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Difference between revisions of "2022 AMC 10A Problems/Problem 15"

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== Solution 2 (Brahmagupta Formula)==
 
== Solution 2 (Brahmagupta Formula)==
  
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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>\frac{625\pi}{4}</math>. So now the area of the region we are trying to find is <math>\frac{625\pi}{4} - 236</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 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>
 
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 $ABCD$ with side lengths $AB=7, BC=24, CD=20, DA=15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{a\pi-b}{c},$ where $a,b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a+b+c?$

$\textbf{(A) } 260 \qquad \textbf{(B) } 855 \qquad \textbf{(C) } 1235 \qquad \textbf{(D) } 1565 \qquad \textbf{(E) } 1997$

Solution 1

DIAGRAM IN PROGRESS.

WILL BE DONE TOMORROW, WAIT FOR ME THANKS.

Opposite angles of every cyclic quadrilateral are supplementary, so \[\angle B + \angle D = 180^{\circ}.\] We claim that $AC=25.$ We can prove it by contradiction:

  • If $AC<25,$ then $\angle B$ and $\angle D$ are both acute angles. This arrives at a contradiction.
  • If $AC>25,$ then $\angle B$ and $\angle D$ are both obtuse angles. This arrives at a contradiction.

By the Inscribed Angle Theorem, we conclude that $\overline{AC}$ is the diameter of the circle. So, the radius of the circle is $r=\frac{AC}{2}=\frac{25}{2}.$

The area of the requested region is \[\pi r^2 - \frac12\cdot AB\cdot BC - \frac12\cdot AD\cdot DC = \frac{625\pi}{4}-\frac{168}{2}-\frac{300}{2}=\frac{625\pi-936}{4}.\] Therefore, the answer is $a+b+c=\boxed{\textbf{(D) } 1565}.$

~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 $A$ and $C$ to make a diameter of $25$. Since the diameter is 25, we can see the area of the circle is just $\frac{625\pi}{4}$ from the formula of the area of the circle with just a diameter. Then we can Brahmagupta Formula, $\sqrt{(s - a)(s - b)(s - c)(s - c)}$ 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 $\sqrt{54756}$ which is $234$. So now the area of the region we are trying to find is $\frac{625\pi}{4} - 234$ or it can be written as $\frac{625\pi-936}{4}$ Therefore, the answer is $a+b+c=\boxed{\textbf{(D) } 1565}.$

~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

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