18-76 If the area of a square is numerically equal to its perimeter, and the area of an equilateral triangle is also numerically equal to its perimeter, then when comparing the radius of the inscribed circle of the former with that of the latter, the radius of the inscribed circle of the former is
(A) equal to that of the latter.
(B) $\frac{4}{3}$ times that of the latter.
(C) $\frac{2}{\sqrt{3}}$ times that of the latter.
(D) $\frac{\sqrt{2}}{\sqrt{3}}$ times that of the latter.
(E) in an indeterminate relationship with that of the latter.
(2nd American High School Mathematics Examination, 1951)
[Solution] Let the side length of the square be $a_{1}$, the distance from the center to a side (apothem) be $p_{1}$; the side length of the equilateral triangle be $a_{2}$, the distance from the center to a side (apothem) be $p_{2}$, and the height be $h$.

Given $a_{1}^{2}=4 a_{1}$, and $2 p_{1}=a_{1}$, thus $\left(2 p_{1}\right)^{2}=4\left(2 p_{1}\right)$, we get $p_{1}=2$.

Also given $\frac{\sqrt{3}}{4} a_{2}^{2}=3 a_{2}$, and $3 p_{2}=h$, while $h=\frac{\sqrt{3}}{2} a_{2}$, hence $a_{2}=2 \sqrt{3} p_{2}$,
thus $\frac{\sqrt{3}}{4}\left(2 \sqrt{3} p_{2}\right)^{2}=3 \cdot 2 \sqrt{3} p_{2}$,
we get $p_{2}=2=p_{1}$.
Therefore, the answer is $(A)$.