Difference between revisions of "2008 AIME I Problems/Problem 14"

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This is a [[quadratic equation]], maximized when <math>\sin\theta = \frac { - 2}{ - 6} = \frac {1}{3}</math>. Thus, <math>m^2 = 9^2[5 + \frac {2}{3} - \frac {1}{3}] = \boxed{432}</math>.
 
This is a [[quadratic equation]], maximized when <math>\sin\theta = \frac { - 2}{ - 6} = \frac {1}{3}</math>. Thus, <math>m^2 = 9^2[5 + \frac {2}{3} - \frac {1}{3}] = \boxed{432}</math>.
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 +
===Solution 3 (Calculus Bash)===
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 +
<center><asy>
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unitsize(3mm);
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pair B=(0,13.5), C=(23.383,0);
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pair O=(7.794, 9), P=(2*7.794,0);
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pair T=(7.794,0), Q=(0,0);
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pair A=(2*7.794,4.5);
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draw(Q--B--C--Q);
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draw(O--T);
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draw(A--P);
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draw(Circle(O,9));
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dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q);
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label("\(B\)",B,NW);
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label("\(A\)",A,NE);
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label("\(\omega\)",O,N);
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label("\(P\)",P,S);
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label("\(T\)",T,S);
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label("\(Q\)",Q,S);
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label("\(C\)",C,E);
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label("\(9\)", (B+O)/2, N);
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label("\(9\)", (O+A)/2, N);
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label("\(9\)", (O+T)/2,W);
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</asy></center>
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(Diagram credit goes to Solution 2)
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 +
We let <math>AC=x</math>. From similar triangles, we have that <math>PC=\frac{x\sqrt{x^2+18x}}{x+9}</math>. Similarly, <math>TP=QT=\frac{9\sqrt{x^2+18x}}{x+9}</math>. Using the Pythagorean Theorem, <math>BQ=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2}</math>. Using the Pythagorean Theorem once again, <math>BP=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2+(\frac{18\sqrt{x^2+18x}}{x+9})^2}</math>. After a large bashful simplification, <math>BP=\sqrt{405+\frac{1458x-6561}{x^2+18x+81}}</math>. The fraction is equivalent to <math>729\frac{2x-9}{(x+9)^2}</math>. Taking the derivative of the fraction and solving for x, we get that <math>x=18</math>. Plugging <math>x=18</math> back into the expression for <math>BP</math> yields <math>\sqrt{432}</math>, so the answer is <math>(\sqrt{432})^2=\boxed{432}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 02:00, 20 March 2019

Problem

Let $\overline{AB}$ be a diameter of circle $\omega$. Extend $\overline{AB}$ through $A$ to $C$. Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$. Point $P$ is the foot of the perpendicular from $A$ to line $CT$. Suppose $AB = 18$, and let $m$ denote the maximum possible length of segment $BP$. Find $m^{2}$.

Solution

Solution 1

[asy] size(250); defaultpen(0.70 + fontsize(10)); import olympiad; pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T); draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P); draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35)); draw(B--P, blue); label("\(A\)",A,NW); label("\(B\)",B,NW); label("\(C\)",C,NW); label("\(O\)",O,NW); label("\(P\)",P,SE); label("\(T\)",T,SE); label("\(9\)",(O+A)/2,N); label("\(9\)",(O+B)/2,N); label("\(x-9\)",(C+A)/2,N); [/asy]

Let $x = OC$. Since $OT, AP \perp TC$, it follows easily that $\triangle APC \sim \triangle OTC$. Thus $\frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}$. By the Law of Cosines on $\triangle BAP$, \begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*} where $\cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}$, so: \begin{align*}BP^2 &= 18^2 + \frac{9^2(x-9)^2}{x^2} + 2(18) \cdot \frac{9(x-9)}{x} \cdot \frac 9x = 405 + 729\left(\frac{2x - 27}{x^2}\right)\end{align*} Let $m = \frac{2x-27}{x^2} \Longrightarrow mx^2 - 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(m)(27) \ge 0 \Longleftrightarrow m \le \frac{1}{27}$. Thus, \[BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}\] Equality holds when $x = 27$.

Solution 2

[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5);  draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9));  dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label("\(B\)",B,NW); label("\(A\)",A,NE); label("\(\omega\)",O,N); label("\(P\)",P,S); label("\(T\)",T,S); label("\(Q\)",Q,S); label("\(C\)",C,E); label("\(\theta\)",C + (-1.7,-0.2), NW); label("\(9\)", (B+O)/2, N); label("\(9\)", (O+A)/2, N); label("\(9\)", (O+T)/2,W); [/asy]

From the diagram, we see that $BQ = \omega T + B\omega\sin\theta = 9 + 9\sin\theta = 9(1 + \sin\theta)$, and that $QP = BA\cos\theta = 18\cos\theta$.

\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \sin\theta)^2 + 18^2\cos^2\theta\\ &= 9^2[1 + 2\sin\theta + \sin^2\theta + 4(1 - \sin^2\theta)]\\ BP^2 &= 9^2[5 + 2\sin\theta - 3\sin^2\theta]\end{align*}

This is a quadratic equation, maximized when $\sin\theta = \frac { - 2}{ - 6} = \frac {1}{3}$. Thus, $m^2 = 9^2[5 + \frac {2}{3} - \frac {1}{3}] = \boxed{432}$.

Solution 3 (Calculus Bash)

[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5);  draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9));  dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label("\(B\)",B,NW); label("\(A\)",A,NE); label("\(\omega\)",O,N); label("\(P\)",P,S); label("\(T\)",T,S); label("\(Q\)",Q,S); label("\(C\)",C,E); label("\(9\)", (B+O)/2, N); label("\(9\)", (O+A)/2, N); label("\(9\)", (O+T)/2,W); [/asy]

(Diagram credit goes to Solution 2)

We let $AC=x$. From similar triangles, we have that $PC=\frac{x\sqrt{x^2+18x}}{x+9}$. Similarly, $TP=QT=\frac{9\sqrt{x^2+18x}}{x+9}$. Using the Pythagorean Theorem, $BQ=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2}$. Using the Pythagorean Theorem once again, $BP=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2+(\frac{18\sqrt{x^2+18x}}{x+9})^2}$. After a large bashful simplification, $BP=\sqrt{405+\frac{1458x-6561}{x^2+18x+81}}$. The fraction is equivalent to $729\frac{2x-9}{(x+9)^2}$. Taking the derivative of the fraction and solving for x, we get that $x=18$. Plugging $x=18$ back into the expression for $BP$ yields $\sqrt{432}$, so the answer is $(\sqrt{432})^2=\boxed{432}$.

See also

2008 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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