Difference between revisions of "2001 AIME I Problems/Problem 7"
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== Solution == | == Solution == | ||
| − | By Heron's formula, the area of the whole triangle is <math> | + | By Heron's formula, the area of the whole triangle is <math>21\sqrt{1311}/2</math>. Since the area of a triangle is the inradius times the semiperimeter, the inradius is <math>\sqrt{1311}/6</math>. The ratio of the heights of triangles ADE and ABC is equal to the ratio between sides DE and BC. Thus, we have <math>(21\sqrt{1311}/40-\sqrt{1311}/6)/(21\sqrt{1311}/40)=x/20</math>. Solving for x gives x=<math>860/63</math>, so the answer is <math>923</math>. |
== See also == | == See also == | ||
{{AIME box|year=2001|n=I|num-b=6|num-a=8}} | {{AIME box|year=2001|n=I|num-b=6|num-a=8}} | ||
Revision as of 00:36, 19 January 2008
Problem
Triangle 
has 
, 
and 
. Points 
and 
are located on 
and 
, respectively, such that 
is parallel to 
and contains the center of the inscribed circle of triangle . Then 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
By Heron's formula, the area of the whole triangle is 
. Since the area of a triangle is the inradius times the semiperimeter, the inradius is 
. The ratio of the heights of triangles ADE and ABC is equal to the ratio between sides DE and BC. Thus, we have 
. Solving for x gives x=
, so the answer is 
.
See also
| 2001 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
