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)
(→‎Video Solution by Math-X (First understand the problem!!!))
 
(71 intermediate revisions by 33 users not shown)
Line 1: Line 1:
 +
==Problem==
 
Rectangle <math>ABCD</math> is inscribed in a semicircle with diameter <math>\overline{FE},</math> as shown in the figure. Let <math>DA=16,</math> and let <math>FD=AE=9.</math> What is the area of <math>ABCD?</math>
 
Rectangle <math>ABCD</math> is inscribed in a semicircle with diameter <math>\overline{FE},</math> as shown in the figure. Let <math>DA=16,</math> and let <math>FD=AE=9.</math> What is the area of <math>ABCD?</math>
  
Line 6: Line 7:
 
<math>\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272</math>
 
<math>\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272</math>
  
−
==Solution==
+
==Solution 1 (Pythagorean Theorem)==
−
First, realize <math>ABCD</math> is not a square. It can easily be seen that the diameter of the semicircle is <math>9+16+9=34</math>, so the radius is <math>\frac{34}{2}=17</math>. Express the area of Rectangle <math>ABCD</math> as <math>16h</math>, where <math>h=AB</math>. Notice that by the Pythagorean theorem <math>8^2+h^{2}=17^{2}\implies h=15</math>. Then, the area of Rectangle <math>ABCD</math> is equal to <math>16\cdot 15=\boxed{\textbf{(A) }240}</math>. ~icematrix
+
<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 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==
+
==Solution 2 (Coordinate Geometry)==
 +
Let the midpoint of segment <math>FE</math> be the origin. Evidently, point <math>D=(-8,0)</math> and <math>A=(8,0)</math>. Since points <math>C</math> and <math>B</math> share <math>x</math>-coordinates with <math>D</math> and <math>A</math> respectively, it suffices to find the <math>y</math>-coordinate of <math>B</math> (which will be the height of the rectangle) and multiply this by <math>DA</math> (which we know is <math>16</math>). The radius of the semicircle is <math>\frac{9+16+9}{2} = 17</math>, so the whole circle has equation <math>x^2+y^2=289</math>; as already stated, <math>B</math> has the same <math>x</math>-coordinate as <math>A</math>, i.e. <math>8</math>, so substituting this into the equation shows that <math>y=\pm15</math>. Since <math>y>0</math> at <math>B</math>, the y-coordinate of <math>B</math> is <math>15</math>. Therefore, the answer is <math>16\cdot 15 = \boxed{\textbf{(A) }240}</math>.
 +
 
 +
(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 <i>Introduction to Algebra</i> book Sidenote: for a semicircle with diameter <math>(1+n)</math>, such that the <math>1</math> part is on one side and the <math>n</math> part is on the other side, the height from the end of the <math>1</math> side (or the start of the <math>n</math> side) is <math>\sqrt{n}</math>. To use this formula, we scale the figure down by <math>9</math>; this will give the height a length of <math>\sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}</math>. Now, scaling back up by <math>9</math>, the height <math>DC</math> is <math>9 \cdot \frac{5}{3} = 15</math>. The answer is then <math>15 \cdot 16 = \boxed{\textbf{(A) }240}</math>.
 +
-[[User:Sweetmango77|SweetMango77]]
 +
 
 +
==Solution 4 (Power of a Point Theorem)==
 +
Draw the other half of the circle as follows:
 +
<asy>
 +
draw(arc((0,0),17,360,0)); 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*SE); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*SW); dot("$E$",(17,0), 1.25*E); dot("$F$",(-17,0), 1.25*W); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); draw((-8,-15)--(-8,0)--(8,0)--(8,-15)--cycle); dot("$B'$",(8,-15), 1.25*S); dot("$C'$",(-8,-15), 1.25*S);
 +
</asy>
 +
By the [[Power of a Point Theorem]], <math>FD\cdot DE = CD\cdot C'D</math>. By symmetry, <math>CD = C'D</math>. We see that <math>FD = 9</math> and <math>DE = 16 + 9 = 25</math>. Substituting in these values, <math>9\cdot 25 = CD^2</math>, giving <math>CD^2 = 225</math> and <math>CD = 15</math>.
 +
The area of the rectangle is therefore <math>15\cdot 16 = \boxed{\textbf{(A) }240}</math>.
 +
 
 +
==Solution 5 (Vertical Theorem)==
 +
<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),S); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$G$", (0,15), SE); dot("$O$", (0,0), NE); draw((0,0)--(0, 15)); draw((-7.5,15)--(0,0));
 +
</asy>
 +
 
 +
According to the Pythagorean Theorem and the Vertical Theorem, we can find out that <math> OG=\sqrt{\left(\frac{2\times9+16}{2}\right)^2 - \left(\frac{16}{2}\right)^2}=15 </math>. Therefore, the answer is <math> 15\times16=\boxed{\textbf{(A) }240} </math>;
 +
 
 +
~[[User:Bloggish|Bloggish]]
 +
 
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=bHNrBwwUCMI
 +
 
 +
~NiuniuMaths
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=_0fveg1r3WFmwmpw&t=3334
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (🚀 Quick 🚀)==
 +
https://youtu.be/pA6Smw91Bc0
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by North America Math Contest Go Go Go==
 +
https://www.youtube.com/watch?v=5Qo4pG3Uk_U
 +
 
 +
~North America Math Contest Go Go Go
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/l9wZS3qGSCg
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution==
 +
https://youtu.be/VnOecUiP-SA
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=852
 +
 
 +
~Interstigation
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/7SwJdAEOeAg?t=23
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=S_CnKCuOA_w
  
−
[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("<math>A</math>",(8,0), 1.25*S); dot("<math>B</math>",(8,15), 1.25*N); dot("<math>C</math>",(-8,15), 1.25*N); dot("<math>D</math>",(-8,0), 1.25*S); dot("<math>E</math>",(17,0), 1.25*S); dot("<math>F</math>",(-17,0), 1.25*S); label("<math>16</math>",(0,0),N); label("<math>9</math>",(12.5,0),N); label("<math>9</math>",(-12.5,0),N); dot("<math>O</math>", (0,0), 1.25*N);[/asy]
 
  
−
We have <math>OC=17</math>, as it is a radius, and <math>OD=8</math> since it is half of <math>AD</math>. This means that <math>CD=\sqrt{17^2-8^2}=15</math>. So <math>16*15=\boxed{\textbf{(A)}240}</math>
 
  
−
~yofro
 
  
−
==See also==
 
 
{{AMC8 box|year=2020|num-b=17|num-a=19}}
 
{{AMC8 box|year=2020|num-b=17|num-a=19}}
−
 
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
>B)
 +
hey ya'lll

Latest revision as of 15:35, 26 January 2024

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 (Pythagorean Theorem)

[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 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 Sidenote: 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 formula, we scale the figure down by $9$; this will give the height a length of $\sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$. Now, scaling back up by $9$, the height $DC$ is $9 \cdot \frac{5}{3} = 15$. The answer is then $15 \cdot 16 = \boxed{\textbf{(A) }240}$. -SweetMango77

Solution 4 (Power of a Point Theorem)

Draw the other half of the circle as follows: [asy] draw(arc((0,0),17,360,0)); 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*SE); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*SW); dot("$E$",(17,0), 1.25*E); dot("$F$",(-17,0), 1.25*W); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); draw((-8,-15)--(-8,0)--(8,0)--(8,-15)--cycle); dot("$B'$",(8,-15), 1.25*S); dot("$C'$",(-8,-15), 1.25*S); [/asy] By the Power of a Point Theorem, $FD\cdot DE = CD\cdot C'D$. By symmetry, $CD = C'D$. We see that $FD = 9$ and $DE = 16 + 9 = 25$. Substituting in these values, $9\cdot 25 = CD^2$, giving $CD^2 = 225$ and $CD = 15$. The area of the rectangle is therefore $15\cdot 16 = \boxed{\textbf{(A) }240}$.

Solution 5 (Vertical Theorem)

[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),S); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$G$", (0,15), SE); dot("$O$", (0,0), NE); draw((0,0)--(0, 15)); draw((-7.5,15)--(0,0)); [/asy]

According to the Pythagorean Theorem and the Vertical Theorem, we can find out that $OG=\sqrt{\left(\frac{2\times9+16}{2}\right)^2 - \left(\frac{16}{2}\right)^2}=15$. Therefore, the answer is $15\times16=\boxed{\textbf{(A) }240}$;

~Bloggish

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=_0fveg1r3WFmwmpw&t=3334

~Math-X

Video Solution (🚀 Quick 🚀)

https://youtu.be/pA6Smw91Bc0

~Education, the Study of Everything

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=5Qo4pG3Uk_U

~North America Math Contest Go Go Go

Video Solution by WhyMath

https://youtu.be/l9wZS3qGSCg

~savannahsolver

Video Solution

https://youtu.be/VnOecUiP-SA

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=852

~Interstigation

Video Solution by OmegaLearn

https://youtu.be/7SwJdAEOeAg?t=23

~ pi_is_3.14

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=S_CnKCuOA_w



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

>B) hey ya'lll

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_18&oldid=213192"