gt;AK=\tfrac mn,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n.&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> unitsize(20);<br /> pair B = (0,0);<br /> pair A = (2,sqrt(45));<br /> pair C = (8,0);<br /> draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));<br /> draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));<br /> draw(B--A--C--cycle);<br /> label(&quot;$A$&quot;,A,dir(105));<br /> label(&quot;$B$&quot;,B,dir(-135));<br /> label(&quot;$C$&quot;,C,dir(-75));<br /> dot((2.68,2.25));<br /> label(&quot;$K$&quot;,(2.68,2.25),dir(-150));<br /> label(&quot;$\omega_1$&quot;,(-6,1));<br /> label(&quot;$\omega_2$&quot;,(14,6));<br /> label(&quot;$7$&quot;,(A+B)/2,dir(140));<br /> label(&quot;$8$&quot;,(B+C)/2,dir(-90));<br /> label(&quot;$9$&quot;,(A+C)/2,dir(60));<br /> &lt;/asy&gt;<br /> -Diagram by Brendanb4321<br /> <br /> <br /> Note that from the tangency condition that the supplement of &lt;math&gt;\angle CAB&lt;/math&gt; with respects to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; are equal to &lt;math&gt;\angle AKB&lt;/math&gt; and &lt;math&gt;\angle AKC&lt;/math&gt;, respectively, so from tangent-chord, &lt;cmath&gt;\angle AKC=\angle AKB=180^{\circ}-\angle BAC&lt;/cmath&gt; Also note that &lt;math&gt;\angle ABK=\angle KAC&lt;/math&gt;, so &lt;math&gt;\triangle AKB\sim \triangle CKA&lt;/math&gt;. Using similarity ratios, we can easily find &lt;cmath&gt;AK^2=BK*KC&lt;/cmath&gt; However, since &lt;math&gt;AB=7&lt;/math&gt; and &lt;math&gt;CA=9&lt;/math&gt;, we can use similarity ratios to get &lt;cmath&gt;BK=\frac{7}{9}AK, CK=\frac{9}{7}AK&lt;/cmath&gt; <br /> <br /> *Now we use Law of Cosines on &lt;math&gt;\triangle AKB&lt;/math&gt;: From reverse Law of Cosines, &lt;math&gt;\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}&lt;/math&gt;. This gives us &lt;cmath&gt;AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;\implies \frac{196}{81}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;AK=\frac{9}{2}&lt;/cmath&gt; so our answer is &lt;math&gt;9+2=\boxed{011}&lt;/math&gt;.<br /> -franchester<br /> <br /> *The motivation for using the Law of Cosines (&quot;LoC&quot;) is after finding the similar triangles it's hard to figure out what to do with &lt;math&gt;BK&lt;/math&gt; and &lt;math&gt;CK&lt;/math&gt; yet we know &lt;math&gt;BC&lt;/math&gt; which somehow has to help us solve the problem--a common theme in solving geometry problems is figuring out how to use what you haven't used yet. We know all three sides of some triangle though, and we're dealing with angles (that's how we found similarity), so why not try the Law of Cosines? This is to help with motivation--the solution is franchester's and I learned about using LoC from reading his solution (I only solved half the problem and got stuck). To anyone in the future reading this, math is beautiful.<br /> -First<br /> <br /> ==Solution 2 (Inversion)==<br /> Consider an inversion with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;r=AK&lt;/math&gt;. Then, we have &lt;math&gt;AB\cdot AB^*=AK^2&lt;/math&gt;, or &lt;math&gt;AB^*=\frac{AK^2}{7}&lt;/math&gt;. Similarly, &lt;math&gt;AC^*=\frac{AK^2}{9}&lt;/math&gt;. Notice that &lt;math&gt;AB^*KC^*&lt;/math&gt; is a parallelogram, since &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; are tangent to &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Thus, &lt;math&gt;AC^*=B^*K&lt;/math&gt;. Now, we get that<br /> &lt;cmath&gt;\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}&lt;/cmath&gt;<br /> so by Law of Cosines on &lt;math&gt;\triangle AB^*K&lt;/math&gt; we have <br /> &lt;cmath&gt;(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{11AK^2}{63\cdot21}&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow AK=\frac{9}{2}&lt;/cmath&gt;<br /> Then, our answer is &lt;math&gt;9+2=\boxed{11}&lt;/math&gt;. <br /> -brianzjk<br /> <br /> <br /> <br /> == Solution 3 (Death By Trig Bash) ==<br /> <br /> 14. Let the centers of the circles be &lt;math&gt;O_{1}&lt;/math&gt; and &lt;math&gt;O_{2}&lt;/math&gt; where the &lt;math&gt;O_{1}&lt;/math&gt; has the side length &lt;math&gt;7&lt;/math&gt; contained in the circle. Now let &lt;math&gt;\angle BAC =x.&lt;/math&gt; This implies &lt;math&gt;\angle AO_{1}B = \angle AO_{2}C = 2x&lt;/math&gt; by the angle by by tangent. Then we also know that &lt;math&gt;\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x.&lt;/math&gt; Now we first find &lt;math&gt;\cos x.&lt;/math&gt; We use law of cosines on &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; to obtain &lt;math&gt;64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x} \implies \cos{x} =\frac{11}{21} \implies \sin{x} =\frac{8\sqrt{5}}{21}.&lt;/math&gt; Then applying law of sines on &lt;math&gt;\bigtriangleup AO_{1}B&lt;/math&gt; we obtain &lt;math&gt;\frac{7}{\sin{2x}} =\frac{OB_{1}}{\sin{90^{\circ}-x}} \implies\frac{7}{2\sin{x}\cos{x}} =\frac{OB_{1}}{\cos{x}} \implies OB_{1} = O_{1}A=\frac{147}{16\sqrt{5}}.&lt;/math&gt; Using similar logic we obtain &lt;math&gt;OA_{1} =\frac{189}{16\sqrt{5}}.&lt;/math&gt; Now we know that &lt;math&gt;\angle O_{1}AO_{2}=180^{\circ}-x.&lt;/math&gt; Thus using law of cosines on &lt;math&gt;\bigtriangleup O_{1}AO_{2}&lt;/math&gt; yields &lt;math&gt;O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}.&lt;/math&gt; While this does look daunting we can write the above expression as &lt;math&gt;\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}.&lt;/math&gt; Then factoring yields &lt;math&gt;\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}.&lt;/math&gt; The area &lt;math&gt;[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}.&lt;/math&gt; Now &lt;math&gt;AK&lt;/math&gt; is twice the length of the altitude of &lt;math&gt;\bigtriangleup O_{1}AO_{2}&lt;/math&gt; so we let the altitude be &lt;math&gt;h&lt;/math&gt; and we have &lt;math&gt;\frac{1}{2} \cdot h \cdot\frac{147\sqrt{5}}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21} \implies h =\frac{9}{4}.&lt;/math&gt; Thus our desired length is &lt;math&gt;\frac{9}{2} \implies n+n = \boxed{11}.&lt;/math&gt;<br /> <br /> ==Solution 4 (Video)==<br /> <br /> Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=10|num-a=12}}<br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_11&diff=141431 2019 AIME II Problems/Problem 11 2021-01-03T19:04:23Z <p>First: </p> <hr /> <div>==Problem==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB=7, BC=8, &lt;/math&gt; and &lt;math&gt;CA=9.&lt;/math&gt; Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and is tangent to line &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;C&lt;/math&gt; and is tangent to line &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Let &lt;math&gt;K&lt;/math&gt; be the intersection of circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; not equal to &lt;math&gt;A.&lt;/math&gt; Then &lt;math&gt;AK=\tfrac mn,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n.&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> unitsize(20);<br /> pair B = (0,0);<br /> pair A = (2,sqrt(45));<br /> pair C = (8,0);<br /> draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));<br /> draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));<br /> draw(B--A--C--cycle);<br /> label(&quot;$A$&quot;,A,dir(105));<br /> label(&quot;$B$&quot;,B,dir(-135));<br /> label(&quot;$C$&quot;,C,dir(-75));<br /> dot((2.68,2.25));<br /> label(&quot;$K$&quot;,(2.68,2.25),dir(-150));<br /> label(&quot;$\omega_1$&quot;,(-6,1));<br /> label(&quot;$\omega_2$&quot;,(14,6));<br /> label(&quot;$7$&quot;,(A+B)/2,dir(140));<br /> label(&quot;$8$&quot;,(B+C)/2,dir(-90));<br /> label(&quot;$9$&quot;,(A+C)/2,dir(60));<br /> &lt;/asy&gt;<br /> -Diagram by Brendanb4321<br /> <br /> <br /> Note that from the tangency condition that the supplement of &lt;math&gt;\angle CAB&lt;/math&gt; with respects to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; are equal to &lt;math&gt;\angle AKB&lt;/math&gt; and &lt;math&gt;\angle AKC&lt;/math&gt;, respectively, so from tangent-chord, &lt;cmath&gt;\angle AKC=\angle AKB=180^{\circ}-\angle BAC&lt;/cmath&gt; Also note that &lt;math&gt;\angle ABK=\angle KAC&lt;/math&gt;, so &lt;math&gt;\triangle AKB\sim \triangle CKA&lt;/math&gt;. Using similarity ratios, we can easily find &lt;cmath&gt;AK^2=BK*KC&lt;/cmath&gt; However, since &lt;math&gt;AB=7&lt;/math&gt; and &lt;math&gt;CA=9&lt;/math&gt;, we can use similarity ratios to get &lt;cmath&gt;BK=\frac{7}{9}AK, CK=\frac{9}{7}AK&lt;/cmath&gt; <br /> <br /> *Now we use Law of Cosines on &lt;math&gt;\triangle AKB&lt;/math&gt;: From reverse Law of Cosines, &lt;math&gt;\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}&lt;/math&gt;. This gives us &lt;cmath&gt;AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;\implies \frac{196}{81}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;AK=\frac{9}{2}&lt;/cmath&gt; so our answer is &lt;math&gt;9+2=\boxed{011}&lt;/math&gt;.<br /> -franchester<br /> <br /> *The motivation for using the Law of Cosines (&quot;LoC&quot;) is after finding the similar triangles it's hard to figure out what to do with &lt;math&gt;BK&lt;/math&gt; and &lt;math&gt;CK&lt;/math&gt; yet we know &lt;math&gt;BC&lt;/math&gt; which somehow has to help us solve the problem--a common theme in solving geometry problems is figuring out how to use what you haven't used yet. We know all three sides of some triangle though, and we're dealign with angles (that's how we found similarity), so why not try the Law of Cosines? This is to help with motivation--the solution is franchester's and I learned about using LoC from reading his solution (I only solved half the problem and got stuck). To anyone in the future reading this, math is beautiful.<br /> -First<br /> <br /> ==Solution 2 (Inversion)==<br /> Consider an inversion with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;r=AK&lt;/math&gt;. Then, we have &lt;math&gt;AB\cdot AB^*=AK^2&lt;/math&gt;, or &lt;math&gt;AB^*=\frac{AK^2}{7}&lt;/math&gt;. Similarly, &lt;math&gt;AC^*=\frac{AK^2}{9}&lt;/math&gt;. Notice that &lt;math&gt;AB^*KC^*&lt;/math&gt; is a parallelogram, since &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; are tangent to &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Thus, &lt;math&gt;AC^*=B^*K&lt;/math&gt;. Now, we get that<br /> &lt;cmath&gt;\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}&lt;/cmath&gt;<br /> so by Law of Cosines on &lt;math&gt;\triangle AB^*K&lt;/math&gt; we have <br /> &lt;cmath&gt;(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{11AK^2}{63\cdot21}&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow AK=\frac{9}{2}&lt;/cmath&gt;<br /> Then, our answer is &lt;math&gt;9+2=\boxed{11}&lt;/math&gt;. <br /> -brianzjk<br /> <br /> <br /> <br /> == Solution 3 (Death By Trig Bash) ==<br /> <br /> 14. Let the centers of the circles be &lt;math&gt;O_{1}&lt;/math&gt; and &lt;math&gt;O_{2}&lt;/math&gt; where the &lt;math&gt;O_{1}&lt;/math&gt; has the side length &lt;math&gt;7&lt;/math&gt; contained in the circle. Now let &lt;math&gt;\angle BAC =x.&lt;/math&gt; This implies &lt;math&gt;\angle AO_{1}B = \angle AO_{2}C = 2x&lt;/math&gt; by the angle by by tangent. Then we also know that &lt;math&gt;\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x.&lt;/math&gt; Now we first find &lt;math&gt;\cos x.&lt;/math&gt; We use law of cosines on &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; to obtain &lt;math&gt;64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x} \implies \cos{x} =\frac{11}{21} \implies \sin{x} =\frac{8\sqrt{5}}{21}.&lt;/math&gt; Then applying law of sines on &lt;math&gt;\bigtriangleup AO_{1}B&lt;/math&gt; we obtain &lt;math&gt;\frac{7}{\sin{2x}} =\frac{OB_{1}}{\sin{90^{\circ}-x}} \implies\frac{7}{2\sin{x}\cos{x}} =\frac{OB_{1}}{\cos{x}} \implies OB_{1} = O_{1}A=\frac{147}{16\sqrt{5}}.&lt;/math&gt; Using similar logic we obtain &lt;math&gt;OA_{1} =\frac{189}{16\sqrt{5}}.&lt;/math&gt; Now we know that &lt;math&gt;\angle O_{1}AO_{2}=180^{\circ}-x.&lt;/math&gt; Thus using law of cosines on &lt;math&gt;\bigtriangleup O_{1}AO_{2}&lt;/math&gt; yields &lt;math&gt;O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}.&lt;/math&gt; While this does look daunting we can write the above expression as &lt;math&gt;\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}.&lt;/math&gt; Then factoring yields &lt;math&gt;\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}.&lt;/math&gt; The area &lt;math&gt;[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}.&lt;/math&gt; Now &lt;math&gt;AK&lt;/math&gt; is twice the length of the altitude of &lt;math&gt;\bigtriangleup O_{1}AO_{2}&lt;/math&gt; so we let the altitude be &lt;math&gt;h&lt;/math&gt; and we have &lt;math&gt;\frac{1}{2} \cdot h \cdot\frac{147\sqrt{5}}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21} \implies h =\frac{9}{4}.&lt;/math&gt; Thus our desired length is &lt;math&gt;\frac{9}{2} \implies n+n = \boxed{11}.&lt;/math&gt;<br /> <br /> ==Solution 4 (Video)==<br /> <br /> Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=10|num-a=12}}<br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_11&diff=141430 2019 AIME II Problems/Problem 11 2021-01-03T19:03:41Z <p>First: </p> <hr /> <div>==Problem==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB=7, BC=8, &lt;/math&gt; and &lt;math&gt;CA=9.&lt;/math&gt; Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and is tangent to line &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;C&lt;/math&gt; and is tangent to line &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Let &lt;math&gt;K&lt;/math&gt; be the intersection of circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; not equal to &lt;math&gt;A.&lt;/math&gt; Then &lt;math&gt;AK=\tfrac mn,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n.&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> unitsize(20);<br /> pair B = (0,0);<br /> pair A = (2,sqrt(45));<br /> pair C = (8,0);<br /> draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));<br /> draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));<br /> draw(B--A--C--cycle);<br /> label(&quot;$A$&quot;,A,dir(105));<br /> label(&quot;$B$&quot;,B,dir(-135));<br /> label(&quot;$C$&quot;,C,dir(-75));<br /> dot((2.68,2.25));<br /> label(&quot;$K$&quot;,(2.68,2.25),dir(-150));<br /> label(&quot;$\omega_1$&quot;,(-6,1));<br /> label(&quot;$\omega_2$&quot;,(14,6));<br /> label(&quot;$7$&quot;,(A+B)/2,dir(140));<br /> label(&quot;$8$&quot;,(B+C)/2,dir(-90));<br /> label(&quot;$9$&quot;,(A+C)/2,dir(60));<br /> &lt;/asy&gt;<br /> -Diagram by Brendanb4321<br /> <br /> <br /> Note that from the tangency condition that the supplement of &lt;math&gt;\angle CAB&lt;/math&gt; with respects to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; are equal to &lt;math&gt;\angle AKB&lt;/math&gt; and &lt;math&gt;\angle AKC&lt;/math&gt;, respectively, so from tangent-chord, &lt;cmath&gt;\angle AKC=\angle AKB=180^{\circ}-\angle BAC&lt;/cmath&gt; Also note that &lt;math&gt;\angle ABK=\angle KAC&lt;/math&gt;, so &lt;math&gt;\triangle AKB\sim \triangle CKA&lt;/math&gt;. Using similarity ratios, we can easily find &lt;cmath&gt;AK^2=BK*KC&lt;/cmath&gt; However, since &lt;math&gt;AB=7&lt;/math&gt; and &lt;math&gt;CA=9&lt;/math&gt;, we can use similarity ratios to get &lt;cmath&gt;BK=\frac{7}{9}AK, CK=\frac{9}{7}AK&lt;/cmath&gt; <br /> <br /> *Now we use Law of Cosines on &lt;math&gt;\triangle AKB&lt;/math&gt;: From reverse Law of Cosines, &lt;math&gt;\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}&lt;/math&gt;. This gives us &lt;cmath&gt;AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;\implies \frac{196}{81}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;AK=\frac{9}{2}&lt;/cmath&gt; so our answer is &lt;math&gt;9+2=\boxed{011}&lt;/math&gt;.<br /> -franchester<br /> <br /> *The motivation for using the Law of Cosines (&quot;LoC&quot;) is after finding the similar triangles it's hard to figure out what to do with &lt;math&gt;BK&lt;/math&gt; and &lt;math&gt;CK&lt;/math&gt; yet we know &lt;math&gt;BC&lt;/math&gt; which somehow has to help us solve the problem--a common theme in solving geometry problems is figuring out how to use what you haven't used yet. We know all three sides of some triangle though, and we're dealign with angles (that's how we found similarity), so why not try the Law of Cosines? This is to help with motivation--the solution is franchester's and I learned about using LoC from reading his solution (I only solved half the problem and got struck). To anyone in the future reading this, math is beautiful.<br /> -First<br /> <br /> ==Solution 2 (Inversion)==<br /> Consider an inversion with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;r=AK&lt;/math&gt;. Then, we have &lt;math&gt;AB\cdot AB^*=AK^2&lt;/math&gt;, or &lt;math&gt;AB^*=\frac{AK^2}{7}&lt;/math&gt;. Similarly, &lt;math&gt;AC^*=\frac{AK^2}{9}&lt;/math&gt;. Notice that &lt;math&gt;AB^*KC^*&lt;/math&gt; is a parallelogram, since &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; are tangent to &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Thus, &lt;math&gt;AC^*=B^*K&lt;/math&gt;. Now, we get that<br /> &lt;cmath&gt;\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}&lt;/cmath&gt;<br /> so by Law of Cosines on &lt;math&gt;\triangle AB^*K&lt;/math&gt; we have <br /> &lt;cmath&gt;(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{11AK^2}{63\cdot21}&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow AK=\frac{9}{2}&lt;/cmath&gt;<br /> Then, our answer is &lt;math&gt;9+2=\boxed{11}&lt;/math&gt;. <br /> -brianzjk<br /> <br /> <br /> <br /> == Solution 3 (Death By Trig Bash) ==<br /> <br /> 14. Let the centers of the circles be &lt;math&gt;O_{1}&lt;/math&gt; and &lt;math&gt;O_{2}&lt;/math&gt; where the &lt;math&gt;O_{1}&lt;/math&gt; has the side length &lt;math&gt;7&lt;/math&gt; contained in the circle. Now let &lt;math&gt;\angle BAC =x.&lt;/math&gt; This implies &lt;math&gt;\angle AO_{1}B = \angle AO_{2}C = 2x&lt;/math&gt; by the angle by by tangent. Then we also know that &lt;math&gt;\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x.&lt;/math&gt; Now we first find &lt;math&gt;\cos x.&lt;/math&gt; We use law of cosines on &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; to obtain &lt;math&gt;64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x} \implies \cos{x} =\frac{11}{21} \implies \sin{x} =\frac{8\sqrt{5}}{21}.&lt;/math&gt; Then applying law of sines on &lt;math&gt;\bigtriangleup AO_{1}B&lt;/math&gt; we obtain &lt;math&gt;\frac{7}{\sin{2x}} =\frac{OB_{1}}{\sin{90^{\circ}-x}} \implies\frac{7}{2\sin{x}\cos{x}} =\frac{OB_{1}}{\cos{x}} \implies OB_{1} = O_{1}A=\frac{147}{16\sqrt{5}}.&lt;/math&gt; Using similar logic we obtain &lt;math&gt;OA_{1} =\frac{189}{16\sqrt{5}}.&lt;/math&gt; Now we know that &lt;math&gt;\angle O_{1}AO_{2}=180^{\circ}-x.&lt;/math&gt; Thus using law of cosines on &lt;math&gt;\bigtriangleup O_{1}AO_{2}&lt;/math&gt; yields &lt;math&gt;O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}.&lt;/math&gt; While this does look daunting we can write the above expression as &lt;math&gt;\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}.&lt;/math&gt; Then factoring yields &lt;math&gt;\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}.&lt;/math&gt; The area &lt;math&gt;[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}.&lt;/math&gt; Now &lt;math&gt;AK&lt;/math&gt; is twice the length of the altitude of &lt;math&gt;\bigtriangleup O_{1}AO_{2}&lt;/math&gt; so we let the altitude be &lt;math&gt;h&lt;/math&gt; and we have &lt;math&gt;\frac{1}{2} \cdot h \cdot\frac{147\sqrt{5}}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21} \implies h =\frac{9}{4}.&lt;/math&gt; Thus our desired length is &lt;math&gt;\frac{9}{2} \implies n+n = \boxed{11}.&lt;/math&gt;<br /> <br /> ==Solution 4 (Video)==<br /> <br /> Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=10|num-a=12}}<br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_11&diff=141429 2019 AIME II Problems/Problem 11 2021-01-03T19:03:12Z <p>First: </p> <hr /> <div>==Problem==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB=7, BC=8, &lt;/math&gt; and &lt;math&gt;CA=9.&lt;/math&gt; Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and is tangent to line &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;C&lt;/math&gt; and is tangent to line &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Let &lt;math&gt;K&lt;/math&gt; be the intersection of circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; not equal to &lt;math&gt;A.&lt;/math&gt; Then &lt;math&gt;AK=\tfrac mn,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n.&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> unitsize(20);<br /> pair B = (0,0);<br /> pair A = (2,sqrt(45));<br /> pair C = (8,0);<br /> draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));<br /> draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));<br /> draw(B--A--C--cycle);<br /> label(&quot;$A$&quot;,A,dir(105));<br /> label(&quot;$B$&quot;,B,dir(-135));<br /> label(&quot;$C$&quot;,C,dir(-75));<br /> dot((2.68,2.25));<br /> label(&quot;$K$&quot;,(2.68,2.25),dir(-150));<br /> label(&quot;$\omega_1$&quot;,(-6,1));<br /> label(&quot;$\omega_2$&quot;,(14,6));<br /> label(&quot;$7$&quot;,(A+B)/2,dir(140));<br /> label(&quot;$8$&quot;,(B+C)/2,dir(-90));<br /> label(&quot;$9$&quot;,(A+C)/2,dir(60));<br /> &lt;/asy&gt;<br /> -Diagram by Brendanb4321<br /> <br /> <br /> Note that from the tangency condition that the supplement of &lt;math&gt;\angle CAB&lt;/math&gt; with respects to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; are equal to &lt;math&gt;\angle AKB&lt;/math&gt; and &lt;math&gt;\angle AKC&lt;/math&gt;, respectively, so from tangent-chord, &lt;cmath&gt;\angle AKC=\angle AKB=180^{\circ}-\angle BAC&lt;/cmath&gt; Also note that &lt;math&gt;\angle ABK=\angle KAC&lt;/math&gt;, so &lt;math&gt;\triangle AKB\sim \triangle CKA&lt;/math&gt;. Using similarity ratios, we can easily find &lt;cmath&gt;AK^2=BK*KC&lt;/cmath&gt; However, since &lt;math&gt;AB=7&lt;/math&gt; and &lt;math&gt;CA=9&lt;/math&gt;, we can use similarity ratios to get &lt;cmath&gt;BK=\frac{7}{9}AK, CK=\frac{9}{7}AK&lt;/cmath&gt; <br /> <br /> *Now we use Law of Cosines on &lt;math&gt;\triangle AKB&lt;/math&gt;: From reverse Law of Cosines, &lt;math&gt;\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}&lt;/math&gt;. This gives us &lt;cmath&gt;AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;\implies \frac{196}{81}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;AK=\frac{9}{2}&lt;/cmath&gt; so our answer is &lt;math&gt;9+2=\boxed{011}&lt;/math&gt;.<br /> -franchester<br /> <br /> *<br /> The motivation for using the Law of Cosines (&quot;LoC&quot;) is after finding the similar triangles it's hard to figure out what to do with &lt;math&gt;BK&lt;/math&gt; and &lt;math&gt;CK&lt;/math&gt; yet we know &lt;math&gt;BC&lt;/math&gt; which somehow has to help us solve the problem--a common theme in solving geometry problems is figuring out how to use what you haven't used yet. We know all three sides of some triangle though, and we're dealign with angles (that's how we found similarity), so why not try the Law of Cosines? This is to help with motivation--the solution is franchester's and I learned about using LoC from reading his solution (I only solved half the problem and got struck). To anyone in the future reading this, math is beautiful.<br /> -First<br /> <br /> ==Solution 2 (Inversion)==<br /> Consider an inversion with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;r=AK&lt;/math&gt;. Then, we have &lt;math&gt;AB\cdot AB^*=AK^2&lt;/math&gt;, or &lt;math&gt;AB^*=\frac{AK^2}{7}&lt;/math&gt;. Similarly, &lt;math&gt;AC^*=\frac{AK^2}{9}&lt;/math&gt;. Notice that &lt;math&gt;AB^*KC^*&lt;/math&gt; is a parallelogram, since &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; are tangent to &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Thus, &lt;math&gt;AC^*=B^*K&lt;/math&gt;. Now, we get that<br /> &lt;cmath&gt;\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}&lt;/cmath&gt;<br /> so by Law of Cosines on &lt;math&gt;\triangle AB^*K&lt;/math&gt; we have <br /> &lt;cmath&gt;(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{11AK^2}{63\cdot21}&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow AK=\frac{9}{2}&lt;/cmath&gt;<br /> Then, our answer is &lt;math&gt;9+2=\boxed{11}&lt;/math&gt;. <br /> -brianzjk<br /> <br /> <br /> <br /> == Solution 3 (Death By Trig Bash) ==<br /> <br /> 14. Let the centers of the circles be &lt;math&gt;O_{1}&lt;/math&gt; and &lt;math&gt;O_{2}&lt;/math&gt; where the &lt;math&gt;O_{1}&lt;/math&gt; has the side length &lt;math&gt;7&lt;/math&gt; contained in the circle. Now let &lt;math&gt;\angle BAC =x.&lt;/math&gt; This implies &lt;math&gt;\angle AO_{1}B = \angle AO_{2}C = 2x&lt;/math&gt; by the angle by by tangent. Then we also know that &lt;math&gt;\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x.&lt;/math&gt; Now we first find &lt;math&gt;\cos x.&lt;/math&gt; We use law of cosines on &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; to obtain &lt;math&gt;64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x} \implies \cos{x} =\frac{11}{21} \implies \sin{x} =\frac{8\sqrt{5}}{21}.&lt;/math&gt; Then applying law of sines on &lt;math&gt;\bigtriangleup AO_{1}B&lt;/math&gt; we obtain &lt;math&gt;\frac{7}{\sin{2x}} =\frac{OB_{1}}{\sin{90^{\circ}-x}} \implies\frac{7}{2\sin{x}\cos{x}} =\frac{OB_{1}}{\cos{x}} \implies OB_{1} = O_{1}A=\frac{147}{16\sqrt{5}}.&lt;/math&gt; Using similar logic we obtain &lt;math&gt;OA_{1} =\frac{189}{16\sqrt{5}}.&lt;/math&gt; Now we know that &lt;math&gt;\angle O_{1}AO_{2}=180^{\circ}-x.&lt;/math&gt; Thus using law of cosines on &lt;math&gt;\bigtriangleup O_{1}AO_{2}&lt;/math&gt; yields &lt;math&gt;O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}.&lt;/math&gt; While this does look daunting we can write the above expression as &lt;math&gt;\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}.&lt;/math&gt; Then factoring yields &lt;math&gt;\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}.&lt;/math&gt; The area &lt;math&gt;[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}.&lt;/math&gt; Now &lt;math&gt;AK&lt;/math&gt; is twice the length of the altitude of &lt;math&gt;\bigtriangleup O_{1}AO_{2}&lt;/math&gt; so we let the altitude be &lt;math&gt;h&lt;/math&gt; and we have &lt;math&gt;\frac{1}{2} \cdot h \cdot\frac{147\sqrt{5}}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21} \implies h =\frac{9}{4}.&lt;/math&gt; Thus our desired length is &lt;math&gt;\frac{9}{2} \implies n+n = \boxed{11}.&lt;/math&gt;<br /> <br /> ==Solution 4 (Video)==<br /> <br /> Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=10|num-a=12}}<br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_13&diff=128352 2010 AIME I Problems/Problem 13 2020-07-15T16:30:42Z <p>First: /* Solution */</p> <hr /> <div>__TOC__<br /> == Problem ==<br /> [[Rectangle]] &lt;math&gt;ABCD&lt;/math&gt; and a [[semicircle]] with diameter &lt;math&gt;AB&lt;/math&gt; are coplanar and have nonoverlapping interiors. Let &lt;math&gt;\mathcal{R}&lt;/math&gt; denote the region enclosed by the semicircle and the rectangle. Line &lt;math&gt;\ell&lt;/math&gt; meets the semicircle, segment &lt;math&gt;AB&lt;/math&gt;, and segment &lt;math&gt;CD&lt;/math&gt; at distinct points &lt;math&gt;N&lt;/math&gt;, &lt;math&gt;U&lt;/math&gt;, and &lt;math&gt;T&lt;/math&gt;, respectively. Line &lt;math&gt;\ell&lt;/math&gt; divides region &lt;math&gt;\mathcal{R}&lt;/math&gt; into two regions with areas in the ratio &lt;math&gt;1: 2&lt;/math&gt;. Suppose that &lt;math&gt;AU = 84&lt;/math&gt;, &lt;math&gt;AN = 126&lt;/math&gt;, and &lt;math&gt;UB = 168&lt;/math&gt;. Then &lt;math&gt;DA&lt;/math&gt; can be represented as &lt;math&gt;m\sqrt {n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> ===Diagram===<br /> &lt;center&gt;&lt;asy&gt; /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */<br /> import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500);<br /> pen zzttqq = rgb(0.6,0.2,0);<br /> pen xdxdff = rgb(0.4902,0.4902,1);<br /> <br /> /* segments and figures */<br /> draw((0,-154.31785)--(0,0));<br /> draw((0,0)--(252,0));<br /> draw((0,0)--(126,0),zzttqq);<br /> draw((126,0)--(63,109.1192),zzttqq);<br /> draw((63,109.1192)--(0,0),zzttqq);<br /> draw((-71.4052,(+9166.01287-109.1192*-71.4052)/21)--(504.60925,(+9166.01287-109.1192*504.60925)/21));<br /> draw((0,-154.31785)--(252,-154.31785));<br /> draw((252,-154.31785)--(252,0));<br /> draw((0,0)--(84,0));<br /> draw((84,0)--(252,0));<br /> draw((63,109.1192)--(63,0));<br /> draw((84,0)--(84,-154.31785));<br /> draw(arc((126,0),126,0,180));<br /> <br /> /* points and labels */<br /> dot((0,0));<br /> label(&quot;$A$&quot;,(-16.43287,-9.3374),NE/2);<br /> dot((252,0));<br /> label(&quot;$B$&quot;,(255.242,5.00321),NE/2);<br /> dot((0,-154.31785));<br /> label(&quot;$D$&quot;,(3.48464,-149.55669),NE/2);<br /> dot((252,-154.31785));<br /> label(&quot;$C$&quot;,(255.242,-149.55669),NE/2);<br /> dot((126,0));<br /> label(&quot;$O$&quot;,(129.36332,5.00321),NE/2);<br /> dot((63,109.1192));<br /> label(&quot;$N$&quot;,(44.91307,108.57427),NE/2);<br /> label(&quot;$126$&quot;,(28.18236,40.85473),NE/2);<br /> dot((84,0));<br /> label(&quot;$U$&quot;,(87.13819,5.00321),NE/2);<br /> dot((113.69848,-154.31785));<br /> label(&quot;$T$&quot;,(116.61611,-149.55669),NE/2);<br /> dot((63,0));<br /> label(&quot;$N'$&quot;,(66.42398,5.00321),NE/2);<br /> label(&quot;$84$&quot;,(41.72627,-12.5242),NE/2);<br /> label(&quot;$168$&quot;,(167.60494,-12.5242),NE/2);<br /> dot((84,-154.31785));<br /> label(&quot;$T'$&quot;,(87.13819,-149.55669),NE/2);<br /> dot((252,0));<br /> label(&quot;$I$&quot;,(255.242,5.00321),NE/2);<br /> clip((-71.4052,-225.24323)--(-71.4052,171.51361)--(504.60925,171.51361)--(504.60925,-225.24323)--cycle);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> === Solution 1 ===<br /> <br /> The center of the semicircle is also the midpoint of &lt;math&gt;AB&lt;/math&gt;. Let this point be O. Let &lt;math&gt;h&lt;/math&gt; be the length of &lt;math&gt;AD&lt;/math&gt;.<br /> <br /> Rescale everything by 42, so &lt;math&gt;AU = 2, AN = 3, UB = 4&lt;/math&gt;. Then &lt;math&gt;AB = 6&lt;/math&gt; so &lt;math&gt;OA = OB = 3&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;ON&lt;/math&gt; is a radius of the semicircle, &lt;math&gt;ON = 3&lt;/math&gt;. Thus &lt;math&gt;OAN&lt;/math&gt; is an equilateral triangle.<br /> <br /> Let &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; be the areas of triangle &lt;math&gt;OUN&lt;/math&gt;, sector &lt;math&gt;ONB&lt;/math&gt;, and trapezoid &lt;math&gt;UBCT&lt;/math&gt; respectively.<br /> <br /> &lt;math&gt;X = \frac {1}{2}(UO)(NO)\sin{O} = \frac {1}{2}(1)(3)\sin{60^\circ} = \frac {3}{4}\sqrt {3}&lt;/math&gt;<br /> <br /> &lt;math&gt;Y = \frac {1}{3}\pi(3)^2 = 3\pi&lt;/math&gt;<br /> <br /> To find &lt;math&gt;Z&lt;/math&gt; we have to find the length of &lt;math&gt;TC&lt;/math&gt;. Project &lt;math&gt;T&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; onto &lt;math&gt;AB&lt;/math&gt; to get points &lt;math&gt;T'&lt;/math&gt; and &lt;math&gt;N'&lt;/math&gt;. Notice that &lt;math&gt;UNN'&lt;/math&gt; and &lt;math&gt;TUT'&lt;/math&gt; are similar. Thus:<br /> <br /> &lt;math&gt;\frac {TT'}{UT'} = \frac {UN'}{NN'} \implies \frac {TT'}{h} = \frac {1/2}{3\sqrt {3}/2} \implies TT' = \frac {\sqrt {3}}{9}h&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;TC = T'C - T'T = UB - TT' = 4 - \frac {\sqrt {3}}{9}h&lt;/math&gt;. So:<br /> <br /> &lt;math&gt;Z = \frac {1}{2}(BU + TC)(CB) = \frac {1}{2}\left(8 - \frac {\sqrt {3}}{9}h\right)h = 4h - \frac {\sqrt {3}}{18}h^2&lt;/math&gt;<br /> <br /> Let &lt;math&gt;L&lt;/math&gt; be the area of the side of line &lt;math&gt;l&lt;/math&gt; containing regions &lt;math&gt;X, Y, Z&lt;/math&gt;. Then<br /> <br /> &lt;math&gt;L = X + Y + Z = \frac {3}{4}\sqrt {3} + 3\pi + 4h - \frac {\sqrt {3}}{18}h^2&lt;/math&gt;<br /> <br /> Obviously, the &lt;math&gt;L&lt;/math&gt; is greater than the area on the other side of line &lt;math&gt;l&lt;/math&gt;. This other area is equal to the total area minus &lt;math&gt;L&lt;/math&gt;. Thus:<br /> <br /> &lt;math&gt;\frac {2}{1} = \frac {L}{6h + \frac {9}{2}{\pi} - L} \implies 12h + 9\pi = 3L&lt;/math&gt;.<br /> <br /> Now just solve for &lt;math&gt;h&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*} 12h + 9\pi &amp; = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\<br /> 0 &amp; = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\<br /> h^2 &amp; = \frac {9}{4}(6) \\<br /> h &amp; = \frac {3}{2}\sqrt {6} \end{align*}&lt;/cmath&gt;<br /> <br /> Don't forget to un-rescale at the end to get &lt;math&gt;AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}&lt;/math&gt;. <br /> <br /> Finally, the answer is &lt;math&gt;63 + 6 = \boxed{069}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the semicircle. It follows that &lt;math&gt;AU + UO = AN = NO = 126&lt;/math&gt;, so triangle &lt;math&gt;ANO&lt;/math&gt; is [[equilateral]].<br /> <br /> Let &lt;math&gt;Y&lt;/math&gt; be the foot of the altitude from &lt;math&gt;N&lt;/math&gt;, such that &lt;math&gt;NY = 63\sqrt{3}&lt;/math&gt; and &lt;math&gt;NU = 21&lt;/math&gt;.<br /> <br /> Finally, denote &lt;math&gt;DT = a&lt;/math&gt;, and &lt;math&gt;AD = x&lt;/math&gt;. Extend &lt;math&gt;U&lt;/math&gt; to point &lt;math&gt;Z&lt;/math&gt; so that &lt;math&gt;Z&lt;/math&gt; is on &lt;math&gt;CD&lt;/math&gt; and &lt;math&gt;UZ&lt;/math&gt; is perpendicular to &lt;math&gt;CD&lt;/math&gt;. It then follows that &lt;math&gt;ZT = a-84&lt;/math&gt;. Since &lt;math&gt;NYU&lt;/math&gt; and &lt;math&gt;UZT&lt;/math&gt; are [[similar]],<br /> <br /> &lt;math&gt;\frac {x}{a-84} = \frac {63\sqrt{3}}{21} = 3\sqrt{3}&lt;/math&gt;<br /> <br /> Given that line &lt;math&gt;NT&lt;/math&gt; divides &lt;math&gt;R&lt;/math&gt; into a ratio of &lt;math&gt;1:2&lt;/math&gt;, we can also say that<br /> <br /> &lt;math&gt;(x)(\frac{84+a}{2}) + \frac {126^2\pi}{6} - (63)(21)(\sqrt{3}) = (\frac{1}{3})(252x + \frac{126^2\pi}{2})&lt;/math&gt;<br /> <br /> where the first term is the area of trapezoid &lt;math&gt;AUTD&lt;/math&gt;, the second and third terms denote the areas of &lt;math&gt;\frac{1}{6}&lt;/math&gt; a full circle, and the area of &lt;math&gt;NUO&lt;/math&gt;, respectively, and the fourth term on the right side of the equation is equal to &lt;math&gt;R&lt;/math&gt;. Cancelling out the &lt;math&gt;\frac{126^2\pi}{6}&lt;/math&gt; on both sides, we obtain<br /> <br /> &lt;math&gt;(x)(\frac{84+a}{2}) - \frac{252x}{3} = (63)(21)(\sqrt{3})&lt;/math&gt;<br /> <br /> By adding and collecting like terms,<br /> &lt;math&gt;\frac{3ax - 252x}{6} = (63)(21)(\sqrt{3})&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{(3x)(a-84)}{6} = (63)(21)(\sqrt{3})&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;a - 84 = \frac{x}{3\sqrt{3}}&lt;/math&gt;,<br /> <br /> &lt;math&gt;\frac {(3x)(\frac{x}{3\sqrt{3}})}{6} = (63)(21)(\sqrt{3})&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac {x^2}{\sqrt{3}} = (63)(126)(\sqrt{3})&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 = (63)(126)(3) = (2)(3^5)(7^2)&lt;/math&gt;<br /> <br /> &lt;math&gt;x = AD = (7)(3^2)(\sqrt{6}) = 63\sqrt{6}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{069}.&lt;/math&gt;<br /> <br /> <br /> === Solution 3 ===<br /> <br /> Note that the total area of &lt;math&gt; \mathcal{R} &lt;/math&gt; is &lt;math&gt;252DA + \frac {126^2 \pi}{2}&lt;/math&gt; and thus one of the regions has area &lt;math&gt;84DA + \frac {126^2 \pi}{6}&lt;/math&gt;<br /> <br /> As in the above solutions we discover that &lt;math&gt;\angle AON = 60^\circ&lt;/math&gt;, thus sector &lt;math&gt;ANO&lt;/math&gt; of the semicircle has &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the semicircle's area.<br /> <br /> Similarly, dropping the &lt;math&gt;N'T'&lt;/math&gt; perpendicular we observe that &lt;math&gt;[AN'T'D] = 84DA&lt;/math&gt;, which is &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the total rectangle.<br /> <br /> Denoting the region to the left of &lt;math&gt;\overline {NT}&lt;/math&gt; as &lt;math&gt;\alpha&lt;/math&gt; and to the right as &lt;math&gt;\beta&lt;/math&gt;, it becomes clear that if &lt;math&gt;[\triangle UT'T] = [\triangle NUO]&lt;/math&gt; then the regions will have the desired ratio.<br /> <br /> Using the 30-60-90 triangle, the slope of &lt;math&gt;NT&lt;/math&gt;, is &lt;math&gt;{-3\sqrt{3}}&lt;/math&gt;, and thus &lt;math&gt;[\triangle UT'T] = \frac {DA^2}{6\sqrt{3}}&lt;/math&gt;.<br /> <br /> &lt;math&gt;[NUO]&lt;/math&gt; is most easily found by &lt;math&gt;\frac{absin(c)}{2}&lt;/math&gt;: &lt;math&gt;[\triangle NUO] = \frac {126*42 * \frac {\sqrt{3}}{2}}{2}&lt;/math&gt;<br /> <br /> Equating, &lt;math&gt;\frac {126*42 * \frac {\sqrt{3}}{2}}{2} = \frac {DA^2}{6\sqrt{3}}&lt;/math&gt;<br /> <br /> Solving, &lt;math&gt; 63 * 21 * 3 * 6 = DA^2&lt;/math&gt;<br /> <br /> &lt;math&gt;DA = 63 \sqrt{6} \longrightarrow \boxed {069}&lt;/math&gt;<br /> <br /> === Solution 4 (Coordinates) ===<br /> Like above solutions, note that &lt;math&gt;ANO&lt;/math&gt; is equilateral with side length &lt;math&gt;126,&lt;/math&gt; where &lt;math&gt;O&lt;/math&gt; is the midpoint of &lt;math&gt;AB.&lt;/math&gt; Then, if we let &lt;math&gt;DA=a&lt;/math&gt; and set origin at &lt;math&gt;D=(0,0),&lt;/math&gt; we get &lt;math&gt;N=(63,a+63\sqrt{3}), U=(84,a).&lt;/math&gt; Line &lt;math&gt;NU&lt;/math&gt; is then &lt;math&gt;y-a=\sqrt{27}(x-84),&lt;/math&gt; so it intersects &lt;math&gt;CA,&lt;/math&gt; the &lt;math&gt;x&lt;/math&gt;-axis, at &lt;math&gt;x=(a/\sqrt{27}+84),&lt;/math&gt; giving us point &lt;math&gt;T.&lt;/math&gt; Now the area of region &lt;math&gt;R&lt;/math&gt; is &lt;math&gt;252a+\pi(126)^2 / 2,&lt;/math&gt; so one third of that is &lt;math&gt;84a+\pi(126)^2 / 6.&lt;/math&gt;<br /> <br /> The area of the smaller piece of &lt;math&gt;R&lt;/math&gt; is &lt;math&gt;[AUTD] + [ANU] + [\text{lune} AN] = \frac{1}{2} \cdot a(84+\frac{a}{\sqrt{27}}+84)+\frac{1}{2} \cdot 84 \cdot 63 \sqrt{3}+ \frac{\pi (126)^2}{6}-\frac{1}{2} \cdot 126 \cdot 63 \sqrt{2}&lt;/math&gt;<br /> &lt;math&gt;=\frac{a^2}{2\sqrt{27}}+84a-21\cdot 63\sqrt{2} + \frac{\pi(126)^2}{6}.&lt;/math&gt;<br /> Setting this equal to &lt;math&gt;84a+\pi(126)^2 / 6&lt;/math&gt; and canceling the &lt;math&gt;84a + \pi(126)^2&lt;/math&gt; yields<br /> &lt;math&gt;\frac{a^2}{2\sqrt{27}}=21 \cdot 63 \sqrt{3},&lt;/math&gt; so &lt;math&gt;a = 63 \sqrt{6}&lt;/math&gt; and the anser is &lt;math&gt;\boxed{069}.&lt;/math&gt;<br /> ~ rzlng<br /> <br /> == See Also ==<br /> *&lt;url&gt;viewtopic.php?t=338915 Discussion&lt;/url&gt;, with a Geogebra diagram.<br /> <br /> {{AIME box|year=2010|num-b=12|num-a=14|n=I}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_12&diff=124281 2005 AIME II Problems/Problem 12 2020-06-07T20:32:03Z <p>First: /* Problem */</p> <hr /> <div>== Problem ==<br /> [[Square]] &lt;math&gt;ABCD &lt;/math&gt; has [[center]] &lt;math&gt; O,\ AB=900,\ E &lt;/math&gt; and &lt;math&gt; F &lt;/math&gt; are on &lt;math&gt; AB &lt;/math&gt; with &lt;math&gt; AE&lt;BF &lt;/math&gt; and &lt;math&gt; E &lt;/math&gt; between &lt;math&gt; A &lt;/math&gt; and &lt;math&gt; F, m\angle EOF =45^\circ, &lt;/math&gt; and &lt;math&gt; EF=400. &lt;/math&gt; Given that &lt;math&gt; BF=p+q\sqrt{r}, &lt;/math&gt; where &lt;math&gt; p,q, &lt;/math&gt; and &lt;math&gt; r &lt;/math&gt; are [[positive]] [[integer]]s and &lt;math&gt; r &lt;/math&gt; is not divisible by the [[square]] of any [[prime]], find &lt;math&gt; p+q+r. &lt;/math&gt;<br /> __TOC__<br /> <br /> == Solutions ==<br /> === Solution 1 (trigonometry) ===<br /> &lt;center&gt;&lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);draw(G--O); dot(A^^B^^C^^D^^E^^F^^G^^O); label(&quot;$$A$$&quot;,A,(-1,1));label(&quot;$$B$$&quot;,B,(1,1));label(&quot;$$C$$&quot;,C,(1,-1));label(&quot;$$D$$&quot;,D,(-1,-1)); label(&quot;$$E$$&quot;,E,(0,1));label(&quot;$$F$$&quot;,F,(1,1));label(&quot;$$G$$&quot;,G,(-1,1));label(&quot;$$O$$&quot;,O,(1,-1)); label(&quot;$$x$$&quot;,E/2+G/2,(0,1));label(&quot;$$y$$&quot;,G/2+F/2,(0,1)); label(&quot;$$450$$&quot;,(O+G)/2,(-1,1)); <br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- Asymptote replacement for Image:AIME_2005II_Solution_12_1.png by Minsoens --&gt;<br /> <br /> Let &lt;math&gt;G&lt;/math&gt; be the foot of the [[perpendicular]] from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt;. Denote &lt;math&gt;x = EG&lt;/math&gt; and &lt;math&gt;y = FG&lt;/math&gt;, and &lt;math&gt;x &gt; y&lt;/math&gt; (since &lt;math&gt;AE &lt; BF&lt;/math&gt; and &lt;math&gt;AG = BG&lt;/math&gt;). Then &lt;math&gt;\tan \angle EOG = \frac{x}{450}&lt;/math&gt;, and &lt;math&gt;\tan \angle FOG = \frac{y}{450}&lt;/math&gt;.<br /> <br /> By the [[trigonometric identity|tangent addition rule]] &lt;math&gt;\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)&lt;/math&gt;, we see that &lt;cmath&gt;\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.&lt;/cmath&gt; Since &lt;math&gt;\tan 45 = 1&lt;/math&gt;, this simplifies to &lt;math&gt;1 - \frac{xy}{450^2} = \frac{x + y}{450}&lt;/math&gt;. We know that &lt;math&gt;x + y = 400&lt;/math&gt;, so we can substitute this to find that &lt;math&gt;1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;x = 400 - y&lt;/math&gt; again, we know have &lt;math&gt;xy = (400 - y)y = 150^2&lt;/math&gt;. This is a quadratic with roots &lt;math&gt;200 \pm 50\sqrt{7}&lt;/math&gt;. Since &lt;math&gt;y &lt; x&lt;/math&gt;, use the smaller root, &lt;math&gt;200 - 50\sqrt{7}&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}&lt;/math&gt;. The answer is &lt;math&gt;250 + 50 + 7 = \boxed{307}&lt;/math&gt;.<br /> <br /> === Solution 2 (synthetic) ===<br /> &lt;center&gt;&lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), O=(4.5,4.5), G=O+(E-O)*dir(-90), J=O+(F-O)*dir(-90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype(&quot;4 4&quot;)); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label(&quot;$$A$$&quot;,A,(-1,1));label(&quot;$$B$$&quot;,B,(1,1));label(&quot;$$C$$&quot;,C,(1,-1));label(&quot;$$D$$&quot;,D,(-1,-1)); label(&quot;$$E$$&quot;,E,(0,1));label(&quot;$$F$$&quot;,F,(1,1));label(&quot;$$G$$&quot;,G,(1,0));label(&quot;$$J$$&quot;,J,(1,0));label(&quot;$$O$$&quot;,O,(1,-1)); label(&quot;$$x$$&quot;,(B+F)/2,(0,1)); label(&quot;$$400$$&quot;,(E+F)/2,(0,1)); label(&quot;$$900$$&quot;,(C+D)/2,(0,-1));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Label &lt;math&gt;BF=x&lt;/math&gt;, so &lt;math&gt;EA =&lt;/math&gt; &lt;math&gt;500 - x&lt;/math&gt;. Rotate &lt;math&gt;\triangle{OEF}&lt;/math&gt; about &lt;math&gt;O&lt;/math&gt; until &lt;math&gt;EF&lt;/math&gt; lies on &lt;math&gt;BC&lt;/math&gt;. Now we know that &lt;math&gt;\angle{EOF}=45^\circ&lt;/math&gt; therefore &lt;math&gt;\angle BOF+\angle AOE=45^\circ&lt;/math&gt; also since &lt;math&gt;O&lt;/math&gt; is the center of the square. Label the new triangle that we created &lt;math&gt;\triangle OGJ&lt;/math&gt;. Now we know that rotation preserves angles and side lengths, so &lt;math&gt;BG=500-x&lt;/math&gt; and &lt;math&gt;JC=x&lt;/math&gt;. Draw &lt;math&gt;GF&lt;/math&gt; and &lt;math&gt;OB&lt;/math&gt;. Notice that &lt;math&gt;\angle BOG =\angle OAE&lt;/math&gt; since rotations preserve the same angles so <br /> &lt;math&gt;\angle{FOG}=45^\circ&lt;/math&gt; too. By SAS we know that &lt;math&gt;\triangle FOE\cong \triangle FOG,&lt;/math&gt; so &lt;math&gt;FG=400&lt;/math&gt;. Now we have a right &lt;math&gt;\triangle BFG&lt;/math&gt; with legs &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;500-x&lt;/math&gt; and hypotenuse &lt;math&gt;400&lt;/math&gt;. By the [[Pythagorean Theorem]], <br /> <br /> &lt;cmath&gt;\begin{align*}<br /> (500-x)^2+x^2&amp;=400^2 \\<br /> 250000-1000x+2x^2&amp;=16000 \\<br /> 90000-1000x+2x^2&amp;=0 \end{align*}&lt;/cmath&gt;<br /> <br /> and applying the [[quadratic formula]] we get that <br /> &lt;math&gt;x=250\pm 50\sqrt{7}&lt;/math&gt;. Since &lt;math&gt;BF &gt; AE,&lt;/math&gt; we take the positive root, and our answer is &lt;math&gt;p+q+r = 250 + 50 + 7 = 307&lt;/math&gt;.<br /> <br /> === Solution 3 (similar triangles)===<br /> &lt;asy&gt;<br /> size(3inch);<br /> pair A, B, C, D, M, O, X, Y;<br /> A = (0,900); B = (900,900); C = (900,0); D = (0,0);<br /> M = (450,900); O = (450,450); X = (250 - 50*sqrt(7),900); Y = (650 - 50*sqrt(7),900);<br /> draw(A--B--C--D--cycle);<br /> draw(X--O--Y);<br /> draw(M--O--A);<br /> label(&quot;$A$&quot;,A,NW); label(&quot;$B$&quot;,B,NE); label(&quot;$C$&quot;,C,SE); label(&quot;$D$&quot;,D,SW); label(&quot;$E$&quot;,X,N); label(&quot;$F$&quot;,Y,NNE); label(&quot;$O$&quot;,O,S); label(&quot;$M$&quot;,M,N);<br /> &lt;/asy&gt;<br /> Let the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt; and let &lt;math&gt;FB = x&lt;/math&gt;, so then &lt;math&gt;MF = 450 - x&lt;/math&gt; and &lt;math&gt;AF = 900 - x&lt;/math&gt;. Drawing &lt;math&gt;\overline{AO}&lt;/math&gt;, we have &lt;math&gt;\triangle OEF\sim\triangle AOF&lt;/math&gt;, so<br /> &lt;cmath&gt;\frac{OF}{EF} = \frac{AF}{OF}\Rightarrow (OF)^2 = 400(900 - x).&lt;/cmath&gt;<br /> By the Pythagorean Theorem on &lt;math&gt;\triangle OMF&lt;/math&gt;,<br /> &lt;cmath&gt;(OF)^2 = 450^2 + (450 - x)^2.&lt;/cmath&gt;<br /> Setting these two expressions for &lt;math&gt;(OF)^2&lt;/math&gt; equal and solving for &lt;math&gt;x&lt;/math&gt; (it is helpful to scale the problem down by a factor of 50 first), we get &lt;math&gt;x = 250\pm 50\sqrt{7}&lt;/math&gt;. Since &lt;math&gt;BF &gt; AE&lt;/math&gt;, we want the value &lt;math&gt;x = 250 + 50\sqrt{7}&lt;/math&gt;, and the answer is &lt;math&gt;250 + 50 + 7 = \boxed{307}&lt;/math&gt;.<br /> <br /> === Solution 4 (Abusing Stewart) ===<br /> Let &lt;math&gt;x = BF&lt;/math&gt;, so &lt;math&gt;AE = 500-x&lt;/math&gt;. Let &lt;math&gt;a = OE&lt;/math&gt;, &lt;math&gt;b = OF&lt;/math&gt;. Applying Stewart's Theorem on triangles &lt;math&gt;AOB&lt;/math&gt; twice, first using &lt;math&gt;E&lt;/math&gt; as the base point and then &lt;math&gt;F&lt;/math&gt;, we arrive at the equations &lt;cmath&gt;(450 \sqrt{2})^2 (900) = 900(500-x)(400+x) + a^2 (900)&lt;/cmath&gt; and &lt;cmath&gt;(450 \sqrt{2})^2 (900) = 900x(900-x) + b^2 (900)&lt;/cmath&gt; Now applying law of sines and law of cosines on &lt;math&gt;\triangle EOF&lt;/math&gt; yields &lt;cmath&gt;\frac{1}{2} ab \sin 45^{\circ} = \frac{4}{9} \times \frac{1}{4} \times 900^2 = 202500&lt;/cmath&gt; and &lt;cmath&gt;a^2+b^2- 2 ab \cos 45^{\circ} = 160000&lt;/cmath&gt; Solving for &lt;math&gt;ab&lt;/math&gt; from the sines equation and plugging into the law of cosines equation yields &lt;math&gt;a^2+b^2 = 290000&lt;/math&gt;. We now finish by adding the two original stewart equations and obtaining: &lt;cmath&gt;2(450\sqrt{2})^2 = (500-x)(400+x)+x(900-x)+520000&lt;/cmath&gt; This is a quadratic which only takes some patience to solve for &lt;math&gt;x = 250 + 50\sqrt{7}&lt;/math&gt;<br /> <br /> === Solution 5 (Complex Numbers) ===<br /> Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with &lt;math&gt;o = 0, a = -450 + 450i, b = 450 + 450i&lt;/math&gt;, and &lt;math&gt;f = x + 450i&lt;/math&gt;. Since &lt;math&gt;EF&lt;/math&gt; = 400, &lt;math&gt;e = (x-400) + 450i&lt;/math&gt;. From &lt;math&gt;\angle{EOF} = 45^{\circ}&lt;/math&gt;, we can deduce that the rotation of point &lt;math&gt;F&lt;/math&gt; 45 degrees counterclockwise, &lt;math&gt;E&lt;/math&gt;, and the origin are collinear. In other words, &lt;cmath&gt;\dfrac{e^{i \frac{\pi}{4}} \cdot (x + 450i)}{(x - 400) + 450i}&lt;/cmath&gt; is a real number. Simplyfying using the fact that &lt;math&gt;e^{i \frac{\pi}{4}} = \dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}&lt;/math&gt;, clearing the denominator, and setting the imaginary part equal to &lt;math&gt;0&lt;/math&gt;, we eventually get the quadratic &lt;cmath&gt;x^2 - 400x + 22500 = 0&lt;/cmath&gt; which has solutions &lt;math&gt;x = 200 \pm 50\sqrt{7}&lt;/math&gt;. It is given that &lt;math&gt;AE &lt; BF&lt;/math&gt;, so &lt;math&gt;x = 200 - 50\sqrt{7}&lt;/math&gt; and &lt;cmath&gt;BF = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7} \Rightarrow \boxed{307}.&lt;/cmath&gt;<br /> <br /> -MP8148<br /> <br /> === Solution 6 ===<br /> &lt;asy&gt;<br /> size(250);<br /> pair A,B,C,D,O,E,F,G,H,K;<br /> A = (0,0);<br /> B = (900,0);<br /> C = (900,900);<br /> D = (0,900);<br /> O = (450,450);<br /> E = (600,0);<br /> F = (150,0);<br /> G = (-600,0);<br /> H = (450,0);<br /> K = (0,270);<br /> draw(A--B--C--D--cycle);<br /> draw(O--E);<br /> draw(O--F);<br /> draw(O--G);<br /> draw(A--G);<br /> draw(O--H);<br /> label(&quot;O&quot;,O,N);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,SE);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,NW);<br /> label(&quot;E&quot;,E,SE);<br /> label(&quot;F&quot;,F,S);<br /> label(&quot;H&quot;,H,SW);<br /> label(&quot;G&quot;,G,SW);<br /> label(&quot;x&quot;,H--E,S);<br /> label(&quot;K&quot;,K,NW);<br /> &lt;/asy&gt;<br /> Let G be a point such that it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB.<br /> Since &lt;math&gt;\triangle GOE \sim \triangle OHE&lt;/math&gt;, &lt;math&gt;\frac{GO}{OE} = \frac{450}{x}&lt;/math&gt;, and by [[Angle Bisector Theorem]], &lt;math&gt;\frac{GF}{FE} = \frac{450}{x}&lt;/math&gt;. Thus, &lt;math&gt;GF = \frac{450 \cdot 400}{x}&lt;/math&gt;. &lt;math&gt;AF = AH-FH = 50+x&lt;/math&gt;, and &lt;math&gt;KA = EB&lt;/math&gt; (90 degree rotation), and now we can bash on 2 similar triangles &lt;math&gt;\triangle GAK \sim \triangle GHO&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\frac{GA}{AK} = \frac{GH}{OH}&lt;/cmath&gt; <br /> &lt;cmath&gt;\frac{\frac{450 \cdot 400}{x}-50-x}{450-x} = \frac{\frac{450 \cdot 400}{x}+400-x}{450}&lt;/cmath&gt;<br /> I hope you like expanding<br /> &lt;cmath&gt;x^2 - 850x + \frac{81000000}{x} = -450x - 22500 + \frac{81000000}{x}&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2 - 400x + 22500 = 0&lt;/cmath&gt;<br /> Quadratic formula gives us<br /> &lt;cmath&gt;x = 200 \pm 50 \sqrt{7}&lt;/cmath&gt;<br /> Since AE &lt; BF <br /> &lt;cmath&gt;x = 200 - 50 \sqrt{7}&lt;/cmath&gt;<br /> Thus, <br /> &lt;cmath&gt;BF = 250 + 50 \sqrt{7}&lt;/cmath&gt;<br /> So, our answer is &lt;math&gt;\boxed{307}&lt;/math&gt;.<br /> <br /> -AlexLikeMath<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_14&diff=121957 1998 AIME Problems/Problem 14 2020-05-03T18:17:26Z <p>First: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> An &lt;math&gt;m\times n\times p&lt;/math&gt; rectangular box has half the volume of an &lt;math&gt;(m + 2)\times(n + 2)\times(p + 2)&lt;/math&gt; rectangular box, where &lt;math&gt;m, n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers, and &lt;math&gt;m\le n\le p.&lt;/math&gt; What is the largest possible value of &lt;math&gt;p&lt;/math&gt;?<br /> <br /> == Solution 1 ==<br /> &lt;cmath&gt;2mnp = (m+2)(n+2)(p+2)&lt;/cmath&gt;<br /> <br /> Let’s solve for &lt;math&gt;p&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)&lt;/cmath&gt;<br /> &lt;cmath&gt;[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)&lt;/cmath&gt;<br /> &lt;cmath&gt;p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}&lt;/cmath&gt;<br /> <br /> Clearly, we want to minimize the denominator, so we test &lt;math&gt;(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9&lt;/math&gt;. The possible pairs of factors of &lt;math&gt;9&lt;/math&gt; are &lt;math&gt;(1,9)(3,3)&lt;/math&gt;. These give &lt;math&gt;m = 3, n = 11&lt;/math&gt; and &lt;math&gt;m = 5, n = 5&lt;/math&gt; respectively. Substituting into the numerator, we see that the first pair gives &lt;math&gt;130&lt;/math&gt;, while the second pair gives &lt;math&gt;98&lt;/math&gt;. We now check that &lt;math&gt;130&lt;/math&gt; is optimal, setting &lt;math&gt;a=m-2&lt;/math&gt;, &lt;math&gt;b=n-2&lt;/math&gt; in order to simplify calculations. Since<br /> &lt;cmath&gt;0 \le (a-1)(b-1) \implies a+b \le ab+1&lt;/cmath&gt;<br /> We have<br /> &lt;cmath&gt;p = \frac{2(a+4)(b+4)}{ab-8} = \frac{2ab+8(a+b)+32}{ab-8} \le \frac{2ab+8(ab+1)+32}{ab-8} = 10 + \frac{120}{ab-8} \le 130&lt;/cmath&gt;<br /> Where we see &lt;math&gt;(m,n)=(3,11)&lt;/math&gt; gives us our maximum value of &lt;math&gt;\boxed{130}&lt;/math&gt;.<br /> <br /> *Note that &lt;math&gt;0 \le (a-1)(b-1)&lt;/math&gt; assumes &lt;math&gt;m,n \ge 3&lt;/math&gt;, but this is clear as &lt;math&gt;\frac{2m}{m+2} = \frac{(n+2)(p+2)}{np} &gt; 1&lt;/math&gt; and similarly for &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Similarly as above, we solve for &lt;math&gt;p,&lt;/math&gt; but we express the denominator differently:<br /> <br /> &lt;cmath&gt;p=\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(m+n+2)}{(m+2)(n+2)}.&lt;/cmath&gt;<br /> Hence, it suffices to maximize &lt;math&gt;\dfrac{m+n+2}{(m+2)(n+2)},&lt;/math&gt; under the conditions that &lt;math&gt;p&lt;/math&gt; is a positive integer.<br /> <br /> Then since &lt;math&gt;\dfrac{m+n+2}{(m+2)(n+2)}&gt;\dfrac{1}{2}&lt;/math&gt; for &lt;math&gt;m=1,2,&lt;/math&gt; we fix &lt;math&gt;m=3.&lt;/math&gt;<br /> &lt;cmath&gt;\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+2)}{5(n+2)}=\dfrac{n-10}{10(n+2)},&lt;/cmath&gt; <br /> where we simply let &lt;math&gt;n=11&lt;/math&gt; to achieve &lt;math&gt;p=\boxed{130}.&lt;/math&gt;<br /> <br /> ~Generic_Username<br /> <br /> == Solution 3 ==<br /> Observe that <br /> &lt;cmath&gt;2 = \left ( 1 + \frac{2}{m} \right ) \left ( 1 + \frac{2}{n} \right ) \left (1 + \frac{2}{p} \right ) \leq \left ( 1 + \frac{2}{m} \right )^3&lt;/cmath&gt; thus &lt;math&gt;m &lt; 7&lt;/math&gt;. <br /> <br /> Now, we can use casework on &lt;math&gt;m&lt;/math&gt; and Simon's Favorite Factoring Trick to check that &lt;math&gt;m = 5,2,1&lt;/math&gt; have no solution and for &lt;math&gt;m = 3,4,6&lt;/math&gt;, we have the corresponding values of &lt;math&gt;p&lt;/math&gt;: &lt;math&gt;130,54,16&lt;/math&gt;. <br /> <br /> Thus, the maximum value is &lt;math&gt;\boxed{130}&lt;/math&gt;. <br /> <br /> ~amplreneo<br /> <br /> == See also ==<br /> {{AIME box|year=1998|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_11&diff=121864 1998 AIME Problems/Problem 11 2020-05-01T00:49:54Z <p>First: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Three of the edges of a [[cube]] are &lt;math&gt;\overline{AB}, \overline{BC},&lt;/math&gt; and &lt;math&gt;\overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{AD}&lt;/math&gt; is an interior [[diagonal]]. Points &lt;math&gt;P, Q,&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; are on &lt;math&gt;\overline{AB}, \overline{BC},&lt;/math&gt; and &lt;math&gt;\overline{CD},&lt;/math&gt; respectively, so that &lt;math&gt;AP = 5, PB = 15, BQ = 15,&lt;/math&gt; and &lt;math&gt;CR = 10.&lt;/math&gt; What is the [[area]] of the [[polygon]] that is the [[intersection]] of [[plane]] &lt;math&gt;PQR&lt;/math&gt; and the cube?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> &lt;span style=&quot;font-size:100%&quot;&gt;For non-asymptote version of image, see [[:Image:1998_AIME-11.png]]&lt;/span&gt;<br /> &lt;center&gt;&lt;asy&gt;<br /> import three;<br /> size(280); defaultpen(linewidth(0.6)+fontsize(9));<br /> currentprojection=perspective(30,-60,40);<br /> triple A=(0,0,0),B=(20,0,0),C=(20,0,20),D=(20,20,20);<br /> triple P=(5,0,0),Q=(20,0,15),R=(20,10,20),Pa=(15,20,20),Qa=(0,20,5),Ra=(0,10,0);<br /> draw(box((0,0,0),(20,20,20)));<br /> draw(P--Q--R--Pa--Qa--Ra--cycle,linewidth(0.7));<br /> label(&quot;$$A\,(0,0,0)$$&quot;,A,SW);<br /> label(&quot;$$B\,(20,0,0)$$&quot;,B,S);<br /> label(&quot;$$C\,(20,0,20)$$&quot;,C,SW);<br /> label(&quot;$$D\,(20,20,20)$$&quot;,D,E);<br /> label(&quot;$$P\,(5,0,0)$$&quot;,P,SW);<br /> label(&quot;$$Q\,(20,0,15)$$&quot;,Q,E);<br /> label(&quot;$$R\,(20,10,20)$$&quot;,R,E);<br /> label(&quot;$$(15,20,20)$$&quot;,Pa,N);<br /> label(&quot;$$(0,20,5)$$&quot;,Qa,W);<br /> label(&quot;$$(0,10,0)$$&quot;,Ra,W);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> This approach uses [[analytical geometry]]. Let &lt;math&gt;A&lt;/math&gt; be at the origin, &lt;math&gt;B&lt;/math&gt; at &lt;math&gt;(20,0,0)&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; at &lt;math&gt;(20,0,20)&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;(20,20,20)&lt;/math&gt;. Thus, &lt;math&gt;P&lt;/math&gt; is at &lt;math&gt;(5,0,0)&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt; is at &lt;math&gt;(20,0,15)&lt;/math&gt;, and &lt;math&gt;R&lt;/math&gt; is at &lt;math&gt;(20,10,20)&lt;/math&gt;. <br /> <br /> Let the plane &lt;math&gt;PQR&lt;/math&gt; have the equation &lt;math&gt;ax + by + cz = d&lt;/math&gt;. Using point &lt;math&gt;P&lt;/math&gt;, we get that &lt;math&gt;5a = d&lt;/math&gt;. Using point &lt;math&gt;Q&lt;/math&gt;, we get &lt;math&gt;20a + 15c = d \Longrightarrow 4d + 15c = d \Longrightarrow d = -5c&lt;/math&gt;. Using point &lt;math&gt;R&lt;/math&gt;, we get &lt;math&gt;20a + 10b + 20c = d \Longrightarrow 4d + 10b - 4d = d \Longrightarrow d = 10b&lt;/math&gt;. Thus plane &lt;math&gt;PQR&lt;/math&gt;’s [[equation]] reduces to &lt;math&gt;\frac{d}{5}x + \frac{d}{10}y - \frac{d}{5}z = d \Longrightarrow 2x + y - 2z = 10&lt;/math&gt;.<br /> <br /> We know need to find the intersection of this plane with that of &lt;math&gt;z = 0&lt;/math&gt;, &lt;math&gt;z = 20&lt;/math&gt;, &lt;math&gt;x = 0&lt;/math&gt;, and &lt;math&gt;y = 20&lt;/math&gt;. After doing a little bit of algebra, the intersections are the lines &lt;math&gt;y = -2x + 10&lt;/math&gt;, &lt;math&gt;y = -2x + 50&lt;/math&gt;, &lt;math&gt;y = 2z + 10&lt;/math&gt;, and &lt;math&gt;z = x + 5&lt;/math&gt;. Thus, there are three more vertices on the polygon, which are at &lt;math&gt;(0,10,0)(0,20,5)(15,20,20)&lt;/math&gt;. <br /> <br /> We can find the lengths of the sides of the polygons now. There are 4 [[right triangle]]s with legs of length 5 and 10, so their [[hypotenuse]]s are &lt;math&gt;5\sqrt{5}&lt;/math&gt;. The other two are of &lt;math&gt;45-45-90 \triangle&lt;/math&gt;s with legs of length 15, so their hypotenuses are &lt;math&gt;15\sqrt{2}&lt;/math&gt;. So we have a [[hexagon]] with sides &lt;math&gt;15\sqrt{2},5\sqrt{5}, 5\sqrt{5},15\sqrt{2}, 5\sqrt{5},5\sqrt{5}&lt;/math&gt; By [[symmetry]], we know that opposite angles of the polygon are congruent. We can also calculate the length of the long diagonal by noting that it is of the same length of a face diagonal, making it &lt;math&gt;20\sqrt{2}&lt;/math&gt;.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(190);<br /> pointpen=black;pathpen=black;<br /> real s=2^.5;<br /> pair P=(0,0),Q=(7.5*s,2.5*s),R=Q+(0,15*s),Pa=(0,20*s),Qa=(-Q.x,Q.y),Ra=(-R.x,R.y);<br /> D(P--Q--R--Pa--Ra--Qa--cycle);D(R--Ra);D(Q--Qa);D(P--Pa);<br /> MP(&quot;15\sqrt{2}&quot;,(Q+R)/2,E);<br /> MP(&quot;5\sqrt{5}&quot;,(P+Q)/2,SE);<br /> MP(&quot;5\sqrt{5}&quot;,(R+Pa)/2,NE);<br /> MP(&quot;20\sqrt{2}&quot;,(P+Pa)/2,W);<br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- asymptote replaced Image:1998_AIME-11b.png by azjps --&gt;<br /> <br /> The height of the triangles at the top/bottom is &lt;math&gt;\frac{20\sqrt{2} - 15\sqrt{2}}{2} = \frac{5}{2}\sqrt{2}&lt;/math&gt;. The [[Pythagorean Theorem]] gives that half of the base of the triangles is &lt;math&gt;\frac{15}{\sqrt{2}}&lt;/math&gt;. We find that the middle [[rectangle]] is actually a [[square]], so the total area is &lt;math&gt;(15\sqrt{2})^2 + 4\left(\frac 12\right)\left(\frac 52\sqrt{2}\right)\left(\frac{15}{\sqrt{2}}\right) = 525&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> First, note that whenever the plane intersects two opposite faces of the cube, the resulting line segments must be parallel. Because they are part of parallel planes (the faces), they must be either parallel or [[skew]]; they are both part of plane &lt;math&gt;PQR&lt;/math&gt;, so they cannot be skew. Therefore, they are [[parallel]].<br /> <br /> Let the cube's vertices be &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;H&lt;/math&gt;, with &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; on the bottom face as before, &lt;math&gt;H&lt;/math&gt; being the other bottom vertex, &lt;math&gt;D&lt;/math&gt; directly above &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt; above &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; above &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; above &lt;math&gt;H&lt;/math&gt;.<br /> <br /> Clearly, the next vertex of the intersection (starting with &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt;) will be somewhere on &lt;math&gt;DG&lt;/math&gt;. Let it be &lt;math&gt;X&lt;/math&gt;, and have a distance of &lt;math&gt;x&lt;/math&gt; from D, and a distance of &lt;math&gt;20 - x&lt;/math&gt; from &lt;math&gt;G&lt;/math&gt;.<br /> <br /> Then, the next vertex will be somewhere on &lt;math&gt;FG&lt;/math&gt;. It must be parallel to &lt;math&gt;PQ&lt;/math&gt;, so this implies that it has a distance of &lt;math&gt;20 - x&lt;/math&gt; from &lt;math&gt;G&lt;/math&gt;, and thus a distance of &lt;math&gt;x&lt;/math&gt; from &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, the next vertex (call it &lt;math&gt;Y&lt;/math&gt;) will be somewhere on &lt;math&gt;AF&lt;/math&gt;. The segment must be parallel to &lt;math&gt;QR&lt;/math&gt;, so &lt;math&gt;FY&lt;/math&gt; must have length &lt;math&gt;2x&lt;/math&gt;, and &lt;math&gt;AY&lt;/math&gt; must be &lt;math&gt;20 - 2x&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DX \parallel AP&lt;/math&gt;, &lt;math&gt;DR \parallel AY&lt;/math&gt;, and &lt;math&gt;RX \parallel PY&lt;/math&gt;, we must have &lt;math&gt;\triangle{APY} \sim \triangle{DXR}&lt;/math&gt;; therefore, &lt;cmath&gt;\frac{AP}{DX}=\frac{AY}{DR}&lt;/cmath&gt; &lt;cmath&gt;\frac{5}{x}=\frac{20-2x}{10}&lt;/cmath&gt; &lt;cmath&gt;x^{2}-10x+25=0&lt;/cmath&gt; &lt;cmath&gt;x=5&lt;/cmath&gt;<br /> <br /> We can now find that the hexagon has side lengths &lt;math&gt;15\sqrt {2}&lt;/math&gt;, &lt;math&gt;5\sqrt {5}&lt;/math&gt;, &lt;math&gt;5\sqrt {5}&lt;/math&gt;, &lt;math&gt;15\sqrt {2}&lt;/math&gt;, &lt;math&gt;5\sqrt {5}&lt;/math&gt;, and &lt;math&gt;5\sqrt {5}&lt;/math&gt;. Moreover, opposite angles of this must be equal (by symmetry), so segment &lt;math&gt;RY&lt;/math&gt; divides the [[hexagon]] into two [[isosceles trapezoid]]s. It is easy to find the length of &lt;math&gt;RY&lt;/math&gt; (they're midpoints of opposite edges, so the distance between the two points is equal to a face diagonal of the cube, or &lt;math&gt;20\sqrt {2}&lt;/math&gt;), so it is now easy to finish the problem. From here, we can continue as in the first solution.<br /> <br /> == See also ==<br /> {{AIME box|year=1998|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_7&diff=121149 1996 AIME Problems/Problem 7 2020-04-19T15:53:11Z <p>First: /* Solution 2 (Casework) */</p> <hr /> <div>== Problem ==<br /> Two [[square]]s of a &lt;math&gt;7\times 7&lt;/math&gt; checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a [[rotation]] in the plane board. How many inequivalent color schemes are possible?<br /> <br /> == Solution 1 (Generalized)==<br /> There are &lt;math&gt;{49 \choose 2}&lt;/math&gt; possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. <br /> <br /> &lt;center&gt;&lt;table&gt;&lt;tr&gt;&lt;td&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;asy&gt;<br /> pathpen = black; pair O = (3.5,3.5); D(O);<br /> for(int i=0;i&lt;7;++i)<br /> for(int j=0;j&lt;7;++j)<br /> D(shift(i,j)*unitsquare);<br /> fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4));<br /> fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5));<br /> fill(shift(2,3)*unitsquare,rgb(.8,.8,.5));fill(shift(2,1)*unitsquare,rgb(.8,.8,.5));<br /> fill(shift(3,2)*unitsquare,rgb(.8,.8,.5));fill(shift(5,2)*unitsquare,rgb(.8,.8,.5));<br /> <br /> D(arc(O,1,280,350),EndArrow(4));<br /> D(arc(O,5^.5,-20,50),EndArrow(4));<br /> D(arc(O,1,10,80),EndArrow(4));<br /> D(arc(O,5^.5,70,140),EndArrow(4));<br /> D(arc(O,1,190,260),EndArrow(4));<br /> D(arc(O,5^.5,250,320),EndArrow(4));<br /> D(arc(O,1,100,170),EndArrow(4));<br /> D(arc(O,5^.5,160,230),EndArrow(4));<br /> &lt;/asy&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;/td&gt;&lt;td&gt;&lt;asy&gt;pathpen = black; pair O = (3.5,3.5); D(O);<br /> for(int i=0;i&lt;7;++i)<br /> for(int j=0;j&lt;7;++j)<br /> D(shift(i,j)*unitsquare);<br /> fill(shift(4,5)*unitsquare,rgb(1,1,.4));<br /> fill(shift(2,1)*unitsquare,rgb(1,1,.4));<br /> fill(shift(1,4)*unitsquare,rgb(.8,.8,.5));<br /> fill(shift(5,2)*unitsquare,rgb(.8,.8,.5));<br /> <br /> D(arc(O,5^.5,-20,50),EndArrow(4));<br /> D(arc(O,5^.5,70,140),EndArrow(4));<br /> D(arc(O,5^.5,250,320),EndArrow(4));<br /> D(arc(O,5^.5,160,230),EndArrow(4));<br /> &lt;/asy&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td&gt;&lt;font style=&quot;font-size:85%&quot;&gt;For most pairs, there will be &lt;br /&gt; three other equivalent boards.&lt;/font&gt;&lt;/td&gt;&lt;td&gt;&lt;font style=&quot;font-size:85%&quot;&gt;For those symmetric about the center, &lt;br /&gt; there is only one other.&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;&lt;/center&gt;<br /> <br /> Note that a pair of yellow squares will only yield &lt;math&gt;2&lt;/math&gt; distinct boards upon rotation [[iff]] the yellow squares are rotationally symmetric about the center square; there are &lt;math&gt;\frac{49-1}{2}=24&lt;/math&gt; such pairs. There are then &lt;math&gt;{49 \choose 2}-24&lt;/math&gt; pairs that yield &lt;math&gt;4&lt;/math&gt; distinct boards upon rotation; in other words, for each of the &lt;math&gt;{49 \choose 2}-24&lt;/math&gt; pairs, there are three other pairs that yield an equivalent board.<br /> <br /> Thus, the number of inequivalent boards is &lt;math&gt;\frac{{49 \choose 2} - 24}{4} + \frac{24}{2} = \boxed{300}&lt;/math&gt;. For a &lt;math&gt;(2n+1) \times (2n+1)&lt;/math&gt; board, this argument generalizes to &lt;math&gt;n(n+1)(2n^2+2n+1)&lt;/math&gt; inequivalent configurations.<br /> <br /> == Solution 2 (Casework) ==<br /> There are 4 cases: <br /> 1. The center square is occupied, in which there are &lt;math&gt;12&lt;/math&gt; cases.<br /> 2. The center square isn't occupied and the two squares are opposite each other with respect to the center square, in which there are &lt;math&gt;12&lt;/math&gt; cases.<br /> 3. The center square isn't occupied and the two squares can rotate to each other with a &lt;math&gt;90^{\circ}&lt;/math&gt; rotation with each other with respect to the center square, in which case there are &lt;math&gt;12&lt;/math&gt; cases.<br /> 4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are &lt;math&gt;\dbinom{12}{2} \cdot \frac{16}{4} = 264&lt;/math&gt; cases. <br /> Add up all the values for each case to get &lt;math&gt;\boxed{300}&lt;/math&gt; as your answer.<br /> <br /> ~First<br /> <br /> == See also ==<br /> {{AIME box|year=1996|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_7&diff=121148 1996 AIME Problems/Problem 7 2020-04-19T15:52:41Z <p>First: /* Solution 2 (Casework) */</p> <hr /> <div>== Problem ==<br /> Two [[square]]s of a &lt;math&gt;7\times 7&lt;/math&gt; checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a [[rotation]] in the plane board. How many inequivalent color schemes are possible?<br /> <br /> == Solution 1 (Generalized)==<br /> There are &lt;math&gt;{49 \choose 2}&lt;/math&gt; possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. <br /> <br /> &lt;center&gt;&lt;table&gt;&lt;tr&gt;&lt;td&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;asy&gt;<br /> pathpen = black; pair O = (3.5,3.5); D(O);<br /> for(int i=0;i&lt;7;++i)<br /> for(int j=0;j&lt;7;++j)<br /> D(shift(i,j)*unitsquare);<br /> fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4));<br /> fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5));<br /> fill(shift(2,3)*unitsquare,rgb(.8,.8,.5));fill(shift(2,1)*unitsquare,rgb(.8,.8,.5));<br /> fill(shift(3,2)*unitsquare,rgb(.8,.8,.5));fill(shift(5,2)*unitsquare,rgb(.8,.8,.5));<br /> <br /> D(arc(O,1,280,350),EndArrow(4));<br /> D(arc(O,5^.5,-20,50),EndArrow(4));<br /> D(arc(O,1,10,80),EndArrow(4));<br /> D(arc(O,5^.5,70,140),EndArrow(4));<br /> D(arc(O,1,190,260),EndArrow(4));<br /> D(arc(O,5^.5,250,320),EndArrow(4));<br /> D(arc(O,1,100,170),EndArrow(4));<br /> D(arc(O,5^.5,160,230),EndArrow(4));<br /> &lt;/asy&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;/td&gt;&lt;td&gt;&lt;asy&gt;pathpen = black; pair O = (3.5,3.5); D(O);<br /> for(int i=0;i&lt;7;++i)<br /> for(int j=0;j&lt;7;++j)<br /> D(shift(i,j)*unitsquare);<br /> fill(shift(4,5)*unitsquare,rgb(1,1,.4));<br /> fill(shift(2,1)*unitsquare,rgb(1,1,.4));<br /> fill(shift(1,4)*unitsquare,rgb(.8,.8,.5));<br /> fill(shift(5,2)*unitsquare,rgb(.8,.8,.5));<br /> <br /> D(arc(O,5^.5,-20,50),EndArrow(4));<br /> D(arc(O,5^.5,70,140),EndArrow(4));<br /> D(arc(O,5^.5,250,320),EndArrow(4));<br /> D(arc(O,5^.5,160,230),EndArrow(4));<br /> &lt;/asy&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td&gt;&lt;font style=&quot;font-size:85%&quot;&gt;For most pairs, there will be &lt;br /&gt; three other equivalent boards.&lt;/font&gt;&lt;/td&gt;&lt;td&gt;&lt;font style=&quot;font-size:85%&quot;&gt;For those symmetric about the center, &lt;br /&gt; there is only one other.&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;&lt;/center&gt;<br /> <br /> Note that a pair of yellow squares will only yield &lt;math&gt;2&lt;/math&gt; distinct boards upon rotation [[iff]] the yellow squares are rotationally symmetric about the center square; there are &lt;math&gt;\frac{49-1}{2}=24&lt;/math&gt; such pairs. There are then &lt;math&gt;{49 \choose 2}-24&lt;/math&gt; pairs that yield &lt;math&gt;4&lt;/math&gt; distinct boards upon rotation; in other words, for each of the &lt;math&gt;{49 \choose 2}-24&lt;/math&gt; pairs, there are three other pairs that yield an equivalent board.<br /> <br /> Thus, the number of inequivalent boards is &lt;math&gt;\frac{{49 \choose 2} - 24}{4} + \frac{24}{2} = \boxed{300}&lt;/math&gt;. For a &lt;math&gt;(2n+1) \times (2n+1)&lt;/math&gt; board, this argument generalizes to &lt;math&gt;n(n+1)(2n^2+2n+1)&lt;/math&gt; inequivalent configurations.<br /> <br /> == Solution 2 (Casework) ==<br /> There are 4 cases: <br /> 1. The center square is occupied, in which there are &lt;math&gt;12&lt;/math&gt; cases.<br /> 2. The center square isn't occupied and the two squares are opposite each other with respect to the center square, in which there are &lt;math&gt;12&lt;/math&gt; cases.<br /> 3. The center square isn't occupied and the two squares can rotate to each other with a &lt;math&gt;90^{\circ}&lt;/math&gt; rotation with each other with respect to the center square, in which case there are &lt;math&gt;12&lt;/math&gt; cases.<br /> 4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are &lt;math&gt;\dbinom{12}{2} \cdot \frac{16}{4} = 264&lt;/math&gt; cases. <br /> Add up all the cases to get &lt;math&gt;\boxed{300}&lt;/math&gt; as your answer.<br /> <br /> ~First<br /> <br /> == See also ==<br /> {{AIME box|year=1996|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_7&diff=121147 1996 AIME Problems/Problem 7 2020-04-19T15:51:33Z <p>First: /* Solution */</p> <hr /> <div>== Problem ==<br /> Two [[square]]s of a &lt;math&gt;7\times 7&lt;/math&gt; checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a [[rotation]] in the plane board. How many inequivalent color schemes are possible?<br /> <br /> == Solution 1 (Generalized)==<br /> There are &lt;math&gt;{49 \choose 2}&lt;/math&gt; possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. <br /> <br /> &lt;center&gt;&lt;table&gt;&lt;tr&gt;&lt;td&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;asy&gt;<br /> pathpen = black; pair O = (3.5,3.5); D(O);<br /> for(int i=0;i&lt;7;++i)<br /> for(int j=0;j&lt;7;++j)<br /> D(shift(i,j)*unitsquare);<br /> fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4));<br /> fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5));<br /> fill(shift(2,3)*unitsquare,rgb(.8,.8,.5));fill(shift(2,1)*unitsquare,rgb(.8,.8,.5));<br /> fill(shift(3,2)*unitsquare,rgb(.8,.8,.5));fill(shift(5,2)*unitsquare,rgb(.8,.8,.5));<br /> <br /> D(arc(O,1,280,350),EndArrow(4));<br /> D(arc(O,5^.5,-20,50),EndArrow(4));<br /> D(arc(O,1,10,80),EndArrow(4));<br /> D(arc(O,5^.5,70,140),EndArrow(4));<br /> D(arc(O,1,190,260),EndArrow(4));<br /> D(arc(O,5^.5,250,320),EndArrow(4));<br /> D(arc(O,1,100,170),EndArrow(4));<br /> D(arc(O,5^.5,160,230),EndArrow(4));<br /> &lt;/asy&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;/td&gt;&lt;td&gt;&lt;asy&gt;pathpen = black; pair O = (3.5,3.5); D(O);<br /> for(int i=0;i&lt;7;++i)<br /> for(int j=0;j&lt;7;++j)<br /> D(shift(i,j)*unitsquare);<br /> fill(shift(4,5)*unitsquare,rgb(1,1,.4));<br /> fill(shift(2,1)*unitsquare,rgb(1,1,.4));<br /> fill(shift(1,4)*unitsquare,rgb(.8,.8,.5));<br /> fill(shift(5,2)*unitsquare,rgb(.8,.8,.5));<br /> <br /> D(arc(O,5^.5,-20,50),EndArrow(4));<br /> D(arc(O,5^.5,70,140),EndArrow(4));<br /> D(arc(O,5^.5,250,320),EndArrow(4));<br /> D(arc(O,5^.5,160,230),EndArrow(4));<br /> &lt;/asy&gt;&lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td&gt;&lt;font style=&quot;font-size:85%&quot;&gt;For most pairs, there will be &lt;br /&gt; three other equivalent boards.&lt;/font&gt;&lt;/td&gt;&lt;td&gt;&lt;font style=&quot;font-size:85%&quot;&gt;For those symmetric about the center, &lt;br /&gt; there is only one other.&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;&lt;/center&gt;<br /> <br /> Note that a pair of yellow squares will only yield &lt;math&gt;2&lt;/math&gt; distinct boards upon rotation [[iff]] the yellow squares are rotationally symmetric about the center square; there are &lt;math&gt;\frac{49-1}{2}=24&lt;/math&gt; such pairs. There are then &lt;math&gt;{49 \choose 2}-24&lt;/math&gt; pairs that yield &lt;math&gt;4&lt;/math&gt; distinct boards upon rotation; in other words, for each of the &lt;math&gt;{49 \choose 2}-24&lt;/math&gt; pairs, there are three other pairs that yield an equivalent board.<br /> <br /> Thus, the number of inequivalent boards is &lt;math&gt;\frac{{49 \choose 2} - 24}{4} + \frac{24}{2} = \boxed{300}&lt;/math&gt;. For a &lt;math&gt;(2n+1) \times (2n+1)&lt;/math&gt; board, this argument generalizes to &lt;math&gt;n(n+1)(2n^2+2n+1)&lt;/math&gt; inequivalent configurations.<br /> <br /> == Solution 2 (Casework) ==<br /> There are 4 cases: <br /> 1. The center square is occupied, in which there are &lt;math&gt;12&lt;/math&gt; cases.<br /> 2. The center square isn't occupied and the two squares are opposite each other with respect to the center square, in which there are &lt;math&gt;12&lt;/math&gt; cases.<br /> 3. The center square isn't occupied and the two squares can rotate to each other with a &lt;math&gt;90^{\circ}&lt;/math&gt; rotation with each other with respect to the center square, in which case there are &lt;math&gt;12&lt;/math&gt; cases.<br /> 4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are &lt;math&gt;\dbinom{12}{2} \cdot 4 = 264&lt;/math&gt; cases. <br /> Add up all the cases to get &lt;math&gt;\boxed{300}&lt;/math&gt; as your answer.<br /> <br /> ~First<br /> <br /> == See also ==<br /> {{AIME box|year=1996|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_3&diff=121146 1996 AIME Problems/Problem 3 2020-04-19T14:31:33Z <p>First: /* Solution FASTER */</p> <hr /> <div>== Problem ==<br /> Find the smallest positive [[integer]] &lt;math&gt;n&lt;/math&gt; for which the expansion of &lt;math&gt;(xy-3x+7y-21)^n&lt;/math&gt;, after like terms have been collected, has at least 1996 terms.<br /> <br /> == Solution ==<br /> Using [[Simon's Favorite Factoring Trick]], we rewrite as &lt;math&gt;[(x+7)(y-3)]^n = (x+7)^n(y-3)^n&lt;/math&gt;. Both [[binomial expansion]]s will contain &lt;math&gt;n+1&lt;/math&gt; non-like terms; their product will contain &lt;math&gt;(n+1)^2&lt;/math&gt; terms, as each term will have an unique power of &lt;math&gt;x&lt;/math&gt; or &lt;math&gt;y&lt;/math&gt; and so none of the terms will need to be collected. Hence &lt;math&gt;(n+1)^2 \ge 1996&lt;/math&gt;, the smallest square after &lt;math&gt;1996&lt;/math&gt; is &lt;math&gt;2025 = 45^2&lt;/math&gt;, so our answer is &lt;math&gt;45 - 1 = \boxed{044}&lt;/math&gt;.<br /> <br /> Alternatively, when &lt;math&gt;n = k&lt;/math&gt;, the exponents of &lt;math&gt;x&lt;/math&gt; or &lt;math&gt;y&lt;/math&gt; in &lt;math&gt;x^i y^i&lt;/math&gt; can be any integer between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; inclusive. Thus, when &lt;math&gt;n=1&lt;/math&gt;, there are &lt;math&gt;(2)(2)&lt;/math&gt; terms and, when &lt;math&gt;n = k&lt;/math&gt;, there are &lt;math&gt;(k+1)^2&lt;/math&gt; terms. Therefore, we need to find the smallest perfect square that is greater than &lt;math&gt;1996&lt;/math&gt;. From trial and error, we get &lt;math&gt;44^2 = 1936&lt;/math&gt; and &lt;math&gt;45^2 = 2025&lt;/math&gt;. Thus, &lt;math&gt;k = 45\rightarrow n = \boxed{044}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1996|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_11&diff=120222 2020 AIME I Problems/Problem 11 2020-03-28T16:55:21Z <p>First: /* Solution 1 (Strategic Casework) */</p> <hr /> <div><br /> == Problem ==<br /> For integers &lt;math&gt;a,b,c&lt;/math&gt; and &lt;math&gt;d,&lt;/math&gt; let &lt;math&gt;f(x)=x^2+ax+b&lt;/math&gt; and &lt;math&gt;g(x)=x^2+cx+d.&lt;/math&gt; Find the number of ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; of integers with absolute values not exceeding &lt;math&gt;10&lt;/math&gt; for which there is an integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;g(f(2))=g(f(4))=0.&lt;/math&gt;<br /> <br /> == Solution 1 (Strategic Casework)==<br /> Either &lt;math&gt;f(2)=f(4)&lt;/math&gt; or not. If it is, note that Vieta's forces &lt;math&gt;a = -6&lt;/math&gt;. Then, &lt;math&gt;b&lt;/math&gt; can be anything. However, &lt;math&gt;c&lt;/math&gt; can also be anything, as we can set the root of &lt;math&gt;g&lt;/math&gt; (not equal to &lt;math&gt;f(2) = f(4)&lt;/math&gt;) to any integer, producing a possible integer value of &lt;math&gt;d&lt;/math&gt;. Therefore there are &lt;math&gt;21^2 = 441&lt;/math&gt; in this case*. If it isn't, then &lt;math&gt;f(2),f(4)&lt;/math&gt; are the roots of &lt;math&gt;g&lt;/math&gt;. This means by Vieta's, that:<br /> <br /> &lt;cmath&gt;f(2)+f(4) = -c \in [-10,10]&lt;/cmath&gt;<br /> &lt;cmath&gt;20 + 6a + 2b \in [-10,10]&lt;/cmath&gt;<br /> &lt;cmath&gt;3a + b \in [-15,-5].&lt;/cmath&gt;<br /> <br /> Solving these inequalities while considering that &lt;math&gt;a \neq -6&lt;/math&gt; to prevent &lt;math&gt;f(2) = f(4)&lt;/math&gt;, we obtain &lt;math&gt;69&lt;/math&gt; possible tuples and adding gives &lt;math&gt;441+69=\boxed{510}&lt;/math&gt;. <br /> ~awang11<br /> <br /> == Solution 2 (Bash) ==<br /> Define &lt;math&gt;h(x)=x^2+cx&lt;/math&gt;. Since &lt;math&gt;g(f(2))=g(f(4))=0&lt;/math&gt;, we know &lt;math&gt;h(f(2))=h(f(4))=-d&lt;/math&gt;. Plugging in &lt;math&gt;f(x)&lt;/math&gt; into &lt;math&gt;h(x)&lt;/math&gt;, we get &lt;math&gt;h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)&lt;/math&gt;. Setting &lt;math&gt;h(f(2))=h(f(4))&lt;/math&gt;, &lt;cmath&gt;16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc&lt;/cmath&gt;. Simplifying and cancelling terms, &lt;cmath&gt;240+112a+24b+12a^2+4ab+12c+2ac=0&lt;/cmath&gt; &lt;cmath&gt;120+56a+12b+6a^2+2ab+6c+ac=0&lt;/cmath&gt; &lt;cmath&gt;6a^2+2ab+ac+56a+12b+6c+120=0&lt;/cmath&gt; &lt;cmath&gt;6a^2+2ab+ac+20a+36a+12b+6c+120=0&lt;/cmath&gt; &lt;cmath&gt;a(6a+2b+c+20)+6(6a+2b+c+20)=0&lt;/cmath&gt; &lt;cmath&gt;(a+6)(6a+2b+c+20)=0&lt;/cmath&gt;<br /> <br /> Therefore, either &lt;math&gt;a+6=0&lt;/math&gt; or &lt;math&gt;6a+2b+c=-20&lt;/math&gt;. The first case is easy: &lt;math&gt;a=-6&lt;/math&gt; and there are &lt;math&gt;441&lt;/math&gt; tuples in that case. In the second case, we simply perform casework on even values of &lt;math&gt;c&lt;/math&gt;, to get &lt;math&gt;77&lt;/math&gt; tuples, subtracting the &lt;math&gt;8&lt;/math&gt; tuples in both cases we get &lt;math&gt;441+77-8=\boxed{510}&lt;/math&gt;.<br /> <br /> -EZmath2006<br /> <br /> == Notes For * ==<br /> In case anyone is confused by this (as I initially was). In the case where &lt;math&gt;f(2)=f(4)&lt;/math&gt;, this does not mean that g has a double root of &lt;math&gt;f(2)=f(4)=c&lt;/math&gt;, ONLY that &lt;math&gt;c&lt;/math&gt; is one of the roots of g. So basically since &lt;math&gt;a=-6&lt;/math&gt; in this case, &lt;math&gt;f(2)=f(4)=b-8&lt;/math&gt;, and we have &lt;math&gt;21&lt;/math&gt; choices for b and we [i]still can[/i] ensure c is an integer with absolute value less than or equal to 10 simply by having another integer root of g that when added to &lt;math&gt;b-8&lt;/math&gt; ensures this, and of course an integer multiplied by an integer is an integer so &lt;math&gt;d&lt;/math&gt; will still be an integer. In other words, you have can have &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; be any integer with absolute value less than or equal to 10 with &lt;math&gt;d&lt;/math&gt; still being an integer. Now refer back to the 1st solution.<br /> ~First<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_11&diff=120220 2020 AIME I Problems/Problem 11 2020-03-28T16:54:52Z <p>First: /* Notes */</p> <hr /> <div><br /> == Problem ==<br /> For integers &lt;math&gt;a,b,c&lt;/math&gt; and &lt;math&gt;d,&lt;/math&gt; let &lt;math&gt;f(x)=x^2+ax+b&lt;/math&gt; and &lt;math&gt;g(x)=x^2+cx+d.&lt;/math&gt; Find the number of ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; of integers with absolute values not exceeding &lt;math&gt;10&lt;/math&gt; for which there is an integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;g(f(2))=g(f(4))=0.&lt;/math&gt;<br /> <br /> == Solution 1 (Strategic Casework)==<br /> Either &lt;math&gt;f(2)=f(4)&lt;/math&gt; or not. If it is, note that Vieta's forces &lt;math&gt;a = -6&lt;/math&gt;. Then, &lt;math&gt;b&lt;/math&gt; can be anything. However, &lt;math&gt;c&lt;/math&gt; can also be anything, as we can set the root of &lt;math&gt;g&lt;/math&gt; (not equal to &lt;math&gt;f(2) = f(4)&lt;/math&gt;) to any integer, producing a possible integer value of &lt;math&gt;d&lt;/math&gt;. Therefore there are &lt;math&gt;21^2 = 441&lt;/math&gt; in this case. If it isn't, then &lt;math&gt;f(2),f(4)&lt;/math&gt; are the roots of &lt;math&gt;g&lt;/math&gt;. This means by Vieta's, that:<br /> <br /> &lt;cmath&gt;f(2)+f(4) = -c \in [-10,10]&lt;/cmath&gt;<br /> &lt;cmath&gt;20 + 6a + 2b \in [-10,10]&lt;/cmath&gt;<br /> &lt;cmath&gt;3a + b \in [-15,-5].&lt;/cmath&gt;<br /> <br /> Solving these inequalities while considering that &lt;math&gt;a \neq -6&lt;/math&gt; to prevent &lt;math&gt;f(2) = f(4)&lt;/math&gt;, we obtain &lt;math&gt;69&lt;/math&gt; possible tuples and adding gives &lt;math&gt;441+69=\boxed{510}&lt;/math&gt;. <br /> ~awang11<br /> <br /> == Solution 2 (Bash) ==<br /> Define &lt;math&gt;h(x)=x^2+cx&lt;/math&gt;. Since &lt;math&gt;g(f(2))=g(f(4))=0&lt;/math&gt;, we know &lt;math&gt;h(f(2))=h(f(4))=-d&lt;/math&gt;. Plugging in &lt;math&gt;f(x)&lt;/math&gt; into &lt;math&gt;h(x)&lt;/math&gt;, we get &lt;math&gt;h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)&lt;/math&gt;. Setting &lt;math&gt;h(f(2))=h(f(4))&lt;/math&gt;, &lt;cmath&gt;16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc&lt;/cmath&gt;. Simplifying and cancelling terms, &lt;cmath&gt;240+112a+24b+12a^2+4ab+12c+2ac=0&lt;/cmath&gt; &lt;cmath&gt;120+56a+12b+6a^2+2ab+6c+ac=0&lt;/cmath&gt; &lt;cmath&gt;6a^2+2ab+ac+56a+12b+6c+120=0&lt;/cmath&gt; &lt;cmath&gt;6a^2+2ab+ac+20a+36a+12b+6c+120=0&lt;/cmath&gt; &lt;cmath&gt;a(6a+2b+c+20)+6(6a+2b+c+20)=0&lt;/cmath&gt; &lt;cmath&gt;(a+6)(6a+2b+c+20)=0&lt;/cmath&gt;<br /> <br /> Therefore, either &lt;math&gt;a+6=0&lt;/math&gt; or &lt;math&gt;6a+2b+c=-20&lt;/math&gt;. The first case is easy: &lt;math&gt;a=-6&lt;/math&gt; and there are &lt;math&gt;441&lt;/math&gt; tuples in that case. In the second case, we simply perform casework on even values of &lt;math&gt;c&lt;/math&gt;, to get &lt;math&gt;77&lt;/math&gt; tuples, subtracting the &lt;math&gt;8&lt;/math&gt; tuples in both cases we get &lt;math&gt;441+77-8=\boxed{510}&lt;/math&gt;.<br /> <br /> -EZmath2006<br /> <br /> == Notes For * ==<br /> In case anyone is confused by this (as I initially was). In the case where &lt;math&gt;f(2)=f(4)&lt;/math&gt;, this does not mean that g has a double root of &lt;math&gt;f(2)=f(4)=c&lt;/math&gt;, ONLY that &lt;math&gt;c&lt;/math&gt; is one of the roots of g. So basically since &lt;math&gt;a=-6&lt;/math&gt; in this case, &lt;math&gt;f(2)=f(4)=b-8&lt;/math&gt;, and we have &lt;math&gt;21&lt;/math&gt; choices for b and we [i]still can[/i] ensure c is an integer with absolute value less than or equal to 10 simply by having another integer root of g that when added to &lt;math&gt;b-8&lt;/math&gt; ensures this, and of course an integer multiplied by an integer is an integer so &lt;math&gt;d&lt;/math&gt; will still be an integer. In other words, you have can have &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; be any integer with absolute value less than or equal to 10 with &lt;math&gt;d&lt;/math&gt; still being an integer. Now refer back to the 1st solution.<br /> ~First<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_11&diff=120219 2020 AIME I Problems/Problem 11 2020-03-28T16:48:46Z <p>First: </p> <hr /> <div><br /> == Problem ==<br /> For integers &lt;math&gt;a,b,c&lt;/math&gt; and &lt;math&gt;d,&lt;/math&gt; let &lt;math&gt;f(x)=x^2+ax+b&lt;/math&gt; and &lt;math&gt;g(x)=x^2+cx+d.&lt;/math&gt; Find the number of ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; of integers with absolute values not exceeding &lt;math&gt;10&lt;/math&gt; for which there is an integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;g(f(2))=g(f(4))=0.&lt;/math&gt;<br /> <br /> == Solution 1 (Strategic Casework)==<br /> Either &lt;math&gt;f(2)=f(4)&lt;/math&gt; or not. If it is, note that Vieta's forces &lt;math&gt;a = -6&lt;/math&gt;. Then, &lt;math&gt;b&lt;/math&gt; can be anything. However, &lt;math&gt;c&lt;/math&gt; can also be anything, as we can set the root of &lt;math&gt;g&lt;/math&gt; (not equal to &lt;math&gt;f(2) = f(4)&lt;/math&gt;) to any integer, producing a possible integer value of &lt;math&gt;d&lt;/math&gt;. Therefore there are &lt;math&gt;21^2 = 441&lt;/math&gt; in this case. If it isn't, then &lt;math&gt;f(2),f(4)&lt;/math&gt; are the roots of &lt;math&gt;g&lt;/math&gt;. This means by Vieta's, that:<br /> <br /> &lt;cmath&gt;f(2)+f(4) = -c \in [-10,10]&lt;/cmath&gt;<br /> &lt;cmath&gt;20 + 6a + 2b \in [-10,10]&lt;/cmath&gt;<br /> &lt;cmath&gt;3a + b \in [-15,-5].&lt;/cmath&gt;<br /> <br /> Solving these inequalities while considering that &lt;math&gt;a \neq -6&lt;/math&gt; to prevent &lt;math&gt;f(2) = f(4)&lt;/math&gt;, we obtain &lt;math&gt;69&lt;/math&gt; possible tuples and adding gives &lt;math&gt;441+69=\boxed{510}&lt;/math&gt;. <br /> ~awang11<br /> <br /> == Solution 2 (Bash) ==<br /> Define &lt;math&gt;h(x)=x^2+cx&lt;/math&gt;. Since &lt;math&gt;g(f(2))=g(f(4))=0&lt;/math&gt;, we know &lt;math&gt;h(f(2))=h(f(4))=-d&lt;/math&gt;. Plugging in &lt;math&gt;f(x)&lt;/math&gt; into &lt;math&gt;h(x)&lt;/math&gt;, we get &lt;math&gt;h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)&lt;/math&gt;. Setting &lt;math&gt;h(f(2))=h(f(4))&lt;/math&gt;, &lt;cmath&gt;16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc&lt;/cmath&gt;. Simplifying and cancelling terms, &lt;cmath&gt;240+112a+24b+12a^2+4ab+12c+2ac=0&lt;/cmath&gt; &lt;cmath&gt;120+56a+12b+6a^2+2ab+6c+ac=0&lt;/cmath&gt; &lt;cmath&gt;6a^2+2ab+ac+56a+12b+6c+120=0&lt;/cmath&gt; &lt;cmath&gt;6a^2+2ab+ac+20a+36a+12b+6c+120=0&lt;/cmath&gt; &lt;cmath&gt;a(6a+2b+c+20)+6(6a+2b+c+20)=0&lt;/cmath&gt; &lt;cmath&gt;(a+6)(6a+2b+c+20)=0&lt;/cmath&gt;<br /> <br /> Therefore, either &lt;math&gt;a+6=0&lt;/math&gt; or &lt;math&gt;6a+2b+c=-20&lt;/math&gt;. The first case is easy: &lt;math&gt;a=-6&lt;/math&gt; and there are &lt;math&gt;441&lt;/math&gt; tuples in that case. In the second case, we simply perform casework on even values of &lt;math&gt;c&lt;/math&gt;, to get &lt;math&gt;77&lt;/math&gt; tuples, subtracting the &lt;math&gt;8&lt;/math&gt; tuples in both cases we get &lt;math&gt;441+77-8=\boxed{510}&lt;/math&gt;.<br /> <br /> -EZmath2006<br /> <br /> == Notes ==<br /> In case anyone is confused by this (as I initially was). Say &lt;math&gt;f(2)=f(4)&lt;/math&gt;. This does not mean that g has a double root of &lt;math&gt;f(2)=f(4)=c&lt;/math&gt;, ONLY that &lt;math&gt;c&lt;/math&gt; is one of the roots of g. <br /> ~First<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_11&diff=120218 2020 AIME I Problems/Problem 11 2020-03-28T15:42:11Z <p>First: /* Solution */</p> <hr /> <div><br /> == Problem ==<br /> For integers &lt;math&gt;a,b,c&lt;/math&gt; and &lt;math&gt;d,&lt;/math&gt; let &lt;math&gt;f(x)=x^2+ax+b&lt;/math&gt; and &lt;math&gt;g(x)=x^2+cx+d.&lt;/math&gt; Find the number of ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; of integers with absolute values not exceeding &lt;math&gt;10&lt;/math&gt; for which there is an integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;g(f(2))=g(f(4))=0.&lt;/math&gt;<br /> <br /> == Solution 1 (Strategic Casework)==<br /> Either &lt;math&gt;f(2)=f(4)&lt;/math&gt; or not. If it is, note that Vieta's forces &lt;math&gt;a = -6&lt;/math&gt;. Then, &lt;math&gt;b&lt;/math&gt; can be anything. However, &lt;math&gt;c&lt;/math&gt; can also be anything, as we can set the root of &lt;math&gt;g&lt;/math&gt; (not equal to &lt;math&gt;f(2) = f(4)&lt;/math&gt;) to any integer, producing a possible integer value of &lt;math&gt;d&lt;/math&gt;. Therefore there are &lt;math&gt;21^2 = 441&lt;/math&gt; in this case. If it isn't, then &lt;math&gt;f(2),f(4)&lt;/math&gt; are the roots of &lt;math&gt;g&lt;/math&gt;. This means by Vieta's, that:<br /> <br /> &lt;cmath&gt;f(2)+f(4) = -c \in [-10,10]&lt;/cmath&gt;<br /> &lt;cmath&gt;20 + 6a + 2b \in [-10,10]&lt;/cmath&gt;<br /> &lt;cmath&gt;3a + b \in [-15,-5].&lt;/cmath&gt;<br /> <br /> Solving these inequalities while considering that &lt;math&gt;a \neq -6&lt;/math&gt; to prevent &lt;math&gt;f(2) = f(4)&lt;/math&gt;, we obtain &lt;math&gt;69&lt;/math&gt; possible tuples and adding gives &lt;math&gt;441+69=\boxed{510}&lt;/math&gt;. <br /> ~awang11<br /> <br /> == Solution 2 (Bash) ==<br /> Define &lt;math&gt;h(x)=x^2+cx&lt;/math&gt;. Since &lt;math&gt;g(f(2))=g(f(4))=0&lt;/math&gt;, we know &lt;math&gt;h(f(2))=h(f(4))=-d&lt;/math&gt;. Plugging in &lt;math&gt;f(x)&lt;/math&gt; into &lt;math&gt;h(x)&lt;/math&gt;, we get &lt;math&gt;h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)&lt;/math&gt;. Setting &lt;math&gt;h(f(2))=h(f(4))&lt;/math&gt;, &lt;cmath&gt;16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc&lt;/cmath&gt;. Simplifying and cancelling terms, &lt;cmath&gt;240+112a+24b+12a^2+4ab+12c+2ac=0&lt;/cmath&gt; &lt;cmath&gt;120+56a+12b+6a^2+2ab+6c+ac=0&lt;/cmath&gt; &lt;cmath&gt;6a^2+2ab+ac+56a+12b+6c+120=0&lt;/cmath&gt; &lt;cmath&gt;6a^2+2ab+ac+20a+36a+12b+6c+120=0&lt;/cmath&gt; &lt;cmath&gt;a(6a+2b+c+20)+6(6a+2b+c+20)=0&lt;/cmath&gt; &lt;cmath&gt;(a+6)(6a+2b+c+20)=0&lt;/cmath&gt;<br /> <br /> Therefore, either &lt;math&gt;a+6=0&lt;/math&gt; or &lt;math&gt;6a+2b+c=-20&lt;/math&gt;. The first case is easy: &lt;math&gt;a=-6&lt;/math&gt; and there are &lt;math&gt;441&lt;/math&gt; tuples in that case. In the second case, we simply perform casework on even values of &lt;math&gt;c&lt;/math&gt;, to get &lt;math&gt;77&lt;/math&gt; tuples, subtracting the &lt;math&gt;8&lt;/math&gt; tuples in both cases we get &lt;math&gt;441+77-8=\boxed{510}&lt;/math&gt;.<br /> <br /> -EZmath2006<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_13&diff=96687 2008 AIME II Problems/Problem 13 2018-08-02T20:00:46Z <p>First: </p> <hr /> <div>== Problem ==<br /> A [[regular polygon|regular]] [[hexagon]] with center at the [[origin]] in the [[complex plane]] has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let &lt;math&gt;R&lt;/math&gt; be the region outside the hexagon, and let &lt;math&gt;S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace&lt;/math&gt;. Then the area of &lt;math&gt;S&lt;/math&gt; has the form &lt;math&gt;a\pi + \sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. Find &lt;math&gt;a + b&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> If you're familiar with inversion, you'll see that the problem's basically asking you to invert the hexagon with respect to the unit circle in the Cartesian Plane using the Inversion Distance Formula. This works because the point in the Cartesian Plane's complex plane's equivalent switches places with it's conjugate but we can do that in the Cartesian plane too (just reflect a point in the Cartesian plane over the x-axis)! If you're familiar with inversion you can go plot the inverted figure's Cartesian Plane Equivalent. Then simply continue on with the figure shown in the below solution.<br /> ~First~<br /> == Solution 2==<br /> If a point &lt;math&gt;z = r\text{cis}\,\theta&lt;/math&gt; is in &lt;math&gt;R&lt;/math&gt;, then the point &lt;math&gt;\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; (where [[cis]] denotes &lt;math&gt;\text{cis}\, \theta = \cos \theta + i \sin \theta&lt;/math&gt;). Since &lt;math&gt;R&lt;/math&gt; is symmetric every &lt;math&gt;60^{\circ}&lt;/math&gt; about the origin, it suffices to consider the area of the result of the transformation when &lt;math&gt;-30 \le \theta \le 30&lt;/math&gt;, and then to multiply by &lt;math&gt;6&lt;/math&gt; to account for the entire area.<br /> <br /> We note that if the region &lt;math&gt;S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace&lt;/math&gt;, where &lt;math&gt;R_2&lt;/math&gt; is the region (in green below) outside the circle of radius &lt;math&gt;1/\sqrt{3}&lt;/math&gt; centered at the origin, then &lt;math&gt;S_2&lt;/math&gt; is simply the region inside a circle of radius &lt;math&gt;\sqrt{3}&lt;/math&gt; centered at the origin. It now suffices to find what happens to the mapping of the region &lt;math&gt;R_2 - R&lt;/math&gt; (in blue below). <br /> <br /> The equation of the hexagon side in that region is &lt;math&gt;x = r \cos \theta = \frac{1}{2}&lt;/math&gt;, which is transformed to &lt;math&gt;\frac{1}{r} \cos (-\theta) = \frac{1}{r} \cos \theta = &lt;/math&gt;2 . Let &lt;math&gt;r\cos \theta = a+bi&lt;/math&gt; where &lt;math&gt;a,b \in \mathbb{R}&lt;/math&gt;; then &lt;math&gt;r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}&lt;/math&gt;, so the equation becomes &lt;math&gt;a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1&lt;/math&gt;. Hence the side is sent to an arc of the unit circle centered at &lt;math&gt;(1,0)&lt;/math&gt;, after considering the restriction that the side of the hexagon is a segment of length &lt;math&gt;1/\sqrt{3}&lt;/math&gt;. <br /> <br /> Including &lt;math&gt;S_2&lt;/math&gt;, we find that &lt;math&gt;S&lt;/math&gt; is the union of six unit circles centered at &lt;math&gt;\text{cis}\, \frac{k\pi}{6}&lt;/math&gt;, &lt;math&gt;k = 0,1,2,3,4,5&lt;/math&gt;, as shown below. <br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype(&quot;4 4&quot;); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1));<br /> draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i &lt; 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label(&quot;$1/\sqrt{3}$&quot;,(0,-0.5),W,fontsize(8)); <br /> &lt;/asy&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;math&gt;\Longrightarrow&lt;/math&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;asy&gt;<br /> defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1));<br /> draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype(&quot;4 4&quot;)); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype(&quot;4 4&quot;)); draw(Circle((0,0),1),linetype(&quot;4 4&quot;)); label(&quot;$\sqrt{3}$&quot;,(0,-0.5),W,fontsize(8));<br /> add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> The area of the regular hexagon is &lt;math&gt;6 \cdot \left( \frac{\left(\sqrt{3}\right)^2 \sqrt{3}}{4} \right) = \frac{9}{2}\sqrt{3}&lt;/math&gt;. The total area of the six &lt;math&gt;120^{\circ}&lt;/math&gt; sectors is &lt;math&gt;6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}&lt;/math&gt;. Their sum is &lt;math&gt;2\pi + \sqrt{27}&lt;/math&gt;, and &lt;math&gt;a+b = \boxed{029}&lt;/math&gt;.<br /> <br /> == Solution 3 (Calculus) ==<br /> One can describe the line parallel to the imaginary axis &lt;math&gt;x=\frac{1}{2}&lt;/math&gt; using polar coordinates as &lt;math&gt;r(\theta)=\dfrac{1}{2\cos{\theta}}&lt;/math&gt;<br /> <br /> so &lt;math&gt;z&lt;/math&gt; is equal to &lt;math&gt;z=(\dfrac{1}{2\cos{\theta}})(cis{\theta})<br /> \rightarrow \frac{1}{z}=2\cos{\theta}cis(-\theta)&lt;/math&gt;<br /> <br /> Dividing the hexagon to 12 equal parts we get that <br /> <br /> &lt;math&gt;Area = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}r^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}(2\cos\theta)^2 d\theta&lt;/math&gt;<br /> <br /> which is a routine computation:<br /> <br /> &lt;math&gt;Area = 12\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta&lt;/math&gt;<br /> <br /> &lt;math&gt;Area = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta = 12[\frac{1}{2}\sin{2\theta}+\theta]_0^{\frac{\pi}{6}}=12(\frac{\sqrt{3}}{4}+\frac{\pi}{6})=2\pi+3\sqrt{3}=2\pi + \sqrt{27}&lt;/math&gt;<br /> <br /> &lt;math&gt;a+b = \boxed{029}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=User:Biusetar&diff=96371 User:Biusetar 2018-07-20T17:17:11Z <p>First: Replaced content with &quot;this page is empty except for this sentence and the title.&quot;</p> <hr /> <div>this page is empty except for this sentence and the title.</div> First https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_9&diff=86202 2017 AIME II Problems/Problem 9 2017-06-28T17:58:11Z <p>First: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> A special deck of cards contains &lt;math&gt;49&lt;/math&gt; cards, each labeled with a number from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;7&lt;/math&gt; and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and &lt;math&gt;\textit{still}&lt;/math&gt; have at least one card of each color and at least one card with each number is &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Without loss of generality, assume that the &lt;math&gt;8&lt;/math&gt; numbers on Sharon's cards are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, and &lt;math&gt;7&lt;/math&gt;, in that order, and assume the &lt;math&gt;8&lt;/math&gt; colors are red, red, and six different arbitrary colors. There are &lt;math&gt;{8\choose2}-1&lt;/math&gt; ways of assigning the two red cards to the &lt;math&gt;8&lt;/math&gt; numbers; we subtract &lt;math&gt;1&lt;/math&gt; because we cannot assign the two reds to the two &lt;math&gt;1&lt;/math&gt;'s.<br /> In order for Sharon to be able to remove at least one card and still have at least one card of each color, one of the reds have to be assigned with one of the &lt;math&gt;1&lt;/math&gt;s. The number of ways for this not to happen is &lt;math&gt;{6\choose2}&lt;/math&gt;, so the number of ways for it to happen is &lt;math&gt;\left({8\choose2}-1\right)-{6\choose2}&lt;/math&gt;. Each of these assignments is equally likely, so the probability that Sharon can discard one of her cards and still have at least one card of each color and at least one card with each number is &lt;math&gt;\frac{\left({8\choose2}-1\right)-{6\choose2}}{{8\choose2}-1}=\frac{4}{9} \implies 4 + 9 = 13 = \boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> There have to be &lt;math&gt;2&lt;/math&gt; of &lt;math&gt;8&lt;/math&gt; cards sharing the same number and &lt;math&gt;2&lt;/math&gt; of them sharing same color.<br /> <br /> &lt;math&gt;2&lt;/math&gt; pairs of cards can't be the same or else there will be &lt;math&gt;2&lt;/math&gt; card which are completely same.<br /> <br /> WLOG the numbers are &lt;math&gt;1,1,2,3,4,5,6,&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; and the colors are &lt;math&gt;a,a,b,c,d,e,f,&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt;<br /> Then we can get &lt;math&gt;2&lt;/math&gt; cases:<br /> <br /> Case One:<br /> &lt;math&gt;1a,1b,2a,3c,4d,5e,6f,&lt;/math&gt; and &lt;math&gt;7g&lt;/math&gt;<br /> in this case, we can discard &lt;math&gt;1a&lt;/math&gt;.<br /> there are &lt;math&gt;2*6=12&lt;/math&gt; situations in this case.<br /> <br /> Case Two:<br /> &lt;math&gt;1b,1c,2a,3a,4d,5e,6f,&lt;/math&gt; and &lt;math&gt;7g&lt;/math&gt;<br /> In this case, we can't discard.<br /> There are &lt;math&gt;\dbinom{6}{2}=15&lt;/math&gt; situations in this case<br /> <br /> So the probability is &lt;math&gt;\frac{12}{12+15}=\frac{4}{9}&lt;/math&gt;<br /> <br /> The answer is &lt;math&gt;4+9=\boxed{013}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> There are &lt;math&gt;7!&lt;/math&gt; ways to choose a set of 7 cards that have all the numbers from 1-7 and all 7 colors. There are then &lt;math&gt;42&lt;/math&gt; cards remaining. Thus, there are &lt;math&gt;7!(42)&lt;/math&gt; desired sets.<br /> <br /> Now, the next thing to find is the number of ways to choose 8 cards where there is not a set of 7 such cards. In this case, one color must have 2 cards and one number must have 2 cards, and they can't be the same number/color card. The number of ways to pick this is equal to a multiplication of &lt;math&gt;\binom{7}{2}&lt;/math&gt; ways to pick 2 numbers, &lt;math&gt;7&lt;/math&gt; colors to assign them to, &lt;math&gt;\binom{6}{2}&lt;/math&gt; ways to pick 2 nonchosen colors, &lt;math&gt;5&lt;/math&gt; ways to pick a number to assign them to, and &lt;math&gt;4!&lt;/math&gt; ways to assign the rest.<br /> <br /> Thus, the answer is &lt;math&gt;\frac{7!(42)}{7!(42) + 21(15)(7)(5!)}&lt;/math&gt;. Dividing out &lt;math&gt;5!&lt;/math&gt; yields &lt;math&gt;\frac{42(42)}{42(42) + 21(15)(7)}&lt;/math&gt; which is equal to &lt;math&gt;\frac{2(42)}{2(42) + 15(7)}&lt;/math&gt; which is equal to &lt;math&gt;\frac{12}{12 + 15}&lt;/math&gt; which is equal to &lt;math&gt;\frac{4}{9}&lt;/math&gt; giving a final answer of &lt;math&gt;\boxed{013}&lt;/math&gt;.<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_12&diff=86198 2017 AIME I Problems/Problem 12 2017-06-28T01:24:00Z <p>First: /* Solution 2(PIE) */</p> <hr /> <div>==Problem 12==<br /> Call a set &lt;math&gt;S&lt;/math&gt; product-free if there do not exist &lt;math&gt;a, b, c \in S&lt;/math&gt; (not necessarily distinct) such that &lt;math&gt;a b = c&lt;/math&gt;. For example, the empty set and the set &lt;math&gt;\{16, 20\}&lt;/math&gt; are product-free, whereas the sets &lt;math&gt;\{4, 16\}&lt;/math&gt; and &lt;math&gt;\{2, 8, 16\}&lt;/math&gt; are not product-free. Find the number of product-free subsets of the set &lt;math&gt;\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}&lt;/math&gt;.<br /> <br /> ==Solution 1(Casework)==<br /> <br /> We shall solve this problem by doing casework on the lowest element of the subset. Note that the number &lt;math&gt;1&lt;/math&gt; cannot be in the subset because &lt;math&gt;1*1=1&lt;/math&gt;. Let &lt;math&gt;S&lt;/math&gt; be a product-free set. If the lowest element of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt;, we consider the set &lt;math&gt;\{3, 6, 9\}&lt;/math&gt;. We see that 5 of these subsets can be a subset of &lt;math&gt;S&lt;/math&gt; (&lt;math&gt;\{3\}&lt;/math&gt;, &lt;math&gt;\{6\}&lt;/math&gt;, &lt;math&gt;\{9\}&lt;/math&gt;, &lt;math&gt;\{6, 9\}&lt;/math&gt;, and the empty set). Now consider the set &lt;math&gt;\{5, 10\}&lt;/math&gt;. We see that 3 of these subsets can be a subset of &lt;math&gt;S&lt;/math&gt; (&lt;math&gt;\{5\}&lt;/math&gt;, &lt;math&gt;\{10\}&lt;/math&gt;, and the empty set). Note that &lt;math&gt;4&lt;/math&gt; cannot be an element of &lt;math&gt;S&lt;/math&gt;, because &lt;math&gt;2&lt;/math&gt; is. Now consider the set &lt;math&gt;\{7, 8\}&lt;/math&gt;. All four of these subsets can be a subset of &lt;math&gt;S&lt;/math&gt;. So if the smallest element of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;5*3*4=60&lt;/math&gt; possible such sets.<br /> <br /> If the smallest element of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;, the only restriction we have is that &lt;math&gt;9&lt;/math&gt; is not in &lt;math&gt;S&lt;/math&gt;. This leaves us &lt;math&gt;2^6=64&lt;/math&gt; such sets.<br /> <br /> If the smallest element of &lt;math&gt;S&lt;/math&gt; is not &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt;, then &lt;math&gt;S&lt;/math&gt; can be any subset of &lt;math&gt;\{4, 5, 6, 7, 8, 9, 10\}&lt;/math&gt;, including the empty set. This gives us &lt;math&gt;2^7=128&lt;/math&gt; such subsets.<br /> <br /> So our answer is &lt;math&gt;60+64+128=\boxed{252}&lt;/math&gt;.<br /> <br /> ==Solution 2(PIE)==<br /> We cannot have the following pairs or triplets: &lt;math&gt;\{2, 4\}, \{3, 9\}, \{2, 3, 6\}, \{2, 5, 10\}&lt;/math&gt;.<br /> Since there are &lt;math&gt;512&lt;/math&gt; subsets(&lt;math&gt;1&lt;/math&gt; isn't needed) we have the following:<br /> &lt;math&gt;(512-(384-160+40-4)) \implies \boxed{252}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_12&diff=86197 2017 AIME I Problems/Problem 12 2017-06-28T01:23:43Z <p>First: </p> <hr /> <div>==Problem 12==<br /> Call a set &lt;math&gt;S&lt;/math&gt; product-free if there do not exist &lt;math&gt;a, b, c \in S&lt;/math&gt; (not necessarily distinct) such that &lt;math&gt;a b = c&lt;/math&gt;. For example, the empty set and the set &lt;math&gt;\{16, 20\}&lt;/math&gt; are product-free, whereas the sets &lt;math&gt;\{4, 16\}&lt;/math&gt; and &lt;math&gt;\{2, 8, 16\}&lt;/math&gt; are not product-free. Find the number of product-free subsets of the set &lt;math&gt;\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}&lt;/math&gt;.<br /> <br /> ==Solution 1(Casework)==<br /> <br /> We shall solve this problem by doing casework on the lowest element of the subset. Note that the number &lt;math&gt;1&lt;/math&gt; cannot be in the subset because &lt;math&gt;1*1=1&lt;/math&gt;. Let &lt;math&gt;S&lt;/math&gt; be a product-free set. If the lowest element of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt;, we consider the set &lt;math&gt;\{3, 6, 9\}&lt;/math&gt;. We see that 5 of these subsets can be a subset of &lt;math&gt;S&lt;/math&gt; (&lt;math&gt;\{3\}&lt;/math&gt;, &lt;math&gt;\{6\}&lt;/math&gt;, &lt;math&gt;\{9\}&lt;/math&gt;, &lt;math&gt;\{6, 9\}&lt;/math&gt;, and the empty set). Now consider the set &lt;math&gt;\{5, 10\}&lt;/math&gt;. We see that 3 of these subsets can be a subset of &lt;math&gt;S&lt;/math&gt; (&lt;math&gt;\{5\}&lt;/math&gt;, &lt;math&gt;\{10\}&lt;/math&gt;, and the empty set). Note that &lt;math&gt;4&lt;/math&gt; cannot be an element of &lt;math&gt;S&lt;/math&gt;, because &lt;math&gt;2&lt;/math&gt; is. Now consider the set &lt;math&gt;\{7, 8\}&lt;/math&gt;. All four of these subsets can be a subset of &lt;math&gt;S&lt;/math&gt;. So if the smallest element of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;5*3*4=60&lt;/math&gt; possible such sets.<br /> <br /> If the smallest element of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;, the only restriction we have is that &lt;math&gt;9&lt;/math&gt; is not in &lt;math&gt;S&lt;/math&gt;. This leaves us &lt;math&gt;2^6=64&lt;/math&gt; such sets.<br /> <br /> If the smallest element of &lt;math&gt;S&lt;/math&gt; is not &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt;, then &lt;math&gt;S&lt;/math&gt; can be any subset of &lt;math&gt;\{4, 5, 6, 7, 8, 9, 10\}&lt;/math&gt;, including the empty set. This gives us &lt;math&gt;2^7=128&lt;/math&gt; such subsets.<br /> <br /> So our answer is &lt;math&gt;60+64+128=\boxed{252}&lt;/math&gt;.<br /> <br /> ==Solution 2(PIE)==<br /> We cannot have the following pairs or triplets: &lt;math&gt;\{2, 4\}, \{3, 9}, \{2, 3, 6\}, \{2, 5, 10\}&lt;/math&gt;.<br /> Since there are &lt;math&gt;512&lt;/math&gt; subsets(&lt;math&gt;1&lt;/math&gt; isn't needed) we have the following:<br /> &lt;math&gt;(512-(384-160+40-4)) \implies \boxed{252}&lt;/math&gt;.<br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_10&diff=86196 2017 AIME I Problems/Problem 10 2017-06-28T00:44:41Z <p>First: /* Solution 3 */</p> <hr /> <div>==Problem 10==<br /> Let &lt;math&gt;z_1=18+83i,~z_2=18+39i,&lt;/math&gt; and &lt;math&gt;z_3=78+99i,&lt;/math&gt; where &lt;math&gt;i=\sqrt{-1}.&lt;/math&gt; Let &lt;math&gt;z&lt;/math&gt; be the unique complex number with the properties that &lt;math&gt;\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}&lt;/math&gt; is a real number and the imaginary part of &lt;math&gt;z&lt;/math&gt; is the greatest possible. Find the real part of &lt;math&gt;z&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> (This solution's quality may be very poor. If one feels that the solution is inadequate, one may choose to improve it.)<br /> <br /> Let us write &lt;math&gt;\frac{z_3 - z_1}{z_2 - z_1}&lt;/math&gt; be some imaginary number with form &lt;math&gt;r_1 (\cos \theta_1 + i \sin \theta_1).&lt;/math&gt; Similarly, we can write &lt;math&gt;\frac{z-z_2}{z-z_3}&lt;/math&gt; as some &lt;math&gt;r_2 (\cos \theta_2 + i \sin \theta_2).&lt;/math&gt;<br /> <br /> The product must be real, so we have that &lt;math&gt;r_1 r_2 (\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2)&lt;/math&gt; is real. Of this, &lt;math&gt;r_1 r_2&lt;/math&gt; must be real, so the imaginary parts only arise from the second part of the product. Thus we have <br /> <br /> &lt;cmath&gt;(\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2) = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i(\cos \theta_1 \sin \theta_2 + \cos \theta_2 \sin \theta_1)&lt;/cmath&gt;<br /> <br /> is real. The imaginary part of this is &lt;math&gt;(\cos \theta_1 \sin \theta_2 + \cos \theta_2 \sin \theta_1),&lt;/math&gt; which we recognize as &lt;math&gt;\sin(\theta_1 + \theta_2).&lt;/math&gt; This is only &lt;math&gt;0&lt;/math&gt; when &lt;math&gt;\theta_1 + \theta_2&lt;/math&gt; is some multiple of &lt;math&gt;\pi.&lt;/math&gt; In this problem, this implies &lt;math&gt;z_1, z_2, z_3&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; must form a cyclic quadrilateral, so the possibilities of &lt;math&gt;z&lt;/math&gt; lie on the circumcircle of &lt;math&gt;z_1, z_2&lt;/math&gt; and &lt;math&gt;z_3.&lt;/math&gt;<br /> <br /> To maximize the imaginary part of &lt;math&gt;z,&lt;/math&gt; it must lie at the top of the circumcircle, which means the real part of &lt;math&gt;z&lt;/math&gt; is the same as the real part of the circumcenter. The center of the circumcircle can be found in various ways, (such as computing the intersection of the perpendicular bisectors of the sides) and when computed gives us that the real part of the circumcenter is &lt;math&gt;56,&lt;/math&gt; so the real part of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;56,&lt;/math&gt; and thus our answer is &lt;math&gt;\boxed{056}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Algebra Bash<br /> <br /> First we calculate &lt;math&gt;\frac{z_3 - z_1}{z_3 - z_2}&lt;/math&gt; , which becomes &lt;math&gt;\frac{15i-4}{11}&lt;/math&gt;.<br /> <br /> Next, we define &lt;math&gt;z&lt;/math&gt; to be &lt;math&gt;a-bi&lt;/math&gt; for some real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Then, &lt;math&gt;\frac {z-z_2}{z-z_3}&lt;/math&gt; can be written as &lt;math&gt;\frac{(a-18)+(b-39)i}{(a-78)+(b-99)i}.&lt;/math&gt; Multiplying both the numerator and denominator by the conjugate of the denominator, we get:<br /> <br /> &lt;math&gt;\frac {[(a-18)(a-78)+(b-39)(b-99)]+[(a-78)(b-39)-(a-18)(b-99)]i}{(a-78)^2+(b-99)^2}&lt;/math&gt;<br /> <br /> In order for the product to be a real number, since both denominators are real numbers, we must have the numerator of &lt;math&gt;\frac {z-z_2}{z-z_3}&lt;/math&gt; be a multiple of the conjugate of &lt;math&gt;15i-4&lt;/math&gt;, namely &lt;math&gt;-15i-4&lt;/math&gt; So, we have &lt;math&gt;(a-18)(a-78)+(b-39)(b-99) = -4k&lt;/math&gt; and &lt;math&gt;(a-78)(b-39)-(a-18)(b-99) = -15k&lt;/math&gt; for some real number &lt;math&gt;k&lt;/math&gt;.<br /> <br /> Then, we get:&lt;math&gt;(a-18)(a-78)+(b-39)(b-99) = \frac{4}{15}[(a-78)(b-39)-(a-18)(b-99)]&lt;/math&gt;<br /> <br /> Expanding both sides and combining like terms, we get:<br /> <br /> &lt;math&gt;a^2 - 112a +b^2 - 122b + \frac{1989}{5} = 0&lt;/math&gt;<br /> <br /> which can be rewritten as:<br /> <br /> &lt;math&gt;(a-56)^2 + (b-61)^2 = \frac{32296}{5}&lt;/math&gt;<br /> <br /> Now, common sense tells us that to maximize &lt;math&gt;b&lt;/math&gt;, we would need to maximize &lt;math&gt;(b-61)^2&lt;/math&gt;. Therefore, we must set &lt;math&gt;(a-56)^2&lt;/math&gt; to its lowest value, namely 0. Therefore, &lt;math&gt;a&lt;/math&gt; must be &lt;math&gt;\boxed{056}.&lt;/math&gt;<br /> <br /> You can also notice that the ab terms cancel out so all you need is the x-coordinate of the center and only expand the a parts of the equation.<br /> <br /> ~stronto<br /> <br /> ==Solution 3==<br /> If you don't already(and can consistently solve at least &lt;math&gt;5&lt;/math&gt; questions on the AIME or more), you should learn how to complex bash for Olympiads. The &lt;math&gt;\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}&lt;/math&gt; just means &lt;math&gt;z&lt;/math&gt; is on the circumcircle of &lt;math&gt;\triangle{z_1 z_2 z_3}&lt;/math&gt; and we just want the highest point on the circle in terms of imaginary part. Convert to Cartesian coordinates and we just need to compute the &lt;math&gt;x&lt;/math&gt;-coordinate of the circumcircle of &lt;math&gt;(18, 83), (18, 39), (78, 99)&lt;/math&gt;(just get the intersection of the perpendicular bisectors) and we get the &lt;math&gt;x&lt;/math&gt;-coordinate of the circumcircle is &lt;math&gt;\boxed{056}&lt;/math&gt;.<br /> <br /> ~First<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_10&diff=86195 2017 AIME I Problems/Problem 10 2017-06-28T00:44:28Z <p>First: </p> <hr /> <div>==Problem 10==<br /> Let &lt;math&gt;z_1=18+83i,~z_2=18+39i,&lt;/math&gt; and &lt;math&gt;z_3=78+99i,&lt;/math&gt; where &lt;math&gt;i=\sqrt{-1}.&lt;/math&gt; Let &lt;math&gt;z&lt;/math&gt; be the unique complex number with the properties that &lt;math&gt;\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}&lt;/math&gt; is a real number and the imaginary part of &lt;math&gt;z&lt;/math&gt; is the greatest possible. Find the real part of &lt;math&gt;z&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> (This solution's quality may be very poor. If one feels that the solution is inadequate, one may choose to improve it.)<br /> <br /> Let us write &lt;math&gt;\frac{z_3 - z_1}{z_2 - z_1}&lt;/math&gt; be some imaginary number with form &lt;math&gt;r_1 (\cos \theta_1 + i \sin \theta_1).&lt;/math&gt; Similarly, we can write &lt;math&gt;\frac{z-z_2}{z-z_3}&lt;/math&gt; as some &lt;math&gt;r_2 (\cos \theta_2 + i \sin \theta_2).&lt;/math&gt;<br /> <br /> The product must be real, so we have that &lt;math&gt;r_1 r_2 (\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2)&lt;/math&gt; is real. Of this, &lt;math&gt;r_1 r_2&lt;/math&gt; must be real, so the imaginary parts only arise from the second part of the product. Thus we have <br /> <br /> &lt;cmath&gt;(\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2) = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i(\cos \theta_1 \sin \theta_2 + \cos \theta_2 \sin \theta_1)&lt;/cmath&gt;<br /> <br /> is real. The imaginary part of this is &lt;math&gt;(\cos \theta_1 \sin \theta_2 + \cos \theta_2 \sin \theta_1),&lt;/math&gt; which we recognize as &lt;math&gt;\sin(\theta_1 + \theta_2).&lt;/math&gt; This is only &lt;math&gt;0&lt;/math&gt; when &lt;math&gt;\theta_1 + \theta_2&lt;/math&gt; is some multiple of &lt;math&gt;\pi.&lt;/math&gt; In this problem, this implies &lt;math&gt;z_1, z_2, z_3&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; must form a cyclic quadrilateral, so the possibilities of &lt;math&gt;z&lt;/math&gt; lie on the circumcircle of &lt;math&gt;z_1, z_2&lt;/math&gt; and &lt;math&gt;z_3.&lt;/math&gt;<br /> <br /> To maximize the imaginary part of &lt;math&gt;z,&lt;/math&gt; it must lie at the top of the circumcircle, which means the real part of &lt;math&gt;z&lt;/math&gt; is the same as the real part of the circumcenter. The center of the circumcircle can be found in various ways, (such as computing the intersection of the perpendicular bisectors of the sides) and when computed gives us that the real part of the circumcenter is &lt;math&gt;56,&lt;/math&gt; so the real part of &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;56,&lt;/math&gt; and thus our answer is &lt;math&gt;\boxed{056}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Algebra Bash<br /> <br /> First we calculate &lt;math&gt;\frac{z_3 - z_1}{z_3 - z_2}&lt;/math&gt; , which becomes &lt;math&gt;\frac{15i-4}{11}&lt;/math&gt;.<br /> <br /> Next, we define &lt;math&gt;z&lt;/math&gt; to be &lt;math&gt;a-bi&lt;/math&gt; for some real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Then, &lt;math&gt;\frac {z-z_2}{z-z_3}&lt;/math&gt; can be written as &lt;math&gt;\frac{(a-18)+(b-39)i}{(a-78)+(b-99)i}.&lt;/math&gt; Multiplying both the numerator and denominator by the conjugate of the denominator, we get:<br /> <br /> &lt;math&gt;\frac {[(a-18)(a-78)+(b-39)(b-99)]+[(a-78)(b-39)-(a-18)(b-99)]i}{(a-78)^2+(b-99)^2}&lt;/math&gt;<br /> <br /> In order for the product to be a real number, since both denominators are real numbers, we must have the numerator of &lt;math&gt;\frac {z-z_2}{z-z_3}&lt;/math&gt; be a multiple of the conjugate of &lt;math&gt;15i-4&lt;/math&gt;, namely &lt;math&gt;-15i-4&lt;/math&gt; So, we have &lt;math&gt;(a-18)(a-78)+(b-39)(b-99) = -4k&lt;/math&gt; and &lt;math&gt;(a-78)(b-39)-(a-18)(b-99) = -15k&lt;/math&gt; for some real number &lt;math&gt;k&lt;/math&gt;.<br /> <br /> Then, we get:&lt;math&gt;(a-18)(a-78)+(b-39)(b-99) = \frac{4}{15}[(a-78)(b-39)-(a-18)(b-99)]&lt;/math&gt;<br /> <br /> Expanding both sides and combining like terms, we get:<br /> <br /> &lt;math&gt;a^2 - 112a +b^2 - 122b + \frac{1989}{5} = 0&lt;/math&gt;<br /> <br /> which can be rewritten as:<br /> <br /> &lt;math&gt;(a-56)^2 + (b-61)^2 = \frac{32296}{5}&lt;/math&gt;<br /> <br /> Now, common sense tells us that to maximize &lt;math&gt;b&lt;/math&gt;, we would need to maximize &lt;math&gt;(b-61)^2&lt;/math&gt;. Therefore, we must set &lt;math&gt;(a-56)^2&lt;/math&gt; to its lowest value, namely 0. Therefore, &lt;math&gt;a&lt;/math&gt; must be &lt;math&gt;\boxed{056}.&lt;/math&gt;<br /> <br /> You can also notice that the ab terms cancel out so all you need is the x-coordinate of the center and only expand the a parts of the equation.<br /> <br /> ~stronto<br /> <br /> ==Solution 3==<br /> If you don't already(and can consistently solve at least &lt;math&gt;5&lt;/math&gt; questions on the AIME or more), you should learn how to complex bash for Olympiads. The &lt;math&gt;\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}&lt;/math&gt; just means &lt;math&gt;z&lt;/math&gt; is on the circumcircle of &lt;math&gt;\triangle{z_1 z_2 z_3}&lt;/math&gt; and we just want the highest point on the circle in terms of imaginary part. Convert to Cartesian coordinates and we just need to compute the &lt;math&gt;x&lt;/math&gt;-coordinate of the circumcircle of &lt;math&gt;(18, 83), (18, 39), (78, 99)&lt;/math&gt;(just get the intersection of the perpendicular bisectors) and we get the &lt;math&gt;x&lt;/math&gt;-coordinate of the circumcircle is &lt;math&gt;\boxed{056}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_15&diff=86171 2014 AIME I Problems/Problem 15 2017-06-25T22:08:48Z <p>First: /* Solution 3 */</p> <hr /> <div>== Problem 15 ==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 3&lt;/math&gt;, &lt;math&gt;BC = 4&lt;/math&gt;, and &lt;math&gt;CA = 5&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; intersects &lt;math&gt;\overline{AB}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt;. Given that &lt;math&gt;EF=DF&lt;/math&gt; and &lt;math&gt;\frac{DG}{EG} = \frac{3}{4}&lt;/math&gt;, length &lt;math&gt;DE=\frac{a\sqrt{b}}{c}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are relatively prime positive integers, and &lt;math&gt;b&lt;/math&gt; is a positive integer not divisible by the square of any prime. Find &lt;math&gt;a+b+c&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> <br /> Since &lt;math&gt;\angle DBE = 90^\circ&lt;/math&gt;, &lt;math&gt;DE&lt;/math&gt; is the diameter of &lt;math&gt;\omega&lt;/math&gt;. Then &lt;math&gt;\angle DFE=\angle DGE=90^\circ&lt;/math&gt;. But &lt;math&gt;DF=FE&lt;/math&gt;, so &lt;math&gt;\triangle DEF&lt;/math&gt; is a 45-45-90 triangle. Letting &lt;math&gt;DG=3x&lt;/math&gt;, we have that &lt;math&gt;EG=4x&lt;/math&gt;, &lt;math&gt;DE=5x&lt;/math&gt;, and &lt;math&gt;DF=EF=\frac{5x}{\sqrt{2}}&lt;/math&gt;. <br /> <br /> Note that &lt;math&gt;\triangle DGE \sim \triangle ABC&lt;/math&gt; by SAS similarity, so &lt;math&gt;\angle BAC = \angle GDE&lt;/math&gt; and &lt;math&gt;\angle ACB = \angle DEG&lt;/math&gt;. Since &lt;math&gt;DEFG&lt;/math&gt; is a cyclic quadrilateral, &lt;math&gt;\angle BAC = \angle GDE=180^\circ-\angle EFG = \angle AFE&lt;/math&gt; and &lt;math&gt;\angle ACB = \angle DEG = \angle GFD&lt;/math&gt;, implying that &lt;math&gt;\triangle AFE&lt;/math&gt; and &lt;math&gt;\triangle CDF&lt;/math&gt; are isosceles. As a result, &lt;math&gt;AE=CD=\frac{5x}{\sqrt{2}}&lt;/math&gt;, so &lt;math&gt;BE=3-\frac{5x}{\sqrt{2}}&lt;/math&gt; and &lt;math&gt;BD =4-\frac{5x}{\sqrt{2}}&lt;/math&gt;. <br /> <br /> Finally, using the Pythagorean Theorem on &lt;math&gt;\triangle BDE&lt;/math&gt;, <br /> &lt;cmath&gt; \left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2&lt;/cmath&gt;<br /> Solving for &lt;math&gt;x&lt;/math&gt;, we get that &lt;math&gt;x=\frac{5\sqrt{2}}{14}&lt;/math&gt;, so &lt;math&gt;DE=5x=\frac{25\sqrt{2}}{14}&lt;/math&gt;. Thus, the answer is &lt;math&gt;25+2+14=\boxed{041}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> &lt;asy&gt;<br /> pair A = (0,3);<br /> pair B = (0,0);<br /> pair C = (4,0);<br /> draw(A--B--C--cycle);<br /> dotfactor = 3;<br /> dot(&quot;$A$&quot;,A,dir(135));<br /> dot(&quot;$B$&quot;,B,dir(215));<br /> dot(&quot;$C$&quot;,C,dir(305));<br /> pair D = (2.21, 0);<br /> pair E = (0, 1.21);<br /> pair F = (1.71, 1.71);<br /> pair G = (2, 1.5);<br /> dot(&quot;$D$&quot;,D,dir(270));<br /> dot(&quot;$E$&quot;,E,dir(180));<br /> dot(&quot;$F$&quot;,F,dir(90));<br /> dot(&quot;$G$&quot;,G,dir(0));<br /> draw(Circle((1.109, 0.609), 1.28));<br /> draw(D--E);<br /> draw(E--F);<br /> draw(D--F);<br /> draw(E--G);<br /> draw(D--G);<br /> draw(B--F);<br /> draw(B--G);<br /> &lt;/asy&gt;<br /> <br /> First we note that &lt;math&gt;\triangle DEF&lt;/math&gt; is an isosceles right triangle with hypotenuse &lt;math&gt;\overline{DE}&lt;/math&gt; the same as the diameter of &lt;math&gt;\omega&lt;/math&gt;. We also note that &lt;math&gt;\triangle DGE \sim \triangle ABC&lt;/math&gt; since &lt;math&gt;\angle EGD&lt;/math&gt; is a right angle and the ratios of the sides are &lt;math&gt;3:4:5&lt;/math&gt;. <br /> <br /> From congruent arc intersections, we know that &lt;math&gt;\angle GED \cong \angle GBC&lt;/math&gt;, and that from similar triangles &lt;math&gt;\angle GED&lt;/math&gt; is also congruent to &lt;math&gt;\angle GCB&lt;/math&gt;. Thus, &lt;math&gt;\triangle BGC&lt;/math&gt; is an isosceles triangle with &lt;math&gt;BG = GC&lt;/math&gt;, so &lt;math&gt;G&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;AG = GC = 5/2&lt;/math&gt;. Similarly, we can find from angle chasing that &lt;math&gt;\angle ABF = \angle EDF = \frac{\pi}4&lt;/math&gt;. Therefore, &lt;math&gt;\overline{BF}&lt;/math&gt; is the angle bisector of &lt;math&gt;\angle B&lt;/math&gt;. From the angle bisector theorem, we have &lt;math&gt;\frac{AF}{AB} = \frac{CF}{CB}&lt;/math&gt;, so &lt;math&gt;AF = 15/7&lt;/math&gt; and &lt;math&gt;CF = 20/7&lt;/math&gt;. <br /> <br /> Lastly, we apply power of a point from points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; with respect to &lt;math&gt;\omega&lt;/math&gt; and have &lt;math&gt;AE \times AB=AF \times AG&lt;/math&gt; and &lt;math&gt;CD \times CB=CG \times CF&lt;/math&gt;, so we can compute that &lt;math&gt;EB = \frac{17}{14}&lt;/math&gt; and &lt;math&gt;DB = \frac{31}{14}&lt;/math&gt;. From the Pythagorean Theorem, we result in &lt;math&gt;DE = \frac{25 \sqrt{2}}{14}&lt;/math&gt;, so &lt;math&gt;a+b+c=25+2+14= \boxed{041}&lt;/math&gt;<br /> <br /> <br /> Also: &lt;math&gt;FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}&lt;/math&gt;. We can also use Ptolemy's Theorem on quadrilateral &lt;math&gt;DEFG&lt;/math&gt; to figure what &lt;math&gt;FG&lt;/math&gt; is in terms of &lt;math&gt;d&lt;/math&gt;:<br /> &lt;cmath&gt;DE\cdot FG+DG\cdot EF=DF\cdot EG&lt;/cmath&gt;<br /> &lt;cmath&gt;d\cdot FG+\frac{3d}{5}\cdot \frac{d}{\sqrt{2}}=\frac{4d}{5}\cdot \frac{d}{\sqrt{2}}&lt;/cmath&gt;<br /> &lt;cmath&gt;d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}&lt;/cmath&gt;<br /> Thus &lt;math&gt;\frac{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}&lt;/math&gt;. &lt;math&gt;a+b+c=25+2+14= \boxed{041}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> Call &lt;math&gt;DE=x&lt;/math&gt; and as a result &lt;math&gt;DG=EF=\frac{x\sqrt{2}}{2}, EF=\frac{3x}{5}, FD=\frac{4x}{5}&lt;/math&gt;. Since &lt;math&gt;EFGD&lt;/math&gt; is cyclic we just need to get &lt;math&gt;DG&lt;/math&gt; and using LoS(for more detail see the &lt;math&gt;2&lt;/math&gt;nd paragraph of Solution &lt;math&gt;2&lt;/math&gt;) we get &lt;math&gt;AG=\frac{5}{2}&lt;/math&gt; and using a similar argument(use LoS again) and subtracting you get &lt;math&gt;FG=\frac{5}{14}&lt;/math&gt; so you can use Ptolemy to get &lt;math&gt;x=\frac{25\sqrt{2}}{14} \implies \boxed{041}&lt;/math&gt;.<br /> ~First<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_15&diff=86170 2014 AIME I Problems/Problem 15 2017-06-25T22:08:23Z <p>First: </p> <hr /> <div>== Problem 15 ==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 3&lt;/math&gt;, &lt;math&gt;BC = 4&lt;/math&gt;, and &lt;math&gt;CA = 5&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; intersects &lt;math&gt;\overline{AB}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt;. Given that &lt;math&gt;EF=DF&lt;/math&gt; and &lt;math&gt;\frac{DG}{EG} = \frac{3}{4}&lt;/math&gt;, length &lt;math&gt;DE=\frac{a\sqrt{b}}{c}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are relatively prime positive integers, and &lt;math&gt;b&lt;/math&gt; is a positive integer not divisible by the square of any prime. Find &lt;math&gt;a+b+c&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> <br /> Since &lt;math&gt;\angle DBE = 90^\circ&lt;/math&gt;, &lt;math&gt;DE&lt;/math&gt; is the diameter of &lt;math&gt;\omega&lt;/math&gt;. Then &lt;math&gt;\angle DFE=\angle DGE=90^\circ&lt;/math&gt;. But &lt;math&gt;DF=FE&lt;/math&gt;, so &lt;math&gt;\triangle DEF&lt;/math&gt; is a 45-45-90 triangle. Letting &lt;math&gt;DG=3x&lt;/math&gt;, we have that &lt;math&gt;EG=4x&lt;/math&gt;, &lt;math&gt;DE=5x&lt;/math&gt;, and &lt;math&gt;DF=EF=\frac{5x}{\sqrt{2}}&lt;/math&gt;. <br /> <br /> Note that &lt;math&gt;\triangle DGE \sim \triangle ABC&lt;/math&gt; by SAS similarity, so &lt;math&gt;\angle BAC = \angle GDE&lt;/math&gt; and &lt;math&gt;\angle ACB = \angle DEG&lt;/math&gt;. Since &lt;math&gt;DEFG&lt;/math&gt; is a cyclic quadrilateral, &lt;math&gt;\angle BAC = \angle GDE=180^\circ-\angle EFG = \angle AFE&lt;/math&gt; and &lt;math&gt;\angle ACB = \angle DEG = \angle GFD&lt;/math&gt;, implying that &lt;math&gt;\triangle AFE&lt;/math&gt; and &lt;math&gt;\triangle CDF&lt;/math&gt; are isosceles. As a result, &lt;math&gt;AE=CD=\frac{5x}{\sqrt{2}}&lt;/math&gt;, so &lt;math&gt;BE=3-\frac{5x}{\sqrt{2}}&lt;/math&gt; and &lt;math&gt;BD =4-\frac{5x}{\sqrt{2}}&lt;/math&gt;. <br /> <br /> Finally, using the Pythagorean Theorem on &lt;math&gt;\triangle BDE&lt;/math&gt;, <br /> &lt;cmath&gt; \left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2&lt;/cmath&gt;<br /> Solving for &lt;math&gt;x&lt;/math&gt;, we get that &lt;math&gt;x=\frac{5\sqrt{2}}{14}&lt;/math&gt;, so &lt;math&gt;DE=5x=\frac{25\sqrt{2}}{14}&lt;/math&gt;. Thus, the answer is &lt;math&gt;25+2+14=\boxed{041}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> &lt;asy&gt;<br /> pair A = (0,3);<br /> pair B = (0,0);<br /> pair C = (4,0);<br /> draw(A--B--C--cycle);<br /> dotfactor = 3;<br /> dot(&quot;$A$&quot;,A,dir(135));<br /> dot(&quot;$B$&quot;,B,dir(215));<br /> dot(&quot;$C$&quot;,C,dir(305));<br /> pair D = (2.21, 0);<br /> pair E = (0, 1.21);<br /> pair F = (1.71, 1.71);<br /> pair G = (2, 1.5);<br /> dot(&quot;$D$&quot;,D,dir(270));<br /> dot(&quot;$E$&quot;,E,dir(180));<br /> dot(&quot;$F$&quot;,F,dir(90));<br /> dot(&quot;$G$&quot;,G,dir(0));<br /> draw(Circle((1.109, 0.609), 1.28));<br /> draw(D--E);<br /> draw(E--F);<br /> draw(D--F);<br /> draw(E--G);<br /> draw(D--G);<br /> draw(B--F);<br /> draw(B--G);<br /> &lt;/asy&gt;<br /> <br /> First we note that &lt;math&gt;\triangle DEF&lt;/math&gt; is an isosceles right triangle with hypotenuse &lt;math&gt;\overline{DE}&lt;/math&gt; the same as the diameter of &lt;math&gt;\omega&lt;/math&gt;. We also note that &lt;math&gt;\triangle DGE \sim \triangle ABC&lt;/math&gt; since &lt;math&gt;\angle EGD&lt;/math&gt; is a right angle and the ratios of the sides are &lt;math&gt;3:4:5&lt;/math&gt;. <br /> <br /> From congruent arc intersections, we know that &lt;math&gt;\angle GED \cong \angle GBC&lt;/math&gt;, and that from similar triangles &lt;math&gt;\angle GED&lt;/math&gt; is also congruent to &lt;math&gt;\angle GCB&lt;/math&gt;. Thus, &lt;math&gt;\triangle BGC&lt;/math&gt; is an isosceles triangle with &lt;math&gt;BG = GC&lt;/math&gt;, so &lt;math&gt;G&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;AG = GC = 5/2&lt;/math&gt;. Similarly, we can find from angle chasing that &lt;math&gt;\angle ABF = \angle EDF = \frac{\pi}4&lt;/math&gt;. Therefore, &lt;math&gt;\overline{BF}&lt;/math&gt; is the angle bisector of &lt;math&gt;\angle B&lt;/math&gt;. From the angle bisector theorem, we have &lt;math&gt;\frac{AF}{AB} = \frac{CF}{CB}&lt;/math&gt;, so &lt;math&gt;AF = 15/7&lt;/math&gt; and &lt;math&gt;CF = 20/7&lt;/math&gt;. <br /> <br /> Lastly, we apply power of a point from points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; with respect to &lt;math&gt;\omega&lt;/math&gt; and have &lt;math&gt;AE \times AB=AF \times AG&lt;/math&gt; and &lt;math&gt;CD \times CB=CG \times CF&lt;/math&gt;, so we can compute that &lt;math&gt;EB = \frac{17}{14}&lt;/math&gt; and &lt;math&gt;DB = \frac{31}{14}&lt;/math&gt;. From the Pythagorean Theorem, we result in &lt;math&gt;DE = \frac{25 \sqrt{2}}{14}&lt;/math&gt;, so &lt;math&gt;a+b+c=25+2+14= \boxed{041}&lt;/math&gt;<br /> <br /> <br /> Also: &lt;math&gt;FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}&lt;/math&gt;. We can also use Ptolemy's Theorem on quadrilateral &lt;math&gt;DEFG&lt;/math&gt; to figure what &lt;math&gt;FG&lt;/math&gt; is in terms of &lt;math&gt;d&lt;/math&gt;:<br /> &lt;cmath&gt;DE\cdot FG+DG\cdot EF=DF\cdot EG&lt;/cmath&gt;<br /> &lt;cmath&gt;d\cdot FG+\frac{3d}{5}\cdot \frac{d}{\sqrt{2}}=\frac{4d}{5}\cdot \frac{d}{\sqrt{2}}&lt;/cmath&gt;<br /> &lt;cmath&gt;d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}&lt;/cmath&gt;<br /> Thus &lt;math&gt;\frac{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}&lt;/math&gt;. &lt;math&gt;a+b+c=25+2+14= \boxed{041}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> Call &lt;math&gt;DE=x&lt;/math&gt; and as a result &lt;math&gt;DG=EF=\frac{x\sqrt{2}}{2}, EF=\frac{3x}{5}, FD=\frac{4x}{5}&lt;/math&gt;. Since &lt;math&gt;EFGD&lt;/math&gt; is cyclic we just need to get &lt;math&gt;DG&lt;/math&gt; and using LoS(for more detail see the &lt;math&gt;2&lt;/math&gt;nd paragraph of Solution &lt;math&gt;2&lt;/math&gt;) we get &lt;math&gt;AG=\frac{5}{2}&lt;/math&gt; and using a similar argument(use LoS again) and subtracting you get &lt;math&gt;FG=\frac{5}{14}&lt;/math&gt; so you can use Ptolemy to get &lt;math&gt;x=\frac{25\sqrt{2}}{14} \implies \boxed{041}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=I|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=86149 2012 AIME II Problems/Problem 15 2017-06-23T21:42:47Z <p>First: /* Solution 3 */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use the Stewart's Theorem to find &lt;math&gt;AD=15/8&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=49/8&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(6cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle CFB=\angle BFE= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=86148 2012 AIME II Problems/Problem 15 2017-06-23T21:42:35Z <p>First: /* Solution 3 */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use the Stewart's Theorem to find &lt;math&gt;AD=15/8&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=49/8&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(5cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle CFB=\angle BFE= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=86147 2012 AIME II Problems/Problem 15 2017-06-23T21:40:30Z <p>First: /* Solution 3 */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use the Stewart's Theorem to find &lt;math&gt;AD=15/8&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=49/8&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle CFB=\angle BFE= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=86146 2012 AIME II Problems/Problem 15 2017-06-23T21:40:13Z <p>First: /* See Also */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use the Stewart's Theorem to find &lt;math&gt;AD=15/8&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=49/8&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(9cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle CFB=\angle BFE= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_7&diff=86096 2009 AIME II Problems/Problem 7 2017-06-18T18:24:53Z <p>First: /* Solution */</p> <hr /> <div>== Problem ==<br /> Define &lt;math&gt;n!!&lt;/math&gt; to be &lt;math&gt;n(n-2)(n-4)\cdots 3\cdot 1&lt;/math&gt; for &lt;math&gt;n&lt;/math&gt; odd and &lt;math&gt;n(n-2)(n-4)\cdots 4\cdot 2&lt;/math&gt; for &lt;math&gt;n&lt;/math&gt; even. When &lt;math&gt;\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}&lt;/math&gt; is expressed as a fraction in lowest terms, its denominator is &lt;math&gt;2^ab&lt;/math&gt; with &lt;math&gt;b&lt;/math&gt; odd. Find &lt;math&gt;\dfrac{ab}{10}&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> <br /> First, note that &lt;math&gt;(2n)!! = 2^n \cdot n!&lt;/math&gt;, and that &lt;math&gt;(2n)!! \cdot (2n-1)!! = (2n)!&lt;/math&gt;. <br /> <br /> We can now take the fraction &lt;math&gt;\dfrac{(2i-1)!!}{(2i)!!}&lt;/math&gt; and multiply both the numerator and the denominator by &lt;math&gt;(2i)!!&lt;/math&gt;. We get that this fraction is equal to &lt;math&gt;\dfrac{(2i)!}{(2i)!!^2} = \dfrac{(2i)!}{2^{2i}(i!)^2}&lt;/math&gt;.<br /> <br /> Now we can recognize that &lt;math&gt;\dfrac{(2i)!}{(i!)^2}&lt;/math&gt; is simply &lt;math&gt;{2i \choose i}&lt;/math&gt;, hence this fraction is &lt;math&gt;\dfrac{{2i\choose i}}{2^{2i}}&lt;/math&gt;, and our sum turns into &lt;math&gt;S=\sum_{i=1}^{2009} \dfrac{{2i\choose i}}{2^{2i}}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}&lt;/math&gt;.<br /> Obviously &lt;math&gt;c&lt;/math&gt; is an integer, and &lt;math&gt;S&lt;/math&gt; can be written as &lt;math&gt;\dfrac{c}{2^{2\cdot 2009}}&lt;/math&gt;.<br /> Hence if &lt;math&gt;S&lt;/math&gt; is expressed as a fraction in lowest terms, its denominator will be of the form &lt;math&gt;2^a&lt;/math&gt; for some &lt;math&gt;a\leq 2\cdot 2009&lt;/math&gt;. <br /> <br /> In other words, we just showed that &lt;math&gt;b=1&lt;/math&gt;.<br /> To determine &lt;math&gt;a&lt;/math&gt;, we need to determine the largest power of &lt;math&gt;2&lt;/math&gt; that divides &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;p(i)&lt;/math&gt; be the largest &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;2^x&lt;/math&gt; that divides &lt;math&gt;i&lt;/math&gt;. <br /> <br /> We can now return to the observation that &lt;math&gt;(2i)! = (2i)!! \cdot (2i-1)!! = 2^i \cdot i! \cdot (2i-1)!!&lt;/math&gt;. Together with the obvious fact that &lt;math&gt;(2i-1)!!&lt;/math&gt; is odd, we get that &lt;math&gt;p((2i)!)=p(i!)+i&lt;/math&gt;.<br /> <br /> It immediately follows that &lt;math&gt;p\left( {2i\choose i} \right) = p((2i)!) - 2p(i!) = i - p(i!)&lt;/math&gt;,<br /> and hence &lt;math&gt;p\left( {2i\choose i} \cdot 2^{2\cdot 2009 - 2i} \right) = 2\cdot 2009 - i - p(i!)&lt;/math&gt;. <br /> <br /> Obviously, for &lt;math&gt;i\in\{1,2,\dots,2009\}&lt;/math&gt; the function &lt;math&gt;f(i)=2\cdot 2009 - i - p(i!)&lt;/math&gt; is is a strictly decreasing function. <br /> Therefore &lt;math&gt;p(c) = p\left( {2\cdot 2009\choose 2009} \right) = 2009 - p(2009!)&lt;/math&gt;.<br /> <br /> We can now compute &lt;math&gt;p(2009!) = \sum_{k=1}^{\infty} \left\lfloor \dfrac{2009}{2^k} \right\rfloor = 1004 + 502 + \cdots + 3 + 1 = 2001&lt;/math&gt;.<br /> Hence &lt;math&gt;p(c)=2009-2001=8&lt;/math&gt;.<br /> <br /> And thus we have &lt;math&gt;a=2\cdot 2009 - p(c) = 4010&lt;/math&gt;, and the answer is &lt;math&gt;\dfrac{ab}{10} = \dfrac{4010\cdot 1}{10} = \boxed{401}&lt;/math&gt;.<br /> <br /> ----<br /> Additionally, once you count the number of factors of &lt;math&gt;2&lt;/math&gt; in the summation, one can consider the fact that, since &lt;math&gt;b&lt;/math&gt; must be odd, it has to take on a value of &lt;math&gt;1,3,5,7,&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt; (Because the number of &lt;math&gt;2&lt;/math&gt;s in the summation is clearly greater than &lt;math&gt;1000&lt;/math&gt;, dividing by &lt;math&gt;10&lt;/math&gt; will yield a number greater than &lt;math&gt;100&lt;/math&gt;, and multiplying this number by any odd number greater than &lt;math&gt;9&lt;/math&gt; will yield an answer &lt;math&gt;&gt;999&lt;/math&gt;, which cannot happen on the AIME.) Once you calculate the value of &lt;math&gt;4010&lt;/math&gt;, and divide by &lt;math&gt;10&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; must be equal to &lt;math&gt;1&lt;/math&gt;, as any other value of &lt;math&gt;b&lt;/math&gt; will result in an answer &lt;math&gt;&gt;999&lt;/math&gt;. This gives &lt;math&gt;\boxed{401}&lt;/math&gt; as the answer.<br /> <br /> ==Solution 2==<br /> Using the steps of the previous solution we get &lt;math&gt;c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}&lt;/math&gt; and if you do the small cases(like &lt;math&gt;1, 2, 3, 4, 5, 6&lt;/math&gt;) you realize that youu can &quot;thin-slice&quot; the problem and simply look at the cases where &lt;math&gt;i=2009, 2008&lt;/math&gt;(they're nearly identical in nature but one has &lt;math&gt;4&lt;/math&gt; with it) since &lt;math&gt;\dbinom{2i}{I}&lt;/math&gt; hardly contains any powers of &lt;math&gt;2&lt;/math&gt; or in other words it's very inefficient and the inefficient cases hold all the power so you can just look at the highest powers of &lt;math&gt;2&lt;/math&gt; in &lt;math&gt;\dbinom{4018}{2009}&lt;/math&gt; and &lt;math&gt;\dbinom{4016}{2008}&lt;/math&gt; and you get the minimum power of &lt;math&gt;2&lt;/math&gt; in either expression is &lt;math&gt;8&lt;/math&gt; so the answer is &lt;math&gt;\frac{4010}{10} \implies \boxed{401}&lt;/math&gt; since it would violate the rules of the AIME and the small cases if &lt;math&gt;b&gt;1&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AIME box|year=2009|n=II|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_15&diff=86009 2008 AIME I Problems/Problem 15 2017-06-11T15:23:35Z <p>First: /* Solution */</p> <hr /> <div>== Problem ==<br /> A square piece of paper has sides of length &lt;math&gt;100&lt;/math&gt;. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance &lt;math&gt;\sqrt{17}&lt;/math&gt; from the corner, and they meet on the diagonal at an angle of &lt;math&gt;60^{\circ}&lt;/math&gt; (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upped edges, can be written in the form &lt;math&gt;\sqrt[n]{m}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;1000&lt;/math&gt;, and &lt;math&gt;m&lt;/math&gt; is not divisible by the &lt;math&gt;n&lt;/math&gt;th power of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;import cse5;<br /> size(200);<br /> pathpen=black;<br /> real s=sqrt(17);<br /> real r=(sqrt(51)+s)/sqrt(2);<br /> D((0,2*s)--(0,0)--(2*s,0));<br /> D((0,s)--r*dir(45)--(s,0));<br /> D((0,0)--r*dir(45));<br /> D((r*dir(45).x,2*s)--r*dir(45)--(2*s,r*dir(45).y));<br /> MP(&quot;30^\circ&quot;,r*dir(45)-(0.25,1),SW);<br /> MP(&quot;30^\circ&quot;,r*dir(45)-(1,0.5),SW);<br /> MP(&quot;\sqrt{17}&quot;,(0,s/2),W);<br /> MP(&quot;\sqrt{17}&quot;,(s/2,0),S);<br /> MP(&quot;\mathrm{cut}&quot;,((0,s)+r*dir(45))/2,N);<br /> MP(&quot;\mathrm{cut}&quot;,((s,0)+r*dir(45))/2,E);<br /> MP(&quot;\mathrm{fold}&quot;,(r*dir(45).x,s+r/2*dir(45).y),E);<br /> MP(&quot;\mathrm{fold}&quot;,(s+r/2*dir(45).x,r*dir(45).y));&lt;/asy&gt;&lt;/center&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> &lt;center&gt;&lt;asy&gt;<br /> import three; import math; import cse5;<br /> size(500);<br /> pathpen=blue;<br /> real r = (51^0.5-17^0.5)/200, h=867^0.25/100;<br /> triple A=(0,0,0),B=(1,0,0),C=(1,1,0),D=(0,1,0);<br /> triple F=B+(r,-r,h),G=(1,-r,h),H=(1+r,0,h),I=B+(0,0,h);<br /> draw(B--F--H--cycle); draw(B--F--G--cycle);<br /> draw(G--I--H); draw(B--I); draw(A--B--C--D--cycle);<br /> triple Fa=A+(-r,-r, h), Fc=C+(r,r, h), Fd=D+(-r,r, h);<br /> triple Ia = A+(0,0,h), Ic = C+(0,0,h), Id = D+(0,0,h);<br /> draw(Ia--I--Ic); draw(Fa--F--Fc--Fd--cycle);<br /> draw(A--Fa); draw(C--Fc); draw(D--Fd);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> === Solution 1 ===<br /> In the original picture, let &lt;math&gt;P&lt;/math&gt; be the corner, and &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be the two points whose distance is &lt;math&gt;\sqrt{17}&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt;. Also, let &lt;math&gt;R&lt;/math&gt; be the point where the two cuts intersect.<br /> <br /> Using &lt;math&gt;\triangle{MNP}&lt;/math&gt; (a 45-45-90 triangle), &lt;math&gt;MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}&lt;/math&gt;. &lt;math&gt;\triangle{MNR}&lt;/math&gt; is [[equilateral triangle|equilateral]], so &lt;math&gt;MR = NR = \sqrt{34}&lt;/math&gt;. (Alternatively, we could find this by the [[Law of Sines]].)<br /> <br /> The length of the perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;MN&lt;/math&gt; in &lt;math&gt;\triangle{MNP}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{17}}{\sqrt{2}}&lt;/math&gt;, and the length of the perpendicular from &lt;math&gt;R&lt;/math&gt; to &lt;math&gt;MN&lt;/math&gt; in &lt;math&gt;\triangle{MNR}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{51}}{\sqrt{2}}&lt;/math&gt;. Adding those two lengths, &lt;math&gt;PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}&lt;/math&gt;. (Alternatively, we could have used that &lt;math&gt;\tan 75^{\circ} = \tan (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}&lt;/math&gt;.)<br /> <br /> Drop a [[perpendicular]] from &lt;math&gt;R&lt;/math&gt; to the side of the square containing &lt;math&gt;M&lt;/math&gt; and let the intersection be &lt;math&gt;G&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}PG&amp;=\frac{PR}{\sqrt{2}}=\frac{\sqrt{17}+\sqrt{51}}{2}\\<br /> MG=PG-PM&amp;=\frac{\sqrt{17}+\sqrt{51}}{2}-\sqrt{17}=\frac{\sqrt{51}-\sqrt{17}}{2}\end{align*}&lt;/cmath&gt;<br /> <br /> &lt;center&gt;&lt;asy&gt;import cse5;<br /> size(200);<br /> pathpen=black;<br /> real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2));<br /> pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45), G=((sqrt(51)+sqrt(17))/2,0);<br /> D(2*N--P--2*M); D(N--R--M); D(P--R);<br /> D((R.x,2*N.y)--R--(2*M.x,R.y));<br /> MP(&quot;30^\circ&quot;,R-(0.25,1),SW);<br /> MP(&quot;30^\circ&quot;,R-(1,0.5),SW);<br /> MP(&quot;\sqrt{17}&quot;,N/2,W);<br /> MP(&quot;\sqrt{17}&quot;,M/2,S);<br /> D(N--M,dashed);<br /> D(G--R,dashed);<br /> MP(&quot;P&quot;,P,SW); MP(&quot;N&quot;,N,SW); MP(&quot;M&quot;,M,SW); MP(&quot;R&quot;,R,NE);<br /> MP(&quot;G&quot;,G,SW);<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be the smaller square base of the tray and let &lt;math&gt;A'B'C'D'&lt;/math&gt; be the larger square, such that &lt;math&gt;AA'&lt;/math&gt;, etc, are edges. Let &lt;math&gt;F&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;A&lt;/math&gt; to plane &lt;math&gt;A'B'C'D'&lt;/math&gt;.<br /> <br /> We know &lt;math&gt;AA'=MR=\sqrt{34}&lt;/math&gt; and &lt;math&gt;A'F=MG\sqrt{2}=\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}&lt;/math&gt;. Now, use the Pythagorean Theorem on triangle &lt;math&gt;AFA'&lt;/math&gt; to find &lt;math&gt;AF&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}\left(\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}\right)^2+AF^2&amp;=\left(\sqrt{34}\right)^2\\ \frac{51-34\sqrt{3}+17}{2}+AF^2&amp;=34\\AF&amp;=\sqrt{34-\frac{68-34\sqrt{3}}{2}}\\AF&amp;=\sqrt{\frac{34\sqrt{3}}{2}}\\AF&amp;=\sqrt{867}\end{align*}&lt;/cmath&gt;<br /> <br /> The answer is &lt;math&gt;867 + 4 = \boxed{871}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> In the final pyramid, let &lt;math&gt;ABCD&lt;/math&gt; be the smaller square and let &lt;math&gt;A'B'C'D'&lt;/math&gt; be the larger square such that &lt;math&gt;AA'&lt;/math&gt;, etc. are edges.<br /> <br /> It is obvious from the diagram that &lt;math&gt;\angle A'AB = \angle A'AD = 105^\circ&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; be the positive &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; axes in a 3-d coordinate system such that &lt;math&gt;A'&lt;/math&gt; has a positive &lt;math&gt;z&lt;/math&gt; coordinate. Let &lt;math&gt;\alpha&lt;/math&gt; be the angle made with the positive &lt;math&gt;x&lt;/math&gt; axis. Define &lt;math&gt;\beta&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; analogously.<br /> <br /> It is easy to see that if &lt;math&gt;P: = (x,y,z)&lt;/math&gt;, then &lt;math&gt;x = AA'\cdot \cos\alpha&lt;/math&gt;. Furthermore, this means that &lt;math&gt;\frac {x^2}{AA'^2} + \frac {y^2}{AA'^2} + \frac {z^2}{AA'^2} = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1&lt;/math&gt;.<br /> <br /> We have that &lt;math&gt;\alpha = \beta = 105^\circ&lt;/math&gt;, so &lt;math&gt;\cos^2 105^\circ + \cos^2105^\circ + \cos^2\gamma = 1\implies \cos\gamma = \sqrt {\frac {3}{4}}&lt;/math&gt;.<br /> <br /> It is easy to see from the [[Law of Sines]] that &lt;math&gt;\frac {AA'}{\sin 45^\circ} = \frac {\sqrt {17}}{\sin 30^\circ}\implies AA' = \sqrt {34}&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;z = AA'\cdot \cos\gamma = \sqrt {34^2\cdot \frac {3}{4}} = \sqrt {867}&lt;/math&gt;.<br /> <br /> It follows that the answer is &lt;math&gt;867 + 4 = \boxed{871}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=I|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_15&diff=86008 2008 AIME I Problems/Problem 15 2017-06-11T15:22:29Z <p>First: /* Solution */</p> <hr /> <div>== Problem ==<br /> A square piece of paper has sides of length &lt;math&gt;100&lt;/math&gt;. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance &lt;math&gt;\sqrt{17}&lt;/math&gt; from the corner, and they meet on the diagonal at an angle of &lt;math&gt;60^{\circ}&lt;/math&gt; (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upped edges, can be written in the form &lt;math&gt;\sqrt[n]{m}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;1000&lt;/math&gt;, and &lt;math&gt;m&lt;/math&gt; is not divisible by the &lt;math&gt;n&lt;/math&gt;th power of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;import cse5;<br /> size(200);<br /> pathpen=black;<br /> real s=sqrt(17);<br /> real r=(sqrt(51)+s)/sqrt(2);<br /> D((0,2*s)--(0,0)--(2*s,0));<br /> D((0,s)--r*dir(45)--(s,0));<br /> D((0,0)--r*dir(45));<br /> D((r*dir(45).x,2*s)--r*dir(45)--(2*s,r*dir(45).y));<br /> MP(&quot;30^\circ&quot;,r*dir(45)-(0.25,1),SW);<br /> MP(&quot;30^\circ&quot;,r*dir(45)-(1,0.5),SW);<br /> MP(&quot;\sqrt{17}&quot;,(0,s/2),W);<br /> MP(&quot;\sqrt{17}&quot;,(s/2,0),S);<br /> MP(&quot;\mathrm{cut}&quot;,((0,s)+r*dir(45))/2,N);<br /> MP(&quot;\mathrm{cut}&quot;,((s,0)+r*dir(45))/2,E);<br /> MP(&quot;\mathrm{fold}&quot;,(r*dir(45).x,s+r/2*dir(45).y),E);<br /> MP(&quot;\mathrm{fold}&quot;,(s+r/2*dir(45).x,r*dir(45).y));&lt;/asy&gt;&lt;/center&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> &lt;center&gt;&lt;asy&gt;<br /> import three; import math; import cse5;<br /> size(750);<br /> pathpen=black;<br /> real r = (51^0.5-17^0.5)/200, h=867^0.25/100;<br /> triple A=(0,0,0),B=(1,0,0),C=(1,1,0),D=(0,1,0);<br /> triple F=B+(r,-r,h),G=(1,-r,h),H=(1+r,0,h),I=B+(0,0,h);<br /> draw(B--F--H--cycle); draw(B--F--G--cycle);<br /> draw(G--I--H); draw(B--I); draw(A--B--C--D--cycle);<br /> triple Fa=A+(-r,-r, h), Fc=C+(r,r, h), Fd=D+(-r,r, h);<br /> triple Ia = A+(0,0,h), Ic = C+(0,0,h), Id = D+(0,0,h);<br /> draw(Ia--I--Ic); draw(Fa--F--Fc--Fd--cycle);<br /> draw(A--Fa); draw(C--Fc); draw(D--Fd);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> === Solution 1 ===<br /> In the original picture, let &lt;math&gt;P&lt;/math&gt; be the corner, and &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be the two points whose distance is &lt;math&gt;\sqrt{17}&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt;. Also, let &lt;math&gt;R&lt;/math&gt; be the point where the two cuts intersect.<br /> <br /> Using &lt;math&gt;\triangle{MNP}&lt;/math&gt; (a 45-45-90 triangle), &lt;math&gt;MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}&lt;/math&gt;. &lt;math&gt;\triangle{MNR}&lt;/math&gt; is [[equilateral triangle|equilateral]], so &lt;math&gt;MR = NR = \sqrt{34}&lt;/math&gt;. (Alternatively, we could find this by the [[Law of Sines]].)<br /> <br /> The length of the perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;MN&lt;/math&gt; in &lt;math&gt;\triangle{MNP}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{17}}{\sqrt{2}}&lt;/math&gt;, and the length of the perpendicular from &lt;math&gt;R&lt;/math&gt; to &lt;math&gt;MN&lt;/math&gt; in &lt;math&gt;\triangle{MNR}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{51}}{\sqrt{2}}&lt;/math&gt;. Adding those two lengths, &lt;math&gt;PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}&lt;/math&gt;. (Alternatively, we could have used that &lt;math&gt;\tan 75^{\circ} = \tan (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}&lt;/math&gt;.)<br /> <br /> Drop a [[perpendicular]] from &lt;math&gt;R&lt;/math&gt; to the side of the square containing &lt;math&gt;M&lt;/math&gt; and let the intersection be &lt;math&gt;G&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}PG&amp;=\frac{PR}{\sqrt{2}}=\frac{\sqrt{17}+\sqrt{51}}{2}\\<br /> MG=PG-PM&amp;=\frac{\sqrt{17}+\sqrt{51}}{2}-\sqrt{17}=\frac{\sqrt{51}-\sqrt{17}}{2}\end{align*}&lt;/cmath&gt;<br /> <br /> &lt;center&gt;&lt;asy&gt;import cse5;<br /> size(200);<br /> pathpen=black;<br /> real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2));<br /> pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45), G=((sqrt(51)+sqrt(17))/2,0);<br /> D(2*N--P--2*M); D(N--R--M); D(P--R);<br /> D((R.x,2*N.y)--R--(2*M.x,R.y));<br /> MP(&quot;30^\circ&quot;,R-(0.25,1),SW);<br /> MP(&quot;30^\circ&quot;,R-(1,0.5),SW);<br /> MP(&quot;\sqrt{17}&quot;,N/2,W);<br /> MP(&quot;\sqrt{17}&quot;,M/2,S);<br /> D(N--M,dashed);<br /> D(G--R,dashed);<br /> MP(&quot;P&quot;,P,SW); MP(&quot;N&quot;,N,SW); MP(&quot;M&quot;,M,SW); MP(&quot;R&quot;,R,NE);<br /> MP(&quot;G&quot;,G,SW);<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be the smaller square base of the tray and let &lt;math&gt;A'B'C'D'&lt;/math&gt; be the larger square, such that &lt;math&gt;AA'&lt;/math&gt;, etc, are edges. Let &lt;math&gt;F&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;A&lt;/math&gt; to plane &lt;math&gt;A'B'C'D'&lt;/math&gt;.<br /> <br /> We know &lt;math&gt;AA'=MR=\sqrt{34}&lt;/math&gt; and &lt;math&gt;A'F=MG\sqrt{2}=\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}&lt;/math&gt;. Now, use the Pythagorean Theorem on triangle &lt;math&gt;AFA'&lt;/math&gt; to find &lt;math&gt;AF&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}\left(\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}\right)^2+AF^2&amp;=\left(\sqrt{34}\right)^2\\ \frac{51-34\sqrt{3}+17}{2}+AF^2&amp;=34\\AF&amp;=\sqrt{34-\frac{68-34\sqrt{3}}{2}}\\AF&amp;=\sqrt{\frac{34\sqrt{3}}{2}}\\AF&amp;=\sqrt{867}\end{align*}&lt;/cmath&gt;<br /> <br /> The answer is &lt;math&gt;867 + 4 = \boxed{871}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> In the final pyramid, let &lt;math&gt;ABCD&lt;/math&gt; be the smaller square and let &lt;math&gt;A'B'C'D'&lt;/math&gt; be the larger square such that &lt;math&gt;AA'&lt;/math&gt;, etc. are edges.<br /> <br /> It is obvious from the diagram that &lt;math&gt;\angle A'AB = \angle A'AD = 105^\circ&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; be the positive &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; axes in a 3-d coordinate system such that &lt;math&gt;A'&lt;/math&gt; has a positive &lt;math&gt;z&lt;/math&gt; coordinate. Let &lt;math&gt;\alpha&lt;/math&gt; be the angle made with the positive &lt;math&gt;x&lt;/math&gt; axis. Define &lt;math&gt;\beta&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; analogously.<br /> <br /> It is easy to see that if &lt;math&gt;P: = (x,y,z)&lt;/math&gt;, then &lt;math&gt;x = AA'\cdot \cos\alpha&lt;/math&gt;. Furthermore, this means that &lt;math&gt;\frac {x^2}{AA'^2} + \frac {y^2}{AA'^2} + \frac {z^2}{AA'^2} = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1&lt;/math&gt;.<br /> <br /> We have that &lt;math&gt;\alpha = \beta = 105^\circ&lt;/math&gt;, so &lt;math&gt;\cos^2 105^\circ + \cos^2105^\circ + \cos^2\gamma = 1\implies \cos\gamma = \sqrt {\frac {3}{4}}&lt;/math&gt;.<br /> <br /> It is easy to see from the [[Law of Sines]] that &lt;math&gt;\frac {AA'}{\sin 45^\circ} = \frac {\sqrt {17}}{\sin 30^\circ}\implies AA' = \sqrt {34}&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;z = AA'\cdot \cos\gamma = \sqrt {34^2\cdot \frac {3}{4}} = \sqrt {867}&lt;/math&gt;.<br /> <br /> It follows that the answer is &lt;math&gt;867 + 4 = \boxed{871}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=I|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_15&diff=86006 2008 AIME I Problems/Problem 15 2017-06-11T15:22:10Z <p>First: /* Solution */</p> <hr /> <div>== Problem ==<br /> A square piece of paper has sides of length &lt;math&gt;100&lt;/math&gt;. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance &lt;math&gt;\sqrt{17}&lt;/math&gt; from the corner, and they meet on the diagonal at an angle of &lt;math&gt;60^{\circ}&lt;/math&gt; (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upped edges, can be written in the form &lt;math&gt;\sqrt[n]{m}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;1000&lt;/math&gt;, and &lt;math&gt;m&lt;/math&gt; is not divisible by the &lt;math&gt;n&lt;/math&gt;th power of any prime. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;import cse5;<br /> size(200);<br /> pathpen=black;<br /> real s=sqrt(17);<br /> real r=(sqrt(51)+s)/sqrt(2);<br /> D((0,2*s)--(0,0)--(2*s,0));<br /> D((0,s)--r*dir(45)--(s,0));<br /> D((0,0)--r*dir(45));<br /> D((r*dir(45).x,2*s)--r*dir(45)--(2*s,r*dir(45).y));<br /> MP(&quot;30^\circ&quot;,r*dir(45)-(0.25,1),SW);<br /> MP(&quot;30^\circ&quot;,r*dir(45)-(1,0.5),SW);<br /> MP(&quot;\sqrt{17}&quot;,(0,s/2),W);<br /> MP(&quot;\sqrt{17}&quot;,(s/2,0),S);<br /> MP(&quot;\mathrm{cut}&quot;,((0,s)+r*dir(45))/2,N);<br /> MP(&quot;\mathrm{cut}&quot;,((s,0)+r*dir(45))/2,E);<br /> MP(&quot;\mathrm{fold}&quot;,(r*dir(45).x,s+r/2*dir(45).y),E);<br /> MP(&quot;\mathrm{fold}&quot;,(s+r/2*dir(45).x,r*dir(45).y));&lt;/asy&gt;&lt;/center&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> &lt;center&gt;&lt;asy&gt;<br /> import three; import math; import cse5;<br /> size(1000);<br /> pathpen=black;<br /> real r = (51^0.5-17^0.5)/200, h=867^0.25/100;<br /> triple A=(0,0,0),B=(1,0,0),C=(1,1,0),D=(0,1,0);<br /> triple F=B+(r,-r,h),G=(1,-r,h),H=(1+r,0,h),I=B+(0,0,h);<br /> draw(B--F--H--cycle); draw(B--F--G--cycle);<br /> draw(G--I--H); draw(B--I); draw(A--B--C--D--cycle);<br /> triple Fa=A+(-r,-r, h), Fc=C+(r,r, h), Fd=D+(-r,r, h);<br /> triple Ia = A+(0,0,h), Ic = C+(0,0,h), Id = D+(0,0,h);<br /> draw(Ia--I--Ic); draw(Fa--F--Fc--Fd--cycle);<br /> draw(A--Fa); draw(C--Fc); draw(D--Fd);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> === Solution 1 ===<br /> In the original picture, let &lt;math&gt;P&lt;/math&gt; be the corner, and &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be the two points whose distance is &lt;math&gt;\sqrt{17}&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt;. Also, let &lt;math&gt;R&lt;/math&gt; be the point where the two cuts intersect.<br /> <br /> Using &lt;math&gt;\triangle{MNP}&lt;/math&gt; (a 45-45-90 triangle), &lt;math&gt;MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}&lt;/math&gt;. &lt;math&gt;\triangle{MNR}&lt;/math&gt; is [[equilateral triangle|equilateral]], so &lt;math&gt;MR = NR = \sqrt{34}&lt;/math&gt;. (Alternatively, we could find this by the [[Law of Sines]].)<br /> <br /> The length of the perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;MN&lt;/math&gt; in &lt;math&gt;\triangle{MNP}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{17}}{\sqrt{2}}&lt;/math&gt;, and the length of the perpendicular from &lt;math&gt;R&lt;/math&gt; to &lt;math&gt;MN&lt;/math&gt; in &lt;math&gt;\triangle{MNR}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{51}}{\sqrt{2}}&lt;/math&gt;. Adding those two lengths, &lt;math&gt;PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}&lt;/math&gt;. (Alternatively, we could have used that &lt;math&gt;\tan 75^{\circ} = \tan (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}&lt;/math&gt;.)<br /> <br /> Drop a [[perpendicular]] from &lt;math&gt;R&lt;/math&gt; to the side of the square containing &lt;math&gt;M&lt;/math&gt; and let the intersection be &lt;math&gt;G&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}PG&amp;=\frac{PR}{\sqrt{2}}=\frac{\sqrt{17}+\sqrt{51}}{2}\\<br /> MG=PG-PM&amp;=\frac{\sqrt{17}+\sqrt{51}}{2}-\sqrt{17}=\frac{\sqrt{51}-\sqrt{17}}{2}\end{align*}&lt;/cmath&gt;<br /> <br /> &lt;center&gt;&lt;asy&gt;import cse5;<br /> size(200);<br /> pathpen=black;<br /> real s=sqrt(17), r=(sqrt(51)+s)/(sqrt(2));<br /> pair P=(0,0), N=(0,sqrt(17)), M=(sqrt(17),0), R=r*dir(45), G=((sqrt(51)+sqrt(17))/2,0);<br /> D(2*N--P--2*M); D(N--R--M); D(P--R);<br /> D((R.x,2*N.y)--R--(2*M.x,R.y));<br /> MP(&quot;30^\circ&quot;,R-(0.25,1),SW);<br /> MP(&quot;30^\circ&quot;,R-(1,0.5),SW);<br /> MP(&quot;\sqrt{17}&quot;,N/2,W);<br /> MP(&quot;\sqrt{17}&quot;,M/2,S);<br /> D(N--M,dashed);<br /> D(G--R,dashed);<br /> MP(&quot;P&quot;,P,SW); MP(&quot;N&quot;,N,SW); MP(&quot;M&quot;,M,SW); MP(&quot;R&quot;,R,NE);<br /> MP(&quot;G&quot;,G,SW);<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be the smaller square base of the tray and let &lt;math&gt;A'B'C'D'&lt;/math&gt; be the larger square, such that &lt;math&gt;AA'&lt;/math&gt;, etc, are edges. Let &lt;math&gt;F&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;A&lt;/math&gt; to plane &lt;math&gt;A'B'C'D'&lt;/math&gt;.<br /> <br /> We know &lt;math&gt;AA'=MR=\sqrt{34}&lt;/math&gt; and &lt;math&gt;A'F=MG\sqrt{2}=\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}&lt;/math&gt;. Now, use the Pythagorean Theorem on triangle &lt;math&gt;AFA'&lt;/math&gt; to find &lt;math&gt;AF&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}\left(\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}\right)^2+AF^2&amp;=\left(\sqrt{34}\right)^2\\ \frac{51-34\sqrt{3}+17}{2}+AF^2&amp;=34\\AF&amp;=\sqrt{34-\frac{68-34\sqrt{3}}{2}}\\AF&amp;=\sqrt{\frac{34\sqrt{3}}{2}}\\AF&amp;=\sqrt{867}\end{align*}&lt;/cmath&gt;<br /> <br /> The answer is &lt;math&gt;867 + 4 = \boxed{871}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> In the final pyramid, let &lt;math&gt;ABCD&lt;/math&gt; be the smaller square and let &lt;math&gt;A'B'C'D'&lt;/math&gt; be the larger square such that &lt;math&gt;AA'&lt;/math&gt;, etc. are edges.<br /> <br /> It is obvious from the diagram that &lt;math&gt;\angle A'AB = \angle A'AD = 105^\circ&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; be the positive &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; axes in a 3-d coordinate system such that &lt;math&gt;A'&lt;/math&gt; has a positive &lt;math&gt;z&lt;/math&gt; coordinate. Let &lt;math&gt;\alpha&lt;/math&gt; be the angle made with the positive &lt;math&gt;x&lt;/math&gt; axis. Define &lt;math&gt;\beta&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; analogously.<br /> <br /> It is easy to see that if &lt;math&gt;P: = (x,y,z)&lt;/math&gt;, then &lt;math&gt;x = AA'\cdot \cos\alpha&lt;/math&gt;. Furthermore, this means that &lt;math&gt;\frac {x^2}{AA'^2} + \frac {y^2}{AA'^2} + \frac {z^2}{AA'^2} = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1&lt;/math&gt;.<br /> <br /> We have that &lt;math&gt;\alpha = \beta = 105^\circ&lt;/math&gt;, so &lt;math&gt;\cos^2 105^\circ + \cos^2105^\circ + \cos^2\gamma = 1\implies \cos\gamma = \sqrt {\frac {3}{4}}&lt;/math&gt;.<br /> <br /> It is easy to see from the [[Law of Sines]] that &lt;math&gt;\frac {AA'}{\sin 45^\circ} = \frac {\sqrt {17}}{\sin 30^\circ}\implies AA' = \sqrt {34}&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;z = AA'\cdot \cos\gamma = \sqrt {34^2\cdot \frac {3}{4}} = \sqrt {867}&lt;/math&gt;.<br /> <br /> It follows that the answer is &lt;math&gt;867 + 4 = \boxed{871}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=I|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_6&diff=85948 2007 AIME II Problems/Problem 6 2017-06-04T17:41:06Z <p>First: /* Solution */</p> <hr /> <div>&lt;noinclude&gt;== Problem ==<br /> &lt;/noinclude&gt;An [[integer]] is called ''parity-monotonic'' if its [[decimal representation]] &lt;math&gt;a_{1}a_{2}a_{3}\cdots a_{k}&lt;/math&gt; satisfies &lt;math&gt;a_{i}&lt;a_{i+1}&lt;/math&gt; if &lt;math&gt;a_{i}&lt;/math&gt; is [[odd]], and &lt;math&gt;a_{i}&gt;a_{i+1}&lt;/math&gt; if &lt;math&gt;a_{i}&lt;/math&gt; is [[even]]. How many four-digit parity-monotonic integers are there?&lt;noinclude&gt;<br /> <br /> == Solution 1==<br /> Let's set up a table of values. Notice that 0 and 9 both cannot appear as any of &lt;math&gt;a_1,\ a_2,\ a_3&lt;/math&gt; because of the given conditions. A clear pattern emerges.<br /> <br /> For example, for &lt;math&gt;3&lt;/math&gt; in the second column, we note that &lt;math&gt;3&lt;/math&gt; is less than &lt;math&gt;4,6,8&lt;/math&gt;, but greater than &lt;math&gt;1&lt;/math&gt;, so there are four possible places to align &lt;math&gt;3&lt;/math&gt; as the second digit.<br /> <br /> {| align=center style=&quot;border:1px solid black;&quot;<br /> |- <br /> | Digit&amp;nbsp;&amp;nbsp; || 1st&amp;nbsp;&amp;nbsp; || 2nd&amp;nbsp;&amp;nbsp; || 3rd&amp;nbsp;&amp;nbsp; || 4th <br /> |-<br /> | 0 || 0 || 0 || 0 || 64<br /> |-<br /> | 1 || 1 || 4 || 16 || 64<br /> |-<br /> | 2 || 1 || 4 || 16 || 64<br /> |-<br /> | 3 || 1 || 4 || 16 || 64<br /> |-<br /> | 4 || 1 || 4 || 16 || 64<br /> |-<br /> | 5 || 1 || 4 || 16 || 64<br /> |-<br /> | 6 || 1 || 4 || 16 || 64<br /> |-<br /> | 7 || 1 || 4 || 16 || 64<br /> |-<br /> | 8 || 1 || 4 || 16 || 64<br /> |-<br /> | 9 || 0 || 0 || 0 || 64<br /> |}<br /> <br /> ==Solution 2: Recursion==<br /> This problem can be solved via recursion since we are &quot;building a string&quot; of numbers with some condition. We want to create a new string by adding a new digit at the front so we avoid complications(&lt;math&gt;0&lt;/math&gt; can't be at the front and do digit is less than &lt;math&gt;9&lt;/math&gt;). There are &lt;math&gt;4&lt;/math&gt; options to add no matter what(try some examples if you want so the recursion is &lt;math&gt;S_n=4S_{n-1}&lt;/math&gt; where &lt;math&gt;S_n&lt;/math&gt; stands for the number of such numbers with &lt;math&gt;n&lt;/math&gt; digits. Since &lt;math&gt;S_1=10&lt;/math&gt; the answer is &lt;math&gt;\boxed{640}&lt;/math&gt;. <br /> <br /> For any number from 1-8, there are exactly 4 numbers from 1-8 that are odd and less than the number or that are even and greater than the number (the same will happen for 0 and 9 in the last column). Thus, the answer is &lt;math&gt;4^{k-1} \cdot 10 = 4^3\cdot10 = 640&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> &lt;/noinclude&gt;<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_6&diff=85947 2007 AIME II Problems/Problem 6 2017-06-04T17:40:45Z <p>First: /* Solution 2 */</p> <hr /> <div>&lt;noinclude&gt;== Problem ==<br /> &lt;/noinclude&gt;An [[integer]] is called ''parity-monotonic'' if its [[decimal representation]] &lt;math&gt;a_{1}a_{2}a_{3}\cdots a_{k}&lt;/math&gt; satisfies &lt;math&gt;a_{i}&lt;a_{i+1}&lt;/math&gt; if &lt;math&gt;a_{i}&lt;/math&gt; is [[odd]], and &lt;math&gt;a_{i}&gt;a_{i+1}&lt;/math&gt; if &lt;math&gt;a_{i}&lt;/math&gt; is [[even]]. How many four-digit parity-monotonic integers are there?&lt;noinclude&gt;<br /> <br /> == Solution ==<br /> Let's set up a table of values. Notice that 0 and 9 both cannot appear as any of &lt;math&gt;a_1,\ a_2,\ a_3&lt;/math&gt; because of the given conditions. A clear pattern emerges.<br /> <br /> For example, for &lt;math&gt;3&lt;/math&gt; in the second column, we note that &lt;math&gt;3&lt;/math&gt; is less than &lt;math&gt;4,6,8&lt;/math&gt;, but greater than &lt;math&gt;1&lt;/math&gt;, so there are four possible places to align &lt;math&gt;3&lt;/math&gt; as the second digit.<br /> <br /> {| align=center style=&quot;border:1px solid black;&quot;<br /> |- <br /> | Digit&amp;nbsp;&amp;nbsp; || 1st&amp;nbsp;&amp;nbsp; || 2nd&amp;nbsp;&amp;nbsp; || 3rd&amp;nbsp;&amp;nbsp; || 4th <br /> |-<br /> | 0 || 0 || 0 || 0 || 64<br /> |-<br /> | 1 || 1 || 4 || 16 || 64<br /> |-<br /> | 2 || 1 || 4 || 16 || 64<br /> |-<br /> | 3 || 1 || 4 || 16 || 64<br /> |-<br /> | 4 || 1 || 4 || 16 || 64<br /> |-<br /> | 5 || 1 || 4 || 16 || 64<br /> |-<br /> | 6 || 1 || 4 || 16 || 64<br /> |-<br /> | 7 || 1 || 4 || 16 || 64<br /> |-<br /> | 8 || 1 || 4 || 16 || 64<br /> |-<br /> | 9 || 0 || 0 || 0 || 64<br /> |}<br /> ==Solution 2: Recursion==<br /> This problem can be solved via recursion since we are &quot;building a string&quot; of numbers with some condition. We want to create a new string by adding a new digit at the front so we avoid complications(&lt;math&gt;0&lt;/math&gt; can't be at the front and do digit is less than &lt;math&gt;9&lt;/math&gt;). There are &lt;math&gt;4&lt;/math&gt; options to add no matter what(try some examples if you want so the recursion is &lt;math&gt;S_n=4S_{n-1}&lt;/math&gt; where &lt;math&gt;S_n&lt;/math&gt; stands for the number of such numbers with &lt;math&gt;n&lt;/math&gt; digits. Since &lt;math&gt;S_1=10&lt;/math&gt; the answer is &lt;math&gt;\boxed{640}&lt;/math&gt;. <br /> <br /> For any number from 1-8, there are exactly 4 numbers from 1-8 that are odd and less than the number or that are even and greater than the number (the same will happen for 0 and 9 in the last column). Thus, the answer is &lt;math&gt;4^{k-1} \cdot 10 = 4^3\cdot10 = 640&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> &lt;/noinclude&gt;<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_6&diff=85946 2007 AIME II Problems/Problem 6 2017-06-04T17:40:30Z <p>First: /* Solution */</p> <hr /> <div>&lt;noinclude&gt;== Problem ==<br /> &lt;/noinclude&gt;An [[integer]] is called ''parity-monotonic'' if its [[decimal representation]] &lt;math&gt;a_{1}a_{2}a_{3}\cdots a_{k}&lt;/math&gt; satisfies &lt;math&gt;a_{i}&lt;a_{i+1}&lt;/math&gt; if &lt;math&gt;a_{i}&lt;/math&gt; is [[odd]], and &lt;math&gt;a_{i}&gt;a_{i+1}&lt;/math&gt; if &lt;math&gt;a_{i}&lt;/math&gt; is [[even]]. How many four-digit parity-monotonic integers are there?&lt;noinclude&gt;<br /> <br /> == Solution ==<br /> Let's set up a table of values. Notice that 0 and 9 both cannot appear as any of &lt;math&gt;a_1,\ a_2,\ a_3&lt;/math&gt; because of the given conditions. A clear pattern emerges.<br /> <br /> For example, for &lt;math&gt;3&lt;/math&gt; in the second column, we note that &lt;math&gt;3&lt;/math&gt; is less than &lt;math&gt;4,6,8&lt;/math&gt;, but greater than &lt;math&gt;1&lt;/math&gt;, so there are four possible places to align &lt;math&gt;3&lt;/math&gt; as the second digit.<br /> <br /> {| align=center style=&quot;border:1px solid black;&quot;<br /> |- <br /> | Digit&amp;nbsp;&amp;nbsp; || 1st&amp;nbsp;&amp;nbsp; || 2nd&amp;nbsp;&amp;nbsp; || 3rd&amp;nbsp;&amp;nbsp; || 4th <br /> |-<br /> | 0 || 0 || 0 || 0 || 64<br /> |-<br /> | 1 || 1 || 4 || 16 || 64<br /> |-<br /> | 2 || 1 || 4 || 16 || 64<br /> |-<br /> | 3 || 1 || 4 || 16 || 64<br /> |-<br /> | 4 || 1 || 4 || 16 || 64<br /> |-<br /> | 5 || 1 || 4 || 16 || 64<br /> |-<br /> | 6 || 1 || 4 || 16 || 64<br /> |-<br /> | 7 || 1 || 4 || 16 || 64<br /> |-<br /> | 8 || 1 || 4 || 16 || 64<br /> |-<br /> | 9 || 0 || 0 || 0 || 64<br /> |}<br /> ==Solution 2==<br /> This problem can be solved via recursion since we are &quot;building a string&quot; of numbers with some condition. We want to create a new string by adding a new digit at the front so we avoid complications(&lt;math&gt;0&lt;/math&gt; can't be at the front and do digit is less than &lt;math&gt;9&lt;/math&gt;). There are &lt;math&gt;4&lt;/math&gt; options to add no matter what(try some examples if you want so the recursion is &lt;math&gt;S_n=4S_{n-1}&lt;/math&gt; where &lt;math&gt;S_n&lt;/math&gt; stands for the number of such numbers with &lt;math&gt;n&lt;/math&gt; digits. Since &lt;math&gt;S_1=10&lt;/math&gt; the answer is &lt;math&gt;\boxed{640}&lt;/math&gt;. <br /> <br /> For any number from 1-8, there are exactly 4 numbers from 1-8 that are odd and less than the number or that are even and greater than the number (the same will happen for 0 and 9 in the last column). Thus, the answer is &lt;math&gt;4^{k-1} \cdot 10 = 4^3\cdot10 = 640&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> &lt;/noinclude&gt;<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_15&diff=85914 2003 AIME II Problems/Problem 15 2017-05-31T23:45:46Z <p>First: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P(x) = x+2x^{2}+3x^{3} \dots 24x^{24}+23x^{25}+22x^{26} \dots x^{47}&lt;/math&gt;. Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> &lt;center&gt;&lt;math&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/math&gt;&lt;/center&gt;<br /> where &lt;math&gt;m, n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;<br /> <br /> == Solution ==<br /> This can quite obviously be factored as:<br /> <br /> &lt;cmath&gt; P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2 &lt;/cmath&gt;<br /> <br /> Note that &lt;math&gt; \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 &lt;/math&gt;.<br /> So the roots of &lt;math&gt;x^{23} + x^{22} + \cdots + x^2 + x + 1&lt;/math&gt; are exactly all &lt;math&gt;24&lt;/math&gt;-th complex roots of &lt;math&gt;1&lt;/math&gt;, except for the root &lt;math&gt;x=1&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;\omega=\cos \frac{360^\circ}{24} + i\sin \frac{360^\circ}{24}&lt;/math&gt;. Then the distinct zeros of &lt;math&gt;P&lt;/math&gt; are &lt;math&gt;0,\omega,\omega^2,\dots,\omega^{23}&lt;/math&gt;.<br /> <br /> We can clearly ignore the root &lt;math&gt;x=0&lt;/math&gt; as it does not contribute to the value that we need to compute. <br /> <br /> The squares of the other roots are &lt;math&gt;\omega^2,~\omega^4,~\dots,~\omega^{24}=1,~\omega^{26}=\omega^2,~\dots,~\omega^{46}=\omega^{22}&lt;/math&gt;.<br /> <br /> Hence we need to compute the following sum:<br /> <br /> &lt;cmath&gt;R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|&lt;/cmath&gt;<br /> <br /> Using basic properties of the sine function, we can simplify this to<br /> <br /> &lt;cmath&gt;R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)&lt;/cmath&gt;<br /> <br /> The five-element sum is just &lt;math&gt;\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ&lt;/math&gt;. <br /> We know that &lt;math&gt;\sin 30^\circ = \sin 150^\circ = \frac 12&lt;/math&gt;, &lt;math&gt;\sin 60^\circ = \sin 120^\circ = \frac{\sqrt 3}2&lt;/math&gt;, and &lt;math&gt;\sin 90^\circ = 1&lt;/math&gt;.<br /> Hence our sum evaluates to:<br /> <br /> &lt;cmath&gt;R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3&lt;/cmath&gt;<br /> <br /> Therefore the answer is &lt;math&gt;8+4+3 = \boxed{015}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that &lt;math&gt;x^k + x^{k-1} + \dots + x + 1 = \frac{x^{k+1} - 1}{x - 1}&lt;/math&gt;. Our sum can be reformed as &lt;cmath&gt;\frac{x(x^{47} - 1) + x^2(x^{45} - 1) + \dots + x^{24}(x - 1)}{x-1}&lt;/cmath&gt;<br /> <br /> So &lt;cmath&gt;\frac{x^{48} + x^{47} + x^{46} + \dots + x^{25} - x^{24} - x^{23} - \dots - x}{x-1} = 0&lt;/cmath&gt;<br /> <br /> &lt;math&gt;x(x^{47} + x^{46} + \dots - x - 1) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{47} + x^{46} + \dots - x - 1 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{47} + x^{46} + \dots + x + 1 = 2(x^{23} + x^{22} + \dots + x + 1)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{x^{48} - 1}{x - 1} = 2\frac{x^{24} - 1}{x - 1}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{48} - 1 - 2x^{24} + 2 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(x^{24} - 1)^2 = 0&lt;/math&gt;<br /> <br /> And we can proceed as above.<br /> <br /> == See also ==<br /> {{AIME box|year=2003|n=II|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_15&diff=85913 2003 AIME II Problems/Problem 15 2017-05-31T22:50:12Z <p>First: /* Problem */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P(x) = x+2x^{2}+3x^{3} \dots 24x^{24}+23x^{25}+22x^{26} \dots x^{47}&lt;/math&gt;. Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> &lt;center&gt;&lt;math&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/math&gt;&lt;/center&gt;<br /> where &lt;math&gt;m, n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;<br /> <br /> == Solution ==<br /> We can rewrite the definition of &lt;math&gt;P(x)&lt;/math&gt; as follows:<br /> <br /> &lt;cmath&gt; P(x) = x^{47} + 2x^{46} + \cdots + 23x^{25} + 24x^{24} + 23x^{23} + \cdots + 2x^2 + x &lt;/cmath&gt;<br /> <br /> This can quite obviously be factored as:<br /> <br /> &lt;cmath&gt; P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2 &lt;/cmath&gt;<br /> <br /> Note that &lt;math&gt; \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 &lt;/math&gt;.<br /> So the roots of &lt;math&gt;x^{23} + x^{22} + \cdots + x^2 + x + 1&lt;/math&gt; are exactly all &lt;math&gt;24&lt;/math&gt;-th complex roots of &lt;math&gt;1&lt;/math&gt;, except for the root &lt;math&gt;x=1&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;\omega=\cos \frac{360^\circ}{24} + i\sin \frac{360^\circ}{24}&lt;/math&gt;. Then the distinct zeros of &lt;math&gt;P&lt;/math&gt; are &lt;math&gt;0,\omega,\omega^2,\dots,\omega^{23}&lt;/math&gt;.<br /> <br /> We can clearly ignore the root &lt;math&gt;x=0&lt;/math&gt; as it does not contribute to the value that we need to compute. <br /> <br /> The squares of the other roots are &lt;math&gt;\omega^2,~\omega^4,~\dots,~\omega^{24}=1,~\omega^{26}=\omega^2,~\dots,~\omega^{46}=\omega^{22}&lt;/math&gt;.<br /> <br /> Hence we need to compute the following sum:<br /> <br /> &lt;cmath&gt;R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|&lt;/cmath&gt;<br /> <br /> Using basic properties of the sine function, we can simplify this to<br /> <br /> &lt;cmath&gt;R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)&lt;/cmath&gt;<br /> <br /> The five-element sum is just &lt;math&gt;\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ&lt;/math&gt;. <br /> We know that &lt;math&gt;\sin 30^\circ = \sin 150^\circ = \frac 12&lt;/math&gt;, &lt;math&gt;\sin 60^\circ = \sin 120^\circ = \frac{\sqrt 3}2&lt;/math&gt;, and &lt;math&gt;\sin 90^\circ = 1&lt;/math&gt;.<br /> Hence our sum evaluates to:<br /> <br /> &lt;cmath&gt;R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3&lt;/cmath&gt;<br /> <br /> Therefore the answer is &lt;math&gt;8+4+3 = \boxed{015}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that &lt;math&gt;x^k + x^{k-1} + \dots + x + 1 = \frac{x^{k+1} - 1}{x - 1}&lt;/math&gt;. Our sum can be reformed as &lt;cmath&gt;\frac{x(x^{47} - 1) + x^2(x^{45} - 1) + \dots + x^{24}(x - 1)}{x-1}&lt;/cmath&gt;<br /> <br /> So &lt;cmath&gt;\frac{x^{48} + x^{47} + x^{46} + \dots + x^{25} - x^{24} - x^{23} - \dots - x}{x-1} = 0&lt;/cmath&gt;<br /> <br /> &lt;math&gt;x(x^{47} + x^{46} + \dots - x - 1) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{47} + x^{46} + \dots - x - 1 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{47} + x^{46} + \dots + x + 1 = 2(x^{23} + x^{22} + \dots + x + 1)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{x^{48} - 1}{x - 1} = 2\frac{x^{24} - 1}{x - 1}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{48} - 1 - 2x^{24} + 2 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(x^{24} - 1)^2 = 0&lt;/math&gt;<br /> <br /> And we can proceed as above.<br /> <br /> == See also ==<br /> {{AIME box|year=2003|n=II|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems&diff=85912 2003 AIME II Problems 2017-05-31T22:49:34Z <p>First: /* Problem 15 */</p> <hr /> <div>{{AIME Problems|year=2003|n=II}}<br /> <br /> == Problem 1 ==<br /> The product &lt;math&gt;N&lt;/math&gt; of three positive integers is 6 times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> [[2003 AIME II Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Let &lt;math&gt;N&lt;/math&gt; be the greatest integer multiple of 8, whose digits are all different. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by 1000?<br /> <br /> [[2003 AIME II Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Define a &lt;math&gt;good~word&lt;/math&gt; as a sequence of letters that consists only of the letters &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; - some of these letters may not appear in the sequence - and in which &lt;math&gt;A&lt;/math&gt; is never immediately followed by &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; is never immediately followed by &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; is never immediately followed by &lt;math&gt;A&lt;/math&gt;. How many seven-letter good words are there?<br /> <br /> [[2003 AIME II Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is &lt;math&gt;m/n&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> [[2003 AIME II Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A cylindrical log has diameter &lt;math&gt;12&lt;/math&gt; inches. A wedge is cut from the log by making two planar cuts that go entirely through the log. The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a &lt;math&gt;45^\circ&lt;/math&gt; angle with the plane of the first cut. The intersection of these two planes has exactly one point in common with the log. The number of cubic inches in the wedge can be expressed as &lt;math&gt;n\pi&lt;/math&gt;, where n is a positive integer. Find &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2003 AIME II Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> In triangle &lt;math&gt;ABC,&lt;/math&gt; &lt;math&gt;AB = 13,&lt;/math&gt; &lt;math&gt;BC = 14,&lt;/math&gt; &lt;math&gt;AC = 15,&lt;/math&gt; and point &lt;math&gt;G&lt;/math&gt; is the intersection of the medians. Points &lt;math&gt;A',&lt;/math&gt; &lt;math&gt;B',&lt;/math&gt; and &lt;math&gt;C',&lt;/math&gt; are the images of &lt;math&gt;A,&lt;/math&gt; &lt;math&gt;B,&lt;/math&gt; and &lt;math&gt;C,&lt;/math&gt; respectively, after a &lt;math&gt;180^\circ&lt;/math&gt; rotation about &lt;math&gt;G.&lt;/math&gt; What is the area of the union of the two regions enclosed by the triangles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;A'B'C'?&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Find the area of rhombus &lt;math&gt;ABCD&lt;/math&gt; given that the radii of the circles circumscribed around triangles &lt;math&gt;ABD&lt;/math&gt; and &lt;math&gt;ACD&lt;/math&gt; are &lt;math&gt;12.5&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt;, respectively.<br /> <br /> [[2003 AIME II Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Find the eighth term of the sequence &lt;math&gt;1440,&lt;/math&gt; &lt;math&gt;1716,&lt;/math&gt; &lt;math&gt;1848,\ldots,&lt;/math&gt; whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.<br /> <br /> [[2003 AIME II Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Consider the polynomials &lt;math&gt;P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x&lt;/math&gt; and &lt;math&gt;Q(x) = x^{4} - x^{3} - x^{2} - 1.&lt;/math&gt; Given that &lt;math&gt;z_{1},z_{2},z_{3},&lt;/math&gt; and &lt;math&gt;z_{4}&lt;/math&gt; are the roots of &lt;math&gt;Q(x) = 0,&lt;/math&gt; find &lt;math&gt;P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Two positive integers differ by &lt;math&gt;60.&lt;/math&gt; The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?<br /> <br /> [[2003 AIME II Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is a right triangle with &lt;math&gt;AC = 7,&lt;/math&gt; &lt;math&gt;BC = 24,&lt;/math&gt; and right angle at &lt;math&gt;C.&lt;/math&gt; Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;AB,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; is on the same side of line &lt;math&gt;AB&lt;/math&gt; as &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;AD = BD = 15.&lt;/math&gt; Given that the area of triangle &lt;math&gt;CDM&lt;/math&gt; may be expressed as &lt;math&gt;\frac {m\sqrt {n}}{p},&lt;/math&gt; where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are relatively prime, and &lt;math&gt;n&lt;/math&gt; is not divisible by the square of any prime, find &lt;math&gt;m + n + p.&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The members of a distinguished committee were choosing a president, and each member gave one vote to one of the &lt;math&gt;27&lt;/math&gt; candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least &lt;math&gt;1&lt;/math&gt; than the number of votes for that candidate. What is the smallest possible number of members of the committee?<br /> <br /> [[2003 AIME II Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Let &lt;math&gt;A = (0,0)&lt;/math&gt; and &lt;math&gt;B = (b,2)&lt;/math&gt; be points on the coordinate plane. Let &lt;math&gt;ABCDEF&lt;/math&gt; be a convex equilateral hexagon such that &lt;math&gt;\angle FAB = 120^\circ,&lt;/math&gt; &lt;math&gt;\overline{AB}\parallel \overline{DE},&lt;/math&gt; &lt;math&gt;\overline{BC}\parallel \overline{EF,}&lt;/math&gt; &lt;math&gt;\overline{CD}\parallel \overline{FA},&lt;/math&gt; and the y-coordinates of its vertices are distinct elements of the set &lt;math&gt;\{0,2,4,6,8,10\}.&lt;/math&gt; The area of the hexagon can be written in the form &lt;math&gt;m\sqrt {n},&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers and n is not divisible by the square of any prime. Find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> [[2003 AIME II Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Let &lt;math&gt;P(x) = x+2x^{2}+3x^{3} \dots 24x^{24}+23x^{25}+22x^{26} \dots x^{47}&lt;/math&gt;. Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> &lt;center&gt;&lt;math&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/math&gt;&lt;/center&gt;<br /> where &lt;math&gt;m, n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;<br /> [[2003 AIME II Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_15&diff=85911 2003 AIME II Problems/Problem 15 2017-05-31T22:48:52Z <p>First: /* Problem */</p> <hr /> <div>== Problem ==<br /> [quote=2003 AIME II #15]Let &lt;math&gt;P(x) = x+2x^{2}+3x^{3} \dots 24x^{24}+23x^{25}+22x^{26} \dots x^{47}&lt;/math&gt;. Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> &lt;center&gt;&lt;math&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/math&gt;&lt;/center&gt;<br /> where &lt;math&gt;m, n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;[/quote]<br /> <br /> == Solution ==<br /> We can rewrite the definition of &lt;math&gt;P(x)&lt;/math&gt; as follows:<br /> <br /> &lt;cmath&gt; P(x) = x^{47} + 2x^{46} + \cdots + 23x^{25} + 24x^{24} + 23x^{23} + \cdots + 2x^2 + x &lt;/cmath&gt;<br /> <br /> This can quite obviously be factored as:<br /> <br /> &lt;cmath&gt; P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2 &lt;/cmath&gt;<br /> <br /> Note that &lt;math&gt; \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 &lt;/math&gt;.<br /> So the roots of &lt;math&gt;x^{23} + x^{22} + \cdots + x^2 + x + 1&lt;/math&gt; are exactly all &lt;math&gt;24&lt;/math&gt;-th complex roots of &lt;math&gt;1&lt;/math&gt;, except for the root &lt;math&gt;x=1&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;\omega=\cos \frac{360^\circ}{24} + i\sin \frac{360^\circ}{24}&lt;/math&gt;. Then the distinct zeros of &lt;math&gt;P&lt;/math&gt; are &lt;math&gt;0,\omega,\omega^2,\dots,\omega^{23}&lt;/math&gt;.<br /> <br /> We can clearly ignore the root &lt;math&gt;x=0&lt;/math&gt; as it does not contribute to the value that we need to compute. <br /> <br /> The squares of the other roots are &lt;math&gt;\omega^2,~\omega^4,~\dots,~\omega^{24}=1,~\omega^{26}=\omega^2,~\dots,~\omega^{46}=\omega^{22}&lt;/math&gt;.<br /> <br /> Hence we need to compute the following sum:<br /> <br /> &lt;cmath&gt;R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|&lt;/cmath&gt;<br /> <br /> Using basic properties of the sine function, we can simplify this to<br /> <br /> &lt;cmath&gt;R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)&lt;/cmath&gt;<br /> <br /> The five-element sum is just &lt;math&gt;\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ&lt;/math&gt;. <br /> We know that &lt;math&gt;\sin 30^\circ = \sin 150^\circ = \frac 12&lt;/math&gt;, &lt;math&gt;\sin 60^\circ = \sin 120^\circ = \frac{\sqrt 3}2&lt;/math&gt;, and &lt;math&gt;\sin 90^\circ = 1&lt;/math&gt;.<br /> Hence our sum evaluates to:<br /> <br /> &lt;cmath&gt;R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3&lt;/cmath&gt;<br /> <br /> Therefore the answer is &lt;math&gt;8+4+3 = \boxed{015}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that &lt;math&gt;x^k + x^{k-1} + \dots + x + 1 = \frac{x^{k+1} - 1}{x - 1}&lt;/math&gt;. Our sum can be reformed as &lt;cmath&gt;\frac{x(x^{47} - 1) + x^2(x^{45} - 1) + \dots + x^{24}(x - 1)}{x-1}&lt;/cmath&gt;<br /> <br /> So &lt;cmath&gt;\frac{x^{48} + x^{47} + x^{46} + \dots + x^{25} - x^{24} - x^{23} - \dots - x}{x-1} = 0&lt;/cmath&gt;<br /> <br /> &lt;math&gt;x(x^{47} + x^{46} + \dots - x - 1) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{47} + x^{46} + \dots - x - 1 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{47} + x^{46} + \dots + x + 1 = 2(x^{23} + x^{22} + \dots + x + 1)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{x^{48} - 1}{x - 1} = 2\frac{x^{24} - 1}{x - 1}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{48} - 1 - 2x^{24} + 2 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(x^{24} - 1)^2 = 0&lt;/math&gt;<br /> <br /> And we can proceed as above.<br /> <br /> == See also ==<br /> {{AIME box|year=2003|n=II|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_15&diff=85910 2003 AIME II Problems/Problem 15 2017-05-31T22:48:16Z <p>First: /* Problem */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P(x) = x+2x^{2}+3x^{3} \dots 24x^{24}+23x^{25}+22x^{26} \dots x^{47}&lt;/math&gt;. Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> &lt;center&gt;&lt;math&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/math&gt;&lt;/center&gt;<br /> where &lt;math&gt;m, n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;<br /> <br /> == Solution ==<br /> We can rewrite the definition of &lt;math&gt;P(x)&lt;/math&gt; as follows:<br /> <br /> &lt;cmath&gt; P(x) = x^{47} + 2x^{46} + \cdots + 23x^{25} + 24x^{24} + 23x^{23} + \cdots + 2x^2 + x &lt;/cmath&gt;<br /> <br /> This can quite obviously be factored as:<br /> <br /> &lt;cmath&gt; P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2 &lt;/cmath&gt;<br /> <br /> Note that &lt;math&gt; \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 &lt;/math&gt;.<br /> So the roots of &lt;math&gt;x^{23} + x^{22} + \cdots + x^2 + x + 1&lt;/math&gt; are exactly all &lt;math&gt;24&lt;/math&gt;-th complex roots of &lt;math&gt;1&lt;/math&gt;, except for the root &lt;math&gt;x=1&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;\omega=\cos \frac{360^\circ}{24} + i\sin \frac{360^\circ}{24}&lt;/math&gt;. Then the distinct zeros of &lt;math&gt;P&lt;/math&gt; are &lt;math&gt;0,\omega,\omega^2,\dots,\omega^{23}&lt;/math&gt;.<br /> <br /> We can clearly ignore the root &lt;math&gt;x=0&lt;/math&gt; as it does not contribute to the value that we need to compute. <br /> <br /> The squares of the other roots are &lt;math&gt;\omega^2,~\omega^4,~\dots,~\omega^{24}=1,~\omega^{26}=\omega^2,~\dots,~\omega^{46}=\omega^{22}&lt;/math&gt;.<br /> <br /> Hence we need to compute the following sum:<br /> <br /> &lt;cmath&gt;R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|&lt;/cmath&gt;<br /> <br /> Using basic properties of the sine function, we can simplify this to<br /> <br /> &lt;cmath&gt;R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)&lt;/cmath&gt;<br /> <br /> The five-element sum is just &lt;math&gt;\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ&lt;/math&gt;. <br /> We know that &lt;math&gt;\sin 30^\circ = \sin 150^\circ = \frac 12&lt;/math&gt;, &lt;math&gt;\sin 60^\circ = \sin 120^\circ = \frac{\sqrt 3}2&lt;/math&gt;, and &lt;math&gt;\sin 90^\circ = 1&lt;/math&gt;.<br /> Hence our sum evaluates to:<br /> <br /> &lt;cmath&gt;R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3&lt;/cmath&gt;<br /> <br /> Therefore the answer is &lt;math&gt;8+4+3 = \boxed{015}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that &lt;math&gt;x^k + x^{k-1} + \dots + x + 1 = \frac{x^{k+1} - 1}{x - 1}&lt;/math&gt;. Our sum can be reformed as &lt;cmath&gt;\frac{x(x^{47} - 1) + x^2(x^{45} - 1) + \dots + x^{24}(x - 1)}{x-1}&lt;/cmath&gt;<br /> <br /> So &lt;cmath&gt;\frac{x^{48} + x^{47} + x^{46} + \dots + x^{25} - x^{24} - x^{23} - \dots - x}{x-1} = 0&lt;/cmath&gt;<br /> <br /> &lt;math&gt;x(x^{47} + x^{46} + \dots - x - 1) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{47} + x^{46} + \dots - x - 1 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{47} + x^{46} + \dots + x + 1 = 2(x^{23} + x^{22} + \dots + x + 1)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{x^{48} - 1}{x - 1} = 2\frac{x^{24} - 1}{x - 1}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{48} - 1 - 2x^{24} + 2 = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(x^{24} - 1)^2 = 0&lt;/math&gt;<br /> <br /> And we can proceed as above.<br /> <br /> == See also ==<br /> {{AIME box|year=2003|n=II|num-b=14|after=Last Question}}<br /> {{MAA Notice}}</div> First