# 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 1

Since $\angle DBE = 90^\circ$, $DE$ is the diameter of $\omega$. Then $\angle DFE=\angle DGE=90^\circ$. But $DF=FE$, so $\triangle DEF$ is a 45-45-90 triangle. Letting $DG=3x$, we have that $EG=4x$, $DE=5x$, and $DF=EF=\frac{5x}{\sqrt{2}}$.

Note that $\triangle DGE \sim \triangle ABC$ by SAS similarity, so $\angle BAC = \angle GDE$ and $\angle ACB = \angle DEG$. Since $DEFG$ is a cyclic quadrilateral, $\angle BAC = \angle GDE=180^\circ-\angle EFG = \angle AFE$ and $\angle ACB = \angle DEG = \angle GFD$, implying that $\triangle AFE$ and $\triangle CDF$ are isosceles. As a result, $AE=CD=\frac{5x}{\sqrt{2}}$, so $BE=3-\frac{5x}{\sqrt{2}}$ and $BD =4-\frac{5x}{\sqrt{2}}$.

Finally, using the Pythagorean Theorem on $\triangle BDE$, $$\left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2$$ Solving for $x$, we get that $x=\frac{5\sqrt{2}}{14}$, so $DE=5x=\frac{25\sqrt{2}}{14}$. Thus, the answer is $25+2+14=\boxed{041}$.

## Solution 2 $[asy] pair A = (0,3); pair B = (0,0); pair C = (4,0); draw(A--B--C--cycle); dotfactor = 3; dot("A",A,dir(135)); dot("B",B,dir(215)); dot("C",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot("D",D,dir(270)); dot("E",E,dir(180)); dot("F",F,dir(90)); dot("G",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]$

First we note that $\triangle DEF$ is an isosceles right triangle with hypotenuse $\overline{DE}$ 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 $\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\angle ABF = \angle EDF = \frac{\pi}4$. Therefore, $\overline{BF}$ 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 $A$ 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 $a+b+c=25+2+14= \boxed{041}$

Also: $FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}$. We can also use Ptolemy's Theorem on quadrilateral $DEFG$ to figure what $FG$ is in terms of $d$: $$DE\cdot FG+DG\cdot EF=DF\cdot EG$$ $$d\cdot FG+\frac{3d}{5}\cdot \frac{d}{\sqrt{2}}=\frac{4d}{5}\cdot \frac{d}{\sqrt{2}}$$ $$d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}$$ Thus $\frac{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}$. $a+b+c=25+2+14= \boxed{041}$

### Solution 3

Call $DE=x$ and as a result $DF=EF=\frac{x\sqrt{2}}{2}, EG=\frac{4x}{5}, GD=\frac{3x}{5}$. Since $EFGD$ is cyclic we just need to get $DG$ and using LoS(for more detail see the $2$nd paragraph of Solution $2$) we get $AG=\frac{5}{2}$ and using a similar argument(use LoS again) and subtracting you get $FG=\frac{5}{14}$ so you can use Ptolemy to get $x=\frac{25\sqrt{2}}{14} \implies \boxed{041}$. ~First

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