Difference between revisions of "2002 AMC 12A Problems/Problem 22"
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Revision as of 14:28, 14 December 2010
Problem
Triangle 
is a right triangle with 
as its right angle, 
, and 
. Let 
be randomly chosen inside , and extend 
to meet 
at 
. What is the probability that 
?
![[asy] import math; unitsize(4mm); defaultpen(fontsize(8pt)+linewidth(0.7)); dotfactor=4; pair A=(10,0); pair C=(0,0); pair B=(0,10.0/sqrt(3)); pair P=(2,2); pair D=extension(A,C,B,P); draw(A--C--B--cycle); draw(B--D); dot(P); label("A",A,S); label("D",D,S); label("C",C,S); label("P",P,NE); label("B",B,N);[/asy]](http://latex.artofproblemsolving.com/0/3/1/031229b02b99c3f8d5235b9e281cbab7abe4250b.png)
Solution
Clearly 
and 
. Choose a 
and get a corresponding 
such that 
and 
. For we need 
. Thus the point may only lie in the triangle 
. The probability of it doing so is the ratio of areas of to , or equivalently, the ratio of 
to 
because the triangles have identical altitudes when taking and as bases. This ratio is equal to 
. Thus the answer is 
.
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
