Difference between revisions of "2008 AIME I Problems/Problem 14"
m (fmt) |
m (→Solution 2: typo fix) |
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Line 31: | Line 31: | ||
===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 17:38, 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
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 , 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 |