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 "2020 AMC 8 Problems/Problem 18"

(→‎Solution 1)
(→‎Solution 1)
Line 10: Line 10:
 
<asy>  draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$O$", (0,0), 1.25*S); draw((0,0)--(-8,15));</asy>
 
<asy>  draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$O$", (0,0), 1.25*S); draw((0,0)--(-8,15));</asy>
  
−
Let <math>O</math> be the center of the semicircle. The diameter of the semicircle is <math>9+16+9=34</math>, so <math>OC = 17</math>. By symmetry, <math>O</math> is in fact the midpoint of <math>DA</math>, so <math>OD=OA=\frac{16}{2}= 8. By the Pythagorean theorem in right-angled triangle </math>ODC<math> (or </math>OBA<math>), we have that </math>CD<math> (or </math>AB<math>) is </math>\sqrt{17^2-8^2}=15<math>. Accordingly, the area of </math>ABCD<math> is </math>16\cdot 15=\boxed{\textbf{(A) }240}$.
+
Let <math>O</math> be the center of the semicircle. The diameter of the semicircle is <math>9+16+9=34</math>, so <math>OC = 17</math>. By symmetry, <math>O</math> is in fact the midpoint of <math>DA</math>, so <math>OD=OA=\frac{16}{2}= 8</math>. By the Pythagorean theorem in right-angled triangle <math>ODC</math> (or <math>OBA</math>), we have that <math>CD</math> (or <math>AB</math>) is <math>\sqrt{17^2-8^2}=15</math>. Accordingly, the area of <math>ABCD</math> is <math>16\cdot 15=\boxed{\textbf{(A) }240}</math>.
  
 
==Solution 2 (coordinate geometry)==
 
==Solution 2 (coordinate geometry)==

Revision as of 20:48, 24 November 2020

Problem

Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$

[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); [/asy] $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$

Solution 1

[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$O$", (0,0), 1.25*S); draw((0,0)--(-8,15));[/asy]

Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$, so $OC = 17$. By symmetry, $O$ is in fact the midpoint of $DA$, so $OD=OA=\frac{16}{2}= 8$. By the Pythagorean theorem in right-angled triangle $ODC$ (or $OBA$), we have that $CD$ (or $AB$) is $\sqrt{17^2-8^2}=15$. Accordingly, the area of $ABCD$ is $16\cdot 15=\boxed{\textbf{(A) }240}$.

Solution 2 (coordinate geometry)

Let the midpoint of segment $FE$ be the origin. Evidently, point $D=(-8,0)$ and $A=(8,0)$. Since points $C$ and $B$ share $x$-coordinates with $D$ and $A$ respectively, it suffices to find the $y$-coordinate of $B$ (which will be the height of the rectangle) and multiply this by $DA$ (which we know is $16$). The radius of the semicircle is $\frac{9+16+9}{2} = 17$, so the whole circle has equation $x^2+y^2=289$; as already stated, $B$ has the same $x$-coordinate as $A$, i.e. $8$, so substituting this into the equation shows that $y=\pm15$. Since $y>0$ at $B$, the y-coordinate of $B$ is $15$. Therefore, the answer is $16\cdot 15 = \boxed{\textbf{(A) }240}$.

(Note that the synthetic solution (Solution 1) is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier strategy.)

Solution 3

We can use a result from the Art of Problem Solving Introduction to Algebra book: for a semicircle with diameter $(1+n)$, such that the $1$ part is on one side and the $n$ part is on the other side, the height from the end of the $1$ side (or the start of the $n$ side) is $\sqrt{n}$. To use this, we scale the figure down by $9$; then the height is $\sqrt{1+\frac{16}{9}} = \sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$. Now, scaling back up by $9$, the height $DC$ is $9 \cdot \frac{5}{3} = 15$. The answer is $15 \cdot 16 = \boxed{\textbf{(A) }240}$.
-SweetMango77; edited by Sevenoptimus, who made it easier to read

Video Solution

https://youtu.be/l9wZS3qGSCg

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 AJHSME/AMC 8 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=2020_AMC_8_Problems/Problem_18&oldid=138429"