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

Revision as of 15:39, 20 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 $a+b+c?$

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

Diagram

$[asy] /* Made by MRENTHUSIASM */ size(200); pair O, A, B, C, D; O = origin; A = (-25/2,0); C = (25/2,0); B = intersectionpoints(Circle(A,7),Circle(C,24))[0]; D = intersectionpoints(Circle(A,15),Circle(C,20))[1]; fill(Circle(O,25/2),yellow); fill(A--B--C--D--cycle,white); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot(O,linewidth(4)); draw(Circle(O,25/2)); draw(A--B--C--D--cycle); label("$7$",midpoint(A--B),rotate(90)*dir(midpoint(A--B)--A)); label("$24$",midpoint(B--C),rotate(-90)*dir(midpoint(B--C)--B)); label("$20$",midpoint(C--D),rotate(-90)*dir(midpoint(C--D)--C)); label("$15$",midpoint(D--A),rotate(90)*dir(midpoint(D--A)--D)); [/asy]$ ~MRENTHUSIASM

Solution 1 (Inscribed Angle Theorem)

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‘s 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 use Brahmagupta Formula, $\sqrt{(s - a)(s - b)(s - c)(s - d)}$ 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 $\sqrt{54756}$ 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

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: $\triangle CDA$ has a $3-4-5$ ratio in the side lengths, and $\triangle ABC$ is a $7-24-25$ triangle.

Next, we can choose one of these triangles and use the circumradius formula to find the radius. Let's choose the $15-20-25$ triangle. The area of the triangle is equal to the product of the side lengths divided by $4$ times the circumradius. Therefore, $150 = \frac{15\cdot20\cdot25}{4r}$ . Solving this simple algebraic equation gives us $r = \frac{25}{2}$ .

Plugging in the values, we have $\frac{25}{2}^2\cdot\pi - (\frac{15\cdot20}{2}+\frac{7\cdot24}{2}) = \frac{625\cdot\pi}{4} - 234$ . Rewriting this gives us $\frac{625\pi-936}{4}$ .

Therefore, adding these values gets us $\boxed{\textbf{(D) } 1565}.$

~orenbad

Video Solution

https://youtu.be/x3DrtvR3sQ8

- Whiz

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 https://youtu.be/x3DrtvR3sQ8
+- Whiz
 == See Also ==

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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