Difference between revisions of "2013 AIME II Problems/Problem 8"
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+ | Using solution 1's diagram, extend line segments <math>AB</math> and <math>CD</math> upwards until they meet at point <math>G</math>. Let point <math>O</math> be the center of the hexagon. By the <math>AA</math> postulate, <math>\Delta ADG \sim \Delta BCG \sim \Delta CDO</math>. This means <math>\frac{DC}{AD} = \frac{22}{2r} = \frac{CO}{AG}</math>, so <math>AG = r \times \frac{2r}{22} = \frac{r^2}{11}</math>. We then solve for <math>AB</math>: <cmath> \frac{AD-BC}{AD} = \frac{2r-20}{2r} = \frac{AB}{AG} </cmath> <cmath>AB = \frac{r^2}{11} \times \frac{2r-20}{2r} = \frac{r^2-10r}{11}</cmath> | ||
+ | Remember that <math>AB=22</math> as well, so <math>22 = \frac{r^2-10r}{11} \Rightarrow r^2-10r-242=0</math>. Solving for <math>r</math> gives <math>r=5+\sqrt{267}</math>. So the solution is <math>5+267=\boxed{272}</math>. | ||
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+ | ~SoilMilk | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2013|n=II|num-b=7|num-a=9}} | {{AIME box|year=2013|n=II|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:33, 27 May 2023
Contents
Problem 8
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
.
Solution 3
Join the diameter of the circle and let the length be
. By Ptolemy's Theorem on trapezoid
,
. Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to
each. Then
Since is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right
:
From the above equations, we have:
Since the radius is half the diameter, it is , so the answer is
.
Solution 4
As we can see this image, it is symmetrical hence the diameter divides the hexagon into two congruent quadrilateral. Now we can apply the Ptolemy's theorem. Denote the radius is r, we can get , after simple factorization, we can get
, it is easy to see that
are two solutions for the equation, so we can factorize that into
so we only need to find the solution for
and we can get
is the desired answer for the problem, and our answer is
.~bluesoul
Solution 5
Using solution 1's diagram, extend line segments and
upwards until they meet at point
. Let point
be the center of the hexagon. By the
postulate,
. This means
, so
. We then solve for
:
Remember that
as well, so
. Solving for
gives
. So the solution is
.
~SoilMilk
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.