10-1. At the base of the pyramid $S A B C D$ lies a trapezoid $A B C D$ with bases $B C$ and $A D$. Points $P_{1}$, $P_{2}, P_{3}$ belong to the side $B C$, such that $B P_{1}<B P_{2}<B P_{3}<B C$. Points $Q_{1}, Q_{2}, Q_{3}$ belong to the side $A D$, such that $A Q_{1}<A Q_{2}<A Q_{3}<A D$. Denote the points of intersection of $B Q_{1}$ with $A P_{1}$, $P_{2} Q_{1}$ with $P_{1} Q_{2}$, $P_{3} Q_{2}$ with $P_{2} Q_{3}$, and $C Q_{3}$ with $P_{3} D$ as $R_{1}, R_{2}, R_{3}$, and $R_{4}$, respectively. It is known that the sum of the volumes of the pyramids $S R_{1} P_{1} R_{2} Q_{1}$ and $S R_{3} P_{3} R_{4} Q_{3}$ is 78. Find the minimum value of the quantity

$$
V_{S A B R_{1}}^{2}+V_{S R_{2} P_{2} R_{3} Q_{2}}^{2}+V_{S C D R_{4}}^{2}
$$

In your answer, specify the integer closest to the found value.
Solution. From the properties of the trapezoid, it follows that the triangles (see Fig. 2), shaded with the same color, have the same area. From this, the equality of the sums of the areas marked

![](https://cdn.mathpix.com/cropped/2024_05_06_997c53c06135e2de9c16g-10.jpg?height=317&width=934&top_left_y=1980&top_left_x=615)

Fig. 2:

with the same color on Fig. 3 follows. The equality of the sums of the areas takes the form

$$
S_{A B R_{1}}+S_{R_{2} P_{2} R_{3} Q_{2}}+S_{C D R_{4}}=S_{R_{1} P_{1} R_{2} Q_{1}}+S_{R_{3} P_{3} R_{4} Q_{3}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_997c53c06135e2de9c16g-11.jpg?height=323&width=934&top_left_y=221&top_left_x=618)

Fig. 3:

From which

$$
V_{S A B R_{1}}+V_{S R_{2} P_{2} R_{3} Q_{2}}+V_{S C D R_{4}}=V_{S R_{1} P_{1} R_{2} Q_{1}}+V_{S R_{3} P_{3} R_{4} Q_{3}}=78
$$

Let $a_{1}=V_{S A B R_{1}}, a_{2}=V_{S R_{2} P_{2} R_{3} Q_{2}}, a_{3}=V_{S C D R_{4}}$. From the problem statement: $a_{1}+a_{2}+a_{3}=78$, and we need to find $a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \rightarrow \min$, under the condition that $a_{1}, a_{2}, a_{3}$ are non-negative. The inequality holds

$$
\left(\frac{a_{1}}{3}+\frac{a_{2}}{3}+\frac{a_{3}}{3}\right)^{2} \leqslant \frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{3}
$$

For school students, this inequality is justified by applying the inequality $2 a b \leqslant a^{2}+b^{2}$ three times. We obtain

$$
a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \geqslant \frac{\left(a_{1}+a_{2}+a_{3}\right)^{2}}{3}=2028
$$

Equality is achieved when $a_{1}=a_{2}=a_{3}=26$.

Answer: 2028.