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

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draw(B--G);
 
draw(B--G);
 
</asy>
 
</asy>
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Solution 1
 
First we note that <math>\triangle DEF</math> is an isosceles right triangle with hypotenuse <math>\overline{DE}</math> the same as the diameter of <math>\omega</math>. We also note that <math>\triangle DGE \sim \triangle ABC</math> since <math>\angle EGD</math> is a right angle and the ratios of the sides are <math>3:4:5</math>.  
 
First we note that <math>\triangle DEF</math> is an isosceles right triangle with hypotenuse <math>\overline{DE}</math> the same as the diameter of <math>\omega</math>. We also note that <math>\triangle DGE \sim \triangle ABC</math> since <math>\angle EGD</math> is a right angle and the ratios of the sides are <math>3:4:5</math>.  
  
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Lastly, we apply power of a point from points <math>A</math> and <math>C</math> with respect to <math>\omega</math> and have <math>AE \times AB=AF \times AG</math> and <math>CD \times CB=CG \times CF</math>, so we can compute that <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math>. From the Pythagorean Theorem, we result in <math>DE = \frac{25 \sqrt{2}}{14}</math>, so <math>a+b+c=25+2+14= \boxed{041}</math>
 
Lastly, we apply power of a point from points <math>A</math> and <math>C</math> with respect to <math>\omega</math> and have <math>AE \times AB=AF \times AG</math> and <math>CD \times CB=CG \times CF</math>, so we can compute that <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math>. From the Pythagorean Theorem, we result in <math>DE = \frac{25 \sqrt{2}}{14}</math>, so <math>a+b+c=25+2+14= \boxed{041}</math>
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 +
Solution 2
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From solution 1, we have CG = 5/2 and <math>\angle EFG</math> = <math>\angle EDG</math> =<math>\angle EAG</math> . Therefore, <math>\triangle EAF</math> is isosceles with EF = EA.
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Let EF = x, then DE = \sqrt{2}x. Therefore <math>EG = \frac{4 \sqrt{2}}{5}x</math>.
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Using Cosine rule on <math>\triangle EGA</math>
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(14x - 25)(2x + 25) = 0 and x = \frac{25}{14}. Hence, <math>DE = \frac{25 \sqrt{2}}{14}</math>, so <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 20:19, 25 July 2015

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] Solution 1 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 $a+b+c=25+2+14= \boxed{041}$

Solution 2 From solution 1, we have CG = 5/2 and $\angle EFG$ = $\angle EDG$ =$\angle EAG$ . Therefore, $\triangle EAF$ is isosceles with EF = EA. Let EF = x, then DE = \sqrt{2}x. Therefore $EG = \frac{4 \sqrt{2}}{5}x$. Using Cosine rule on $\triangle EGA$ (14x - 25)(2x + 25) = 0 and x = \frac{25}{14}. Hence, $DE = \frac{25 \sqrt{2}}{14}$, so $a+b+c=25+2+14= \boxed{041}$

See also

2014 AIME I (Problems • Answer Key • Resources)
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