Difference between revisions of "2019 AIME I Problems/Problem 13"
(→Problem 13) |
|||
Line 1: | Line 1: | ||
− | + | ==Problem 13== | |
− | |||
Triangle <math>ABC</math> has side lengths <math>AB=4</math>, <math>BC=5</math>, and <math>CA=6</math>. Points <math>D</math> and <math>E</math> are on ray <math>AB</math> with <math>AB<AD<AE</math>. The point <math>F \neq C</math> is a point of intersection of the circumcircles of <math>\triangle ACD</math> and <math>\triangle EBC</math> satisfying <math>DF=2</math> and <math>EF=7</math>. Then <math>BE</math> can be expressed as <math>\tfrac{a+b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers such that <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | Triangle <math>ABC</math> has side lengths <math>AB=4</math>, <math>BC=5</math>, and <math>CA=6</math>. Points <math>D</math> and <math>E</math> are on ray <math>AB</math> with <math>AB<AD<AE</math>. The point <math>F \neq C</math> is a point of intersection of the circumcircles of <math>\triangle ACD</math> and <math>\triangle EBC</math> satisfying <math>DF=2</math> and <math>EF=7</math>. Then <math>BE</math> can be expressed as <math>\tfrac{a+b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers such that <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | ||
Revision as of 04:09, 15 March 2019
Problem 13
Triangle has side lengths , , and . Points and are on ray with . The point is a point of intersection of the circumcircles of and satisfying and . Then can be expressed as , where , , , and are positive integers such that and are relatively prime, and is not divisible by the square of any prime. Find .
Solution
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.