[ [correct pyramid ]

The side of the base of a regular triangular pyramid is equal to $a$, the lateral edge is equal to $b$. Find the height of the pyramid and the dihedral angle between the lateral faces.

#
Let $O$ be the center of the base $ABC$ of a regular triangular pyramid $ABCD$, $M$ be the midpoint of $BC$, $F$ be the foot of the perpendicular dropped from point $M$ to the lateral edge $AD$, $AB = BC = AC = a$, and $AD = BD = CD = b$. From the right triangle $DAO$, we find that

$$
\begin{gathered}
DO = \sqrt{AD^2 - AO^2} = \sqrt{b^2 - \left(\frac{a \sqrt{3}}{3}\right)^2} = \frac{\sqrt{8b^2 - a^2}}{\sqrt{3}} \\
\sin \angle DAO = \frac{DO}{AD} = \frac{\sqrt{3b^2 - a^2}}{b \sqrt{3}}
\end{gathered}
$$

The line $AM$ is the orthogonal projection of the inclined line $AD$ onto the plane $ABC$. Since $AM \perp BC$, by the theorem of three perpendiculars, $AD \perp BC$. Therefore, the line $AD$ is perpendicular to two intersecting lines $MF$ (by construction) and $BC$. Thus, $BFC$ is the linear angle of the dihedral angle between the planes of the lateral faces $ADB$ and $ADC$. From the right triangles $AFM$ and $BMF$, we find that

$$
\begin{gathered}
MF = AM \sin \angle DAO = \frac{a \sqrt{3}}{2} \cdot \frac{\sqrt{3b^2 - a^2}}{b \sqrt{3}} = \frac{a \sqrt{3b^2 - a^2}}{2b} \\
\tan \frac{1}{2} \angle BFC = \frac{BM}{MF} = \frac{\frac{1}{2} \sqrt{2a^2 - a^2}}{2b} = \frac{b}{\sqrt{3b^2 - a^2}}
\end{gathered}
$$

## Answer

$\frac{\sqrt{3b^2 - a^2}}{\sqrt{3}}, 2 \operatorname{arctan} \frac{b}{\sqrt{3b^2 - a^2}}$.

On the faces of a regular tetrahedron with edge $a$, regular tetrahedra are constructed. Prove that the new vertices of the constructed tetrahedra are the vertices of a regular tetrahedron. Find its edge.

## Solution

Let $ABCD$ be a regular tetrahedron with side $a$; $D_1, A_1, B_1$, and $C_1$ be the centers of its faces $ABC, DBC, ACD$, and $ABD$ respectively; $D_2, A_2, B_2$, and $C_2$ be the vertices of the regular tetrahedra $D_2ABC, A_2BCD, B_2ACD$, and $C_2BAD$. The segments $DD_1$ and $D_2D_1$ are the heights of the regular tetrahedra with the common face $ABC$. Therefore, points $D, D_1$, and $D_2$ lie on the same line, and $D_1$ is the midpoint of $DD_2$. Similarly, $A_1$ is the midpoint of segment $AA_2$. The lines $AA_1$ and $DD_1$ lie in the plane $AMD$, where $M$ is the midpoint of $BC$, and $\frac{MA_1}{DM} = \frac{MD_1}{AM} = \frac{1}{3}$, since $A_1$ and $D_1$ are the points of intersection of the medians of triangles $BCD$ and $ABC$. Therefore,

$$
A_1D_1 \parallel AD, \quad A_1D_1 = \frac{1}{3} AD
$$

Moreover, points $A_1$ and $D_1$ are the midpoints of the diagonals $AA_2$ and $DD_2$ of the trapezoid $AD A_2 D_2$. Thus,

$$
A_1D_1 = \frac{1}{2}(A_2D_2 - AD),
$$

from which we find that

$$
A_2D_2 = 2A_1D_1 + AD = \frac{2}{3}a + a = \frac{5}{3}a
$$

Similarly, we find that the other edges of the triangular pyramid $A_2B_2C_2D_2$ are equal to $\frac{5}{3}a$.

## Answer

$\frac{5a}{3}$