Difference between revisions of "2008 AIME I Problems/Problem 14"
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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>. | 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>. | ||
+ | |||
+ | ===Solution 4=== | ||
+ | |||
+ | <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) | ||
+ | |||
+ | Let <math>AC=x</math>. The only constraint on <math>x</math> is that it must be greater than <math>0</math>. Using similar triangles, we can deduce that <math>PA=\frac{9x}{x+9}</math>. Now, apply law of cosines on <math>\triangle PAB</math>. <cmath>BP^2=\left(\frac{9x^2}{x+9}\right)^2+18^2-2(18)\left(\frac{9x}{x+9}\right)\cos(\angle PAB).</cmath> We can see that <math>\cos(\angle PAB)=\cos(180^{\circ}-\angle PAC)=\cos(\angle PAC -90^{\circ})=-\sin(\angle PCA)</math>. We can find <math>-\sin(\angle PCA)=-\frac{9}{x+9}</math>. Plugging this into our equation, we get: | ||
+ | <cmath>BP^2=\left(\frac{9x^2}{x+9}\right)^2+18^2-2(18)\left(\frac{9x}{x+9}\right)\left(-\frac{9}{x+9}\right).</cmath> Eventually, <cmath>BP^2 = 81\left(\frac{x^2+36x}{(x+9)^2}+4\right).</cmath> We want to maximize <math>\frac{x^2+36x}{(x+9)^2}</math>. There are many ways to maximize this expression, discussed here: https://artofproblemsolving.com/community/c4h2292700_maximization. The maximum result of that expression is <math>\frac{4}{3}</math>. Finally, evaluating <math>BP^2</math> for this value <math>81\left(\frac{4}{3}+4\right) = \boxed{432}</math>. | ||
+ | |||
+ | |||
+ | ~superagh | ||
== See also == | == See also == |
Latest revision as of 22:07, 5 October 2020
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 1.1
Proceed as follows for Solution 1.
Once you approach the function , find the maximum value by setting .
Simplifying to take the derivative, we have , so . Setting , we have .
Solving, we obtain as the critical value. Hence, has the maximum value of . Since , the maximum value of occurs at , so has a maximum value of .
Note: Please edit this solution if it feels inadequate.
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 .
Solution 4
(Diagram credit goes to Solution 2)
Let . The only constraint on is that it must be greater than . Using similar triangles, we can deduce that . Now, apply law of cosines on . We can see that . We can find . Plugging this into our equation, we get: Eventually, We want to maximize . There are many ways to maximize this expression, discussed here: https://artofproblemsolving.com/community/c4h2292700_maximization. The maximum result of that expression is . Finally, evaluating for this value .
~superagh
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.