Difference between revisions of "2005 AIME I Problems/Problem 15"
m (→See also: aime box) |
|||
| Line 19: | Line 19: | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2005|n=I|num-b=14|after=Last Question}} | |
| − | |||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 18:28, 21 March 2007
Problem
Triangle 
has 
The incircle of the triangle evenly trisects the median 
If the area of the triangle is 
where 
and 
are integers and is not divisible by the square of a prime, find 
Solution
Let 
, 
and 
be the points of tangency of the incircle with 
, 
and 
, respectively. Without loss of generality, let 
, so that is between 
and 
. Let the length of the median be 
. Then by two applications of the Power of a Point Theorem, 
, so 
. Now, 
and 
are two tangents to a circle from the same point, so 
and thus 
. Then 
so 
and thus 
.
Now, by Stewart's Theorem in triangle 
with cevian 
, we have
. Our earlier result from Power of a Point was that 
, so we combine these two results to solve for 
and we get
or equivalently
.
Thus 
or 
. We discard the value as extraneous (it gives us an equilateral triangle) and are left with , so our triangle has sides of length 10, 20 and 26. Applying Heron's formula or the equivalent gives that the area is 
and so the answer is 
.
See also
| 2005 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
