11.170. From the midpoint of the height of a regular triangular pyramid, perpendiculars are dropped to a lateral edge and to a lateral face. The lengths of these perpendiculars are $a$ and $b$, respectively. Find the volume of the pyramid. Does the problem have a solution for any $a$ and $b$?
## Solution.

Let $O_{1}$ be the midpoint of the height $S \dot{O}$ of the regular pyramid $S A B C$ (Fig. 11.63), $S D$ be the apothem of the lateral face $A S B$. $O_{1} N$ and $O_{1} M$ are perpendiculars dropped from $O_{1}$ to $S C$ and $S D$, respectively, with $O_{1} N=a$.

Since the pyramid $S A B C$ is regular, the planes $A S B$ and $S D O$ are perpendicular, and thus the segment $O_{1} M$ is perpendicular to the plane $A S B$, with $O_{1} M=b$. Drop perpendiculars $O Q$ and $O P$ to $S C$ and $S D$, respectively. Then, since $O P \parallel O_{1} M$, $O Q \parallel O_{1} N$, and $O_{1}$ is the midpoint of $S O$, we have $O P=2 \cdot O_{1} M=2 b$, $O Q=2 \cdot O_{1} N=2 a$. Let $\angle S C O=\alpha$, $\angle S D O=\beta$, and $S O=H$. Then from $\triangle O P D$ ($\angle O P D=90^{\circ}$) we have $O D=\frac{2 b}{\sin \beta}$, and from $\triangle O Q C$ ($\angle O Q C=90^{\circ}$) we have $O C=\frac{2 a}{\sin \alpha}$. Since $O C=2 \cdot O D$, we get $\frac{2 a}{\sin \alpha}=2 \cdot \frac{2 b}{\sin \beta}$; $\sin \alpha=\frac{a \sin \beta}{2 b}$. Let $O P$ be the height of the right $\triangle S O D$, dropped to the hypotenuse.

Then $\angle S O P=\angle S D O=\beta$. Similarly, $\angle S O Q=\angle S C O=\alpha$. From $\triangle S P O$ ($\angle S P O=90^{\circ}$): $H=\frac{O P}{\cos \angle S O P}=\frac{2 b}{\cos \beta}$.

$$
\text { From } \triangle S Q O\left(\angle S Q O=90^{\circ}\right): H=\frac{O Q}{\cos \angle S O Q}=\frac{2 a}{\cos \alpha} \text {. }
$$

Then $\frac{2 b}{\cos \beta}=\frac{2 a}{\cos \alpha} ; \cos \alpha=\frac{a \cos \beta}{b}$. Since $\sin ^{2} \alpha+\cos ^{2} \alpha=1$, we have $\frac{a^{2} \sin ^{2} \beta}{4 b^{2}}+\frac{a^{2} \cos ^{2} \beta}{b^{2}}=1 ; \frac{a^{2}}{b^{2}}\left(\frac{1}{4} \sin ^{2} \beta+\cos ^{2} \beta\right)=1 ; \frac{1}{4} \sin ^{2} \beta+1-\sin ^{2} \beta=\frac{b^{2}}{a^{2}}$; $\frac{3}{4} \sin ^{2} \beta=\frac{a^{2}-b^{2}}{a^{2}} ; \sin \beta=\frac{2 \sqrt{a^{2}-b^{2}}}{a \sqrt{3}} ; \cos \beta=\sqrt{1-\sin ^{2} \beta}=\sqrt{1-\frac{4\left(a^{2}-b^{2}\right)}{3 a^{2}}}=$ $=\frac{\sqrt{4 b^{2}-a^{2}}}{a \sqrt{3}}$. Then $H=\frac{2 b}{\cos \beta}=\frac{2 a b \sqrt{3}}{\sqrt{4 b^{2}-a^{2}}}$.

The area of the base $S=\frac{A B^{2} \sqrt{3}}{4}=\frac{(2 \sqrt{3} \cdot D O)^{2} \sqrt{3}}{4}=3 \sqrt{3} \cdot D O^{2}=3 \sqrt{3} \cdot\left(\frac{2 b}{\sin \beta}\right)^{2}=3 \sqrt{3} \cdot \frac{12 a^{2} b^{2}}{4\left(a^{2}-b^{2}\right)}=$ $=\frac{9 a^{2} b^{2} \sqrt{3}}{a^{2}-b^{2}}$.

The volume of the pyramid

$V=\frac{1}{3} S H=\frac{1}{3} \cdot \frac{9 a^{2} b^{2} \sqrt{3}}{a^{2}-b^{2}} \cdot \frac{2 a b \sqrt{3}}{\sqrt{4 b^{2}-a^{2}}}=\frac{18 a^{3} b^{3}}{\left(a^{2}-b^{2}\right) \sqrt{4 b^{2}-a^{2}}}$.

The problem has a solution when $\left\{\begin{array}{l}a^{2}-b^{2}>0 ; \\ 4 b^{2}-a^{2}>0,\end{array}\right.$ that is, when $b<a<2 b$.

Answer: $\frac{18 a^{3} b^{3}}{\left(a^{2}-b^{2}\right) \sqrt{4 b^{2}-a^{2}}} ; b<a<2 b$.