What is the ratio of the area of a square inscribed in a semicircle with radius $r$ to the area of a square inscribed in a circle with radius $r$? Express your answer as a common fraction.
Let $s_1$ be the side length of the square inscribed in the semicircle of radius $r$.  Applying the Pythagorean theorem to the right triangle shown in the diagram, we have $(s_1/2)^2+s_1^2=r^2$, which implies $s_1^2=\frac{4}{5}r^2$.  Let $s_2$ be the side length of the square inscribed in the circle of radius $r$.  Applying the Pythagorean theorem to the right triangle shown in the diagram, we have $(s_2/2)^2+(s_2/2)^2=r^2$, which implies $s_2^2=2r^2$.  Therefore, the ratio of  the areas of the two squares is $\dfrac{s_1^2}{s_2^2}=\dfrac{\frac{4}{5}r^2}{2r^2}=\boxed{\dfrac{2}{5}}$. [asy]
import olympiad;
import graph; size(200); dotfactor=3;
defaultpen(linewidth(0.8)+fontsize(10));
draw(Arc((0,0),1,0,180));
draw(dir(0)--dir(180));
real s=1/sqrt(5);
draw((s,0)--(s,2s)--(-s,2s)--(-s,0));
draw((0,0)--(s,2s),linetype("2 3"));
label("$r$",(s/2,s),unit((-2,1)));
draw(rightanglemark((0,0),(s,0),(s,2s),3.0));

picture pic1;

draw(pic1,Circle((0,0),1));
draw(pic1,(1/sqrt(2),1/sqrt(2))--(-1/sqrt(2),1/sqrt(2))--(-1/sqrt(2),-1/sqrt(2))--(1/sqrt(2),-1/sqrt(2))--cycle);
draw(pic1,(0,0)--(1/sqrt(2),1/sqrt(2)),linetype("2 3"));
label(pic1,"$r$",(1/sqrt(2),1/sqrt(2))/2,unit((-1,1)));
dot(pic1,(0,0));
draw(pic1,(0,0)--(1/sqrt(2),0));
draw(pic1,rightanglemark((0,0),(1/sqrt(2),0),(1/sqrt(2),1/sqrt(2)),3.0));

add(shift((2.5,0))*pic1);[/asy]
\dfrac{2}{5}
Level 5
Geometry
Problem: What is the ratio of the area of a square inscribed in a semicircle with radius $r$ to the area of a square inscribed in a circle with radius $r$? Express your answer as a common fraction.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the ratio of the areas of the two squares, we first need to determine the area of the square inscribed in the semicircle. The semicircle has a radius $r$, and the square will have its base on the diameter of the semicircle. Let the side length of the square be $s_1$. The top corners of the square will touch the arc of the semicircle. The height of the square is $s_1$, and the distance from the center of the semicircle to the top corner of the square is $r$. By the Pythagorean theorem, we have:

\[(\frac{s_1}{2})^2 + s_1^2 = r^2.\]

This simplifies to:

\[\frac{s_1^2}{4} + s_1^2 = r^2,\]
\[\frac{5s_1^2}{4} = r^2,\]
\[s_1^2 = \frac{4r^2}{5}.\]

Thus, the area of the square inscribed in the semicircle is:

\[A_1 = s_1^2 = \frac{4r^2}{5}.\]

Next, we calculate the area of the square inscribed in the full circle with radius $r$. Let the side length of this square be $s_2$. The diagonal of the square equals the diameter of the circle, which is $2r$. The relationship between the side length and the diagonal of a square is given by:

\[s_2\sqrt{2} = 2r.\]

Solving for $s_2$, we find:

\[s_2 = \frac{2r}{\sqrt{2}} = r\sqrt{2}.\]

The area of the square inscribed in the circle is:

\[A_2 = s_2^2 = (r\sqrt{2})^2 = 2r^2.\]

Now, we can find the ratio of the areas of the two squares:

\[\text{Ratio} = \frac{A_1}{A_2} = \frac{\frac{4r^2}{5}}{2r^2} = \frac{4/5}{2} = \frac{4}{10} = \frac{2}{5}.\]

Thus, the final answer is $\dfrac{2}{5}$.<|END_OF_LATENT|> The final answer is $\dfrac{2}{5}$.