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

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<cmath>\frac{3\omega}{5}\cdot \frac{\omega\sqrt{2}}{2}+\omega\cdot FG=\frac{4\omega}{5}\cdot \frac{\omega\sqrt{2}}{2}</cmath>
 
<cmath>\frac{3\omega}{5}\cdot \frac{\omega\sqrt{2}}{2}+\omega\cdot FG=\frac{4\omega}{5}\cdot \frac{\omega\sqrt{2}}{2}</cmath>
 
<cmath>\frac{3\omega\sqrt{2}}{10}+EF\left(\frac{4\omega\sqrt{2}}{10}\right)\implies FG=\frac{\omega\sqrt{2}}{10}</cmath>
 
<cmath>\frac{3\omega\sqrt{2}}{10}+EF\left(\frac{4\omega\sqrt{2}}{10}\right)\implies FG=\frac{\omega\sqrt{2}}{10}</cmath>
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Thus <math>\frac{\omega\sqrt{2}}{10}=\frac{5}{14}\rightarrow \omega=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}. </math>a+b+c=25+2+14= \boxed{041}$
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Thus <math>\frac{\omega\sqrt{2}}{10}=\frac{5}{14}\rightarrow \omega=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}</math>. <math>a+b+c=25+2+14= \boxed{041}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=14|after=Last Question}}
 
{{AIME box|year=2014|n=I|num-b=14|after=Last Question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:00, 14 February 2016

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 $a+b+c=25+2+14= \boxed{041}$

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 $\omega$: \[DG\cdot EF+EF\cdot DG=DF\cdot EG\] \[\frac{3\omega}{5}\cdot \frac{\omega\sqrt{2}}{2}+\omega\cdot FG=\frac{4\omega}{5}\cdot \frac{\omega\sqrt{2}}{2}\] \[\frac{3\omega\sqrt{2}}{10}+EF\left(\frac{4\omega\sqrt{2}}{10}\right)\implies FG=\frac{\omega\sqrt{2}}{10}\] Thus $\frac{\omega\sqrt{2}}{10}=\frac{5}{14}\rightarrow \omega=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}$. $a+b+c=25+2+14= \boxed{041}$

See also

2014 AIME I (Problems • Answer Key • Resources)
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