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

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<math>\textbf{(A) } 260 \qquad \textbf{(B) } 855 \qquad \textbf{(C) } 1235 \qquad \textbf{(D) } 1565 \qquad \textbf{(E) } 1997</math>
 
<math>\textbf{(A) } 260 \qquad \textbf{(B) } 855 \qquad \textbf{(C) } 1235 \qquad \textbf{(D) } 1565 \qquad \textbf{(E) } 1997</math>
  
−
==Solution 1==
+
==Diagram==
−
<b>DIAGRAM IN PROGRESS.
+
<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
  
−
WILL BE DONE TOMORROW, WAIT FOR ME THANKS.</b>
+
==Solution 1 (Inscribed Angle Theorem)==
  
 
Opposite angles of every cyclic quadrilateral are supplementary, so <cmath>\angle B + \angle D = 180^{\circ}.</cmath>
 
Opposite angles of every cyclic quadrilateral are supplementary, so <cmath>\angle B + \angle D = 180^{\circ}.</cmath>
Line 24: Line 47:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
== Solution 2 (Brahmagupta Formula)==
+
== Solution 2 (Brahmagupta‘s Formula)==
 +
 
 +
When we look at the side lengths of the quadrilateral we see <math>7</math> and <math>24,</math> which screams out <math>25</math> 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> See Solution 1 for a rigorous proof.
 +
 
 +
This can also be shown using the Law of Cosines: Since <math>7^2+24^2-2\cdot7\cdot24\cdot\cos B=15^2+20^2-2\cdot15\cdot20\cdot\cos D</math> and <math>\cos B + \cos D = 0,</math> it follows that <math>\cos B = \cos D = 0.</math>
 +
 
 +
Since the diameter is <math>25,</math> 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 - d)}</math> where <math>a,b,c,d</math> are side lengths, and <math>s</math> is semi-perimeter to find the area of the quadrilateral.
 +
 
 +
You can find the answer to this quickly by noting that <math>\sqrt{(33-7)(33-24)(33-20)(33-15)} = \sqrt{(26)(9)(13)(18)} = \sqrt{2^2 \cdot 3^4 \cdot 13^2} = 2 \cdot 9 \cdot 13 = 234</math>. So now the area of the region we are trying to find is <math>\frac{625\pi}{4} - 234 = \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, surd <math>(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>surd 54756</math> which is \frac{625\pi-936}{4}. 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>
 
 
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>
  
−
~Gdking
+
~Gdking ~Oinava ~[https://artofproblemsolving.com/wiki/index.php/User:South South]
 +
 
 +
== Solution 3 (Circumradius's Formula) ==
 +
 
 +
We can guess that this quadrilateral is actually made of two right triangles: <math>\triangle CDA</math> has a <math>3 \text{-} 4 \text{-} 5</math> ratio in the side lengths, and <math>\triangle ABC</math> is a <math>7 \text{-} 24 \text{-} 25</math> 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 <math>15-20-25</math> triangle. The area of the triangle is equal to the product of the side lengths divided by <math>4</math> times the circumradius. Therefore, <math>150 = \frac{15\cdot20\cdot25}{4r}</math>. Solving this simple algebraic equation gives us <math>r = \frac{25}{2}</math>.
 +
 
 +
Plugging in the values, we have <math>\frac{25}{2}^2\cdot\pi - \left(\frac{15\cdot20}{2}+\frac{7\cdot24}{2}\right) = \frac{625\cdot\pi}{4} - 234</math>. Rewriting this gives us <math>\frac{625\pi-936}{4}</math>.
 +
 
 +
Therefore, adding these values gets us <math>\boxed{\textbf{(D) } 1565}.</math>
 +
 
 +
~[[OrenSH|orenbad]]
 +
 
 +
==Video Solution 1==
 +
https://youtu.be/ZHuInvG82PY
 +
 
 +
~Education, the Study of Everything
 +
 
 +
== Video Solution 2==
 +
 
 +
https://youtu.be/Ov9AA7veKKk
 +
 
 +
== Video Solution 3==
 +
 
 +
https://youtu.be/x3DrtvR3sQ8
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/0kkc4-y8TkU?t=1632
  
 +
~IceMatrix
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2022|ab=A|num-b=14|num-a=16}}
 
{{AMC10 box|year=2022|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:06, 10 October 2024

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.

This can also be shown using the Law of Cosines: Since $7^2+24^2-2\cdot7\cdot24\cdot\cos B=15^2+20^2-2\cdot15\cdot20\cdot\cos D$ and $\cos B + \cos D = 0,$ it follows that $\cos B = \cos D = 0.$

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.

You can find the answer to this quickly by noting that $\sqrt{(33-7)(33-24)(33-20)(33-15)} = \sqrt{(26)(9)(13)(18)} = \sqrt{2^2 \cdot 3^4 \cdot 13^2} = 2 \cdot 9 \cdot 13 = 234$. 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 ~Oinava ~South

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/Ov9AA7veKKk

Video Solution 3

https://youtu.be/x3DrtvR3sQ8

Video Solution by TheBeautyofMath

https://youtu.be/0kkc4-y8TkU?t=1632

~IceMatrix

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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