Difference between revisions of "2013 AIME II Problems/Problem 8"
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− | Using the trapezoid <math>ABCD</math> mentioned above, draw an altitude of the trapezoid passing through point <math>B</math> onto <math>AD</math> at point <math>E</math>. Now, we can use the pythagorean theorem: <math>(22^2-(r-10)^2)+10^2=r^2</math>. Expanding and combining like terms gives us the quadratic <cmath>r^2-10r-242</cmath> and solving for <math>r</math> gives <math>r=5+\sqrt{267}</math>. So the solution is <math>5+267=\boxed{272}</math> | + | Using the trapezoid <math>ABCD</math> mentioned above, draw an altitude of the trapezoid passing through point <math>B</math> onto <math>AD</math> at point <math>E</math>. Now, we can use the pythagorean theorem: <math>(22^2-(r-10)^2)+10^2=r^2</math>. Expanding and combining like terms gives us the quadratic <cmath>r^2-10r-242=0</cmath> and solving for <math>r</math> gives <math>r=5+\sqrt{267}</math>. So the solution is <math>5+267=\boxed{272}</math> |
Revision as of 21:43, 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
Solution 1
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
Solution 2
Using the trapezoid mentioned above, draw an altitude of the trapezoid passing through point
onto
at point
. Now, we can use the pythagorean theorem:
. Expanding and combining like terms gives us the quadratic
and solving for
gives
. So the solution is