2008 AIME I Problems/Problem 14
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
![[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 (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
![[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("\(O\)",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/3/f/1/3f1f0f6ecfec2a185ec153b2b42b8b91e59ffd0c.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
(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
![[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)
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 |
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