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

(Created page with "==Problem== Quadrilateral <math>ABCD</math> with side lengths <math>AB=7, BC=24, CD=20, DA=15</math> is inscribed in a circle. The area interior to the circle but exterior to...")
 
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==Solution==
 
==Solution==
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<b>DIAGRAM IN PROGRESS.
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 +
WILL BE DONE TOMORROW, WAIT FOR ME THANKS.</b>
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 +
Opposite angles of every cyclic quadrilateral are supplementary, so <math>\angle B + \angle D = 180^{\circ}.</math>
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 +
We claim that <math>\angle B = \angle D = 90^\circ,</math> so <math>AC=25</math> by the Pythagorean Theorem. We can prove the claim by contradiction: Suppose that <math>\angle B = \angle D = 90^\circ</math> is false. Then, we can apply the Hinge Theorem to <math>\triangle ABC</math> and <math>\triangle ADC:</math> We get <math>AC>25</math> in one triangle, and <math>AC<25</math> in the other triangle, arriving at a contradiction. Therefore, the claim must be true.
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By the Inscribed Angle Theorem, we conclude that <math>\overline{AC}</math> is the diameter of the circle. So, the radius of the circle is <math>r=\frac{AC}{2}=\frac{25}{2}.</math>
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The area of the requested region is <cmath>\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}.</cmath>
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Therefore, the answer is <math>a+b+c=\boxed{\textbf{(D) } 1565}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 23:34, 11 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

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 $\angle B = \angle D = 90^\circ,$ so $AC=25$ by the Pythagorean Theorem. We can prove the claim by contradiction: Suppose that $\angle B = \angle D = 90^\circ$ is false. Then, we can apply the Hinge Theorem to $\triangle ABC$ and $\triangle ADC:$ We get $AC>25$ in one triangle, and $AC<25$ in the other triangle, arriving at a contradiction. Therefore, the claim must be true.

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

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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