Difference between revisions of "2008 AIME I Problems/Problem 14"
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Let <math>\overline{AB}</math> be a diameter of circle <math>\omega</math>. Extend <math>\overline{AB}</math> through <math>A</math> to <math>C</math>. Point <math>T</math> lies on <math>\omega</math> so that line <math>CT</math> is tangent to <math>\omega</math>. Point <math>P</math> is the foot of the perpendicular from <math>A</math> to line <math>CT</math>. Suppose <math>AB = 18</math>, and let <math>m</math> denote the maximum possible length of segment <math>BP</math>. Find <math>m^{2}</math>. | Let <math>\overline{AB}</math> be a diameter of circle <math>\omega</math>. Extend <math>\overline{AB}</math> through <math>A</math> to <math>C</math>. Point <math>T</math> lies on <math>\omega</math> so that line <math>CT</math> is tangent to <math>\omega</math>. Point <math>P</math> is the foot of the perpendicular from <math>A</math> to line <math>CT</math>. Suppose <math>AB = 18</math>, and let <math>m</math> denote the maximum possible length of segment <math>BP</math>. Find <math>m^{2}</math>. | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
<center><asy> | <center><asy> | ||
size(250); defaultpen(0.70 + fontsize(10)); import olympiad; | size(250); defaultpen(0.70 + fontsize(10)); import olympiad; | ||
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Equality holds when <math>x = 27</math>. | Equality holds when <math>x = 27</math>. | ||
− | == | + | ===Solution 2=== |
<asy> | <asy> | ||
− | unitsize( | + | unitsize(3mm); |
pair B=(0,13.5), C=(23.383,0); | pair B=(0,13.5), C=(23.383,0); | ||
pair O=(7.794, 9), P=(2*7.794,0); | pair O=(7.794, 9), P=(2*7.794,0); | ||
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</asy> | </asy> | ||
− | <math>BQ = \omega T + B\omega\sin\theta = 9 + 9\sin\theta = 9(1 + \sin\theta)</math> | + | From the diagram, we see that <math>BQ = \omega T + B\omega\sin\theta = 9 + 9\sin\theta = 9(1 + \sin\theta)</math>. <math>QP = BA\cos\theta = 18\cos\theta</math> |
− | < | + | <cmath>\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*}</cmath> | ||
− | + | 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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− | <math>m^2 = 9^2[5 + \frac {2}{3} - \frac {1}{3}] = \boxed{432}</math> | ||
== See also == | == See also == |
Revision as of 18:37, 23 March 2008
Problem
Let be a diameter of circle
. Extend
through
to
. Point
lies on
so that line
is tangent to
. Point
is the foot of the perpendicular from
to line
. Suppose
, and let
denote the maximum possible length of segment
. Find
.
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]](http://latex.artofproblemsolving.com/e/6/b/e6b29aed6992abeced47a59aa7fe8bf49e0cf057.png)
Let . Since
, it follows easily that
. Thus
. By the Law of Cosines on
,
where
, so:
Let
; this is a quadratic, and its discriminant must be nonnegative:
. Thus,
Equality holds when
.
Solution 2
From the diagram, we see that .
This is a quadratic equation, maximized when . Thus,
.
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 |