2022 AMC 10A Problems/Problem 15
Contents
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
Diagram
~MRENTHUSIASM
Solution 1 (Inscribed Angle Theorem)
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
Solution 2 (Brahmagupta‘s Formula)
When we look at the side lengths of the quadrilateral we see and which screams out because of Pythagorean triplets. As a result, we can draw a line through points and to make a diameter of See Solution 1 for a rigorous proof.
This can also be shown using the Law of Cosines: Since and it follows that
Since the diameter is we can see the area of the circle is just from the formula of the area of the circle with just a diameter.
Then we can use Brahmagupta Formula where are side lengths, and is semi-perimeter to find the area of the quadrilateral.
If we just plug the values in, we get So now the area of the region we are trying to find is
Therefore, the answer is
~Gdking ~Oinava
Solution 3 (Circumradius's Formula)
We can guess that this quadrilateral is actually made of two right triangles: has a ratio in the side lengths, and is a triangle. (See Solution 1 for a proof.)
Next, we can choose one of these triangles and use the circumradius formula to find the radius. Let's choose the triangle. The area of the triangle is equal to the product of the side lengths divided by times the circumradius. Therefore, . Solving this simple algebraic equation gives us .
Plugging in the values, we have . Rewriting this gives us .
Therefore, adding these values gets us
Video Solution 1
~Education, the Study of Everything
Video Solution 2
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.