3.62. Find the sine of the angle at the vertex of an isosceles triangle, given that the perimeter of any rectangle inscribed in it, with two vertices lying on the base, is a constant value.
3.62. Given that $A B=A C$, $D E F L$ is a rectangle inscribed in $\triangle A B C$, the perimeter of which does not depend on the choice of point $E$ on $A B$ (Fig. 3.64); we need to find $\sin A$. Draw $A A_{1} \perp B C$ and let $A A_{1}=h, B C=a$. Construct another rectangle inscribed in the same way, one of whose sides passes through point $H$, where $A H=H A_{1}$; in this case, $P_{K N Q R}=P_{D E P L}$ (according to the condition). Since $\triangle A K R \sim \triangle A B C$, we have $\frac{K R}{B C}=\frac{A H}{A A_{1}}$, i.e., $K R=\frac{a h}{2 h}=\frac{a}{2}$. Further, $P_{K N Q R}=2(N K+K R)=2\left(\frac{h}{2}+\frac{a}{2}\right)=a+h$.

Let $D E=x$; considering that $\triangle A E F \sim \triangle A B C$, we get $\frac{E F}{B C}=\frac{A M}{A A_{1}}$,

i.e., $E F=\frac{a(h-x)}{h}$. Then

$$
P_{D E F L}=2(D E+E F)=2\left(x+\frac{a h-a x}{h}\right)=\frac{2(x h+a h-a x)}{h}
$$

Since $P_{\text {KNQR }}=P_{\text {DEPE }}$, we have

$$
\frac{2(x h+a h-a x)}{h}=a+h, \text { or } 2 x h+a h-2 a x-h^{2}=0
$$

from which $(2 x-h)(a-h)=0$. This equality must be valid for any value of $x$; hence, $a-h=0$, i.e., $h=a$.

Finally, from $\triangle A A_{1} C$ we find $\operatorname{tg} \frac{A}{2}=\frac{A_{1} C}{A A_{1}}=\frac{a}{2 h}=\frac{a}{2 a}=\frac{1}{2}$, from which

$$
\sin A=\frac{2 \operatorname{tg} \frac{A}{2}}{1+\operatorname{tg}^{2} \frac{A}{2}}=\frac{1}{1+\frac{1}{4}}=\frac{4}{5}
$$