Difference between revisions of "2022 AMC 10A Problems/Problem 15"
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== Solution 2 (Brahmagupta Formula)== | == Solution 2 (Brahmagupta Formula)== | ||
− | When we look at the side lengths of the quadrilateral we see 7 and 24, which screams out 25 because of Pythagorean triplets. As a result, we can draw a line through points <math>A</math> and <math>C</math> to make a diameter of <math>25</math>. Since the diameter is 25, we can see the area of the circle is just <math>\frac{625\pi}{4}</math> from the formula of the area of the circle with just a diameter. Then we can Brahmagupta Formula, <math>\sqrt{(s - a)(s - b)(s - c)(s - c)}</math> where a,b,c,d side lengths, and s is semi-perimeter to find the area of the quadrilateral. If we just plug the values in, we get <math>\sqrt{54756}</math> which is <math>234</math>. So now the area of the region we are trying to find is <math>\frac{625\pi}{4} - 234</math> or it can be written as <math>\frac{625\pi-936}{4}</math> | + | When we look at the side lengths of the quadrilateral we see 7 and 24, which screams out 25 because of Pythagorean triplets. As a result, we can draw a line through points <math>A</math> and <math>C</math> to make a diameter of <math>25</math>. Since the diameter is 25, we can see the area of the circle is just <math>\frac{625\pi}{4}</math> from the formula of the area of the circle with just a diameter. Then we can use Brahmagupta Formula, <math>\sqrt{(s - a)(s - b)(s - c)(s - c)}</math> where a,b,c,d are side lengths, and s is semi-perimeter to find the area of the quadrilateral. If we just plug the values in, we get <math>\sqrt{54756}</math> which is <math>234</math>. So now the area of the region we are trying to find is <math>\frac{625\pi}{4} - 234</math> or it can be written as <math>\frac{625\pi-936}{4}</math> |
Therefore, the answer is <math>a+b+c=\boxed{\textbf{(D) } 1565}.</math> | Therefore, the answer is <math>a+b+c=\boxed{\textbf{(D) } 1565}.</math> | ||
Revision as of 16:01, 13 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 1
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
Solution 2 (Brahmagupta Formula)
When we look at the side lengths of the 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 of the area of the circle with just a diameter. Then we can use Brahmagupta Formula,
where a,b,c,d 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
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.