2014 AIME I Problems/Problem 15

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

First we note that $\triangle DEF$ is an isosceles right triangle with hypotenuse $DE$ the same as the diameter of $\omega$ . We also note that $\triangle DGE \sim \triangle ABC$ since $\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$ .

From congruent arc intersections, we know that $\angle GED \cong \angle GBC$ , and that from similar triangles $\angle GED$ is also congruent to $\angle GCB$ . Thus, $\triangle BGC$ is an isosceles triangle with $BG = GC$ , so G is the midpoint of $AC$ and $AG = GC = 5/2$ . Similarly, we can find from angle chasing that $BF$ is the angle bisector of $\angle B$ . From the angle bisector theorem, we have $\frac{AF}{AB} = \frac{CF}{CB}, so$ AF = 15/7 $and$ CF = 20/7$.

Lastly, we apply power of a point from points$ (Error compiling LaTeX. Unknown error_msg)A $and$ C $with respect to$ \omega $and have$ AE \times AB=AF \times AG $and$ CD \times CB=CG \times CF $, so we can compute that$ EB = \frac{17}{14} $and$ DB = \frac{31}{14} $. From the Pythagorean Theorem, we result in$ DE = \frac{25 \sqrt{2}}{14} $, so$ a+b+c=25+2+14= \boxed{041}$

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2014 AIME I Problems/Problem 15

Problem 15

Solution

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