Difference between revisions of "2016 AIME I Problems/Problem 6"
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==Solution== | ==Solution== | ||
It is well known that <math>DA = DI = DB</math> and so we have <math>DA = DB = 5</math>. Then <math>\triangle DLB \sim \triangle ALC</math> and so <math>\frac{AL}{AC} = \frac{3}{5}</math> and from the angle bisector theorem <math>\frac{CI}{IL} = \frac{5}{3}</math> so <math>CI = \frac{10}{3}</math> and our answer is <math>\boxed{013}</math> | It is well known that <math>DA = DI = DB</math> and so we have <math>DA = DB = 5</math>. Then <math>\triangle DLB \sim \triangle ALC</math> and so <math>\frac{AL}{AC} = \frac{3}{5}</math> and from the angle bisector theorem <math>\frac{CI}{IL} = \frac{5}{3}</math> so <math>CI = \frac{10}{3}</math> and our answer is <math>\boxed{013}</math> | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2016|n=I|num-b=5|after=7}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:08, 4 March 2016
Problem
In 
let 
be the center of the inscribed circle, and let the bisector of 
intersect 
at 
. The line through 
and intersects the circumscribed circle of at the two points and 
. If 
and 
, then 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
It is well known that 
and so we have 
. Then 
and so 
and from the angle bisector theorem 
so 
and our answer is 
See also
| 2016 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by 7 | |
| 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. 
