Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2012 AIME II Problems/Problem 15"

m (→‎Solution)
(→‎Solution)
Line 13: Line 13:
 
<math>EF = AF \cdot cos \angle AFE + AE \cdot cos \angle AEF</math> (2).  Ah, but <math>\angle AFE = \angle ACE</math> (since F lies on the same circle), and we can find <math>cos \angle ACE</math> using the law of cosines:
 
<math>EF = AF \cdot cos \angle AFE + AE \cdot cos \angle AEF</math> (2).  Ah, but <math>\angle AFE = \angle ACE</math> (since F lies on the same circle), and we can find <math>cos \angle ACE</math> using the law of cosines:
  
−
<math>AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot cos \angle ACE</math>, and plugging in <math>AE = 8, AC = 3, BE = BC = 7,</math> we get <math>cos \angle ACE = -1/7</math>.
+
<math>AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot cos \angle ACE</math>, and plugging in <math>AE = 8, AC = 3, BE = BC = 7,</math> we get <math>cos \angle ACE = -1/7 = cos \angle AFE</math>.
  
 
Also, <math>\angle AEF = \angle DEF</math>, and <math>\angle DFE = \pi/2</math> (since <math>F</math> is on the circle with diameter <math>DE</math>), so <math>cos \angle AEF = EF/DE = 8 \cdot EF/49</math>.   
 
Also, <math>\angle AEF = \angle DEF</math>, and <math>\angle DFE = \pi/2</math> (since <math>F</math> is on the circle with diameter <math>DE</math>), so <math>cos \angle AEF = EF/DE = 8 \cdot EF/49</math>.   

Revision as of 23:59, 17 April 2012

Problem 15

Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Use angle bisector theorem to find $CD=21/8$, $BD=35/8$, and $AD=15/8$. Use Power of the Point to find $ED=49/8$, and so $AE=8$. Use law of cosines to find $\angle CAD = \pi /3$, hence $\angle BAD = \pi /3$ as well, and $\triangle BCE$ is equilateral.

I'm sure there is a more elegant solution from here, but instead we do'll some hairy law of cosines:

$AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot cos \angle AFE.$ (1)

$AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot cos \angle AEF.$ Adding these two and simplifying we get:

$EF = AF \cdot cos \angle AFE + AE \cdot cos \angle AEF$ (2). Ah, but $\angle AFE = \angle ACE$ (since F lies on the same circle), and we can find $cos \angle ACE$ using the law of cosines:

$AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot cos \angle ACE$, and plugging in $AE = 8, AC = 3, BE = BC = 7,$ we get $cos \angle ACE = -1/7 = cos \angle AFE$.

Also, $\angle AEF = \angle DEF$, and $\angle DFE = \pi/2$ (since $F$ is on the circle with diameter $DE$), so $cos \angle AEF = EF/DE = 8 \cdot EF/49$.

Plugging in all our values into equation (2), we get:

$EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}$, or $EF = \frac{7}{15} \cdot AF$.

Finally, we plug this into equation (1), yielding:

$8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot EF \cdot \frac{-1}{7}$ Thus,

$64 = \frac{AF^2}{225} \cdot (225+49+30),$ or $AF^2 = \frac{900}{19}.$ The answer is $919$.

See Also

2012 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&oldid=46452"