10. The base of the pyramid $T A B C D$ is a rectangle $A B C D$. The height of the pyramid, equal to $h$, coincides with the lateral edge $T A$, and the lateral edge $T C$ is inclined to the base plane at an angle of $30^{\circ}$. The plane passing through the edge $T C$ and parallel to the diagonal of the base $B D$,
forms an angle of $60^{\circ}$ with the base plane. What is the smallest area that the section of the pyramid by a plane passing through the diagonal of the base $B D$ can have?
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-30.jpg?height=1059&width=1223&top_left_y=407&top_left_x=385)

Let $\alpha = \angle TCA$, then $AC = BD = TA \operatorname{ctg} \angle TCA = h \operatorname{ctg} \alpha$. Draw $EC \parallel BD$, $AK \perp EC$, $K \in EC$, and let $\beta = \angle TKA$. Let $P = AK \cap BD$, then $AP = PK$. Draw $AF \perp TK$, $PQ \perp TK$, $QM \parallel (EC)$, $M \in TC$ and $MN \perp BD$; then $MN$ is the height of the section $BMD$ with the shortest length, and $MN = PQ$. Find the area of the section: $PQ = 0.5 AF = 0.5 AT \cos \beta = 0.5 h \cos \beta$, $S_{\triangle BMD} = 0.5 BD \cdot MN = 0.25 h^2 \operatorname{ctg} \alpha \cos \beta$.

Answer: $\alpha = 30^\circ$, $\beta = 60^\circ$, $S = h^2 \sqrt{3} / 8$.

## First (Preliminary) Stage of the Academic Competition

## "Step into the Future" School Students' Olympiad in the Subject of

"Mathematics", Fall 2016

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