160. Given a triangle $A B C$, with area $S$, and the radius of the circumscribed circle is $R$. Perpendiculars are erected at vertices $A$, $B$, and $C$ to the plane of the triangle, and points $A_{1}, B_{1}$, and $C_{1}$ are taken on these perpendiculars such that the segments $A A_{1}, B B_{1}, C C_{1}$ are equal in length to the corresponding altitudes of the triangle dropped from vertices $A, B$, and $C$. Find the volume of the pyramid bounded by the planes $A_{1} B_{1} C$, $A_{1} B C_{1}$, $A B_{1} C_{1}$, and $A B C$.
160. Let's introduce the usual notations: $a, b, c$ - the sides of the triangle, $h_{a}, h_{b}, h_{c}$ - its altitudes, $p$ - the semiperimeter, $r$ - the radius of the inscribed circle. Let $M$ be the point of intersection of the planes $A_{1} B_{1} C, A_{1} B C_{\hat{1}}$ and $A B_{1} C_{\hat{1}}, O_{a}, O_{b}, O_{c}$ - the centers of the excircles ( $O_{a}$ - the center of the circle tangent to side $B C$ and the extensions of $A B$ and $A C$ and so on). We need to prove that the desired figure is the pyramid $O_{a} O_{b} O_{c} M$, and that the height dropped from point $M$ passes through $O$ - the center of the inscribed circle, and $|M O|=2 r$.

Consider, for example, the plane $A_{1} B_{1} C$. Let $K$ be the point of intersection of this plane with the line $A B$,

$$
\frac{|K A|}{|K B|}=\frac{\left|A A_{1}\right|}{\left|B B_{1}\right|}=\frac{h_{a}}{h_{b}}=\frac{b}{a}=\frac{|A C|}{|B C|}
$$

i.e., $K$ is the point of intersection of the line $A B$ with the bisector of the external angle $C$. From this, it follows that the base of our pyramid is indeed the triangle $O_{a} O_{b} O_{c}$ and that point $M$ projects onto point $O$. Let's find $|M O|:$

$$
\frac{|M O|}{h_{a}}=\frac{\left|O O_{a}\right|}{\left|A O_{a}\right|}=\frac{r_{a}-r}{r_{a}}
$$

where $r_{a}$ is the radius of the excircle with center at $O_{a}$; $r_{a}=\frac{S}{p-a}, r=\frac{S}{p}, h_{a}=\frac{2 S}{a}, \quad$ therefore,

$$
|M O|=h_{a} \frac{r_{a}-r}{r_{a}}=\frac{2 S}{a} \frac{\frac{1}{p-a}-\frac{1}{p}}{\frac{1}{p-a}}=\frac{2 S}{p}=2 r
$$

Let's find the area of $\triangle O_{a} O_{b} O_{c}^{-}$. Note that $O_{a} A, O_{b} B, O_{c} C$ are the altitudes of this triangle. The angles of $\triangle O_{a} O_{b} O_{c}$ are easily found, for example, $\widehat{O_{c} O_{a} O_{b}}=\widehat{B O_{a} C}=180^{\circ}-\left(90^{\circ}-\frac{\widehat{B}}{2}\right)-\left(90^{\circ}-\frac{\widehat{C}}{2}\right)=90^{\circ}-\frac{\hat{A}}{2}$.

Similarly, the other angles are found. The circle with diameter $O_{b} O_{c}$ passes through $B$ and $C$, therefore,

$$
\left|O_{b} O_{c}\right|=\frac{|B C|}{\sin \widehat{B O_{b} C}}=\frac{a}{\sin \frac{\hat{A}}{2}}
$$

Similarly, $\left|O_{b} O_{a}\right|=\frac{c}{\sin \frac{\hat{C}}{2}}$, hence

$$
\left|O_{a} A\right|=\left|O_{a} O_{b}\right| \sin \widehat{O_{a} O_{b} A}=\frac{c}{\sin \frac{\hat{C}}{2}} \cos \frac{\hat{B}}{2}
$$

Thus, the area of $\Delta O_{a} O_{b} O_{c}$ (denote it by $Q$) will be

$$
\begin{aligned}
& Q=\frac{1}{2} \frac{a c}{\sin \frac{\hat{A}}{2} \sin \frac{\hat{C}}{2}} \cos \frac{\hat{B}}{2}= \\
& \quad=\frac{a c \sin \hat{B}}{4} \cdot \frac{1}{\sin \frac{\hat{A}}{2} \sin \frac{\hat{B}}{2} \sin \frac{\hat{C}}{2}}
\end{aligned}
$$

Let's find $\sin \frac{\hat{A}}{2}$ :

$$
\begin{aligned}
& \sin \frac{\hat{A}}{2}=\sqrt{\frac{1-\cos \hat{A}}{2}}=\sqrt{\frac{1}{2}\left(1-\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)}= \\
& =\sqrt{\frac{(p-b)(p-c)}{b c}} .
\end{aligned}
$$

Similarly, $\sin \frac{\hat{B}}{2}$ and $\sin \frac{\hat{C}}{2}$ are found. Replacing them in (1), we get

$$
Q=S \frac{a b c}{2(p-a)(p-b)(p-c)}
$$

and the volume of the pyramid $M O_{a} O_{b} O_{c}$ will be

$$
V=\frac{S a b c r}{3(p-a)(p-b)(p-c)}=\frac{1}{3} a b c=\frac{4}{3} S R
$$