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 ![\[\angle B + \angle D = 180^{\circ}.\]](http://latex.artofproblemsolving.com/1/0/b/10b835ac4f345b1691b1f847eb0bd5445af3efd0.png)
We claim that 
We can prove it by contradiction:
- If
then 
and 
are both acute angles. This arrives at a contradiction.
then 
and 
are both acute angles. This arrives at a contradiction.- If
then and are both obtuse angles. This arrives at a contradiction.
then 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 ![\[\pi r^2 - \frac12\cdot AB\cdot BC - \frac12\cdot AD\cdot DC = \frac{625\pi}{4}-\frac{168}{2}-\frac{300}{2}=\frac{625\pi-936}{4}.\]](http://latex.artofproblemsolving.com/e/a/b/eabc501f3d3709f5610b40fa8eef1209001f1b06.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
