# Problem 6.

In-1 In an irregular pyramid $A B C D$, the sum of the planar angles at vertex $A$ is $180^{\circ}$. Find the surface area of this pyramid if the area of face $B C D$ is $s$ and $A B=$ $C D, A D=B C$.
Answer: $4 s$.

Solution. We will prove that the faces of the pyramid are equal triangles. For this, consider the net $A D^{\prime \prime \prime} B D^{\prime} C D^{\prime \prime}$ of the pyramid $A B C D$, where $A D^{\prime \prime}=A D^{\prime \prime \prime}=A D, B D^{\prime \prime \prime}=B D^{\prime}=B D, C D^{\prime \prime}=$ $C D^{\prime}=C D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_eb9911c05e0d5d3c03abg-11.jpg?height=431&width=762&top_left_y=727&top_left_x=290)

Let $\angle B A C=\alpha, \angle B A D^{\prime \prime \prime}=\beta, \angle C A D^{\prime \prime}=\gamma$. Since $\alpha+\beta+\gamma=180^{\circ}$, the points $D^{\prime \prime}, A, D^{\prime \prime \prime}$ lie on the same line. Since $A D^{\prime \prime}=B C, A B=D^{\prime \prime} C$, then $A B C D^{\prime \prime}$ is a parallelogram and $B C \| A D^{\prime \prime}$. Similarly, $A D^{\prime \prime \prime} B C$ is also a parallelogram. Triangle $A C D^{\prime \prime}$ is equal to triangle $B C D^{\prime}$. Therefore, the faces of the pyramid are equal triangles.

V-2 In an irregular pyramid $A B C D$, the sum of the planar angles at vertex $A$ is $180^{\circ}$. Find the surface area of this pyramid if the area of face $A B C$ is $s$ and $A B=$ $C D, A D=B C$.

Answer: $4 s$.

V-3 In an irregular pyramid $A B C D$, the sum of the planar angles at vertex $A$ is $180^{\circ}$. Find the area of face $B C D$ if the surface area of this pyramid is $s$ and $A B=$ $C D, A D=B C$.

Answer: $\frac{s}{4}$.

V-4 In an irregular pyramid $A B C D$, the sum of the planar angles at vertex $A$ is $180^{\circ}$. Find the area of face $A B C$ if the surface area of this pyramid is $s$ and $A B=$ $C D, A D=B C$.

Answer: $\frac{s}{4}$.

B-5 In an irregular pyramid $A B C D$, the sums of the planar angles at vertices $A$ and $B$ are the same, the sums of the planar angles at vertices $C$ and $D$ are also the same, and the sum of the areas of faces $A B D$ and $A C D$ is $s$. Find the surface area of this pyramid.

Answer: $2 s$.

Solution. Let $\Sigma_{A}, \Sigma_{B}, \Sigma_{C}, \Sigma_{D}$ be the sums of the planar angles at vertices $A, B, C, D$ respectively. Then $\Sigma_{A}+\Sigma_{B}+\Sigma_{C}+\Sigma_{D}=4 \pi$ and $\Sigma_{A}+\Sigma_{C}=2 \pi$. If $\Sigma_{A} \leq \Sigma_{C}$, then $\Sigma_{A} \leq \pi \leq \Sigma_{C}$.

We will prove that $\triangle A B C=\triangle A B D$ and $\triangle C D A=\triangle C D B$ (from which the answer will follow). For this, consider the net $A D_{1} B D_{2} C D_{3}$ of the pyramid $A B C D$, where $A D_{1}=A D_{3}=A D, B D_{1}=$ $B D_{2}=B D, C D_{2}=C D_{3}=C D$. Then $\Sigma_{A}=\angle D_{1} A D_{3}=\angle D_{1} B D_{2}=2 \pi-\Sigma_{C}=\angle D_{2} C D_{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_eb9911c05e0d5d3c03abg-12.jpg?height=571&width=814&top_left_y=80&top_left_x=244)

Thus, the isosceles triangles $\Delta D_{1} A D_{3}$ and $\Delta D_{2} C D_{3}$ with equal angles at the vertex are similar, from which it follows that $\angle A D_{3} D_{1}=\angle C D_{3} D_{2}, \angle A D_{3} C=\angle D_{1} D_{3} D_{2}$ and $\frac{A D_{3}}{D_{3} C}=\frac{D_{1} D_{3}}{D_{2} D_{3}}$. Therefore, $\triangle A C D_{3} \sim \Delta D_{1} D_{2} D_{3}$, and the similarity ratio is $\frac{A D_{3}}{D_{1} D_{3}}=\frac{A C}{D_{1} D_{2}}$, but $\frac{A D_{3}}{D_{1} D_{3}}=\frac{D_{1} B}{D_{1} D_{2}}$ (again from the similarity of isosceles triangles $\triangle A D_{1} D_{3}$ and $\triangle B D_{1} D_{2}$), so $A C=D_{1} B$. Similarly, it can be shown that $B C=A D_{1}$, from which $\triangle A B C=\triangle A B D_{1}=\triangle A B D$ and, similarly, $\triangle C D A=\triangle C D B$.

V-6 In an irregular pyramid $A B C D$, the sums of the planar angles at vertices $A$ and $B$ are the same, the sums of the planar angles at vertices $C$ and $D$ are also the same, and the sum of the areas of faces $A B C$ and $A C D$ is $s$. Find the surface area of this pyramid.

Answer: $2 s$.

V-7 In an irregular pyramid $A B C D$, the sums of the planar angles at vertices $A$ and $B$ are the same and the sums of the planar angles at vertices $C$ and $D$ are the same. Find the sum of the areas of faces $A B D$ and $A C D$ if the surface area of this pyramid is $s$.

Answer: $\frac{s}{2}$.

B-8 In an irregular pyramid $A B C D$, the sums of the planar angles at vertices $A$ and $B$ are the same and the sums of the planar angles at vertices $C$ and $D$ are the same. Find the sum of the areas of faces $A B C$ and $B C D$ if the surface area of this pyramid is $s$.

Answer: $\frac{s}{2}$.

## Lomonosov Olympiad for Schoolchildren in Mathematics

Final stage 2020/21 academic year for 10-11 grades

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