Let the perpendicular projections of vertex $A$ of parallelogram $ABCD$ onto lines $CB$ and $CD$ be denoted by $E$ and $F$, respectively, and let the orthocenter of triangle $AEF$ be $M$. Determine the length of segment $AM$, given the lengths of segments $AC$ and $EF$, as well as the angle between lines $AC$ and $EF$.
I. solution. The altitude $M E$ of the triangle $A E F$ is perpendicular to $A F$, so it is parallel to $C D$. Similarly, $M F \| C B$, so the quadrilateral $M E C F$ is a parallelogram (in the two variations of the diagram, $\angle B A D$ is acute, or obtuse).
![](https://cdn.mathpix.com/cropped/2024_05_02_0a6a4dc5541d43ef4665g-1.jpg?height=556&width=1296&top_left_y=299&top_left_x=404)

Let $K$ and $O$ be the centers of the parallelograms $M E C F$ and $A B C D$, respectively. $K$ bisects $M C$, and $O$ bisects $A C$, so $K O$ is the midline of the triangle $A C M$ parallel to $A M$, hence $A M=2 \cdot K O$.

By construction, $E$ and $F$ lie on the Thales circle with diameter $A C$ (i.e., centered at $O$), so $O E=O F=A C / 2$, and since $K$ also bisects $E F$, $K O$ is the altitude of the isosceles triangle $O E F$. Therefore,

$$
A M^{2}=4 \cdot K O^{2}=4\left(O E^{2}-K E^{2}\right)=A C^{2}-E F^{2}
$$

Thus, the segment can be calculated without using the angle between the lines $A C$ and $E F$, provided that $A C \geqq E F$. If $E F>A C$, then there is no parallelogram that satisfies the given conditions.

II. solution. Translate the segment $A M$ so that the endpoint $A$ coincides with $E$, and let the new position of $M$ be $M^{\prime}$, so $M^{\prime} E$ is equal to $M A$ and parallel to it. Since the altitude $M A$ is perpendicular to $E F$, the triangle $M^{\prime} E F$ has a right angle at $E$:

$$
M A^{2}=M^{\prime} E^{2}=M^{\prime} F^{2}-E F^{2}
$$

We have seen in the first solution that $M E C F$ is a parallelogram, so $M F=E C$. By construction, $M M^{\prime}$ is parallel to $A E$, hence perpendicular to $M F$. Therefore, the right triangle $M M^{\prime} F$ is congruent to the triangle $E A C$ (congruent by reflection), so $M^{\prime} F=A C$, and substituting this into (1) we get

$$
M A^{2}=A C^{2}-E F^{2}
$$

Tibor Schvarcz (Debrecen, Fazekas M. Gym., III. grade)

Remarks. 1. To answer the question of the problem, it was not necessary to use the given angle $\varphi$ between the lines $A C$ and $E F$. This angle did not even appear in the marked source, the organizing committee attached it to the problem to somehow more precisely determine the parallelogram in question, since a parallelogram is determined by 3 independent data. We recommend that the reader construct the parallelogram using $\varphi$ and the two length data.

2. Denoting the angle $\angle B A D$ of our parallelogram by $\alpha$, we have $A E=A B \sin \alpha, A F=A D \sin \alpha$ and $\angle F A E=180^{\circ}-\alpha=\angle A B C$ (complementary angles), so the triangles $A E F$ and $B A C$ are similar, and thus $E F=A C \sin \alpha$. From this, it also follows that the parallelogram only exists if $E F \leqq A C$.