Difference between revisions of "2005 AIME II Problems/Problem 14"
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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> | ||
| − | + | By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have | |
| − | == | + | <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> | ||
| − | {{ | + | 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 18:18, 25 July 2008
Problem
In triangle 
and 
Point 
is on 
with 
Point 
is on such that 
Given that 
where 
and 
are relatively prime positive integers, find 
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]](http://latex.artofproblemsolving.com/5/e/b/5eb6a853aed7859b36cd02006f4f16fa5f383ea2.png)
By the Law of Sines and since 
, we have
\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 
. The answer is 
.
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 | ||
