In the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, a plane intersects the edges $D D^{\prime}, A A^{\prime}, A B, B C, C C^{\prime}$ at points $P, Q, R, S, T$ respectively. We know that $A R: R B=C S: S B=7: 3$. Can the point $P$ be chosen such that $2 Q R=P Q+R S$?
Let the edge of the cube be 10 units, and $DP = x$. The answer is affirmative if there is an $x$ that satisfies the given equation, and for which $0 < x < 10$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a7737631b94a61d664d1g-1.jpg?height=664&width=959&top_left_y=253&top_left_x=564)

From the requirement, $BR = BS = 3$, so $RS$ is perpendicular to the diagonal $BD$ and to the symmetry plane $BDD'$ of the cube. Therefore, the same holds for the intersecting plane. Let the intersection point of the lines $RS$ and $AD$ be $A_1$, then $A_1A = RA = 7$, $A_1D = 17$. The intersecting plane intersects the plane $ADD'$ at $PA_1$, on which $Q$ is such that

$$
PQ : PA_1 = DA : DA_1 = 10 : 17, \quad \text{and} \quad QR : PA_1 = QA_1 : PA_1 = 7 : 17
$$

since the right triangles $QAR$ and $QAA_1$ are congruent.

Substituting these into the requirement:

$$
\begin{gathered}
\frac{14}{17} PA_1 = \frac{10}{17} PA_1 + 3\sqrt{2} \\
PA_1 = \sqrt{x^2 + 17^2} = \frac{51\sqrt{2}}{4} \\
x = \frac{17\sqrt{2}}{4} \approx 6.01
\end{gathered}
$$

Thus, the answer to the question is: yes, it is solvable.

Ildikó Csapó (Sopron, Széchenyi I. Gymnasium, 4th grade)

Remarks. 1. According to the requirement, the sides $SR$, $RQ$, and $QP$ of the intersection form an arithmetic sequence.

2. The same holds for the sides $RS$, $ST$, and $TP$, because the intersection domain is symmetric with respect to the line connecting the midpoint of segment $RS$ and point $P$.

However, the solution did not require this!