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

Revision as of 20:54, 20 December 2014

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*}$ (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 $q = \boxed{463}$ .

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

Revision as of 20:54, 20 December 2014

Problem

Solution

See also

@@ Line 8: / Line 8: @@
 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);
+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></center>