Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (Brahmagupta Formula))
(→‎Solution 2 (Brahmaguptas Formula))
Line 24: Line 24:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
== Solution 2 (Brahmaguptas Formula)==
+
== 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 <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.  
+
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>. 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 a,b,c,d are side lengths, and s is semi-perimeter to find the area of the quadrilateral.  
+
*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 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>
 
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>

Revision as of 14:08, 14 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‘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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_15&oldid=181428"