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

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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>.  
 
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>
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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, <math>\triangle BGC</math> is an isosceles triangle with <math>BG = GC</math>, so G is the midpoint of <math>AC</math> and <math>AG  = GC = 5/2</math>. Similarly, we can find from angle chasing that <math>BF</math> is the angle bisector of <math>\angle B</math>. From the angle bisector theorem, we have <math>\frac{AF}{AB} = \frac{CF}{CB}, so </math>AF = 15/7<math> and </math>CF = 20/7<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}$
  
 
== 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 01:02, 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 $\angle B$. 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$ (Error compiling LaTeX. Unknown error_msg)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}$

See also

2014 AIME I (Problems • Answer Key • Resources)
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