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

m (→‎Solution 2: typo fix)
m (→‎Solution 1)
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Let <math>x = OC</math>. Since <math>OT, AP \perp TC</math>, it follows easily that <math>\triangle APC \sim \triangle OTC</math>. Thus <math>\frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}</math>. By the [[Law of Cosines]] on <math>\triangle BAP</math>,  
 
Let <math>x = OC</math>. Since <math>OT, AP \perp TC</math>, it follows easily that <math>\triangle APC \sim \triangle OTC</math>. Thus <math>\frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}</math>. By the [[Law of Cosines]] on <math>\triangle BAP</math>,  
 
<cmath>\begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*}</cmath>
 
<cmath>\begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*}</cmath>
−
where <math>\cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}</math>, so:  
+
where <math>\cos \angle BAP = -\cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}</math>, so:  
 
<cmath>\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*}</cmath>
 
<cmath>\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*}</cmath>
 
Let <math>m = \frac{2x-27}{x^2} \Longrightarrow mx^2 - 2x + 27 = 0</math>; this is a quadratic, and its [[discriminant]] must be nonnegative: <math>(-2)^2 - 4(m)(27) \ge 0 \Longleftrightarrow m \le \frac{1}{27}</math>. Thus,
 
Let <math>m = \frac{2x-27}{x^2} \Longrightarrow mx^2 - 2x + 27 = 0</math>; this is a quadratic, and its [[discriminant]] must be nonnegative: <math>(-2)^2 - 4(m)(27) \ge 0 \Longleftrightarrow m \le \frac{1}{27}</math>. Thus,

Revision as of 22:41, 18 October 2008

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, $m^2 = 9^2[5 + \frac {2}{3} - \frac {1}{3}] = \boxed{432}$.

See also

2008 AIME I (Problems • Answer Key • Resources)
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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