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

m (→‎Solution 1 (Inscribed Angle Theorem): added law of cosines)
(→‎Solution 1 (Inscribed Angle Theorem))
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* If <math>AC>25,</math> then <math>\angle B</math> and <math>\angle D</math> are both obtuse angles. This arrives at a contradiction.
 
* If <math>AC>25,</math> then <math>\angle B</math> and <math>\angle D</math> are both obtuse angles. This arrives at a contradiction.
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(This can also be shown using Law of Cosines: Since the sum of the squares of (7 and 24) and of (20 and 15) are equal, and the cosines of the corner angles are additive opposites, the cosines of the angles must 0. --pinava)
 
  
 
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>
 
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>

Revision as of 00:43, 3 September 2023

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 $25.$ See Solution 1 for a rigorous proof.

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 $\sqrt{54756}=234.$ So now the area of the region we are trying to find is $\frac{625\pi}{4} - 234 = \frac{625\pi-936}{4}.$

Therefore, the answer is $a+b+c=\boxed{\textbf{(D) } 1565}.$

~Gdking

Solution 3 (Circumradius's Formula)

We can guess that this quadrilateral is actually made of two right triangles: $\triangle CDA$ has a $3 \text{-} 4 \text{-} 5$ ratio in the side lengths, and $\triangle ABC$ is a $7 \text{-} 24 \text{-} 25$ triangle. (See Solution 1 for a proof.)

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 $\frac{25}{2}^2\cdot\pi - \left(\frac{15\cdot20}{2}+\frac{7\cdot24}{2}\right) = \frac{625\cdot\pi}{4} - 234$. Rewriting this gives us $\frac{625\pi-936}{4}$.

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

~orenbad

Video Solution 1

https://youtu.be/ZHuInvG82PY

~Education, the Study of Everything

Video Solution 2

https://youtu.be/x3DrtvR3sQ8

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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