12.187. In an acute isosceles triangle, the angle at the base is $\alpha$, and the area is $S$. Find the area of the triangle whose vertices are the feet of the altitudes of the given triangle.
Solution.

In $\triangle ABC$ (Fig. 12.55) $AB=BC, \angle BAC=\alpha; BD, AF, CE$ are altitudes.

Let $AB=a$, then from $\triangle ADB\left(\angle ADB=90^{\circ}\right): AD=a \cos \alpha, BD=a \sin \alpha$. $AC=2AD=2a \cos \alpha$.

In $\triangle AEC\left(\angle AEC=90^{\circ}\right)$:

$AE=AC \cos \angle BAC=2a \cos^2 \alpha \Rightarrow BE=AB-AE=a\left(1-2 \cos^2 \alpha\right)=-a \cos 2\alpha$. Let $L$ be the intersection point of $EF$ and $BD$. Since $\triangle EBF \sim \triangle ABC$, then

$\frac{EF}{AC}=\frac{BL}{BD}=\frac{BE}{AB} \Rightarrow EF=\frac{AC \cdot BE}{AB}=\frac{2a \cos \alpha \cdot(-a \cos 2\alpha)}{a}=$

$=-2a \cos \alpha \cos 2\alpha; BL=\frac{BD \cdot BE}{AB}=-a \sin \alpha \cos 2\alpha$.

$DL=BD-BL=a \sin \alpha+a \sin \alpha \cos 2\alpha=a \sin \alpha(1+\cos 2\alpha)=$ $=2a \sin \alpha \cos^2 \alpha$.
$S_{\triangle EDF}=\frac{1}{2} EF \cdot DL=-2a^2 \sin \alpha \cos^3 \alpha \cos 2\alpha=-a^2 \sin 2\alpha \cos 2\alpha \cos^2 \alpha=$ $=-\frac{1}{2} a^2 \sin 4\alpha \cos^2 \alpha$.

$$
S=S_{\triangle ABC}=\frac{1}{2} AB^2 \sin \angle ABC=\frac{1}{2} a^2 \sin^2 \alpha
$$

Thus $a^2=\frac{2S}{\sin 2\alpha}=\frac{S}{\sin \alpha \cos \alpha}$.

$$
S_{\triangle EDF}=-\frac{1}{2} \cdot \frac{S}{\sin \alpha \cos \alpha} \cdot \sin 4\alpha \cos^2 \alpha=-\frac{1}{2} S \sin 4\alpha \operatorname{ctg} \alpha
$$

Answer: $-\frac{1}{2} S \sin 4\alpha \operatorname{ctg} \alpha$.