101. Inscribed in the triangle $A B C$ is a triangle whose sides are perpendicular to the radii $O A, O B, O C$ of the circumscribed circle. Derive from this construction another solution to problem 99 b) for an acute-angled triangle.

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101. If $D, E, F$ are the feet of the altitudes of triangle $ABC$, then $\triangle ACD \sim \triangle BCE$ (see Fig. 220); therefore, $\frac{CE}{CD} = \frac{CB}{CA}$. Consequently, $\triangle ABC \approx \triangle DEC$ " $\angle CED = \angle CBA$. Let $MN$ be the tangent to the circumcircle at point $C$. Clearly, $\angle MCA = \angle CBA \left(= \frac{1}{2} - AC\right)$. Thus, $\angle MCE = \angle CED$, i.e., $ED \parallel MN \perp OC$. Similarly, it can be shown that $EF \perp OA$, $DF \perp OB$, i.e., triangle $DEF$ is the desired one.

If triangle $ABC$ is acute-angled, then the center $O$ of the circumcircle lies inside it (see Fig. 220); therefore, $S_{\triangle ABC} = S_{ODCE} + S_{OEAF} + S_{OFBD}$. But the diagonals of the last three quadrilaterals are mutually perpendicular; hence, their areas are equal to half the product of the diagonals. Thus,

\[
\begin{aligned}
S_{\triangle ABC} & = \frac{1}{2} OC \cdot DE + \frac{1}{2} OA \cdot EF + \frac{1}{2} OB \cdot FD = \\
& = \frac{1}{2} R(DE + EF + FD)
\end{aligned}
\]

where $R$ is the radius of the circumcircle.

If $F'D'E'$ is any triangle inscribed in triangle $ABC$, then

\[
S_{\triangle ABC} = S_{OD'CE'} + S_{OE'AF'} + S_{OF'BD'} =
\]

\[
= \frac{1}{2} OC \cdot D'E' \sin \gamma + \frac{1}{2} OA \cdot E'F' \sin \alpha + \frac{1}{2} OB \cdot F'D' \sin \beta,
\]

where $\gamma, \alpha$, and $\beta$ are the angles between the diagonals of the quadrilaterals $OD'CE'$, $OE'AF'$, and $OF'BD'$; therefore,

\[
S_{\triangle ABC} \leq \frac{1}{2} R(D'E' + E'F' + F'D')
\]

Comparing the last relation with relation (*), we find that

\[
DE + EF + FD \leq D'E' + E'F' + F'D'
\]

Thus, among all triangles inscribed in a given acute-angled triangle, the one with the smallest perimeter has vertices at the feet of the altitudes of the given triangle.