https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=First&feedformat=atom AoPS Wiki - User contributions [en] 2021-08-03T09:01:52Z User contributions MediaWiki 1.31.1 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 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_14&diff=85877 2001 AIME I Problems/Problem 14 2017-05-28T19:34:30Z <p>First: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?<br /> <br /> == Solutions==<br /> ==Solution 1==<br /> Let &lt;math&gt;0&lt;/math&gt; represent a house that does not receive mail and &lt;math&gt;1&lt;/math&gt; represent a house that does receive mail. This problem is now asking for the number of &lt;math&gt;19&lt;/math&gt;-digit strings of &lt;math&gt;0&lt;/math&gt;'s and &lt;math&gt;1&lt;/math&gt;'s such that there are no two consecutive &lt;math&gt;1&lt;/math&gt;'s and no three consecutive &lt;math&gt;0&lt;/math&gt;'s. <br /> <br /> The last two digits of any &lt;math&gt;n&lt;/math&gt;-digit string can't be &lt;math&gt;11&lt;/math&gt;, so the only possibilities are &lt;math&gt;00&lt;/math&gt;, &lt;math&gt;01&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;00&lt;/math&gt;, &lt;math&gt;b_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;01&lt;/math&gt;, and &lt;math&gt;c_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;10&lt;/math&gt;. <br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;00&lt;/math&gt;, then the previous digit must be a &lt;math&gt;1&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring will be &lt;math&gt;10&lt;/math&gt;. So <br /> &lt;cmath&gt;a_{n} = c_{n-1}.&lt;/cmath&gt;<br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;01&lt;/math&gt;, then the previous digit can be either a &lt;math&gt;0&lt;/math&gt; or a &lt;math&gt;1&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring can be either &lt;math&gt;00&lt;/math&gt; or &lt;math&gt;10&lt;/math&gt;. So <br /> &lt;cmath&gt;b_{n} = a_{n-1} + c_{n-1}.&lt;/cmath&gt;<br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;10&lt;/math&gt;, then the previous digit must be a &lt;math&gt;0&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring will be &lt;math&gt;01&lt;/math&gt;. So <br /> &lt;cmath&gt;c_{n} = b_{n-1}.&lt;/cmath&gt;<br /> <br /> Clearly, &lt;math&gt;a_2=b_2=c_2=1&lt;/math&gt;. Using the [[recursion|recursive]] equations and initial values: <br /> &lt;cmath&gt;\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}<br /> \multicolumn{19}{c}{}\\\hline<br /> n&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9&amp;10&amp;11&amp;12&amp;13&amp;14&amp;15&amp;16&amp;17&amp;18&amp;19\\\hline<br /> a_n&amp;1&amp;1&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86\\\hline<br /> b_n&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86&amp;114&amp;151\\\hline<br /> c_n&amp;1&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86&amp;114\\\hline<br /> \end{array}&lt;/cmath&gt;<br /> <br /> As a result &lt;math&gt;a_19+b_19+c_19=\boxed{351}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=13|num-a=15|t=384210}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_14&diff=85876 2001 AIME I Problems/Problem 14 2017-05-28T19:33:56Z <p>First: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?<br /> <br /> == Solutions==<br /> ==Solution 1==<br /> Let &lt;math&gt;0&lt;/math&gt; represent a house that does not receive mail and &lt;math&gt;1&lt;/math&gt; represent a house that does receive mail. This problem is now asking for the number of &lt;math&gt;19&lt;/math&gt;-digit strings of &lt;math&gt;0&lt;/math&gt;'s and &lt;math&gt;1&lt;/math&gt;'s such that there are no two consecutive &lt;math&gt;1&lt;/math&gt;'s and no three consecutive &lt;math&gt;0&lt;/math&gt;'s. <br /> <br /> The last two digits of any &lt;math&gt;n&lt;/math&gt;-digit string can't be &lt;math&gt;11&lt;/math&gt;, so the only possibilities are &lt;math&gt;00&lt;/math&gt;, &lt;math&gt;01&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;00&lt;/math&gt;, &lt;math&gt;b_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;01&lt;/math&gt;, and &lt;math&gt;c_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;10&lt;/math&gt;. <br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;00&lt;/math&gt;, then the previous digit must be a &lt;math&gt;1&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring will be &lt;math&gt;10&lt;/math&gt;. So <br /> &lt;cmath&gt;a_{n} = c_{n-1}.&lt;/cmath&gt;<br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;01&lt;/math&gt;, then the previous digit can be either a &lt;math&gt;0&lt;/math&gt; or a &lt;math&gt;1&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring can be either &lt;math&gt;00&lt;/math&gt; or &lt;math&gt;10&lt;/math&gt;. So <br /> &lt;cmath&gt;b_{n} = a_{n-1} + c_{n-1}.&lt;/cmath&gt;<br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;10&lt;/math&gt;, then the previous digit must be a &lt;math&gt;0&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring will be &lt;math&gt;01&lt;/math&gt;. So <br /> &lt;cmath&gt;c_{n} = b_{n-1}.&lt;/cmath&gt;<br /> <br /> Clearly, &lt;math&gt;a_2=b_2=c_2=1&lt;/math&gt;. Using the [[recursion|recursive]] equations and initial values: <br /> &lt;cmath&gt;\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}<br /> \multicolumn{19}{c}{}\\\hline<br /> n&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9&amp;10&amp;11&amp;12&amp;13&amp;14&amp;15&amp;16&amp;17&amp;18&amp;19\\\hline<br /> a_n&amp;1&amp;1&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86\\\hline<br /> b_n&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86&amp;114&amp;151\\\hline<br /> c_n&amp;1&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86&amp;114\\\hline<br /> \end{array}&lt;/cmath&gt;<br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=13|num-a=15|t=384210}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_14&diff=85875 2001 AIME I Problems/Problem 14 2017-05-28T19:28:26Z <p>First: /* Solution */</p> <hr /> <div>== Problem ==<br /> A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?<br /> <br /> == Solutions==<br /> ==Solution 1==<br /> Let &lt;math&gt;0&lt;/math&gt; represent a house that does not receive mail and &lt;math&gt;1&lt;/math&gt; represent a house that does receive mail. This problem is now asking for the number of &lt;math&gt;19&lt;/math&gt;-digit strings of &lt;math&gt;0&lt;/math&gt;'s and &lt;math&gt;1&lt;/math&gt;'s such that there are no two consecutive &lt;math&gt;1&lt;/math&gt;'s and no three consecutive &lt;math&gt;0&lt;/math&gt;'s. <br /> <br /> The last two digits of any &lt;math&gt;n&lt;/math&gt;-digit string can't be &lt;math&gt;11&lt;/math&gt;, so the only possibilities are &lt;math&gt;00&lt;/math&gt;, &lt;math&gt;01&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;00&lt;/math&gt;, &lt;math&gt;b_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;01&lt;/math&gt;, and &lt;math&gt;c_n&lt;/math&gt; be the number of &lt;math&gt;n&lt;/math&gt;-digit strings ending in &lt;math&gt;10&lt;/math&gt;. <br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;00&lt;/math&gt;, then the previous digit must be a &lt;math&gt;1&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring will be &lt;math&gt;10&lt;/math&gt;. So <br /> &lt;cmath&gt;a_{n} = c_{n-1}.&lt;/cmath&gt;<br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;01&lt;/math&gt;, then the previous digit can be either a &lt;math&gt;0&lt;/math&gt; or a &lt;math&gt;1&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring can be either &lt;math&gt;00&lt;/math&gt; or &lt;math&gt;10&lt;/math&gt;. So <br /> &lt;cmath&gt;b_{n} = a_{n-1} + c_{n-1}.&lt;/cmath&gt;<br /> <br /> If an &lt;math&gt;n&lt;/math&gt;-digit string ends in &lt;math&gt;10&lt;/math&gt;, then the previous digit must be a &lt;math&gt;0&lt;/math&gt;, and the last two digits of the &lt;math&gt;n-1&lt;/math&gt; digits substring will be &lt;math&gt;01&lt;/math&gt;. So <br /> &lt;cmath&gt;c_{n} = b_{n-1}.&lt;/cmath&gt;<br /> <br /> Clearly, &lt;math&gt;a_2=b_2=c_2=1&lt;/math&gt;. Using the [[recursion|recursive]] equations and initial values: <br /> &lt;cmath&gt;\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}<br /> \multicolumn{19}{c}{}\\\hline<br /> n&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9&amp;10&amp;11&amp;12&amp;13&amp;14&amp;15&amp;16&amp;17&amp;18&amp;19\\\hline<br /> a_n&amp;1&amp;1&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86\\\hline<br /> b_n&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86&amp;114&amp;151\\\hline<br /> c_n&amp;1&amp;1&amp;2&amp;2&amp;3&amp;4&amp;5&amp;7&amp;9&amp;12&amp;16&amp;21&amp;28&amp;37&amp;49&amp;65&amp;86&amp;114\\\hline<br /> \end{array}&lt;/cmath&gt;<br /> ==Solution 2==<br /> Let's call the number of sequences that start with a house getting mail and of length &lt;math&gt;n&lt;/math&gt; as &lt;math&gt;A_n&lt;/math&gt; and call &lt;math&gt;B_n&lt;/math&gt; the opposite(first house gets no mail). We get &lt;math&gt;A_n=B_{n-1}&lt;/math&gt; and &lt;math&gt;B_{n}=B_{n-1}+<br /> <br /> Therefore, the number of &lt;/math&gt;19&lt;math&gt;-digit strings is &lt;/math&gt;a_{19}+b_{19}+c_{19} = 86+151+114 = \boxed{351}.$<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=13|num-a=15|t=384210}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_6&diff=85869 2000 AIME I Problems/Problem 6 2017-05-27T22:00:02Z <p>First: /* Solution 3(2-Liner) */</p> <hr /> <div>== Problem ==<br /> For how many [[ordered pair]]s &lt;math&gt;(x,y)&lt;/math&gt; of [[integer]]s is it true that &lt;math&gt;0 &lt; x &lt; y &lt; 10^{6}&lt;/math&gt; and that the [[arithmetic mean]] of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is exactly &lt;math&gt;2&lt;/math&gt; more than the [[geometric mean]] of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;?<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \frac{x+y}{2} &amp;=&amp; \sqrt{xy} + 2\\<br /> x+y-4 &amp;=&amp; 2\sqrt{xy}\\<br /> y - 2\sqrt{xy} + x &amp;=&amp; 4\\<br /> \sqrt{y} - \sqrt{x} &amp;=&amp; \pm 2\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;y &gt; x&lt;/math&gt;, we only consider &lt;math&gt;+2&lt;/math&gt;.<br /> <br /> For simplicity, we can count how many valid pairs of &lt;math&gt;(\sqrt{x},\sqrt{y})&lt;/math&gt; that satisfy our equation.<br /> <br /> The maximum that &lt;math&gt;\sqrt{y}&lt;/math&gt; can be is &lt;math&gt;10^3 - 1 = 999&lt;/math&gt; because &lt;math&gt;\sqrt{y}&lt;/math&gt; must be an integer (this is because &lt;math&gt;\sqrt{y} - \sqrt{x} = 2&lt;/math&gt;, an integer). Then &lt;math&gt;\sqrt{x} = 997&lt;/math&gt;, and we continue this downward until &lt;math&gt;\sqrt{y} = 3&lt;/math&gt;, in which case &lt;math&gt;\sqrt{x} = 1&lt;/math&gt;. The number of pairs of &lt;math&gt;(\sqrt{x},\sqrt{y})&lt;/math&gt;, and so &lt;math&gt;(x,y)&lt;/math&gt; is then &lt;math&gt;\boxed{997}&lt;/math&gt;.<br /> &lt;!-- solution lost in edit conflict - azjps -<br /> Since &lt;math&gt;y&gt;x&lt;/math&gt;, it follows that each ordered pair &lt;math&gt;(x,y) = (n^2, (n+2)^2)&lt;/math&gt; satisfies this equation. The minimum value of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; and the maximum value of &lt;math&gt;y = 999^2&lt;/math&gt; which would make &lt;math&gt;x = 997^2&lt;/math&gt;. Thus &lt;math&gt;x&lt;/math&gt; can be any of the squares between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;997^2&lt;/math&gt; inclusive and the answer is &lt;math&gt;\boxed{997}&lt;/math&gt;.<br /> --&gt;<br /> <br /> === Solution 2 ===<br /> <br /> <br /> Let &lt;math&gt;a^2&lt;/math&gt; = &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;b^2&lt;/math&gt; = &lt;math&gt;y&lt;/math&gt;<br /> <br /> Then &lt;cmath&gt;\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2 + b^2 = 2ab + 4&lt;/cmath&gt;<br /> &lt;cmath&gt;(a-b)^2 = 4&lt;/cmath&gt;<br /> &lt;cmath&gt;(a-b) = \pm 2&lt;/cmath&gt;<br /> <br /> This makes counting a lot easier since now we just have to find all pairs &lt;math&gt;(a,b)&lt;/math&gt; that differ by 2.<br /> <br /> <br /> Because &lt;math&gt;\sqrt{10^6} = 10^3&lt;/math&gt;, then we can use all positive integers less than 1000 for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;.<br /> <br /> <br /> [[Without loss of generality]], let's say &lt;math&gt;a &lt; b&lt;/math&gt;.<br /> <br /> <br /> We can count even and odd pairs separately to make things easier*:<br /> <br /> <br /> Odd: &lt;cmath&gt;(1,3) , (3,5) , (5,7) . . . (997,999)&lt;/cmath&gt;<br /> <br /> <br /> Even: &lt;cmath&gt;(2,4) , (4,6) , (6,8) . . . (996,998)&lt;/cmath&gt;<br /> <br /> <br /> This makes 499 odd pairs and 498 even pairs, for a total of &lt;math&gt;\boxed{997}&lt;/math&gt; pairs.<br /> <br /> <br /> <br /> &lt;math&gt;*&lt;/math&gt;Note: We are counting the pairs for the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, which, when squared, translate to the pairs of &lt;math&gt;(x,y)&lt;/math&gt; we are trying to find.<br /> ===Solution 3(2-Liner)===<br /> Our equation is &lt;math&gt;x+y-4=2\sqrt{xy} \implies \sqrt{y}-\sqrt{x}=2&lt;/math&gt; since &lt;math&gt;y&gt;x&lt;/math&gt;. As a result &lt;math&gt;y&lt;/math&gt; must be a perfect square and cannot be &lt;math&gt;10^6, 4, 1&lt;/math&gt; so the answer is &lt;math&gt;\boxed{997}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_6&diff=85868 2000 AIME I Problems/Problem 6 2017-05-27T21:59:42Z <p>First: /* Solution */</p> <hr /> <div>== Problem ==<br /> For how many [[ordered pair]]s &lt;math&gt;(x,y)&lt;/math&gt; of [[integer]]s is it true that &lt;math&gt;0 &lt; x &lt; y &lt; 10^{6}&lt;/math&gt; and that the [[arithmetic mean]] of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is exactly &lt;math&gt;2&lt;/math&gt; more than the [[geometric mean]] of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;?<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> &lt;cmath&gt;\begin{eqnarray*}<br /> \frac{x+y}{2} &amp;=&amp; \sqrt{xy} + 2\\<br /> x+y-4 &amp;=&amp; 2\sqrt{xy}\\<br /> y - 2\sqrt{xy} + x &amp;=&amp; 4\\<br /> \sqrt{y} - \sqrt{x} &amp;=&amp; \pm 2\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;y &gt; x&lt;/math&gt;, we only consider &lt;math&gt;+2&lt;/math&gt;.<br /> <br /> For simplicity, we can count how many valid pairs of &lt;math&gt;(\sqrt{x},\sqrt{y})&lt;/math&gt; that satisfy our equation.<br /> <br /> The maximum that &lt;math&gt;\sqrt{y}&lt;/math&gt; can be is &lt;math&gt;10^3 - 1 = 999&lt;/math&gt; because &lt;math&gt;\sqrt{y}&lt;/math&gt; must be an integer (this is because &lt;math&gt;\sqrt{y} - \sqrt{x} = 2&lt;/math&gt;, an integer). Then &lt;math&gt;\sqrt{x} = 997&lt;/math&gt;, and we continue this downward until &lt;math&gt;\sqrt{y} = 3&lt;/math&gt;, in which case &lt;math&gt;\sqrt{x} = 1&lt;/math&gt;. The number of pairs of &lt;math&gt;(\sqrt{x},\sqrt{y})&lt;/math&gt;, and so &lt;math&gt;(x,y)&lt;/math&gt; is then &lt;math&gt;\boxed{997}&lt;/math&gt;.<br /> &lt;!-- solution lost in edit conflict - azjps -<br /> Since &lt;math&gt;y&gt;x&lt;/math&gt;, it follows that each ordered pair &lt;math&gt;(x,y) = (n^2, (n+2)^2)&lt;/math&gt; satisfies this equation. The minimum value of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; and the maximum value of &lt;math&gt;y = 999^2&lt;/math&gt; which would make &lt;math&gt;x = 997^2&lt;/math&gt;. Thus &lt;math&gt;x&lt;/math&gt; can be any of the squares between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;997^2&lt;/math&gt; inclusive and the answer is &lt;math&gt;\boxed{997}&lt;/math&gt;.<br /> --&gt;<br /> <br /> === Solution 2 ===<br /> <br /> <br /> Let &lt;math&gt;a^2&lt;/math&gt; = &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;b^2&lt;/math&gt; = &lt;math&gt;y&lt;/math&gt;<br /> <br /> Then &lt;cmath&gt;\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2 + b^2 = 2ab + 4&lt;/cmath&gt;<br /> &lt;cmath&gt;(a-b)^2 = 4&lt;/cmath&gt;<br /> &lt;cmath&gt;(a-b) = \pm 2&lt;/cmath&gt;<br /> <br /> This makes counting a lot easier since now we just have to find all pairs &lt;math&gt;(a,b)&lt;/math&gt; that differ by 2.<br /> <br /> <br /> Because &lt;math&gt;\sqrt{10^6} = 10^3&lt;/math&gt;, then we can use all positive integers less than 1000 for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;.<br /> <br /> <br /> [[Without loss of generality]], let's say &lt;math&gt;a &lt; b&lt;/math&gt;.<br /> <br /> <br /> We can count even and odd pairs separately to make things easier*:<br /> <br /> <br /> Odd: &lt;cmath&gt;(1,3) , (3,5) , (5,7) . . . (997,999)&lt;/cmath&gt;<br /> <br /> <br /> Even: &lt;cmath&gt;(2,4) , (4,6) , (6,8) . . . (996,998)&lt;/cmath&gt;<br /> <br /> <br /> This makes 499 odd pairs and 498 even pairs, for a total of &lt;math&gt;\boxed{997}&lt;/math&gt; pairs.<br /> <br /> <br /> <br /> &lt;math&gt;*&lt;/math&gt;Note: We are counting the pairs for the values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, which, when squared, translate to the pairs of &lt;math&gt;(x,y)&lt;/math&gt; we are trying to find.<br /> ===Solution 3(2-Liner)===<br /> Our equation is &lt;math&gt;x+y-4=2\sqrt{xy} \implies &lt;/math&gt;\sqrt{y}-\sqrt{x}=2&lt;math&gt; since &lt;/math&gt;y&gt;x&lt;math&gt;. As a result &lt;/math&gt;y&lt;math&gt; must be a perfect square and cannot be &lt;/math&gt;10^6, 4, 1&lt;math&gt; so the answer is &lt;/math&gt;\boxed{997}$.<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=1999_AIME_Problems&diff=85865 1999 AIME Problems 2017-05-27T16:41:47Z <p>First: /* Problem 13 */</p> <hr /> <div>{{AIME Problems|year=1999}}<br /> <br /> == Problem 1 ==<br /> Find the smallest prime that is the fifth term of an increasing arithmetic sequence, all four preceding terms also being prime.<br /> <br /> [[1999 AIME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> Consider the parallelogram with vertices &lt;math&gt;(10,45),&lt;/math&gt; &lt;math&gt;(10,114),&lt;/math&gt; &lt;math&gt;(28,153),&lt;/math&gt; and &lt;math&gt;(28,84).&lt;/math&gt; A line through the origin cuts this figure into two congruent polygons. The slope of the line 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 /> [[1999 AIME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;n^2-19n+99&lt;/math&gt; is a perfect square.<br /> <br /> [[1999 AIME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> The two squares shown share the same center &lt;math&gt;O_{}&lt;/math&gt; and have sides of length 1. The length of &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;43/99&lt;/math&gt; and the area of octagon &lt;math&gt;ABCDEFGH&lt;/math&gt; 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 /> [[Image:AIME_1999_Problem_4.png]]<br /> <br /> [[1999 AIME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> For any positive integer &lt;math&gt;x_{}&lt;/math&gt;, let &lt;math&gt;S(x)&lt;/math&gt; be the sum of the digits of &lt;math&gt;x_{}&lt;/math&gt;, and let &lt;math&gt;T(x)&lt;/math&gt; be &lt;math&gt;|S(x+2)-S(x)|.&lt;/math&gt; For example, &lt;math&gt;T(199)=|S(201)-S(199)|=|3-19|=16.&lt;/math&gt; How many values of &lt;math&gt;T(x)&lt;/math&gt; do not exceed 1999?<br /> <br /> [[1999 AIME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> A transformation of the first quadrant of the coordinate plane maps each point &lt;math&gt;(x,y)&lt;/math&gt; to the point &lt;math&gt;(\sqrt{x},\sqrt{y}).&lt;/math&gt; The vertices of quadrilateral &lt;math&gt;ABCD&lt;/math&gt; are &lt;math&gt;A=(900,300), B=(1800,600), C=(600,1800),&lt;/math&gt; and &lt;math&gt;D=(300,900).&lt;/math&gt; Let &lt;math&gt;k_{}&lt;/math&gt; be the area of the region enclosed by the image of quadrilateral &lt;math&gt;ABCD.&lt;/math&gt; Find the greatest integer that does not exceed &lt;math&gt;k_{}.&lt;/math&gt;<br /> <br /> [[1999 AIME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> There is a set of 1000 switches, each of which has four positions, called &lt;math&gt;A, B, C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;. When the position of any switch changes, it is only from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;, from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;D&lt;/math&gt;, or from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt;. Initially each switch is in position &lt;math&gt;A&lt;/math&gt;. The switches are labeled with the 1000 different integers &lt;math&gt;(2^{x})(3^{y})(5^{z})&lt;/math&gt;, where &lt;math&gt;x, y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; take on the values &lt;math&gt;0, 1, \ldots, 9&lt;/math&gt;. At step i of a 1000-step process, the &lt;math&gt;i&lt;/math&gt;-th switch is advanced one step, and so are all the other switches whose labels divide the label on the &lt;math&gt;i&lt;/math&gt;-th switch. After step 1000 has been completed, how many switches will be in position &lt;math&gt;A&lt;/math&gt;?<br /> <br /> [[1999 AIME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Let &lt;math&gt;\mathcal{T}&lt;/math&gt; be the set of ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of nonnegative real numbers that lie in the plane &lt;math&gt;x+y+z=1.&lt;/math&gt; Let us say that &lt;math&gt;(x,y,z)&lt;/math&gt; supports &lt;math&gt;(a,b,c)&lt;/math&gt; when exactly two of the following are true: &lt;math&gt;x\ge a, y\ge b, z\ge c.&lt;/math&gt; Let &lt;math&gt;\mathcal{S}&lt;/math&gt; consist of those triples in &lt;math&gt;\mathcal{T}&lt;/math&gt; that support &lt;math&gt;\left(\frac 12,\frac 13,\frac 16\right).&lt;/math&gt; The area of &lt;math&gt;\mathcal{S}&lt;/math&gt; divided by the area of &lt;math&gt;\mathcal{T}&lt;/math&gt; 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 /> [[1999 AIME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> A function &lt;math&gt;f&lt;/math&gt; is defined on the complex numbers by &lt;math&gt;f(z)=(a+bi)z,&lt;/math&gt; where &lt;math&gt;a_{}&lt;/math&gt; and &lt;math&gt;b_{}&lt;/math&gt; are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that &lt;math&gt;|a+bi|=8&lt;/math&gt; and that &lt;math&gt;b^2=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 /> [[1999 AIME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Ten points in the plane are given, with no three collinear. Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points 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 /> [[1999 AIME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Given that &lt;math&gt;\sum_{k=1}^{35}\sin 5k=\tan \frac mn,&lt;/math&gt; where angles are measured in degrees, and &lt;math&gt;m_{}&lt;/math&gt; and &lt;math&gt;n_{}&lt;/math&gt; are relatively prime positive integers that satisfy &lt;math&gt;\frac mn&lt;90,&lt;/math&gt; find &lt;math&gt;m+n.&lt;/math&gt;<br /> <br /> [[1999 AIME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The inscribed circle of triangle &lt;math&gt;ABC&lt;/math&gt; is tangent to &lt;math&gt;\overline{AB}&lt;/math&gt; at &lt;math&gt;P_{},&lt;/math&gt; and its radius is 21. Given that &lt;math&gt;AP=23&lt;/math&gt; and &lt;math&gt;PB=27,&lt;/math&gt; find the perimeter of the triangle.<br /> <br /> [[1999 AIME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> Forty teams play a tournament in which every team plays every other(&lt;math&gt;39&lt;/math&gt; different opponents) team exactly once. No ties occur, and each team has a &lt;math&gt;50 \%&lt;/math&gt; chance of winning any game it plays. The probability that no two teams win the same number of games 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;\log_2 n.&lt;/math&gt;<br /> <br /> [[1999 AIME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Point &lt;math&gt;P_{}&lt;/math&gt; is located inside triangle &lt;math&gt;ABC&lt;/math&gt; so that angles &lt;math&gt;PAB, PBC,&lt;/math&gt; and &lt;math&gt;PCA&lt;/math&gt; are all congruent. The sides of the triangle have lengths &lt;math&gt;AB=13, BC=14,&lt;/math&gt; and &lt;math&gt;CA=15,&lt;/math&gt; and the tangent of angle &lt;math&gt;PAB&lt;/math&gt; 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 /> [[1999 AIME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> Consider the paper triangle whose vertices are &lt;math&gt;(0,0), (34,0),&lt;/math&gt; and &lt;math&gt;(16,24).&lt;/math&gt; The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?<br /> <br /> [[1999 AIME 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=1997_AIME_Problems/Problem_10&diff=85842 1997 AIME Problems/Problem 10 2017-05-24T22:30:33Z <p>First: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:<br /> <br /> i. Either each of the three cards has a different shape or all three of the card have the same shape.<br /> <br /> ii. Either each of the three cards has a different color or all three of the cards have the same color.<br /> <br /> iii. Either each of the three cards has a different shade or all three of the cards have the same shade.<br /> <br /> How many different complementary three-card sets are there?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> *'''Case 1''': All three attributes are the same. This is impossible since sets contain distinct cards.<br /> *'''Case 2''': Two of the three attributes are the same. There are &lt;math&gt;{3\choose 2}&lt;/math&gt; ways to pick the two attributes in question. Then there are &lt;math&gt;3&lt;/math&gt; ways to pick the value of the first attribute, &lt;math&gt;3&lt;/math&gt; ways to pick the value of the second attribute, and &lt;math&gt;1&lt;/math&gt; way to arrange the positions of the third attribute, giving us &lt;math&gt;{3\choose 2} \cdot 3 \cdot 3 = 27&lt;/math&gt; ways.<br /> *'''Case 3''': One of the three attributes are the same. There are &lt;math&gt;{3\choose 1}&lt;/math&gt; ways to pick the one attribute in question, and then &lt;math&gt;3&lt;/math&gt; ways to pick the value of that attribute. Then there are &lt;math&gt;3!&lt;/math&gt; ways to arrange the positions of the next two attributes, giving us &lt;math&gt;{3\choose 1} \cdot 3 \cdot 3! = 54&lt;/math&gt; ways.<br /> *'''Case 4''': None of the three attributes are the same. We fix the order of the first attribute, and then there are &lt;math&gt;3!&lt;/math&gt; ways to pick the ordering of the second attribute and &lt;math&gt;3!&lt;/math&gt; ways to pick the ordering of the third attribute. This gives us &lt;math&gt;(3!)^2 = 36&lt;/math&gt; ways.<br /> <br /> Adding the cases up, we get &lt;math&gt;27 + 54 + 36 = \boxed{117}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different there is only one option- choose the only value that is remaining. In this way, every two card pick corresponds to exactly one set, for a total of &lt;math&gt;\binom{27}{2} = 27*13 = 351&lt;/math&gt; possibilities. Note, however, that each set is generated by &lt;math&gt;{3\choose 2} = 3&lt;/math&gt; pairs, so we've overcounted by a multiple of 3 and the answer is &lt;math&gt;\frac{351}{3} = \boxed{117}&lt;/math&gt;.<br /> <br /> === Solution 3===<br /> Treat the sets as ordered. Then for each of the three criterion, there are &lt;math&gt;3!=6&lt;/math&gt; choices if the attribute is different and there are &lt;math&gt;3&lt;/math&gt; choices is the attribute is the same. Thus all three attributes combine to a total of &lt;math&gt;(6+3)^3=729&lt;/math&gt; possibilities. However if all three attributes are the same then the set must be composed of three cards that are the same, which is impossible. This takes out &lt;math&gt;3^3=27&lt;/math&gt; possibilities. Notice that we have counted every set &lt;math&gt;3!=6&lt;/math&gt; times by treating the set as ordered. the final solution is then &lt;math&gt;\frac{729-27}{6}=\boxed{117}&lt;/math&gt;<br /> <br /> {{AIME box|year=1997|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_10&diff=85841 1997 AIME Problems/Problem 10 2017-05-24T22:30:22Z <p>First: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:<br /> <br /> i. Either each of the three cards has a different shape or all three of the card have the same shape.<br /> <br /> ii. Either each of the three cards has a different color or all three of the cards have the same color.<br /> <br /> iii. Either each of the three cards has a different shade or all three of the cards have the same shade.<br /> <br /> How many different complementary three-card sets are there?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> *'''Case 1''': All three attributes are the same. This is impossible since sets contain distinct cards.<br /> *'''Case 2''': Two of the three attributes are the same. There are &lt;math&gt;{3\choose 2}&lt;/math&gt; ways to pick the two attributes in question. Then there are &lt;math&gt;3&lt;/math&gt; ways to pick the value of the first attribute, &lt;math&gt;3&lt;/math&gt; ways to pick the value of the second attribute, and &lt;math&gt;1&lt;/math&gt; way to arrange the positions of the third attribute, giving us &lt;math&gt;{3\choose 2} \cdot 3 \cdot 3 = 27&lt;/math&gt; ways.<br /> *'''Case 3''': One of the three attributes are the same. There are &lt;math&gt;{3\choose 1}&lt;/math&gt; ways to pick the one attribute in question, and then &lt;math&gt;3&lt;/math&gt; ways to pick the value of that attribute. Then there are &lt;math&gt;3!&lt;/math&gt; ways to arrange the positions of the next two attributes, giving us &lt;math&gt;{3\choose 1} \cdot 3 \cdot 3! = 54&lt;/math&gt; ways.<br /> *'''Case 4''': None of the three attributes are the same. We fix the order of the first attribute, and then there are &lt;math&gt;3!&lt;/math&gt; ways to pick the ordering of the second attribute and &lt;math&gt;3!&lt;/math&gt; ways to pick the ordering of the third attribute. This gives us &lt;math&gt;(3!)^2 = 36&lt;/math&gt; ways.<br /> <br /> Adding the cases up, we get &lt;math&gt;27 + 54 + 36 = \boxed{117}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different there is only one option- choose the only value that is remaining. In this way, every two card pick corresponds to exactly one set, for a total of &lt;math&gt;\binom{27}{2} = 27*13 = 351&lt;/math&gt; possibilities. Note, however, that each set is generated by &lt;math&gt;{3\choose 2} = 3&lt;/math&gt; pairs, so we've overcounted by a multiple of 3 and the answer is &lt;math&gt;\frac{351}{3} = \boxed{117}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Treat the sets as ordered. Then for each of the three criterion, there are &lt;math&gt;3!=6&lt;/math&gt; choices if the attribute is different and there are &lt;math&gt;3&lt;/math&gt; choices is the attribute is the same. Thus all three attributes combine to a total of &lt;math&gt;(6+3)^3=729&lt;/math&gt; possibilities. However if all three attributes are the same then the set must be composed of three cards that are the same, which is impossible. This takes out &lt;math&gt;3^3=27&lt;/math&gt; possibilities. Notice that we have counted every set &lt;math&gt;3!=6&lt;/math&gt; times by treating the set as ordered. the final solution is then &lt;math&gt;\frac{729-27}{6}=\boxed{117}&lt;/math&gt;<br /> <br /> {{AIME box|year=1997|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> First https://artofproblemsolving.com/wiki/index.php?title=De_Moivre%27s_Theorem&diff=85047 De Moivre's Theorem 2017-03-28T02:28:16Z <p>First: /* Proof */</p> <hr /> <div>'''DeMoivre's Theorem''' is a very useful theorem in the mathematical fields of [[complex numbers]]. It allows complex numbers in [[polar form]] to be easily raised to certain powers. It states that for &lt;math&gt;x\in\mathbb{R}&lt;/math&gt; and &lt;math&gt;n\in\mathbb{Z}&lt;/math&gt;, &lt;math&gt;\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)&lt;/math&gt;.<br /> <br /> == Proof ==<br /> This is one proof of De Moivre's theorem by [[induction]].<br /> <br /> *If &lt;math&gt;n&gt;0&lt;/math&gt;, for &lt;math&gt;n=1&lt;/math&gt;, the case is obviously true.<br /> <br /> :Assume true for the case &lt;math&gt;n=k&lt;/math&gt;. Now, the case of &lt;math&gt;n=k+1&lt;/math&gt;:<br /> <br /> :[[Image:DeMoivreInductionP1.gif]]<br /> <br /> :Therefore, the result is true for all positive integers &lt;math&gt;n&lt;/math&gt;.<br /> <br /> *If &lt;math&gt;n=0&lt;/math&gt;, the formula holds true because &lt;math&gt;\cos(0x)+i\sin (0x)=1+i0=1&lt;/math&gt;. Since &lt;math&gt;z^0=1&lt;/math&gt;, the equation holds true.<br /> <br /> *If &lt;math&gt;n&lt;0&lt;/math&gt;, one must consider &lt;math&gt;n=-m&lt;/math&gt; when &lt;math&gt;m&lt;/math&gt; is a positive integer.<br /> <br /> :[[Image:DeMoivreInductionP2.gif]]<br /> <br /> And thus, the formula proves true for all integral values of &lt;math&gt;n&lt;/math&gt;. &lt;math&gt;\Box&lt;/math&gt;<br /> <br /> Note that from the functional equation &lt;math&gt;f(x)^n = f(nx)&lt;/math&gt; where &lt;math&gt;f(x) = \cos x + i\sin x&lt;/math&gt;, we see that &lt;math&gt;f(x)&lt;/math&gt; behaves like an exponential function. Indeed, [[Euler's formula]] states that &lt;math&gt;e^{ix} = \cos x+i\sin x&lt;/math&gt;. This extends De Moivre's theorem to all &lt;math&gt;n\in \mathbb{R}&lt;/math&gt;.<br /> <br /> ==Generalization==<br /> <br /> <br /> [[Category:Theorems]]<br /> [[Category:Complex numbers]]</div> First