2022 AMC 10A Problems/Problem 15
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 1
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
Solution 2 (Brahmagupta Formula)
When we look at the sidelengths of quadrilateral we see 7 and 24, which screams out 25 because of Pythagorean triplets. As a result, we can draw a line through points 
and 
to make a diameter of 
. Since the diameter is 25, we can see the area of the circle is just 
from the formula doe the area of the circle with just a diameter. Then using Brahmagupta Formula,
See Also
| 2022 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
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