Difference between revisions of "2008 AIME I Problems/Problem 14"
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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>. | ||
+ | |||
+ | ===Solution 3 (Calculus Bash)=== | ||
+ | |||
+ | <center><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("\(9\)", (B+O)/2, N); | ||
+ | label("\(9\)", (O+A)/2, N); | ||
+ | label("\(9\)", (O+T)/2,W); | ||
+ | </asy></center> | ||
+ | |||
+ | (Diagram credit goes to Solution 2) | ||
+ | |||
+ | We let <math>AC=x</math>. From similar triangles, we have that <math>PC=\frac{x\sqrt{x^2+18x}}{x+9}</math>. Similarly, <math>TP=QT=\frac{9\sqrt{x^2+18x}}{x+9}</math>. Using the Pythagorean Theorem, <math>BQ=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2}</math>. Using the Pythagorean Theorem once again, <math>BP=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2+(\frac{18\sqrt{x^2+18x}}{x+9})^2}</math>. After a large bashful simplification, <math>BP=\sqrt{405+\frac{1458x-6561}{x^2+18x+81}}</math>. The fraction is equivalent to <math>729\frac{2x-9}{(x+9)^2}</math>. Taking the derivative of the fraction and solving for x, we get that <math>x=18</math>. Plugging <math>x=18</math> back into the expression for <math>BP</math> yields <math>\sqrt{432}</math>, so the answer is <math>(\sqrt{432})^2=\boxed{432}</math>. | ||
== See also == | == See also == |
Revision as of 02:00, 20 March 2019
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 .
Contents
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, .
Solution 3 (Calculus Bash)
(Diagram credit goes to Solution 2)
We let . From similar triangles, we have that . Similarly, . Using the Pythagorean Theorem, . Using the Pythagorean Theorem once again, . After a large bashful simplification, . The fraction is equivalent to . Taking the derivative of the fraction and solving for x, we get that . Plugging back into the expression for yields , so the answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.