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2013 AMC 8 Problems/Problem 23: Difference between revisions

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==Problem==
==Problem==
Angle <math>ABC</math> of <math>\triangle ABC</math> is a right angle. The sides of <math>\triangle ABC</math> are the diameters of semicircles as shown. The area of the semicircle on <math>\overline{AB}</math> equals <math>8\pi</math>, and the arc of the semicircle on <math>\overline{AC}</math> has length <math>8.5\pi</math>. What is the radius of the semicircle on <math>\overline{BC}</math>?
<asy>
import graph;
pair A,B,C;
A=(0,8);
B=(0,0);
C=(15,0);
draw((0,8)..(-4,4)..(0,0)--(0,8));
draw((0,0)..(7.5,-7.5)..(15,0)--(0,0));
real theta = aTan(8/15);
draw(arc((15/2,4),17/2,-theta,180-theta));
draw((0,8)--(15,0));
dot(A);
dot(C);
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);</asy>


==Solution==
<math>\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9</math>
 
==Video Solution==
https://youtu.be/crR3uNwKjk0 ~savannahsolver
 
==Solution 1==
If the semicircle on <math>\overline{AB}</math> were a full circle, the area would be <math>16\pi</math>.
 
<math>\pi r^2=16 \pi \Rightarrow r^2=16 \Rightarrow r=+4</math>, therefore the diameter of the first circle is <math>8</math>.
 
The arc of the largest semicircle is <math>8.5 \pi</math>, so if it were a full circle, the circumference would be <math>17 \pi</math>. So the <math>\text{diameter}=17</math>.
 
 
 
By the Pythagorean theorem, the other side has length <math>15</math>, so the radius is <math>\boxed{\textbf{(B)}\ 7.5}</math>
 
~Edited by Theraccoon to correct typos.
 
==Brief Explanation==
SavannahSolver got a diameter of <math>17</math> because the given arc length of the semicircle was
<math>8.5\pi</math>. The arc length of a semicircle can be calculated using the formula
<math>\pi r</math>, where
<math>r</math> is the radius. let’s use the full circumference formula for a circle, which is
<math>2\pi r</math>. Since the semicircle is half of a circle, its arc length is
<math>\pi r</math>, which was given as
<math>8.5\pi</math>. Solving for
<math>r</math>, we get
<math>r=8.5</math>
. Therefore, the diameter, which is
<math>2r</math>, is
<math>2\cdot8.5=17</math>
Then, the other steps to solve the problem will be the same as mentioned above by SavannahSolver
the answer is <math>\boxed{\textbf{(B)}\ 7.5}</math>
 
 
. - TheNerdWhoIsNerdy.
Minor edits by -Coin1
 
==Solution 2==
We go as in Solution 1, finding the diameter of the circle on <math>\overline{AC}</math> and <math>\overline{AB}</math>. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is <math>\frac{289\pi}{8}</math>, and the middle one is <math>\frac{289\pi}{8}-\frac{64\pi}{8}=\frac{225\pi}{8}</math>, so the radius is <math>\frac{15}{2}=\boxed{\textbf{(B)}\ 7.5}</math>.
 
~Note by Theraccoon: The person who posted this did not include their name.
 
==Video Solution by OmegaLearn==
https://youtu.be/abSgjn4Qs34?t=2584
 
~ pi_is_3.14
 
 
===Answer (B) 7.5===
~ Mia Wang the Author
~skibidi


==See Also==
==See Also==
{{AMC8 box|year=2013|before=First Problem|num-a=2}}
{{AMC8 box|year=2013|num-b=22|num-a=24}}
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 14:06, 24 August 2025

Problem

Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$, and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$. What is the radius of the semicircle on $\overline{BC}$? [asy] import graph; pair A,B,C; A=(0,8); B=(0,0); C=(15,0); draw((0,8)..(-4,4)..(0,0)--(0,8)); draw((0,0)..(7.5,-7.5)..(15,0)--(0,0)); real theta = aTan(8/15); draw(arc((15/2,4),17/2,-theta,180-theta)); draw((0,8)--(15,0)); dot(A); dot(C); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE);[/asy]

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 7.5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 8.5 \qquad \textbf{(E)}\ 9$

Video Solution

https://youtu.be/crR3uNwKjk0 ~savannahsolver

Solution 1

If the semicircle on $\overline{AB}$ were a full circle, the area would be $16\pi$.

$\pi r^2=16 \pi \Rightarrow r^2=16 \Rightarrow r=+4$, therefore the diameter of the first circle is $8$.

The arc of the largest semicircle is $8.5 \pi$, so if it were a full circle, the circumference would be $17 \pi$. So the $\text{diameter}=17$.


By the Pythagorean theorem, the other side has length $15$, so the radius is $\boxed{\textbf{(B)}\ 7.5}$

~Edited by Theraccoon to correct typos.

Brief Explanation

SavannahSolver got a diameter of $17$ because the given arc length of the semicircle was $8.5\pi$. The arc length of a semicircle can be calculated using the formula $\pi r$, where $r$ is the radius. let’s use the full circumference formula for a circle, which is $2\pi r$. Since the semicircle is half of a circle, its arc length is $\pi r$, which was given as $8.5\pi$. Solving for $r$, we get $r=8.5$ . Therefore, the diameter, which is $2r$, is $2\cdot8.5=17$ Then, the other steps to solve the problem will be the same as mentioned above by SavannahSolver the answer is $\boxed{\textbf{(B)}\ 7.5}$


. - TheNerdWhoIsNerdy. Minor edits by -Coin1

Solution 2

We go as in Solution 1, finding the diameter of the circle on $\overline{AC}$ and $\overline{AB}$. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is $\frac{289\pi}{8}$, and the middle one is $\frac{289\pi}{8}-\frac{64\pi}{8}=\frac{225\pi}{8}$, so the radius is $\frac{15}{2}=\boxed{\textbf{(B)}\ 7.5}$.

~Note by Theraccoon: The person who posted this did not include their name.

Video Solution by OmegaLearn

https://youtu.be/abSgjn4Qs34?t=2584

~ pi_is_3.14


Answer (B) 7.5

~ Mia Wang the Author ~skibidi

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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