2025 AMC 10A Problems/Problem 10: Difference between revisions

Latest revision as of 16:51, 11 November 2025

Problem

A semicircle has diameter $\overline{AB}$ and chord $\overline{CD}$ of length $16$ parallel to $\overline{AB}$ . A smaller semicircle with diameter on $\overline{AB}$ and tangent to $\overline{CD}$ is cut from the larger semicircle, as shown below.

$[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A = (-3,0); pair B = (3,0); fill(Arc((0,0),3,0,180)--cycle,palered); fill(Arc((-1.125,0),0.75,0,180)--cycle,white); draw(Arc((0,0),3,0,180),black); draw(Arc((-1.125,0),0.75,0,180),black); draw((-3,0) -- (-1.875,0),black); draw((-0.375,0) -- (3,0),black); draw((-2.895, 0.75) -- (2.895,0.75), black); dot((-3,0)); dot((3,0)); dot((-2.925, 0.75)); dot((2.925, 0.75)); label("$16$",midpoint((-2.925, 0.75)--(2.925, 0.75)),N); label("$A$",A,S); label("$B$",B,S); label("$C$",(-2.925, 0.75),W); label("$D$",(2.925, 0.75),E); [/asy]$

What is the area of the resulting figure, shown shaded?

$\textbf{(A) } 16\pi \qquad\textbf{(B) } 24\pi \qquad\textbf{(C) } 32\pi \qquad\textbf{(D) } 48\pi \qquad\textbf{(E) } 64\pi$

Video Solution

https://youtu.be/l1RY_C20Q2M

Note

This question is similar to 2004 AMC 12B 10/AMC 10B 12.

Lets call the radius of the big semi-circle $R$ and the radius of the small semi-circle $r$ . The $c$ in the 2004 AMC 12B 10/AMC 10B 12, corresponds to $r$ and the $b$ corresponds to $R$ . If we cut the chord by half, we get $8$ as a result, and the $8$ corresponds to $a$ . The only difference is this problem contains two semi-circles while the 2004 AMC 12B 10/AMC 10B 12 contains two circles. From that problem, we know that the area of the annulus is $\pi a^2$ , which means the area in this problem should be $\frac{\pi(8)^2}{2}$ which is $\boxed{\mathrm{(C)\ }32\pi}$ . Basically, the 2004 AMC 12B 10/AMC 10B 12 was just a generalized problem while this one was a more specific one

~Neo

Not a cheeky solution

This solution is a general solution if MAA stated the diagram is not to be altered in any way (i.e. the chord is not the diameter, the chord can't be moved, the semicircle can't be moved, shrunk, etc.)

Solution 2 is the only other solution (at the moment) that satisfies those prerequisites.

Call the center of the big semicircle $ \omega $. Also, call the radius of the big semicircle $ R $ and the smaller semicircle $ r $. We then see that the distance from $ \omega $ to the chord is simply $ r $. We also see that if we draw a line from the intersection of the bigger semicircle and the chord to $ \omega $, that is just $ R $. Finally, it is a given fact that drawing a line from $ \omega $ to the top of the semicircle perpendicularly bisects the chord into two lengths of 8. Thus, our first equation is $ r^2 + 64 = R^2 $.

For our second equation, we take our line we drew from $ \omega $ to the top of the semicircle, calling it $ l $. The distance from the intersection of the chord and $ l $, call it $ E $, to the top of the semicircle is $ R-r $. We take the midpoint of this line, call it $ M $, splitting The distance of $ E $ to the top of the semicircle into two equal parts of length $ \frac{R-r}{2} $. We can clearly see that lines $ \omega M $ and $ CD $ intersect $ E $ at $ 90^\circ $, therefore quadrilateral $ C\omega DM $ is a rhombus. With that in mind it isnt hard to prove that triangles $ \omega EC $ and $ MEC $ are congruent through ASA, and then we see that $ \frac{R-r}{2} = r $, and $ R = 3r $.

Through substitution, we see $ r^2+64=R^2 \implies r^2 + 64 = 9r^2 \implies 64 = 8r^2 \implies 8 = r^2 \implies r = \sqrt{8} $, therefore the radius of the smaller semicircle is $ \sqrt{8} $.

We have $ R=3r $, and so the radius of the bigger semicircle is $ 3\sqrt{8} $.

We want the area of the bigger semicircle minus the area of the smaller semicircle, which is $ \frac{1}{2} \cdot (3\sqrt{8})^2 \pi - \frac{1}{2} \cdot (\sqrt{8})^2 \pi = 36 \pi - 4\pi = 32 \pi $

Thus the answer is $\boxed{\text{(C) }32\pi}$

The other solutions are quite interesting. Go check em' out.

~Pinotation

Solution 1 (Somewhat Cheeky)

Notice that the size of the smaller semicircle is not specified, and there is no additional information that hints at any specific size for it. Hence, we can shrink the small semicircle until its area is arbitrarily small and negligible, leaving us with a semicircle with a diameter of $16$ . The area of the semicircle is given by $\frac{\pi r^2}{2}$ , so we have $r=\frac{16}{2}=8\Rightarrow$ $A=\frac{\pi(8)^2}{2}=\boxed{\text{(C) }32\pi}$

~Bocabulary142857

Solution 2

Let the radius of the larger semicircle be $R$ and that of the smaller one be $r.$ We are looking for $\dfrac{1}{2}\pi(R^2-r^2).$ If we connect the center of the large semicircle to one endpoint of the chord and to the center of the chord, we get a right triangle with legs $8$ and $r$ and hypotenuse $R.$ Hence, $R^2=r^2+8^2\implies R^2-r^2=64\implies\dfrac{1}{2}\pi(R^2-r^2)=\boxed{\text{(C) }32\pi}.!$

~Nioronean, $\LaTeX$ and writing by Tacos_are_yummy_1

Solution 3

The problem doesn't restrict where the smaller semicircle is along the larger semicircle's diameter. Therefore, we can assume that the two semicircles are concentric. Let the center of both semicircles be $O$ , and let $CD$ be tangent to the smaller semicircle at $T$ . Let the radius of the smaller semicircle be $x$ , and let the radius of the larger semicircle be $r$ . If we mirror the diagram over $AB$ , we can see that we have two concentric circles. We are trying to find $\pi(\frac{r^2-x^2}{2})$ . By Power of a Point on $T$ , we can see that $64 = (r + x)(r - x) = r^2 - x^2.$ Thus, $\pi(\frac{r^2 - x^2}{2}) = 32\pi \implies \boxed{\text{(C) }32\pi}.$ ~vinceS

Solution 4

The problem doesn't define where the chord is within the circle. So, let's place the chord such that when lines are drawn from its ends to the center of the larger semicircle, an equilateral triangle with side length $16$ is formed.

$[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; fill(Arc((0,0),3,0,180)--cycle,palered); fill(Arc((0,0),2.598,0,180)--cycle,white); draw(Arc((0,0),3,0,180),black); draw(Arc((0,0),2.598,0,180),black); draw((-2.598,0) -- (2.598,0), white); draw((-1.5,2.598) -- (1.5,2.598),black); draw((0,0) -- (-1.5, 2.598), black); draw((0,0) -- (1.5, 2.598), black); draw((0,0) -- (0, 2.598), black); draw((-3,0) -- (-2.598, 0), black); draw((3,0) -- (2.598, 0), black); dot((0,0)); dot((0,2.598)); dot((-1.5,2.598)); dot((1.5,2.598)); dot((3,0)); dot((-3,0)); label("$16$",midpoint((0, 2.598)--(0, 2.598)),N); label("$8\sqrt{3}$",midpoint((0, 1.4)--(0, 1.4)),E); label("$16$",midpoint((-0.9, 1.3)--(-0.9, 1.3)),W); label("$16$",midpoint((0.9, 1.3)--(0.9, 1.3)),E); label("$A$",midpoint((-3, 0)--(-3, 0)),W); label("$B$",midpoint((3, 0)--(3, 0)),E); label("$C$",midpoint((-1.5,2.7)--(-1.5,2.7)),N); label("$D$",midpoint((1.5,2.7)--(1.5,2.7)),N); [/asy]$

In this definition, the radius of the larger semicircle is $16,$ giving it an area of $\frac{\pi\cdot16^2}{2} = 128\pi.$

The height of the equilateral triangle is $8\sqrt{3},$ which is equal to the radius of the smaller semicircle. This gives the smaller semicircle an area of $\frac{\pi\cdot(8\sqrt{3})^2}{2} = 96\pi.$

The total shaded area is the difference of these semicircles, or $128\pi - 96\pi = \boxed{\text{(C) } 32\pi}.$

~chisps

Solution 5

Let the radius of the larger semicircle be $R$ and the smaller one $r.$ We are asked to compute $\frac{R^2-r^2}{2} \cdot \pi$ . Since a diameter is always perpendicular and bisects a chord, by drawing a diameter and applying Power of a Point Theorem, this yields $(R+r)(R-r)=8^2$ $\implies$ $R^2-r^2=64$ . Therefore, the answer is $\frac{64}{2} \pi = \boxed{\text{(C) } 32\pi}.$ ~hxve

Solution 6

Since the problem does not restrict where the chord is, WLOG we can simply let the chord be the base of the semicircle. Therefore, the area is simply $\frac{\pi\cdot8^2}{2} = 32\pi.$

~metrixgo

Solution 7

We can move this small semicircle to the middle of the big semicircle. Let $r$ be the radius of the small semicircle. By Pythagorean Theorem, the line draw from the midpoint of the diameter of the big semicircle to the big semicircle at the height of the small semicircle (see the diagram) has length $\sqrt{r^2 + 64}$ . This is a radius of the big semicircle. So the shaded area would be $\frac{\pi \cdot (\sqrt{r^2+64})^2 - \pi r^2}{2} = \frac{\pi(r^2 + 64 - r^2)}{2} = \frac{64\pi}{2} = \boxed{\text{(C) } 32\pi}.$ Diagram:

$[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real R = 3; pair A = (-R,0), B = (R,0); fill(Arc((0,0),R,0,180)--cycle,palered); draw(Arc((0,0),R,0,180),black); real h = 0.75; pair C = (-sqrt(R^2 - h^2), h); pair D = ( sqrt(R^2 - h^2), h); draw(C--D,black); label("$C$",C,W); label("$D$",D,E); label("$16$", midpoint(C--D), N); real r = h; pair O = (0,0); pair Q = (0,r); fill(Arc(O,r,0,180)--cycle,white); draw(Arc(O,r,0,180),black); draw(A--B,black); draw(O--Q); draw(O--D); draw(Q--D, dashed); label("$A$",A,S); label("$B$",B,S); label("$r$", midpoint(O--Q), W); label("$\sqrt{r^{2}+64}$", midpoint(O--D), S); label("$8$", midpoint(Q--D), N); [/asy]$

~JerryZYang

Chinese Video Solution

https://www.bilibili.com/video/BV1bj2uBxEkU/

~metrixgo

Video Solution (In 1 Min)

https://youtu.be/8VWMNAx55g0?si=IqLseEtKLl2joUNU ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution by Daily Dose of Math

https://youtu.be/gPh9w3X3QSw

~Thesmartgreekmathdude

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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

Page

Toolbox

Search

2025 AMC 10A Problems/Problem 10: Difference between revisions

Latest revision as of 16:51, 11 November 2025

Contents

Problem

Video Solution

Note

Not a cheeky solution

Solution 1 (Somewhat Cheeky)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Chinese Video Solution

Video Solution (In 1 Min)

Video Solution

Video Solution by SpreadTheMathLove

Video Solution by Daily Dose of Math

See Also

@@ Line 4: / Line 4: @@
 <asy>
-import graph;
+import graph; unitsize(14mm);
-unitsize(14mm);
+defaultpen(linewidth(.8pt)+fontsize(10pt));
-defaultpen(linewidth(.8pt)+fontsize(10pt));
+dotfactor=4;
-dotfactor=4;
 pair A = (-3,0);
 pair B = (3,0);
-pair centerBig = (0,0);
+fill(Arc((0,0),3,0,180)--cycle,palered);
-real Rbig = 3;
+fill(Arc((-1.125,0),0.75,0,180)--cycle,white);
-pair centerSmall = (-1.125,0);
+draw(Arc((0,0),3,0,180),black);
-real Rsmall = 0.75;
+draw(Arc((-1.125,0),0.75,0,180),black);
-real ychord = 0.75;
+draw((-3,0) -- (-1.875,0),black);
-real xchord = sqrt(Rbig*Rbig - ychord*ychord);
+draw((-0.375,0) -- (3,0),black);
-pair C = (-xchord, ychord);
+draw((-2.895, 0.75) -- (2.895,0.75), black);
-pair D = ( xchord, ychord);
+dot((-3,0)); dot((3,0)); dot((-2.925, 0.75));
-fill(Arc(centerBig, Rbig, 0, 180)--cycle, palered);
+dot((2.925, 0.75));
-fill(Arc(centerSmall, Rsmall, 0, 180)--cycle, white);
+label("$16$",midpoint((-2.925, 0.75)--(2.925, 0.75)),N);
-draw(Arc(centerBig, Rbig, 0, 180), black);
+label("$A$",A,S);
-draw(Arc(centerSmall, Rsmall, 0, 180), black);
+label("$B$",B,S);
-draw(A -- (-1.875,0), black);    // -1.125 - 0.75 = -1.875
+label("$C$",(-2.925, 0.75),W);
-draw((-0.375,0) -- B, black);    // -1.125 + 0.75 = -0.375
+label("$D$",(2.925, 0.75),E);
-draw(C -- D, black);
-dot(A);
-dot(B);
-dot(C);
-dot(D);
-label("$16$", midpoint(C--D), N);
-label("$A$", A, S);
-label("$B$", B, S);
-label("$C$", C, W);
-label("$D$", D, E);
 </asy>
 What is the area of the resulting figure, shown shaded?
@@ Line 41: / Line 29: @@
 <imath>\textbf{(A) } 16\pi \qquad\textbf{(B) } 24\pi \qquad\textbf{(C) } 32\pi \qquad\textbf{(D) } 48\pi \qquad\textbf{(E) } 64\pi</imath>
-==Solution 1 (Somewhat Cheese)==
+==Video Solution==
-Notice that the size of the smaller semicircle is not specified, and there is no additional information that hints at any specific size for it. Hence, we can shrink the small semicircle until its area is arbitrarily small and negligible, leaving us with a semicircle with a diameter of <imath>16</imath>. The area of the semicircle is given by <imath>\frac{\pi r^2}{2}</imath>, so we have <imath>r=\frac{16}{2}=8\Rightarrow</imath><imath>A=\frac{\pi(8)^2}{2}=\boxed{\textbf{(C) }32\pi}</imath>
+https://youtu.be/l1RY_C20Q2M
+==Note==
+This question is similar to [https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_10 2004 AMC 12B 10/AMC 10B 12].
+Lets call the radius of the big semi-circle <imath>R</imath> and the radius of the small semi-circle <imath>r</imath>. The <imath>c</imath> in the 2004 AMC 12B 10/AMC 10B 12, corresponds to <imath>r</imath> and the <imath>b</imath> corresponds to <imath>R</imath>. If we cut the chord by half, we get <imath>8</imath> as a result, and the <imath>8</imath> corresponds to <imath>a</imath>. The only difference is this problem contains two semi-circles while the 2004 AMC 12B 10/AMC 10B 12 contains two circles.
+From that problem, we know that the area of the annulus is <imath>\pi a^2</imath>, which means the area in this problem should be <imath>\frac{\pi(8)^2}{2}</imath> which is <imath>\boxed{\mathrm{(C)\ }32\pi}</imath>.
+Basically, the 2004 AMC 12B 10/AMC 10B 12 was just a generalized problem while this one was a more specific one
+~Neo
+==Not a cheeky solution==
+This solution is a general solution if MAA stated the diagram is not to be altered in any way (i.e. the chord is not the diameter, the chord can't be moved, the semicircle can't be moved, shrunk, etc.)
+Solution 2 is the only other solution (at the moment) that satisfies those prerequisites.
+Call the center of the big semicircle \( \omega \). Also, call the radius of the big semicircle \( R \) and the smaller semicircle \( r \). We then see that the distance from \( \omega \) to the chord is simply \( r \). We also see that if we draw a line from the intersection of the bigger semicircle and the chord to \( \omega \), that is just \( R \). Finally, it is a given fact that drawing a line from \( \omega \) to the top of the semicircle perpendicularly bisects the chord into two lengths of 8. Thus, our first equation is \( r^2 + 64 = R^2 \).
+For our second equation, we take our line we drew from \( \omega \) to the top of the semicircle, calling it \( l \). The distance from the intersection of the chord and \( l \), call it \( E \), to the top of the semicircle is \( R-r \). We take the midpoint of this line, call it \( M \), splitting The distance of \( E \) to the top of the semicircle into two equal parts of length \( \frac{R-r}{2} \). We can clearly see that lines \( \omega M \) and \( CD \) intersect \( E \) at \( 90^\circ \), therefore quadrilateral \( C\omega DM \) is a rhombus. With that in mind it isnt hard to prove that triangles \( \omega EC \) and \( MEC \) are congruent through ASA, and then we see that \( \frac{R-r}{2} = r \), and \( R = 3r \).
+Through substitution, we see \( r^2+64=R^2 \implies r^2 + 64 = 9r^2 \implies 64 = 8r^2 \implies 8 = r^2 \implies r = \sqrt{8} \), therefore the radius of the smaller semicircle is \( \sqrt{8} \).
+We have \( R=3r \), and so the radius of the bigger semicircle is \( 3\sqrt{8} \).
+We want the area of the bigger semicircle minus the area of the smaller semicircle, which is \( \frac{1}{2} \cdot (3\sqrt{8})^2 \pi - \frac{1}{2} \cdot (\sqrt{8})^2 \pi = 36 \pi - 4\pi = 32 \pi \)
+Thus the answer is <imath>\boxed{\text{(C) }32\pi}</imath>
+The other solutions are quite interesting. Go check em' out.
+~Pinotation
+==Solution 1 (Somewhat Cheeky)==
+Notice that the size of the smaller semicircle is not specified, and there is no additional information that hints at any specific size for it. Hence, we can shrink the small semicircle until its area is arbitrarily small and negligible, leaving us with a semicircle with a diameter of <imath>16</imath>. The area of the semicircle is given by <imath>\frac{\pi r^2}{2}</imath>, so we have <imath>r=\frac{16}{2}=8\Rightarrow</imath><imath>A=\frac{\pi(8)^2}{2}=\boxed{\text{(C) }32\pi}</imath>
 ~Bocabulary142857
 ==Solution 2==
-Let the radius of the larger semicircle be <imath>R</imath> and that of the smaller one be <imath>r.</imath> We are looking for <imath>\dfrac{1}{2}\pi(R^2-r^2).</imath> If we connect the center of the large semicircle to one endpoint of the chord and to the center of the chord, we get a right triangle with legs <imath>8</imath> and <imath>r</imath> and hypotenuse <imath>R.</imath> Hence, <imath>R^2=r^2+8^2\implies R^2-r^2=64\implies\dfrac{1}{2}\pi(R^2-r^2)=\boxed{\text{(C) }32\pi}.</imath>
+Let the radius of the larger semicircle be <imath>R</imath> and that of the smaller one be <imath>r.</imath> We are looking for <imath>\dfrac{1}{2}\pi(R^2-r^2).</imath> If we connect the center of the large semicircle to one endpoint of the chord and to the center of the chord, we get a right triangle with legs <imath>8</imath> and <imath>r</imath> and hypotenuse <imath>R.</imath> Hence, <imath>R^2=r^2+8^2\implies R^2-r^2=64\implies\dfrac{1}{2}\pi(R^2-r^2)=\boxed{\text{(C) }32\pi}.!</imath>
 ~Nioronean, <imath>\LaTeX</imath> and writing by Tacos_are_yummy_1
@@ Line 60: / Line 83: @@
 ==Solution 4==
 The problem doesn't define where the chord is within the circle. So, let's place the chord such that when lines are drawn from its ends to the center of the larger semicircle, an equilateral triangle with side length <imath>16</imath> is formed.
+<asy>
+import graph; unitsize(14mm);
+defaultpen(linewidth(.8pt)+fontsize(10pt));
+dotfactor=4;
+fill(Arc((0,0),3,0,180)--cycle,palered);
+fill(Arc((0,0),2.598,0,180)--cycle,white);
+draw(Arc((0,0),3,0,180),black);
+draw(Arc((0,0),2.598,0,180),black);
+draw((-2.598,0) -- (2.598,0), white);
+draw((-1.5,2.598) -- (1.5,2.598),black);
+draw((0,0) -- (-1.5, 2.598), black);
+draw((0,0) -- (1.5, 2.598), black);
+draw((0,0) -- (0, 2.598), black);
+draw((-3,0) -- (-2.598, 0), black);
+draw((3,0) -- (2.598, 0), black);
+dot((0,0));
+dot((0,2.598));
+dot((-1.5,2.598));
+dot((1.5,2.598));
+dot((3,0));
+dot((-3,0));
+label("$16$",midpoint((0, 2.598)--(0, 2.598)),N);
+label("$8\sqrt{3}$",midpoint((0, 1.4)--(0, 1.4)),E);
+label("$16$",midpoint((-0.9, 1.3)--(-0.9, 1.3)),W);
+label("$16$",midpoint((0.9, 1.3)--(0.9, 1.3)),E);
+label("$A$",midpoint((-3, 0)--(-3, 0)),W);
+label("$B$",midpoint((3, 0)--(3, 0)),E);
+label("$C$",midpoint((-1.5,2.7)--(-1.5,2.7)),N);
+label("$D$",midpoint((1.5,2.7)--(1.5,2.7)),N);
+</asy>
 In this definition, the radius of the larger semicircle is <imath>16,</imath> giving it an area of <imath>\frac{\pi\cdot16^2}{2} = 128\pi.</imath>
@@ Line 73: / Line 127: @@
 ==Solution 6==
-Since the problem does not restrict where the chord is, we can simply let the chord be the base of the semicircle. Therefore, the area is simply <imath>\frac{\pi\cdot8^2}{2} = 32\pi.</imath>
+Since the problem does not restrict where the chord is, WLOG we can simply let the chord be the base of the semicircle. Therefore, the area is simply <imath>\frac{\pi\cdot8^2}{2} = 32\pi.</imath>
+~metrixgo
+==Solution 7==
+We can move this small semicircle to the middle of the big semicircle. Let <imath>r</imath> be the radius of the small semicircle. By Pythagorean Theorem, the line draw from the midpoint of the diameter of the big semicircle to the big semicircle at the height of the small semicircle (see the diagram) has length <imath>\sqrt{r^2 + 64}</imath>. This is a radius of the big semicircle. So the shaded area would be
+<cmath>\frac{\pi \cdot (\sqrt{r^2+64})^2 - \pi r^2}{2} = \frac{\pi(r^2 + 64 - r^2)}{2} = \frac{64\pi}{2} = \boxed{\text{(C) } 32\pi}.</cmath>
+Diagram:
+<asy>
+import graph;
+unitsize(14mm);
+defaultpen(linewidth(.8pt)+fontsize(10pt));
+dotfactor=4;
+real R = 3;
+pair A = (-R,0), B = (R,0);
+fill(Arc((0,0),R,0,180)--cycle,palered);
+draw(Arc((0,0),R,0,180),black);
+real h = 0.75;
+pair C = (-sqrt(R^2 - h^2), h);
+pair D = ( sqrt(R^2 - h^2), h);
+draw(C--D,black);
+label("$C$",C,W);
+label("$D$",D,E);
+label("$16$", midpoint(C--D), N);
+real r = h;
+pair O = (0,0);
+pair Q = (0,r);
+fill(Arc(O,r,0,180)--cycle,white);
+draw(Arc(O,r,0,180),black);
+draw(A--B,black);
+draw(O--Q);
+draw(O--D);
+draw(Q--D, dashed);
+label("$A$",A,S);
+label("$B$",B,S);
+label("$r$", midpoint(O--Q), W);
+label("$\sqrt{r^{2}+64}$", midpoint(O--D), S);
+label("$8$", midpoint(Q--D), N);
+</asy>
+~JerryZYang
+==Chinese Video Solution==
+https://www.bilibili.com/video/BV1bj2uBxEkU/
 ~metrixgo
-==See Also==
+== Video Solution (In 1 Min) ==
-{{AMC10 box|year=2025|ab=A|before=[[2024 AMC 10B Problems]]|after=[[2025 AMC 10B Problems]]}}
+https://youtu.be/8VWMNAx55g0?si=IqLseEtKLl2joUNU ~ Pi Academy
-* [[AMC 10]]
-* [[AMC 10 Problems and Solutions]]
+==Video Solution==
-* [[Mathematics competitions]]
+https://youtu.be/gWSZeCKrOfU
-* [[Mathematics competition resources]]
-{{MAA Notice}}
+~MK
+==Video Solution by SpreadTheMathLove==
+https://www.youtube.com/watch?v=dAeyV60Hu5c
+==Video Solution by Daily Dose of Math==
+https://youtu.be/gPh9w3X3QSw
+~Thesmartgreekmathdude
 ==See Also==
-{{AMC10 box|year=2025|ab=A|before=[[2024 AMC 10B Problems]]|after=[[2025 AMC 10B Problems]]}}
+{{AMC10 box|year=2025|ab=A|num-b=9|num-a=11}}
 * [[AMC 10]]
 * [[AMC 10 Problems and Solutions]]