2008 AIME I Problems/Problem 14

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

$[asy] size(250); defaultpen(0.70 + fontsize(10)); import olympiad; pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T); draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P); draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35)); draw(B--P, blue); label("$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 $x = 27$ .

2008 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2008 AIME I Problems/Problem 14

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