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

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== Solution ==
 
== Solution ==
  
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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 similar to 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 congruent to angle GBC, and that from similar triangles angle GED 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 <math>CD \times CB=CG \times CF</math>, so we can compute that EB = 17/14 and DB = 31/14. From 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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First we note that <math>\triangle DEF</math> is an isosceles right triangle with hypotenuse <math>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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From congruent arc intersections, we know that <math>\angle GED \cong \angle GBC</math>, and that from similar triangles <math>\angle GED</math> is also congruent to <math>\angle GCB</math>. 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 <math>CD \times CB=CG \times CF</math>, so we can compute that EB = 17/14 and DB = 31/14. From Pythagorean Theorem, we result in <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 00:59, 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

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

See also

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

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