Difference between revisions of "2005 AIME II Problems/Problem 14"
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pair C = (0,0), B=(15,0), A=IP(CR(B,13), CR(C,14)), D=(6,0), E = (4410/463,0); | 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)); | 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(" | + | 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)); | D(anglemark(C,A,D,50)); D(anglemark(E,A,B,50)); | ||
</asy></center> | </asy></center> |
Revision as of 20:54, 20 December 2014
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("",(D+B)/2,N,f); D(anglemark(C,A,D,50)); D(anglemark(E,A,B,50)); [/asy]](http://latex.artofproblemsolving.com/8/4/3/843eac690811f538587c302061d460701e05dff2.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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.