Difference between revisions of "2013 AIME II Problems/Problem 8"
(→Solution) |
(→Solution) |
||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
+ | ===Solution 1=== | ||
Let us call the hexagon <math>ABCDEF</math>, where <math>AB=CD=DE=AF=22</math>, and <math>BC=EF=20</math>. | Let us call the hexagon <math>ABCDEF</math>, where <math>AB=CD=DE=AF=22</math>, and <math>BC=EF=20</math>. | ||
We can just consider one half of the hexagon, <math>ABCD</math>, to make matters simpler. | We can just consider one half of the hexagon, <math>ABCD</math>, to make matters simpler. | ||
Line 15: | Line 16: | ||
Multiplying both sides by <math>r^2</math> and reordering gives us the quadratic <cmath>r^2-10r-242=0</cmath> | Multiplying both sides by <math>r^2</math> and reordering gives us the quadratic <cmath>r^2-10r-242=0</cmath> | ||
Using the quadratic equation to solve, we get that <math>r=5+\sqrt{267}</math> (because <math>5-\sqrt{267}</math> gives a negative value), so the answer is <math>5+267=\boxed{272}</math> | Using the quadratic equation to solve, we get that <math>r=5+\sqrt{267}</math> (because <math>5-\sqrt{267}</math> gives a negative value), so the answer is <math>5+267=\boxed{272}</math> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | 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> |
Revision as of 21:42, 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