Difference between revisions of "2022 AMC 10A Problems/Problem 15"
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Opposite angles of every cyclic quadrilateral are supplementary, so <cmath>\angle B + \angle D = 180^{\circ}.</cmath> | Opposite angles of every cyclic quadrilateral are supplementary, so <cmath>\angle B + \angle D = 180^{\circ}.</cmath> | ||
− | We claim that <math>AC=25.</math> | + | We claim that <math>AC=25.</math> We can prove it by contradiction: |
* If <math>AC<25,</math> then <math>\angle B</math> and <math>\angle D</math> are both acute angles. This arrives at a contradiction. | * If <math>AC<25,</math> then <math>\angle B</math> and <math>\angle D</math> are both acute angles. This arrives at a contradiction. |
Revision as of 00:54, 12 November 2022
Problem
Quadrilateral with side lengths
is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form
where
and
are positive integers such that
and
have no common prime factor. What is
Solution
DIAGRAM IN PROGRESS.
WILL BE DONE TOMORROW, WAIT FOR ME THANKS.
Opposite angles of every cyclic quadrilateral are supplementary, so
We claim that
We can prove it by contradiction:
- If
then
and
are both acute angles. This arrives at a contradiction.
- If
then
and
are both obtuse angles. This arrives at a contradiction.
By the Inscribed Angle Theorem, we conclude that is the diameter of the circle. So, the radius of the circle is
The area of the requested region is
Therefore, the answer is
~MRENTHUSIASM
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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