7. Consider quadrilateral pyramids $MABCD$ with the following properties: the base of the pyramid is a convex quadrilateral $ABCD$, where $AB = BC = 1$, $CD = DA = 2$, and each of the planes of the lateral faces $MAB$, $MBC$, $MCD$, $MDA$ forms an angle of $45^{\circ}$ with the plane of the base.

a) Find the volume of such a pyramid if its height, dropped from vertex $M$, is $\frac{9}{5}$.

b) For what height is the volume of the considered pyramids maximized, and what is this maximum volume?
Answer. a) $\frac{27}{25} ;$ b) $\frac{4}{3}$.

Let $M H$ be the height of the pyramid, $(M H=h), P$ be the projection of $M$ onto the line $A B$. Then $M H P$ is a right triangle with angle $\angle M P H=45^{\circ}$, from which $H P=h \cdot \operatorname{ctg} 45^{\circ}=h$. Similarly, it can be proved that the point $H$ is at a distance $r=h$ from each of the lines $B C, C D, D A$ (in other words, the circle of radius $r$ with center $H$ touches the lines $A B, B C, C D, D A$).

Triangles $B A D$ and $B C D$ are equal by three sides, so the quadrilateral $A B C D$ is symmetric with respect to the diagonal $B D$. Its area $S$ is equal to $2 S_{B A D}=A B \cdot A D \sin \angle B A D$, so $S \leqslant A B \cdot A D=2$. Equality is achieved when $\angle B A D=90^{\circ}$, so $S_{\max }=2$.

The point $H$ lies on the internal or external bisector of each of the angles of the quadrilateral $A B C D . B D$ is the internal bisector of angles $B$ and $D$. The external bisectors of angles $B$ and $D$ are parallel, so $H$ must lie on $B D$. Let $I$ and $J$ be the points of intersection of the internal and external bisectors of angle $A$ with the line $B D$. Then $I$ is the center of the inscribed circle of the quadrilateral $A B C D$ (let its radius be $r_{1}$); $J$ is the center of the circle tangent to the extensions of the sides of the quadrilateral $A B C D$ (the exscribed circle, let its radius be $r_{2}$). The area of a quadrilateral in which a circle is inscribed can be given by the formula $S=\frac{(A B+B C+C D+D A) r_{1}}{2}$, from which $r_{1}=\frac{S}{3}$. Also, $S=S_{A D J}+S_{C D J}-S_{A B J}-S_{B C J}=\frac{(A D+C D-A B-B C) r_{2}}{2}$, from which $r_{2}=S$.

The pyramid satisfies the condition of the problem if and only if (1) the height passes through the center of the inscribed circle in the base (i.e., $H=I$) and its length is $h=r_{1}=\frac{S}{3}$ or (2) the height passes through the center of the exscribed circle (i.e., $H=J$) and $h=r_{2}=S$.

a) When $h=\frac{9}{5}$, the first case is impossible ($S=3 r_{1}=3 h=\frac{27}{5}>2$). Therefore, only the second case remains, and then $S=r_{2}=h=\frac{9}{5}$. The volume is $V=\frac{S h}{3}=\frac{27}{25}$.

b) The volume in the first and second cases is $V_{1}=\frac{S h}{3}=\frac{S^{2}}{9}$ and $V_{2}=\frac{S h}{3}=\frac{S^{2}}{3}$. The maximum volume $V_{\max }=\frac{S_{\max }^{2}}{3}=\frac{4}{3}$.

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