Difference between revisions of "2008 AIME I Problems/Problem 14"
m (fmt) |
m (→Solution 2: typo fix) |
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===Solution 2=== | ===Solution 2=== | ||
− | <asy> | + | <center><asy> |
unitsize(3mm); | unitsize(3mm); | ||
pair B=(0,13.5), C=(23.383,0); | pair B=(0,13.5), C=(23.383,0); | ||
Line 51: | Line 51: | ||
label("\(Q\)",Q,S); | label("\(Q\)",Q,S); | ||
label("\(C\)",C,E); | label("\(C\)",C,E); | ||
− | label("\(\theta\)",C + (-1. | + | label("\(\theta\)",C + (-1.7,-0.2), NW); |
label("\(9\)", (B+O)/2, N); | label("\(9\)", (B+O)/2, N); | ||
label("\(9\)", (O+A)/2, N); | label("\(9\)", (O+A)/2, N); | ||
label("\(9\)", (O+T)/2,W); | label("\(9\)", (O+T)/2,W); | ||
− | </asy> | + | </asy></center> |
− | From the diagram, we see that <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>, and that <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\\ | <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)]\\ | &= 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> | + | 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>. | 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>. |
Revision as of 18:38, 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
![[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]](http://latex.artofproblemsolving.com/f/b/8/fb879b761e356cf98b5a05c4eaeb066425612f15.png)
From the diagram, we see that , and 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 |