Problem 4. Let $S A B C D$ be a regular quadrilateral pyramid with base $A B C D$. On the segment $A C$, there is a point $M$ such that $S M = M B$ and the planes $S B M$ and $S A B$ are perpendicular. Find the ratio $A M: A C$.
Answer: $3: 4$.

Solution. Let the center of the base $A B C D$ be $O$. Then $S O \perp A C$ and $B D \perp A C$, which means $S B \perp A C$. This implies that the perpendicular bisector of the segment $S B$ (let's denote this plane as $\alpha$) is parallel to $A C$. On the other hand, from $S M = M B$, it follows that point $M$ lies in $\alpha$. Therefore, the entire line $A C$ must be contained in $\alpha$. From this, we get $S O = O B$.

Let the midpoint of $S B$ be $R$. Note that $R M$ lies in two planes, both perpendicular to $S A B - \alpha$ and $S M B$. This means that $R M$ is itself perpendicular to the plane $S A B$, and also to the segment $R A$.

Let the length $O S = O A = O B = O C = O D$ be 1. Then $O R$ is the median of the right isosceles triangle $S O B$; since the legs of this triangle are both 1, we have $O R = 1 / \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5f2891ebbe9fabbee508g-3.jpg?height=274&width=593&top_left_y=1514&top_left_x=430)

Fig. 1: to the solution of problem 4

Finally, consider the section of the tetrahedron by the plane $\alpha$ (Fig. 1). The triangle $M R A$ is a right triangle, with $R O$ being the altitude to the hypotenuse. We have $R O^{2} = M O \cdot A O$, that is, $(1 / \sqrt{2})^{2} = M O \cdot 1$, from which $M O = 1 / 2$. We get $A M = 3 / 2$ and $A M: A C = 3: 4$.

## Criteria

7 points. Any complete solution to the problem.

5 points. A solution with correct geometric constructions and proofs, but with a computational error at the end of the solution.

5 points. A generally correct geometric solution with the correct answer, in which the proof of the equality of the edges of the pyramid contains inaccuracies.

3 points. A geometric solution with the correct answer, in which the equality of the edges of the pyramid is used but not proven.

6 points. A generally correct coordinate-vector solution with an incorrect answer due to an insignificant arithmetic error at the final stage of calculations.

3 points. A conceptually correct coordinate-vector solution with an incorrect answer due to a computational error at the beginning or middle of the solution.

1 point. It is proven that the height dropped from point $M$ to the triangle $S B M$ on $B S$ is perpendicular to the plane $A B S$, but the rest of the solution is incorrect.

3 points. It is proven that the height of the pyramid is half the diagonal of the base of the pyramid, but the rest of the solution is incorrect.

4 points. It is proven that all edges of the pyramid are equal, but the rest of the solution is incorrect or the progress is insignificant.

The statement that a lateral edge of the pyramid is perpendicular to the diagonal intersecting it is considered obvious; no points are deducted for the lack of its proof.