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

Revision as of 23:45, 18 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

$[asy] pair A = (0,3); pair B = (0,0); pair C = (4,0); draw(A--B--C--cycle); dotfactor = 3; dot("$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 $a+b+c=25+2+14= \boxed{041}$

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

Revision as of 23:45, 18 March 2014

Problem 15

Solution

See also

@@ Line 22: / Line 22: @@
 dot("$F$",F,dir(90));
 dot("$G$",G,dir(0));
-draw(Circle((1.107, 0.607), 1.26));
+draw(Circle((1.109, 0.609), 1.28));
 draw(D--E);
 draw(E--F);