Difference between revisions of "2014 AIME I Problems/Problem 15"

(Problem 15)
(Problem 15)
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== Problem 15 ==
 
== Problem 15 ==
  
In <math>\triangle ABC</math>, <math>AB = 3</math>, <math>BC = 4</math>, and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B</math>, <math>\overline{BC}</math> at <math>B</math> and <math>D</math>, and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4}</math>, <math>DE=\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer not divisible by the square of any prime. Find <math>a+b+c</math>.
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In <math>\triangle ABC</math>, <math>AB = 3</math>, <math>BC = 4</math>, and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B</math>, <math>\overline{BC}</math> at <math>B</math> and <math>D</math>, and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4}</math>, length <math>DE=\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer not divisible by the square of any prime. Find <math>a+b+c</math>.
  
 
== Solution ==
 
== Solution ==

Revision as of 18:20, 14 March 2014

Problem 15

In $\triangle ABC$, $AB = 3$, $BC = 4$, and $CA = 5$. Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$, $\overline{BC}$ at $B$ and $D$, and $\overline{AC}$ at $F$ and $G$. Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$, length $DE=\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$.

Solution