Difference between revisions of "2008 AIME I Problems/Problem 14"
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Once you approach the function m=(2x-27)/x^2 find the maximum value by setting dm/dx=0. | Once you approach the function m=(2x-27)/x^2 find the maximum value by setting dm/dx=0. |
Revision as of 02:07, 25 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
![[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 1.1
Proceed as follows for solution 1.
________________________________________________________________________________________________________________________
Once you approach the function m=(2x-27)/x^2 find the maximum value by setting dm/dx=0.
Simplifying m to take the derivative, we have 2/x-27/x^2, so dm/dx=-2/x^2+54/x^3.
Setting dm/dx=0, we have 2/x^2=54/x^3.
Solving, we obtain x=27 as the critical value.
Hence, m has the maximum value of (2*27-27)/27^2=1/27.
Since BP^2=405+729m, the maximum value of BP occurs at m=1/27, so BP^2 has a maximum value of 405+729/27=432
Note: Please edit this solution if it feels inadequate.
I am not familiar with LaTeX and would greatly appreciate it if someone converted the "solution" into LaTeX.
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,
.
Solution 3 (Calculus Bash)
![[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]](http://latex.artofproblemsolving.com/f/c/f/fcff59bc4258c8afe32a6b740c2fffdbb8bbb395.png)
(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.