Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 14"
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ALTERNATE SOLUTION: | ALTERNATE SOLUTION: | ||
Let AB=x. Use power of a point: | Let AB=x. Use power of a point: | ||
− | (1/18)/(1/18+2r)=(x)/(2x)=1/2 | + | $(1/18)/(1/18+2r)=(x)/(2x)=1/2 |
2/18=1/18+2r | 2/18=1/18+2r | ||
r=1/36 | r=1/36 |
Revision as of 16:21, 20 November 2009
Problem 14
Points and
lie on
, with radius
, so that
is acute. Extend
to point
so that
. Let
be the intersection of
and
such that
and
. If
can be written as
, where
and
are relatively prime and
is not divisible by the square of any prime, find
.
Solution
By the cosine double-angle formula,
![[asy] size(220); defaultpen(fontsize(10)); real r = (2+26^0.5)/198; pair O = (0,0), A=(r,0), B= A*dir(112.02432), C=2*B-A; pair[] D = intersectionpoints(C--O,Circle(O,r)); dot(O); dot(A); dot(B); dot(C); dot(D[0]); draw(O--A--C--cycle); draw(Circle(O,r)); draw(O--B,dashed); draw(anglemark(B,A,O,0.15)); label("\(A\)",A,NE); label("\(B\)",B,N); label("\(C\)",C,N); label("\(D\)",D[0],W); label("\(\frac{1}{18}\)",(C+D[0])/2,W); label("\(O\)",O,SW); [/asy]](http://latex.artofproblemsolving.com/4/c/1/4c1b94039cec9e7ac3ff8f368819afe9284d8251.png)
The Law of Cosines on with respect to
yields
\begin{align*}r^2 &= r^2 + AB^2 - 2 \cdot AB \cdot r \cos \angle BAO \\ AB^2 &= 2 \cdot AB \cdot r \cdot \frac{\sqrt{11}}{4}\\ AB &= \frac{r\sqrt{11}}{2} (Error compiling LaTeX. Unknown error_msg)
Now, . The Law of Cosines on
with respect to
yields
The answer is thus
.
ALTERNATE SOLUTION: Let AB=x. Use power of a point: $(1/18)/(1/18+2r)=(x)/(2x)=1/2 2/18=1/18+2r r=1/36
See also
Mock AIME 1 2007-2008 (Problems, Source) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |