https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=I-+-i&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-20T15:20:55Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_17&diff=161263 2003 AMC 12A Problems/Problem 17 2021-08-31T16:04:15Z <p>I- -i: </p> <hr /> <div>== Problem ==<br /> Square &lt;math&gt;ABCD&lt;/math&gt; has sides of length &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{CD}&lt;/math&gt;. A circle with radius &lt;math&gt;2&lt;/math&gt; and center &lt;math&gt;M&lt;/math&gt; intersects a circle with radius &lt;math&gt;4&lt;/math&gt; and center &lt;math&gt;A&lt;/math&gt; at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. What is the distance from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;\overline{AD}&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,M,P;<br /> D=(0,0);<br /> C=(10,0);<br /> B=(10,10);<br /> A=(0,10);<br /> M=(5,0);<br /> P=(8,4);<br /> dot(M);<br /> dot(P);<br /> draw(A--B--C--D--cycle,linewidth(0.7));<br /> draw((5,5)..D--C..cycle,linewidth(0.7));<br /> draw((7.07,2.93)..B--A--D..cycle,linewidth(0.7));<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$M$&quot;,M,S);<br /> label(&quot;$P$&quot;,P,N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}&lt;/math&gt;<br /> <br /> == Solutions ==<br /> === Solution 1===<br /> <br /> Let &lt;math&gt;D&lt;/math&gt; be the origin. &lt;math&gt;A&lt;/math&gt; is the point &lt;math&gt;(0,4)&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; is the point &lt;math&gt;(2,0)&lt;/math&gt;. We are given the radius of the quarter circle and semicircle as &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;, respectively, so their equations, respectively, are:<br /> <br /> &lt;math&gt;x^2 + (y-4)^2 = 4^2&lt;/math&gt;<br /> <br /> &lt;math&gt;(x-2)^2 + y^2 = 2^2&lt;/math&gt;<br /> <br /> Subtract the second equation from the first:<br /> <br /> &lt;math&gt;x^2 + (y - 4)^2 - (x - 2)^2 - y^2 = 12&lt;/math&gt;<br /> <br /> &lt;math&gt;4x - 8y + 12 = 12&lt;/math&gt;<br /> <br /> &lt;math&gt;x = 2y.&lt;/math&gt;<br /> <br /> Then substitute:<br /> <br /> &lt;math&gt;(2y)^2 + (y - 4)^2 = 16&lt;/math&gt;<br /> <br /> &lt;math&gt;4y^2 + y^2 - 8y + 16 = 16&lt;/math&gt;<br /> <br /> &lt;math&gt;5y^2 - 8y = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;y(5y - 8) = 0.&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;y = 0&lt;/math&gt; and &lt;math&gt;y = \frac{8}{5}&lt;/math&gt; making &lt;math&gt;x = 0&lt;/math&gt; and &lt;math&gt;x = \frac{16}{5}&lt;/math&gt;.<br /> <br /> The first value of &lt;math&gt;0&lt;/math&gt; is obviously referring to the x-coordinate of the point where the circles intersect at the origin, &lt;math&gt;D&lt;/math&gt;, so the second value must be referring to the x coordinate of &lt;math&gt;P&lt;/math&gt;. Since &lt;math&gt;\overline{AD}&lt;/math&gt; is the y-axis, the distance to it from &lt;math&gt;P&lt;/math&gt; is the same as the x-value of the coordinate of &lt;math&gt;P&lt;/math&gt;, so the distance from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;\overline{AD}&lt;/math&gt; is &lt;math&gt;\frac{16}{5} \Rightarrow \boxed{B}.&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> &lt;math&gt;APMD&lt;/math&gt; obviously forms a kite. Let the intersection of the diagonals be &lt;math&gt;E&lt;/math&gt;. &lt;math&gt;AE+EM=AM=2\sqrt{5}&lt;/math&gt; Let &lt;math&gt;AE=x&lt;/math&gt;. Then, &lt;math&gt;EM=2\sqrt{5}-x&lt;/math&gt;.<br /> <br /> <br /> By Pythagorean Theorem, &lt;math&gt;DE^2=4^2-AE^2=2^2-EM^2&lt;/math&gt;. Thus, &lt;math&gt;16-x^2=4-(2\sqrt{5}-x)^2&lt;/math&gt;. Simplifying, &lt;math&gt;x=\frac{8}{\sqrt{5}}&lt;/math&gt;. By Pythagoras again, &lt;math&gt;DE=\frac{4}{\sqrt{5}}&lt;/math&gt;. Then, the area of &lt;math&gt;ADP&lt;/math&gt; is &lt;math&gt;DE\cdot AE=\frac{32}{5}&lt;/math&gt;. <br /> <br /> <br /> Using &lt;math&gt;4&lt;/math&gt; instead as the base, we can drop a altitude from P. &lt;math&gt;\frac{32}{5}=\frac{bh}{2}&lt;/math&gt;. &lt;math&gt;\frac{32}{5}=\frac{4h}{2}&lt;/math&gt;. Thus, the horizontal distance is &lt;math&gt;\frac{16}{5} \implies \boxed{\textbf{(B)}\frac{16}{5}}&lt;/math&gt;<br /> <br /> <br /> <br /> ~BJHHar<br /> <br /> === Solution 3 ===<br /> <br /> Note that &lt;math&gt;P&lt;/math&gt; is merely a reflection of &lt;math&gt;D&lt;/math&gt; over &lt;math&gt;AM&lt;/math&gt;. Call the intersection of &lt;math&gt;AM&lt;/math&gt; and &lt;math&gt;DP&lt;/math&gt; &lt;math&gt;X&lt;/math&gt;. Drop perpendiculars from &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt;, and denote their respective points of intersection by &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;K&lt;/math&gt;. We then have &lt;math&gt;\triangle DXJ\sim\triangle DPK&lt;/math&gt;, with a scale factor of 2. Thus, we can find &lt;math&gt;XJ&lt;/math&gt; and double it to get our answer. With some analytical geometry, we find that &lt;math&gt;XJ=\frac{8}{5}&lt;/math&gt;, implying that &lt;math&gt;PK=\frac{16}{5}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> As in Solution 2, draw in &lt;math&gt;DP&lt;/math&gt; and &lt;math&gt;AM&lt;/math&gt; and denote their intersection point &lt;math&gt;X&lt;/math&gt;. Next, drop a perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; and denote the foot as &lt;math&gt;Z&lt;/math&gt;. &lt;math&gt;AP \cong AD&lt;/math&gt; as they are both radii and similarly &lt;math&gt;DM \cong MP&lt;/math&gt; so &lt;math&gt;APMD&lt;/math&gt; is a kite and &lt;math&gt;DX \perp XM&lt;/math&gt; by a well-known theorem. <br /> <br /> Pythagorean theorem gives us &lt;math&gt;AM=2 \sqrt{5}&lt;/math&gt;. Clearly &lt;math&gt;\triangle XMD \sim \triangle XDA \sim \triangle DMA \sim \triangle ZDP&lt;/math&gt; by angle-angle and &lt;math&gt;\triangle XMD \cong \triangle XMP&lt;/math&gt; by Hypotenuse Leg.<br /> Manipulating similar triangles gives us &lt;math&gt;PZ=\frac{16}{5}&lt;/math&gt;<br /> <br /> === Solution 5 ===<br /> Using the double-angle formula for sine, what we need to find is &lt;math&gt;AP\cdot \sin(DAP) = AP\cdot 2\sin( DAM) \cos(DAM) = 4\cdot 2\cdot \frac{2}{\sqrt{20}}\cdot\frac{4}{\sqrt{20}} = \frac{16}{5}&lt;/math&gt;.<br /> ==Solution 6(LoC)==<br /> We use the Law of Cosines:<br /> <br /> &lt;math&gt;32-32 \cos \theta = 8 + 8 \cos \theta &lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{3}{5} = \cos \theta &lt;/math&gt;<br /> <br /> &lt;math&gt;2 + 2*\frac{3}{5} = \frac{16}{5}&lt;/math&gt;<br /> <br /> === Solution 7 ===<br /> <br /> Let &lt;math&gt;H&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;CD&lt;/math&gt;, and let &lt;math&gt;HD = x&lt;/math&gt;. Then we have &lt;math&gt;HC = 4-x&lt;/math&gt;, and &lt;math&gt;PH = 4 - \sqrt{16 - x^2}&lt;/math&gt;. Since &lt;math&gt;\triangle DHP \sim \triangle PHC&lt;/math&gt;, we have &lt;math&gt;HP^2 = DH \cdot HC&lt;/math&gt;, or &lt;math&gt;-x^2 + 4x = 16 - 8\sqrt{16-x^2}&lt;/math&gt;. Solving gives &lt;math&gt;x = \frac{16}{5}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2003|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> I- -i https://artofproblemsolving.com/wiki/index.php?title=File:Awesome_Face.png&diff=150328 File:Awesome Face.png 2021-03-26T02:30:57Z <p>I- -i: </p> <hr /> <div></div> I- -i https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_14&diff=148962 2020 AIME I Problems/Problem 14 2021-03-09T03:03:31Z <p>I- -i: /* Solution 6 */</p> <hr /> <div><br /> == Problem ==<br /> Let &lt;math&gt;P(x)&lt;/math&gt; be a quadratic polynomial with complex coefficients whose &lt;math&gt;x^2&lt;/math&gt; coefficient is &lt;math&gt;1.&lt;/math&gt; Suppose the equation &lt;math&gt;P(P(x))=0&lt;/math&gt; has four distinct solutions, &lt;math&gt;x=3,4,a,b.&lt;/math&gt; Find the sum of all possible values of &lt;math&gt;(a+b)^2.&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Either &lt;math&gt;P(3) = P(4)&lt;/math&gt; or not. We first see that if &lt;math&gt;P(3) = P(4)&lt;/math&gt; it's easy to obtain by Vieta's that &lt;math&gt;(a+b)^2 = 49&lt;/math&gt;. Now, take &lt;math&gt;P(3) \neq P(4)&lt;/math&gt; and WLOG &lt;math&gt;P(3) = P(a), P(4) = P(b)&lt;/math&gt;. Now, consider the parabola formed by the graph of &lt;math&gt;P&lt;/math&gt;. It has vertex &lt;math&gt;\frac{3+a}{2}&lt;/math&gt;. Now, say that &lt;math&gt;P(x) = x^2 - (3+a)x + c&lt;/math&gt;. We note &lt;math&gt;P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}&lt;/math&gt;. Now, we note &lt;math&gt;P(4) = \frac{7}{2}&lt;/math&gt; by plugging in again. Now, it's easy to find that &lt;math&gt;a = -2.5, b = -3.5&lt;/math&gt;, yielding a value of &lt;math&gt;36&lt;/math&gt;. Finally, we add &lt;math&gt;49 + 36 = \boxed{085}&lt;/math&gt;. ~awang11, charmander3333<br /> <br /> &lt;b&gt;Remark&lt;/b&gt;: We know that &lt;math&gt;c=\frac{8a-1}{2}&lt;/math&gt; from &lt;math&gt;P(3)+P(4)=3+a&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let the roots of &lt;math&gt;P(x)&lt;/math&gt; be &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then we can write &lt;math&gt;P(x)=x^2-(m+n)x+mn&lt;/math&gt;. The fact that &lt;math&gt;P(P(x))=0&lt;/math&gt; has solutions &lt;math&gt;x=3,4,a,b&lt;/math&gt; implies that some combination of &lt;math&gt;2&lt;/math&gt; of these are the solution to &lt;math&gt;P(x)=m&lt;/math&gt;, and the other &lt;math&gt;2&lt;/math&gt; are the solution to &lt;math&gt;P(x)=n&lt;/math&gt;. It's fairly easy to see there are only &lt;math&gt;2&lt;/math&gt; possible such groupings: &lt;math&gt;P(3)=P(4)=m&lt;/math&gt; and &lt;math&gt;P(a)=P(b)=n&lt;/math&gt;, or &lt;math&gt;P(3)=P(a)=m&lt;/math&gt; and &lt;math&gt;P(4)=P(b)=n&lt;/math&gt; (Note that &lt;math&gt;a,b&lt;/math&gt; are interchangeable, and so are &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;). We now casework: <br /> If &lt;math&gt;P(3)=P(4)=m&lt;/math&gt;, then <br /> &lt;cmath&gt;9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7&lt;/cmath&gt;<br /> so this gives &lt;math&gt;(a+b)^2=7^2=49&lt;/math&gt;. <br /> Next, if &lt;math&gt;P(3)=P(a)=m&lt;/math&gt;, then <br /> &lt;cmath&gt;9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n&lt;/cmath&gt;<br /> &lt;cmath&gt;16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n&lt;/cmath&gt;<br /> Subtracting the first part of the first equation from the first part of the second equation gives <br /> &lt;cmath&gt;7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3&lt;/cmath&gt;<br /> Hence, &lt;math&gt;a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6&lt;/math&gt;, and so &lt;math&gt;(a+b)^2=(-6)^2=36&lt;/math&gt;. <br /> Therefore, the solution is &lt;math&gt;49+36=\boxed{085}&lt;/math&gt; ~ktong<br /> <br /> == Solution 3 ==<br /> Write &lt;math&gt;P(x) = x^2+wx+z&lt;/math&gt;. Split the problem into two cases: &lt;math&gt;P(3)\ne P(4)&lt;/math&gt; and &lt;math&gt;P(3) = P(4)&lt;/math&gt;.<br /> <br /> Case 1: We have &lt;math&gt;P(3) \ne P(4)&lt;/math&gt;. We must have <br /> &lt;cmath&gt;w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.&lt;/cmath&gt;<br /> Rearrange and divide through by &lt;math&gt;8&lt;/math&gt; to obtain<br /> &lt;cmath&gt;w = \frac{-25-2z}{8}.&lt;/cmath&gt;<br /> Now, note that<br /> &lt;cmath&gt;z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =&lt;/cmath&gt;<br /> &lt;cmath&gt;\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.&lt;/cmath&gt;<br /> Now, rearrange to get<br /> &lt;cmath&gt;\frac{z}{8} = -\frac{21}{16}&lt;/cmath&gt;<br /> and thus<br /> &lt;cmath&gt;z = -\frac{21}{2}.&lt;/cmath&gt;<br /> Substituting this into our equation for &lt;math&gt;w&lt;/math&gt; yields &lt;math&gt;w = -\frac{1}{2}&lt;/math&gt;. Then, it is clear that &lt;math&gt;P&lt;/math&gt; does not have a double root at &lt;math&gt;P(3)&lt;/math&gt;, so we must have &lt;math&gt;P(a) = P(3)&lt;/math&gt; and &lt;math&gt;P(b) = P(4)&lt;/math&gt; or vice versa. This gives &lt;math&gt;3+a = \frac{1}{2}&lt;/math&gt; and &lt;math&gt;4+b = \frac{1}{2}&lt;/math&gt; or vice versa, implying that &lt;math&gt;a+b = 1-3-4 = -6&lt;/math&gt; and &lt;math&gt;(a+b)^2 = 6&lt;/math&gt;.<br /> <br /> Case 2: We have &lt;math&gt;P(3) = P(4)&lt;/math&gt;. Then, we must have &lt;math&gt;w = -7&lt;/math&gt;. It is clear that &lt;math&gt;P(a) = P(b)&lt;/math&gt; (we would otherwise get &lt;math&gt;P(a)=P(3)=P(4)&lt;/math&gt; implying &lt;math&gt;a \in \{3,4\}&lt;/math&gt; or vice versa), so &lt;math&gt;a+b=-w=7&lt;/math&gt; and &lt;math&gt;(a+b)^2 = 49&lt;/math&gt;.<br /> <br /> Thus, our final answer is &lt;math&gt;49+36=\boxed{085}&lt;/math&gt;. ~GeronimoStilton<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;P(x)=(x-r)(x-s)&lt;/math&gt;. There are two cases: in the first case, &lt;math&gt;(3-r)(3-s)=(4-r)(4-s)&lt;/math&gt; equals &lt;math&gt;r&lt;/math&gt; (without loss of generality), and thus &lt;math&gt;(a-r)(a-s)=(b-r)(b-s)=s&lt;/math&gt;. By Vieta's formulas &lt;math&gt;a+b=r+s=3+4=7&lt;/math&gt;.<br /> <br /> In the second case, say without loss of generality &lt;math&gt;(3-r)(3-s)=r&lt;/math&gt; and &lt;math&gt;(4-r)(4-s)=s&lt;/math&gt;. Subtracting gives &lt;math&gt;-7+r+s=r-s&lt;/math&gt;, so &lt;math&gt;s=7/2&lt;/math&gt;. From this, we have &lt;math&gt;r=-3&lt;/math&gt;.<br /> <br /> Note &lt;math&gt;r+s=1/2&lt;/math&gt;, so by Vieta's, we have &lt;math&gt;\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}&lt;/math&gt;. In this case, &lt;math&gt;a+b=-6&lt;/math&gt;.<br /> <br /> The requested sum is &lt;math&gt;36+49=85&lt;/math&gt;.~TheUltimate123<br /> <br /> ==Solution 5 (Official MAA)==<br /> Note that because &lt;math&gt;P\big(P(3)\big)=P\big(P(4)\big)= 0&lt;/math&gt;, &lt;math&gt;P(3)&lt;/math&gt; and &lt;math&gt;P(4)&lt;/math&gt; are roots of &lt;math&gt;P(x)&lt;/math&gt;. There are two cases.<br /> CASE 1: &lt;math&gt;P(3) = P(4)&lt;/math&gt;. Then &lt;math&gt;P(x)&lt;/math&gt; is symmetric about &lt;math&gt;x=\tfrac72&lt;/math&gt;; that is to say, &lt;math&gt;P(r) = P(7-r)&lt;/math&gt; for all &lt;math&gt;r&lt;/math&gt;. Thus the remaining two roots must sum to &lt;math&gt;7&lt;/math&gt;. Indeed, the polynomials &lt;math&gt;P(x) = \left(x-\frac72\right)^2 + \frac{11}4 \pm i\sqrt3&lt;/math&gt; satisfy the conditions.<br /> CASE 2: &lt;math&gt;P(3)\neq P(4)&lt;/math&gt;. Then &lt;math&gt;P(3)&lt;/math&gt; and &lt;math&gt;P(4)&lt;/math&gt; are the two distinct roots of &lt;math&gt;P(x)&lt;/math&gt;, so&lt;cmath&gt;P(x) = \big(x-P(3)\big)\big(x-P(4)\big)&lt;/cmath&gt;for all &lt;math&gt;x&lt;/math&gt;. Note that any solution to &lt;math&gt;P\big(P(x)\big) = 0&lt;/math&gt; must satisfy either &lt;math&gt;P(x) = P(3)&lt;/math&gt; or &lt;math&gt;P(x) = P(4)&lt;/math&gt;. Because &lt;math&gt;P(x)&lt;/math&gt; is quadratic, the polynomials &lt;math&gt;P(x) - P(3)&lt;/math&gt; and &lt;math&gt;P(x) - P(4)&lt;/math&gt; each have the same sum of roots as the polynomial &lt;math&gt;P(x)&lt;/math&gt;, which is &lt;math&gt;P(3) + P(4)&lt;/math&gt;. Thus the answer in this case is &lt;math&gt;2\big(P(3) + P(4)\big)-7&lt;/math&gt;, and so it suffices to compute the value of &lt;math&gt;P(3)+P(4)&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P(3)=u&lt;/math&gt; and &lt;math&gt;P(4) = v&lt;/math&gt;. Substituting &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;x=4&lt;/math&gt; into the above quadratic polynomial yields the system of equations<br /> &lt;cmath&gt;\begin{align*}<br /> u &amp;= (3-u)(3-v) = 9 - 3u - 3v + uv\\<br /> v &amp;= (4-u)(4-v) = 16 - 4u - 4v + uv.<br /> \end{align*}&lt;/cmath&gt;Subtracting the first equation from the second gives &lt;math&gt;v - u = 7 - u - v&lt;/math&gt;, yielding &lt;math&gt;v = \frac72.&lt;/math&gt; Substituting this value into the second equation gives&lt;cmath&gt;\dfrac72 = \left(4 - u\right)\left(4 - \dfrac72\right),&lt;/cmath&gt;yielding &lt;math&gt;u = -3.&lt;/math&gt; The sum of the two solutions is &lt;math&gt;2\left(\tfrac72-3\right)-7 = -6&lt;/math&gt;. In this case, &lt;math&gt;P(x)= (x+3)\left(x-\frac72\right)&lt;/math&gt;.<br /> <br /> The requested sum of squares is &lt;math&gt;7^2+(-6)^2 = {85}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> <br /> Let &lt;math&gt;P(x) = (x-c)^2 - d&lt;/math&gt; for some &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Then, we can write &lt;math&gt;P(P(x)) = ((x-c)^2 - d - c)^2 - d&lt;/math&gt;. Setting the expression equal to &lt;math&gt;0&lt;/math&gt; and solving for &lt;math&gt;x&lt;/math&gt; gives:<br /> <br /> &lt;math&gt;x = \pm \sqrt{ \pm \sqrt{d} + d + c} + c&lt;/math&gt;<br /> <br /> Therefore, we have that &lt;math&gt;x&lt;/math&gt; takes on the four values &lt;math&gt;\sqrt{\sqrt{d} + d + c} + c&lt;/math&gt;, &lt;math&gt;-\sqrt{\sqrt{d} + d + c} + c&lt;/math&gt;, &lt;math&gt;\sqrt{-\sqrt{d} + d + c} + c&lt;/math&gt;, and &lt;math&gt;-\sqrt{-\sqrt{d} + d + c} + c&lt;/math&gt;. Two of these values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, and the other two are &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;.<br /> <br /> We can split these four values into two &quot;groups&quot; based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values.<br /> <br /> &lt;math&gt;\textbf{Case 1}&lt;/math&gt;: Both the 3 and 4 values are from the same group.<br /> <br /> In this case, the &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; values are both from the other group. The sum of this is just &lt;math&gt;2c&lt;/math&gt; because the radical cancels out. Because of this, we can see that &lt;math&gt;c&lt;/math&gt; is just the average of &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, so we have &lt;math&gt;2c = 3 + 4 = 7&lt;/math&gt;, so &lt;math&gt;(a+b)^2 = 7^2 = 49&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2}&lt;/math&gt;: The 3 and 4 values come from different groups.<br /> <br /> It is easy to see that all possibilities in this case are basically symmetric and yield the same value for &lt;math&gt;(a+b)^2&lt;/math&gt;. Without loss of generality, assume that &lt;math&gt;\sqrt{\sqrt{d} + d + c} + c = 4&lt;/math&gt; and &lt;math&gt;\sqrt{-\sqrt{d} + d + c} + c = 3&lt;/math&gt;. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger.<br /> <br /> We can write &lt;math&gt;\sqrt{\sqrt{d} + d + c} + c = 1 + \sqrt{-\sqrt{d} + d + c} + c&lt;/math&gt;.<br /> <br /> Moving most terms to the left side and simplifying gives &lt;math&gt;\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} = 1&lt;/math&gt;.<br /> <br /> We can square both sides and simplify:<br /> <br /> &lt;math&gt;\sqrt{d} + d + c - \sqrt{d} + d + c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;2d + 2c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = (d+c) - \frac{1}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sqrt{(d+c)^2 - (\sqrt{d})^2} = (d+c) - \frac{1}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sqrt{d^2 + 2dc + c^2 - d} = (d+c) - \frac{1}{2}&lt;/math&gt;<br /> <br /> Squaring both sides again gives the following:<br /> <br /> &lt;math&gt;d^2 + 2dc + c^2 - d = d^2 + 2dc + c^2 - d - c + \frac{1}{4}&lt;/math&gt;<br /> <br /> Nearly all terms cancel out, yielding &lt;math&gt;c = \frac{1}{4}&lt;/math&gt;.<br /> <br /> By substituting this back in, we obtain &lt;math&gt;\sqrt{\sqrt{d} + d + c} = \frac{15}{4}&lt;/math&gt; and &lt;math&gt;\sqrt{-\sqrt{d} + d + c} = \frac{11}{4}&lt;/math&gt;.<br /> <br /> The sum of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; is equal to &lt;math&gt;-\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} + 2c = -\frac{15}{4} - \frac{11}{4} + \frac{1}{2} = -6&lt;/math&gt;, so &lt;math&gt;(a+b)^2 = 36&lt;/math&gt;.<br /> <br /> Adding up both values gives &lt;math&gt;49 + 36 = \boxed{085}&lt;/math&gt; as our final answer.<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> I- -i https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_14&diff=148961 2020 AIME I Problems/Problem 14 2021-03-09T02:56:49Z <p>I- -i: /* Solution 5 (Official MAA) */</p> <hr /> <div><br /> == Problem ==<br /> Let &lt;math&gt;P(x)&lt;/math&gt; be a quadratic polynomial with complex coefficients whose &lt;math&gt;x^2&lt;/math&gt; coefficient is &lt;math&gt;1.&lt;/math&gt; Suppose the equation &lt;math&gt;P(P(x))=0&lt;/math&gt; has four distinct solutions, &lt;math&gt;x=3,4,a,b.&lt;/math&gt; Find the sum of all possible values of &lt;math&gt;(a+b)^2.&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Either &lt;math&gt;P(3) = P(4)&lt;/math&gt; or not. We first see that if &lt;math&gt;P(3) = P(4)&lt;/math&gt; it's easy to obtain by Vieta's that &lt;math&gt;(a+b)^2 = 49&lt;/math&gt;. Now, take &lt;math&gt;P(3) \neq P(4)&lt;/math&gt; and WLOG &lt;math&gt;P(3) = P(a), P(4) = P(b)&lt;/math&gt;. Now, consider the parabola formed by the graph of &lt;math&gt;P&lt;/math&gt;. It has vertex &lt;math&gt;\frac{3+a}{2}&lt;/math&gt;. Now, say that &lt;math&gt;P(x) = x^2 - (3+a)x + c&lt;/math&gt;. We note &lt;math&gt;P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}&lt;/math&gt;. Now, we note &lt;math&gt;P(4) = \frac{7}{2}&lt;/math&gt; by plugging in again. Now, it's easy to find that &lt;math&gt;a = -2.5, b = -3.5&lt;/math&gt;, yielding a value of &lt;math&gt;36&lt;/math&gt;. Finally, we add &lt;math&gt;49 + 36 = \boxed{085}&lt;/math&gt;. ~awang11, charmander3333<br /> <br /> &lt;b&gt;Remark&lt;/b&gt;: We know that &lt;math&gt;c=\frac{8a-1}{2}&lt;/math&gt; from &lt;math&gt;P(3)+P(4)=3+a&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let the roots of &lt;math&gt;P(x)&lt;/math&gt; be &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then we can write &lt;math&gt;P(x)=x^2-(m+n)x+mn&lt;/math&gt;. The fact that &lt;math&gt;P(P(x))=0&lt;/math&gt; has solutions &lt;math&gt;x=3,4,a,b&lt;/math&gt; implies that some combination of &lt;math&gt;2&lt;/math&gt; of these are the solution to &lt;math&gt;P(x)=m&lt;/math&gt;, and the other &lt;math&gt;2&lt;/math&gt; are the solution to &lt;math&gt;P(x)=n&lt;/math&gt;. It's fairly easy to see there are only &lt;math&gt;2&lt;/math&gt; possible such groupings: &lt;math&gt;P(3)=P(4)=m&lt;/math&gt; and &lt;math&gt;P(a)=P(b)=n&lt;/math&gt;, or &lt;math&gt;P(3)=P(a)=m&lt;/math&gt; and &lt;math&gt;P(4)=P(b)=n&lt;/math&gt; (Note that &lt;math&gt;a,b&lt;/math&gt; are interchangeable, and so are &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;). We now casework: <br /> If &lt;math&gt;P(3)=P(4)=m&lt;/math&gt;, then <br /> &lt;cmath&gt;9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7&lt;/cmath&gt;<br /> so this gives &lt;math&gt;(a+b)^2=7^2=49&lt;/math&gt;. <br /> Next, if &lt;math&gt;P(3)=P(a)=m&lt;/math&gt;, then <br /> &lt;cmath&gt;9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n&lt;/cmath&gt;<br /> &lt;cmath&gt;16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n&lt;/cmath&gt;<br /> Subtracting the first part of the first equation from the first part of the second equation gives <br /> &lt;cmath&gt;7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3&lt;/cmath&gt;<br /> Hence, &lt;math&gt;a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6&lt;/math&gt;, and so &lt;math&gt;(a+b)^2=(-6)^2=36&lt;/math&gt;. <br /> Therefore, the solution is &lt;math&gt;49+36=\boxed{085}&lt;/math&gt; ~ktong<br /> <br /> == Solution 3 ==<br /> Write &lt;math&gt;P(x) = x^2+wx+z&lt;/math&gt;. Split the problem into two cases: &lt;math&gt;P(3)\ne P(4)&lt;/math&gt; and &lt;math&gt;P(3) = P(4)&lt;/math&gt;.<br /> <br /> Case 1: We have &lt;math&gt;P(3) \ne P(4)&lt;/math&gt;. We must have <br /> &lt;cmath&gt;w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.&lt;/cmath&gt;<br /> Rearrange and divide through by &lt;math&gt;8&lt;/math&gt; to obtain<br /> &lt;cmath&gt;w = \frac{-25-2z}{8}.&lt;/cmath&gt;<br /> Now, note that<br /> &lt;cmath&gt;z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =&lt;/cmath&gt;<br /> &lt;cmath&gt;\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.&lt;/cmath&gt;<br /> Now, rearrange to get<br /> &lt;cmath&gt;\frac{z}{8} = -\frac{21}{16}&lt;/cmath&gt;<br /> and thus<br /> &lt;cmath&gt;z = -\frac{21}{2}.&lt;/cmath&gt;<br /> Substituting this into our equation for &lt;math&gt;w&lt;/math&gt; yields &lt;math&gt;w = -\frac{1}{2}&lt;/math&gt;. Then, it is clear that &lt;math&gt;P&lt;/math&gt; does not have a double root at &lt;math&gt;P(3)&lt;/math&gt;, so we must have &lt;math&gt;P(a) = P(3)&lt;/math&gt; and &lt;math&gt;P(b) = P(4)&lt;/math&gt; or vice versa. This gives &lt;math&gt;3+a = \frac{1}{2}&lt;/math&gt; and &lt;math&gt;4+b = \frac{1}{2}&lt;/math&gt; or vice versa, implying that &lt;math&gt;a+b = 1-3-4 = -6&lt;/math&gt; and &lt;math&gt;(a+b)^2 = 6&lt;/math&gt;.<br /> <br /> Case 2: We have &lt;math&gt;P(3) = P(4)&lt;/math&gt;. Then, we must have &lt;math&gt;w = -7&lt;/math&gt;. It is clear that &lt;math&gt;P(a) = P(b)&lt;/math&gt; (we would otherwise get &lt;math&gt;P(a)=P(3)=P(4)&lt;/math&gt; implying &lt;math&gt;a \in \{3,4\}&lt;/math&gt; or vice versa), so &lt;math&gt;a+b=-w=7&lt;/math&gt; and &lt;math&gt;(a+b)^2 = 49&lt;/math&gt;.<br /> <br /> Thus, our final answer is &lt;math&gt;49+36=\boxed{085}&lt;/math&gt;. ~GeronimoStilton<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;P(x)=(x-r)(x-s)&lt;/math&gt;. There are two cases: in the first case, &lt;math&gt;(3-r)(3-s)=(4-r)(4-s)&lt;/math&gt; equals &lt;math&gt;r&lt;/math&gt; (without loss of generality), and thus &lt;math&gt;(a-r)(a-s)=(b-r)(b-s)=s&lt;/math&gt;. By Vieta's formulas &lt;math&gt;a+b=r+s=3+4=7&lt;/math&gt;.<br /> <br /> In the second case, say without loss of generality &lt;math&gt;(3-r)(3-s)=r&lt;/math&gt; and &lt;math&gt;(4-r)(4-s)=s&lt;/math&gt;. Subtracting gives &lt;math&gt;-7+r+s=r-s&lt;/math&gt;, so &lt;math&gt;s=7/2&lt;/math&gt;. From this, we have &lt;math&gt;r=-3&lt;/math&gt;.<br /> <br /> Note &lt;math&gt;r+s=1/2&lt;/math&gt;, so by Vieta's, we have &lt;math&gt;\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}&lt;/math&gt;. In this case, &lt;math&gt;a+b=-6&lt;/math&gt;.<br /> <br /> The requested sum is &lt;math&gt;36+49=85&lt;/math&gt;.~TheUltimate123<br /> <br /> ==Solution 5 (Official MAA)==<br /> Note that because &lt;math&gt;P\big(P(3)\big)=P\big(P(4)\big)= 0&lt;/math&gt;, &lt;math&gt;P(3)&lt;/math&gt; and &lt;math&gt;P(4)&lt;/math&gt; are roots of &lt;math&gt;P(x)&lt;/math&gt;. There are two cases.<br /> CASE 1: &lt;math&gt;P(3) = P(4)&lt;/math&gt;. Then &lt;math&gt;P(x)&lt;/math&gt; is symmetric about &lt;math&gt;x=\tfrac72&lt;/math&gt;; that is to say, &lt;math&gt;P(r) = P(7-r)&lt;/math&gt; for all &lt;math&gt;r&lt;/math&gt;. Thus the remaining two roots must sum to &lt;math&gt;7&lt;/math&gt;. Indeed, the polynomials &lt;math&gt;P(x) = \left(x-\frac72\right)^2 + \frac{11}4 \pm i\sqrt3&lt;/math&gt; satisfy the conditions.<br /> CASE 2: &lt;math&gt;P(3)\neq P(4)&lt;/math&gt;. Then &lt;math&gt;P(3)&lt;/math&gt; and &lt;math&gt;P(4)&lt;/math&gt; are the two distinct roots of &lt;math&gt;P(x)&lt;/math&gt;, so&lt;cmath&gt;P(x) = \big(x-P(3)\big)\big(x-P(4)\big)&lt;/cmath&gt;for all &lt;math&gt;x&lt;/math&gt;. Note that any solution to &lt;math&gt;P\big(P(x)\big) = 0&lt;/math&gt; must satisfy either &lt;math&gt;P(x) = P(3)&lt;/math&gt; or &lt;math&gt;P(x) = P(4)&lt;/math&gt;. Because &lt;math&gt;P(x)&lt;/math&gt; is quadratic, the polynomials &lt;math&gt;P(x) - P(3)&lt;/math&gt; and &lt;math&gt;P(x) - P(4)&lt;/math&gt; each have the same sum of roots as the polynomial &lt;math&gt;P(x)&lt;/math&gt;, which is &lt;math&gt;P(3) + P(4)&lt;/math&gt;. Thus the answer in this case is &lt;math&gt;2\big(P(3) + P(4)\big)-7&lt;/math&gt;, and so it suffices to compute the value of &lt;math&gt;P(3)+P(4)&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P(3)=u&lt;/math&gt; and &lt;math&gt;P(4) = v&lt;/math&gt;. Substituting &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;x=4&lt;/math&gt; into the above quadratic polynomial yields the system of equations<br /> &lt;cmath&gt;\begin{align*}<br /> u &amp;= (3-u)(3-v) = 9 - 3u - 3v + uv\\<br /> v &amp;= (4-u)(4-v) = 16 - 4u - 4v + uv.<br /> \end{align*}&lt;/cmath&gt;Subtracting the first equation from the second gives &lt;math&gt;v - u = 7 - u - v&lt;/math&gt;, yielding &lt;math&gt;v = \frac72.&lt;/math&gt; Substituting this value into the second equation gives&lt;cmath&gt;\dfrac72 = \left(4 - u\right)\left(4 - \dfrac72\right),&lt;/cmath&gt;yielding &lt;math&gt;u = -3.&lt;/math&gt; The sum of the two solutions is &lt;math&gt;2\left(\tfrac72-3\right)-7 = -6&lt;/math&gt;. In this case, &lt;math&gt;P(x)= (x+3)\left(x-\frac72\right)&lt;/math&gt;.<br /> <br /> The requested sum of squares is &lt;math&gt;7^2+(-6)^2 = {85}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> <br /> Let &lt;math&gt;P(x) = (x-c)^2 - d&lt;/math&gt; for some &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;.<br /> <br /> Then, we can write &lt;math&gt;P(P(x)) = ((x-c)^2 - d - c)^2 - d&lt;/math&gt;. Setting the expression equal to &lt;math&gt;0&lt;/math&gt; and solving for &lt;math&gt;x&lt;/math&gt; gives:<br /> <br /> &lt;math&gt;x = \pm \sqrt{ \pm \sqrt{d} + d + c} + c&lt;/math&gt;<br /> <br /> Therefore, we have that &lt;math&gt;x&lt;/math&gt; takes on the four values &lt;math&gt;\sqrt{\sqrt{d} + d + c} + c&lt;/math&gt;, &lt;math&gt;-\sqrt{\sqrt{d} + d + c} + c&lt;/math&gt;, &lt;math&gt;\sqrt{-\sqrt{d} + d + c} + c&lt;/math&gt;, and &lt;math&gt;-\sqrt{-\sqrt{d} + d + c} + c&lt;/math&gt;. Two of these values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, and the other two are &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;.<br /> <br /> We can split these four values into two &quot;groups&quot; based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values.<br /> <br /> &lt;math&gt;\textbf{Case 1}&lt;/math&gt;: Both the 3 and 4 values are from the same group.<br /> <br /> In this case, the &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; values are both from the other group. The sum of this is just &lt;math&gt;2c&lt;/math&gt; because the radical cancels out. Because of this, we can see that &lt;math&gt;c&lt;/math&gt; is just the average of &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, so we have &lt;math&gt;2c = 3 + 4 = 7&lt;/math&gt;, so &lt;math&gt;(a+b)^2 = 7^2 = 49&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2}&lt;/math&gt;: The 3 and 4 values come from different groups.<br /> <br /> It is easy to see that all possibilities in this case are basically symmetric and yield the same value for &lt;math&gt;(a+b)^2&lt;/math&gt;. Without loss of generality, assume that &lt;math&gt;\sqrt{\sqrt{d} + d + c} + c = 4&lt;/math&gt; and &lt;math&gt;\sqrt{-\sqrt{d} + d + c} + c = 3&lt;/math&gt;. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger.<br /> <br /> We can write &lt;math&gt;\sqrt{\sqrt{d} + d + c} + c = 1 + \sqrt{-\sqrt{d} + d + c} + c&lt;/math&gt;.<br /> <br /> Moving most terms to the left side and simplifying gives &lt;math&gt;\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} = 1&lt;/math&gt;.<br /> <br /> We can square both sides and simplify:<br /> <br /> &lt;math&gt;\sqrt{d} + d + c - \sqrt{d} + d + c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;2d + 2c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = (d+c) - \frac{1}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sqrt{(d+c)^2 - (\sqrt{d})^2} = (d+c) - \frac{1}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;\sqrt{d^2 + 2dc + c^2 - d} = (d+c) - \frac{1}{2}&lt;/math&gt;<br /> <br /> Squaring both sides again gives the following:<br /> <br /> &lt;math&gt;d^2 + 2dc + c^2 - d = d^2 + 2dc + c^2 - d - c + \frac{1}{4}&lt;/math&gt;<br /> <br /> Nearly all terms cancel out, yielding &lt;math&gt;c = \frac{1}{4}&lt;/math&gt;.<br /> <br /> By substituting this back in, we obtain &lt;math&gt;\sqrt{\sqrt{d} + d + c} = \frac{15}{4}&lt;/math&gt; and &lt;math&gt;\sqrt{-\sqrt{d} + d + c} = \frac{11}{4}&lt;/math&gt;.<br /> <br /> The sum of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; is equal to &lt;math&gt;-\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} + 2c = -\frac{15}{4} - \frac{11}{4} + \frac{1}{2} = -6&lt;/math&gt;, so &lt;math&gt;(a+b)^2 = 36&lt;/math&gt;.<br /> <br /> Adding up both values gives &lt;math&gt;49 + 36 = \boxed{085}&lt;/math&gt; as our final answer. ~ I-_-I<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> I- -i https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_9&diff=113182 2013 AMC 10A Problems/Problem 9 2019-12-22T20:48:13Z <p>I- -i: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the number of attempted three-point shots made be &lt;math&gt;x&lt;/math&gt; and the number of attempted two-point shots be &lt;math&gt;y&lt;/math&gt;. We know that &lt;math&gt;x+y=30&lt;/math&gt;, and we need to evaluate &lt;math&gt;(0.2\cdot3)x +(0.3\cdot2)y&lt;/math&gt;, as we know that the three-point shots are worth 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them.<br /> <br /> Simplifying, we see that this is equal to &lt;math&gt;0.6x + 0.6y = 0.6(x+y)&lt;/math&gt;. Plugging in &lt;math&gt;x+y=30&lt;/math&gt;, we get &lt;math&gt;0.6(30) = \boxed{\textbf{(B) }18}&lt;/math&gt;<br /> <br /> ==Solution 2 (cheap)== <br /> <br /> The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes &lt;math&gt;0.2 \cdot 30 = 6&lt;/math&gt; shots, which are worth &lt;math&gt;6 \cdot 3 = \boxed{\textbf{(B) }18}&lt;/math&gt; points.<br /> <br /> -- I-_-I<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=8|num-a=10}}<br /> {{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> I- -i https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_9&diff=113181 2013 AMC 10A Problems/Problem 9 2019-12-22T20:47:13Z <p>I- -i: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the number of attempted three-point shots made be &lt;math&gt;x&lt;/math&gt; and the number of attempted two-point shots be &lt;math&gt;y&lt;/math&gt;. We know that &lt;math&gt;x+y=30&lt;/math&gt;, and we need to evaluate &lt;math&gt;(0.2\cdot3)x +(0.3\cdot2)y&lt;/math&gt;, as we know that the three-point shots are worth 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them.<br /> <br /> Simplifying, we see that this is equal to &lt;math&gt;0.6x + 0.6y = 0.6(x+y)&lt;/math&gt;. Plugging in &lt;math&gt;x+y=30&lt;/math&gt;, we get &lt;math&gt;0.6(30) = \boxed{\textbf{(B) }18}&lt;/math&gt;<br /> <br /> ==Solution 2== <br /> <br /> The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes &lt;math&gt;0.2 \cdot 30 = 6&lt;/math&gt; shots, which are worth &lt;math&gt;6 \cdot 3 = \boxed{\textbf{(B) }18}&lt;/math&gt; points.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=8|num-a=10}}<br /> {{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> I- -i https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_9&diff=113180 2013 AMC 10A Problems/Problem 9 2019-12-22T20:46:53Z <p>I- -i: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the number of attempted three-point shots made be &lt;math&gt;x&lt;/math&gt; and the number of attempted two-point shots be &lt;math&gt;y&lt;/math&gt;. We know that &lt;math&gt;x+y=30&lt;/math&gt;, and we need to evaluate &lt;math&gt;(0.2\cdot3)x +(0.3\cdot2)y&lt;/math&gt;, as we know that the three-point shots are worth 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them.<br /> <br /> Simplifying, we see that this is equal to &lt;math&gt;0.6x + 0.6y = 0.6(x+y)&lt;/math&gt;. Plugging in &lt;math&gt;x+y=30&lt;/math&gt;, we get &lt;math&gt;0.6(30) = \boxed{\textbf{(B) }18}&lt;/math&gt;<br /> <br /> ==Solution 2== <br /> <br /> The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes &lt;math&gt;0.2 \cdot 30 = 6&lt;/math&gt; shots, which are worth &lt;math&gt;6 \cdot 3 = \boxed{\textbf{(B) }18&lt;/math&gt; points.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=8|num-a=10}}<br /> {{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> I- -i https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=110429 2006 AIME II Problems/Problem 11 2019-10-19T17:44:34Z <p>I- -i: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined as follows &lt;math&gt; a_1=a_2=a_3=1, &lt;/math&gt; and, for all positive integers &lt;math&gt; n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. &lt;/math&gt; Given that &lt;math&gt; a_{28}=6090307, a_{29}=11201821, &lt;/math&gt; and &lt;math&gt; a_{30}=20603361, &lt;/math&gt; find the [[remainder]] when &lt;math&gt;\sum^{28}_{k=1} a_k &lt;/math&gt; is divided by 1000.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Define the sum as &lt;math&gt;s&lt;/math&gt;. Since &lt;math&gt;a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} &lt;/math&gt;, the sum will be:<br /> &lt;center&gt;&lt;math&gt;s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\<br /> s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\<br /> s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\<br /> s = -s + a_{28} + a_{30}<br /> &lt;/math&gt;&lt;/center&gt;<br /> <br /> Thus &lt;math&gt;s = \frac{a_{28} + a_{30}}{2}&lt;/math&gt;, and &lt;math&gt;a_{28},\,a_{30}&lt;/math&gt; are both given; the last four digits of their sum is &lt;math&gt;3668&lt;/math&gt;, and half of that is &lt;math&gt;1834&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{834}&lt;/math&gt;.<br /> <br /> === Solution 2 (bash) ===<br /> <br /> Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:<br /> <br /> &lt;math&gt;a_{1}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{2}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{3}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{4}\equiv 3 \pmod {1000}<br /> \newline<br /> a_{5}\equiv 5 \pmod {1000}<br /> \newline<br /> \cdots <br /> \newline<br /> a_{25} \equiv 793 \pmod {1000}<br /> \newline<br /> a_{26} \equiv 281 \pmod {1000}<br /> \newline<br /> a_{27} \equiv 233 \pmod {1000}<br /> \newline<br /> a_{28} \equiv 307 \pmod {1000}&lt;/math&gt;<br /> <br /> Adding all the residues shows the sum is congruent to &lt;math&gt;\boxed{834}&lt;/math&gt; mod 1000.<br /> <br /> ~ I-_-I<br /> <br /> === Solution 3 (some guessing involved) ===<br /> All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given &lt;math&gt;a_{28}, a_{29}, &lt;/math&gt; and &lt;math&gt;a_{30}&lt;/math&gt;, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some &lt;math&gt;p, q, r&lt;/math&gt; such that &lt;math&gt;\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}&lt;/math&gt;. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that &lt;math&gt;(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})&lt;/math&gt;, at least for the first few terms. From this, we have that &lt;math&gt;\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)&lt;/math&gt;.<br /> <br /> Solution by zeroman<br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> I- -i https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=110428 2006 AIME II Problems/Problem 11 2019-10-19T17:44:19Z <p>I- -i: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined as follows &lt;math&gt; a_1=a_2=a_3=1, &lt;/math&gt; and, for all positive integers &lt;math&gt; n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. &lt;/math&gt; Given that &lt;math&gt; a_{28}=6090307, a_{29}=11201821, &lt;/math&gt; and &lt;math&gt; a_{30}=20603361, &lt;/math&gt; find the [[remainder]] when &lt;math&gt;\sum^{28}_{k=1} a_k &lt;/math&gt; is divided by 1000.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Define the sum as &lt;math&gt;s&lt;/math&gt;. Since &lt;math&gt;a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} &lt;/math&gt;, the sum will be:<br /> &lt;center&gt;&lt;math&gt;s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\<br /> s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\<br /> s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\<br /> s = -s + a_{28} + a_{30}<br /> &lt;/math&gt;&lt;/center&gt;<br /> <br /> Thus &lt;math&gt;s = \frac{a_{28} + a_{30}}{2}&lt;/math&gt;, and &lt;math&gt;a_{28},\,a_{30}&lt;/math&gt; are both given; the last four digits of their sum is &lt;math&gt;3668&lt;/math&gt;, and half of that is &lt;math&gt;1834&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{834}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:<br /> <br /> &lt;math&gt;a_{1}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{2}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{3}\equiv 1 \pmod {1000}<br /> \newline<br /> a_{4}\equiv 3 \pmod {1000}<br /> \newline<br /> a_{5}\equiv 5 \pmod {1000}<br /> \newline<br /> \cdots <br /> \newline<br /> a_{25} \equiv 793 \pmod {1000}<br /> \newline<br /> a_{26} \equiv 281 \pmod {1000}<br /> \newline<br /> a_{27} \equiv 233 \pmod {1000}<br /> \newline<br /> a_{28} \equiv 307 \pmod {1000}&lt;/math&gt;<br /> <br /> Adding all the residues shows the sum is congruent to &lt;math&gt;\boxed{834}&lt;/math&gt; mod 1000.<br /> <br /> ~ I-_-I<br /> <br /> === Solution 3 (some guessing involved) ===<br /> All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given &lt;math&gt;a_{28}, a_{29}, &lt;/math&gt; and &lt;math&gt;a_{30}&lt;/math&gt;, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some &lt;math&gt;p, q, r&lt;/math&gt; such that &lt;math&gt;\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}&lt;/math&gt;. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that &lt;math&gt;(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})&lt;/math&gt;, at least for the first few terms. From this, we have that &lt;math&gt;\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)&lt;/math&gt;.<br /> <br /> Solution by zeroman<br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> I- -i https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=110423 2006 AIME II Problems/Problem 11 2019-10-19T17:40:34Z <p>I- -i: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined as follows &lt;math&gt; a_1=a_2=a_3=1, &lt;/math&gt; and, for all positive integers &lt;math&gt; n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. &lt;/math&gt; Given that &lt;math&gt; a_{28}=6090307, a_{29}=11201821, &lt;/math&gt; and &lt;math&gt; a_{30}=20603361, &lt;/math&gt; find the [[remainder]] when &lt;math&gt;\sum^{28}_{k=1} a_k &lt;/math&gt; is divided by 1000.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Define the sum as &lt;math&gt;s&lt;/math&gt;. Since &lt;math&gt;a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} &lt;/math&gt;, the sum will be:<br /> &lt;center&gt;&lt;math&gt;s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\<br /> s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\<br /> s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\<br /> s = -s + a_{28} + a_{30}<br /> &lt;/math&gt;&lt;/center&gt;<br /> <br /> Thus &lt;math&gt;s = \frac{a_{28} + a_{30}}{2}&lt;/math&gt;, and &lt;math&gt;a_{28},\,a_{30}&lt;/math&gt; are both given; the last four digits of their sum is &lt;math&gt;3668&lt;/math&gt;, and half of that is &lt;math&gt;1834&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{834}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:<br /> <br /> &lt;math&gt;a_{1}\equiv 1 \pmod 1000&lt;/math&gt;<br /> &lt;math&gt;a_{2}\equiv 1 \pmod 1000&lt;/math&gt;<br /> &lt;math&gt;a_{3}\equiv 1 \pmod 1000&lt;/math&gt;<br /> &lt;math&gt;a_{4}\equiv 3 \pmod 1000&lt;/math&gt;<br /> &lt;math&gt;a_{5}\equiv 5 \pmod 1000&lt;/math&gt;<br /> &lt;math&gt;\cdots&lt;/math&gt;<br /> &lt;math&gt;a_{25} \equiv 793 \pmod 1000&lt;/math&gt;<br /> &lt;math&gt;a_{26} \equiv 281 \pmod 1000&lt;/math&gt;<br /> &lt;math&gt;a_{27} \equiv 233 \pmod 1000&lt;/math&gt;<br /> &lt;math&gt;a_{28} \equiv 307 \pmod 1000&lt;/math&gt;<br /> <br /> Adding all the residues shows the sum is congruent to &lt;math&gt;\boxed{834}&lt;/math&gt; mod 1000.<br /> <br /> ~ I-_-I<br /> <br /> === Solution 3 (some guessing involved) ===<br /> All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given &lt;math&gt;a_{28}, a_{29}, &lt;/math&gt; and &lt;math&gt;a_{30}&lt;/math&gt;, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some &lt;math&gt;p, q, r&lt;/math&gt; such that &lt;math&gt;\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}&lt;/math&gt;. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that &lt;math&gt;(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})&lt;/math&gt;, at least for the first few terms. From this, we have that &lt;math&gt;\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)&lt;/math&gt;.<br /> <br /> Solution by zeroman<br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> I- -i