Difference between revisions of "2013 AIME II Problems/Problem 8"
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Let us call <math>\angle BFO=\theta</math>. Therefore, <math>\angle AOB=2\theta</math>, and so <math>\angle AOE=90-2\theta</math>. Let us label the radius of the circle <math>r</math>. This means <cmath>\sin{\theta}=\frac{BF}{r}=\frac{11}{r}</cmath> <cmath>\sin{(90-2\theta)}=\frac{BE}{r}=\frac{10}{r}</cmath> | Let us call <math>\angle BFO=\theta</math>. Therefore, <math>\angle AOB=2\theta</math>, and so <math>\angle AOE=90-2\theta</math>. Let us label the radius of the circle <math>r</math>. This means <cmath>\sin{\theta}=\frac{BF}{r}=\frac{11}{r}</cmath> <cmath>\sin{(90-2\theta)}=\frac{BE}{r}=\frac{10}{r}</cmath> | ||
Now we can use simple trigonometry to solve for <math>r</math>. | Now we can use simple trigonometry to solve for <math>r</math>. | ||
− | Recall that <math>\sin{(90-\alpha)}=\cos(\alpha)</math>: That means <math>\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}</math> | + | Recall that <math>\sin{(90-\alpha)}=\cos(\alpha)</math>: That means <math>\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}</math>. |
Recall that <math>\cos{2\alpha}=1-2\sin^2{\alpha}</math>: That means <math>\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}</math>. | Recall that <math>\cos{2\alpha}=1-2\sin^2{\alpha}</math>: That means <math>\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}</math>. | ||
Let <math>\sin{\theta}=x</math>. | Let <math>\sin{\theta}=x</math>. |
Revision as of 21:23, 4 April 2013
A hexagon that is inscribed in a circle has side lengths ,
,
,
,
, and
in that order. The radius of the circle can be written as
, where
and
are positive integers. Find
.
Solution
Let us call the hexagon , where
, and
.
We can just consider one half of the hexagon,
, to make matters simpler.
Draw a line from the center of the circle,
, to the midpoint of
,
. Now, draw a line from
to the midpoint of
,
. Clearly,
, because
, and
, for similar reasons. Also notice that
.
Let us call
. Therefore,
, and so
. Let us label the radius of the circle
. This means
Now we can use simple trigonometry to solve for
.
Recall that
: That means
.
Recall that
: That means
.
Let
.
Substitute to get
and
Now substitute the first equation into the second equation:
Multiplying both sides by
and reordering gives us the quadratic
Using the quadratic equation to solve, we get that
(because
gives a negative value), so the answer is