Difference between revisions of "2005 AIME II Problems/Problem 14"

m (Solution)
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By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have
 
By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have
  
<center><math>\begin{align*}
+
<cmath>\begin{align*}
  \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC}   
+
  \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 BAD}{\sin ADB \sin AEB} \\
= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}
+
&= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}
 
\end{align*}
 
\end{align*}
</math></center>
+
</cmath>
  
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>.  
+
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 ==
 
== See also ==

Revision as of 19:59, 13 March 2015

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("",(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*}

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 (ProblemsAnswer KeyResources)
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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