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

m (→‎Solution)
(→‎Solution)
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== Solution ==
 
== Solution ==
 
<center><asy>
 
<center><asy>
−
pointpen = black; pathpen = black + linewidth(0.7); pen f = fontsize(10);
+
import olympiad; import cse5; import geometry; size(150);
−
pair C = (0,0), B=(15,0), A=IP(CR(B,13), CR(C,14)), D=(6,0), E = (4410/463,0);
+
defaultpen(fontsize(10pt));
−
D(MP("A",A,N,f)--MP("B",B,f)--MP("C",C,f)--A--MP("D",D,f)--A--MP("E",E,f));
+
defaultpen(0.8);
−
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);
+
dotfactor = 4;
−
D(anglemark(C,A,D,50)); D(anglemark(E,A,B,50));
+
pair A = origin;
 +
pair C = rotate(15,A)*(A+dir(-50));
 +
pair B = rotate(15,A)*(A+dir(-130));
 +
pair D = extension(A,A+dir(-68),B,C);
 +
pair E = extension(A,A+dir(-82),B,C);
 +
label("$A$",A,N);
 +
label("$B$",B,SW);
 +
label("$D$",D,SE);
 +
label("$E$",E,S);
 +
label("$C$",C,SE);
 +
draw(A--B--C--cycle);
 +
draw(A--E);
 +
draw(A--D);
 +
draw(anglemark(B,A,E,5));
 +
draw(anglemark(D,A,C,5));
 
</asy></center>
 
</asy></center>
  

Revision as of 20:12, 16 April 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] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/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 (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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