Difference between revisions of "2022 AMC 10A Problems/Problem 15"
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+ | == Video Solution == | ||
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+ | https://youtu.be/x3DrtvR3sQ8 | ||
== See Also == | == See Also == |
Revision as of 14:39, 20 November 2022
Contents
[hide]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 . 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 s is semi-perimeter to find the area of the quadrilateral.
If we just plug the values in, we get which is . So now the area of the region we are trying to find is or it can be written as
Therefore, the answer is
~Gdking
Solution 3 (if you didn't realize the diagonal was the diameter)
We can observe that this quadrilateral is actually made of two right triangles: has a ratio in the side lengths, and is a triangle.
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
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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