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Difference between revisions of "2005 AIME II Problems/Problem 14"
(solution / asy)
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== Problem ==
 
== Problem ==
  
−
In triangle <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math>
+
In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math>
  
 
== Solution ==
 
== Solution ==
 +
<center><asy>
 +
pointpen = black; pathpen = black + linewidth(0.7); pen f = fontsize(10);
 +
pair C = (0,0), B=(15,0), A=IP(CR(B,13), CR(C,14)), D=(6,0), E = (4410/463,0);
 +
D(MP("A",A,N,f)--MP("B",B,f)--MP("C",C,f)--A--MP("D",D,f)--A--MP("E",E,f));
 +
MP("6",(D+C)/2,f);MP("13",(A+B)/2,NE,f);MP("14",(A+C)/2,NW,f);MP("9",(D+B)/2,N,f);
 +
D(anglemark(C,A,D,50)); D(anglemark(E,A,B,50));
 +
</asy></center>
  
−
{{solution}}
+
By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have
  
−
== See also ==
+
<center><math>\begin{align*}
 +
\frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} 
 +
= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB}
 +
= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}
 +
\end{align*}
 +
</math></center>
  
−
{{AIME box|year=2005|n=II|num-b=13|num-a=15}}
+
Substituting our knowns, we have <math>\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>.
  
 +
== See also ==
 +
{{AIME box|year=2005|n=II|num-b=13|num-a=15|t=368562}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 17:18, 25 July 2008

Problem

In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$

Solution

[asy] pointpen = black; pathpen = black + linewidth(0.7); pen f = fontsize(10); pair C = (0,0), B=(15,0), A=IP(CR(B,13), CR(C,14)), D=(6,0), E = (4410/463,0); D(MP("A",A,N,f)--MP("B",B,f)--MP("C",C,f)--A--MP("D",D,f)--A--MP("E",E,f)); MP("6",(D+C)/2,f);MP("13",(A+B)/2,NE,f);MP("14",(A+C)/2,NW,f);MP("9",(D+B)/2,N,f); D(anglemark(C,A,D,50)); D(anglemark(E,A,B,50)); [/asy]

By the Law of Sines and since $\angle BAE = \angle CAD, \angle BAD = \angle CAE$, we have

$\begin{align*} 
\frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC}

= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} = \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}

\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Substituting our knowns, we have $\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}$. The answer is $q = \boxed{463}$.

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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