Difference between revisions of "2008 AIME I Problems/Problem 14"
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== Problem == | == Problem == | ||
− | 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>\overline{AB} = 18</math>, and let <math>m</math> denote the maximum possible length of segment <math>BP</math>. Find <math>m^{2}</math>. |
− | |||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
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where <math>\cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}</math>, so: | where <math>\cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}</math>, so: | ||
<cmath>\begin{align*}BP^2 &= 18^2 + \frac{9^2(x-9)^2}{x^2} + 2(18) \cdot \frac{9(x-9)}{x} \cdot \frac 9x = 405 + 729\left(\frac{2x - 27}{x^2}\right)\end{align*}</cmath> | <cmath>\begin{align*}BP^2 &= 18^2 + \frac{9^2(x-9)^2}{x^2} + 2(18) \cdot \frac{9(x-9)}{x} \cdot \frac 9x = 405 + 729\left(\frac{2x - 27}{x^2}\right)\end{align*}</cmath> | ||
− | Let <math> | + | Let <math>k = \frac{2x-27}{x^2} \Longrightarrow kx^2 - 2x + 27 = 0</math>; this is a quadratic, and its [[discriminant]] must be nonnegative: <math>(-2)^2 - 4(k)(27) \ge 0 \Longleftrightarrow k \le \frac{1}{27}</math>. Thus, |
<cmath>BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}</cmath> | <cmath>BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}</cmath> | ||
− | Equality holds when <math>x = 27</math>. | + | Equality holds when <math>x = 27</math>.~Shen Kislay Kai |
+ | |||
+ | ==== Solution 1.1 (Calculus) ==== | ||
− | === | + | Proceed as follows for Solution 1. |
+ | |||
+ | Once you approach the function <math>k=(2x-27)/x^2</math>, find the maximum value by setting <math>dk/dx=0</math>. | ||
+ | |||
+ | Simplifying <math>k</math> to take the derivative, we have <math>2/x-27/x^2</math>, so <math>dk/dx=-2/x^2+54/x^3</math>. Setting <math>dk/dx=0</math>, we have <math>2/x^2=54/x^3</math>. | ||
+ | |||
+ | Solving, we obtain <math>x=27</math> as the critical value. Hence, <math>k</math> has the maximum value of <math>(2*27-27)/27^2=1/27</math>. Since <math>BP^2=405+729k</math>, the maximum value of <math>\overline {BP}</math> occurs at <math>k=1/27</math>, so <math>BP^2</math> has a maximum value of <math>405+729/27=\fbox{432}</math>. | ||
+ | |||
+ | Note: Please edit this solution if it feels inadequate. | ||
+ | ~Shen Kislay Kai | ||
===Solution 2=== | ===Solution 2=== | ||
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label("\(B\)",B,NW); | label("\(B\)",B,NW); | ||
label("\(A\)",A,NE); | label("\(A\)",A,NE); | ||
− | label("\( | + | label("\(O\)",O,N); |
label("\(P\)",P,S); | label("\(P\)",P,S); | ||
label("\(T\)",T,S); | label("\(T\)",T,S); | ||
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</asy></center> | </asy></center> | ||
− | From the diagram, we see that <math>BQ = | + | From the diagram, we see that <math>BQ = OT + BO \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\\ | ||
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(Diagram credit goes to Solution 2) | (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 | + | We let <math>AC=x</math>. From similar triangles, we have that <math>PC=\frac{x\sqrt{x^2+18x}}{x+9}</math> (Use Pythagorean on <math>\triangle\omega TC</math> and then using <math>\triangle\omega CT\sim\triangle ACP</math>). Similarly, <math>TP=QT=\frac{9\sqrt{x^2+18x}}{x+9}</math>. Using the Pythagorean Theorem again and <math>\triangle CAP\sim\triangle CBQ</math>, <math>BQ=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2}</math>. Using the Pythagorean Theorem <math>\bold{again}</math>, <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 | ||
+ | |||
+ | ===Solution 5 (Clean)=== | ||
+ | Let <math>h</math> be the distance from <math>A</math> to <math>CT</math>. Observe that <math>h</math> takes any value from <math>0</math> to <math>2r</math>, where <math>r</math> is the radius of the circle. | ||
+ | |||
+ | Let <math>Q</math> be the foot of the altitude from <math>B</math> to <math>CT</math>. It is clear that <math>T</math> is the midpoint of <math>PQ</math>, and so the length <math>OT</math> is the average of <math>AP</math> and <math>BQ</math>. It follows thus that <math>BQ = 2r - h</math>. | ||
+ | |||
+ | We compute <math>PT = \sqrt{r^2 - (r - h)^2} = \sqrt{h(2r - h)},</math> | ||
+ | and so <math>BP^2 = PQ^2 + BQ^2 = 4PT^2 + BQ^2 = 4h(2r - h) + (2r-h)^2 = (2r-h)(2r + 3h)</math>. | ||
+ | This is <math>\frac{1}{3}(6r - 3h)(2r + 3h) \le \frac{1}{3} \cdot \left( \frac{8r}{2} \right)^2</math>. Equality is attained, so thus we extract the answer of <math>\frac{16 \cdot 9^2}{3} = 27 \cdot 16 = \boxed{432}.</math> | ||
== See also == | == See also == |
Latest revision as of 12:29, 3 September 2024
Contents
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 .~Shen Kislay Kai
Solution 1.1 (Calculus)
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. ~Shen Kislay Kai
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 (Use Pythagorean on and then using ). Similarly, . Using the Pythagorean Theorem again and , . Using the Pythagorean Theorem , . 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
Solution 5 (Clean)
Let be the distance from to . Observe that takes any value from to , where is the radius of the circle.
Let be the foot of the altitude from to . It is clear that is the midpoint of , and so the length is the average of and . It follows thus that .
We compute and so . This is . Equality is attained, so thus we extract the answer of
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.