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 B, so from the angle bisector theorem we have AF/AB = CF/CB, so AF = 15/7 and CF = 20/7. Lastly, we apply power of a point from points A and C with respect to "omega" and have AE*AB=AF*AG and $CD \times CB=CG \times CF$ , so we can compute that EB = 17/14 and DB = 31/14. From Pythagorean Theorem, we result in $DE = \frac{25 \sqrt{2}}{14}$ , so $a+b+c=25+2+14= \boxed{041}$

2014 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2014 AIME I Problems/Problem 15

Problem 15

Solution

See also