Difference between revisions of "2016 AIME I Problems"
(→Problem 6) |
(→Problem 6) |
||
| Line 18: | Line 18: | ||
==Problem 6== | ==Problem 6== | ||
| − | In <math>\triangle ABC</math> let <math>I</math> be the center of the inscribed circle, and let the bisector of <math>\angle ACB</math> intersect <math>\ | + | In <math>\triangle ABC</math> let <math>I</math> be the center of the inscribed circle, and let the bisector of <math>\angle ACB</math> intersect <math>\overline{AB}</math> at <math>L</math>. The line through <math>C</math> and <math>L</math> intersects the circumscribed circle of <math>\triangle ABC</math> at the two points <math>C</math> and <math>D</math>. If <math>LI=2</math> and <math>LD=3</math>, then <math>IC= \frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. |
[[2016 AIME I Problems/Problem 6 | Solution]] | [[2016 AIME I Problems/Problem 6 | Solution]] | ||
Revision as of 13:35, 4 March 2016
| 2016 AIME I (Answer Key) | AoPS Contest Collections • PDF | ||
|
Instructions
| ||
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
to 
, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.Contents
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
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 
.
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Let 
be a nonzero polynomial such that 
for every real 
, and 
. Then 
, where 
and 
are relatively prime positive integers. Find 
.
Problem 12
Problem 13
Problem 14
Problem 15
| 2016 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by 2015 AIME II |
Followed by 2016 AIME II | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
