Construct the three altitudes of a given triangle with as few construction steps as possible! Count as one step the drawing of a visible line: a circle, an arc, a straight line; do not count as a step the selection of the compass width, the placement of the compass point, or the appropriate alignment of the ruler (although these do require attention, they are not visible).





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I. Solution: Using the usual notation, let us construct, for example, the reflection of vertex $B, C$ over the sides $C A$ and $A B$, respectively, denoted as $B_{1}$ and $C_{1}$. Thus, the intersection point $M$ of $B B_{1}$ and $C C_{1}$, and the line connecting the orthocenter with $A$ gives the altitudes. For $B_{1}$, we draw 2 arcs: one around $C$ with radius $a$ and one around $A$ with radius $c$; similarly for $C_{1}$: one around $A$ with radius $b$ and one around $B$ with radius $a$. Finally, the above 3 lines result in 7 steps.

If we know that the triangle is right-angled at, for example, $C$, then we only need the altitude $C C_{1}$, which can be obtained with 2 arcs and 1 line, i.e., in 3 steps. (Otherwise, the identity of $M$ and $C$ would indicate that $\angle A C B = 90^{\circ}$.) - If we know that the triangle is isosceles, for example, $A C = B C$, then by saving 1 arc around $C$, we can reach the goal in 6 steps. - The equality of all three sides does not provide further savings.

Zalán Péter (Aszód, Petőfi S. g. I. o. t.)

II. Solution: The circle over the diameter $B C$ (Thales' circle) intersects the other two sides (or their extensions) at the feet of the altitudes $B B^{\prime}$ and $C C^{\prime}$, denoted as $B^{\prime}$ and $C^{\prime}$. From these, we can proceed similarly as from $B_{1}, C_{1}$, and the number of steps is the same. Indeed, preparing the Thales' circle requires 2 equal-radius arcs (around $B$ and $C$) and drawing the perpendicular bisector, which is the 4th step. Here, we draw 3 arcs and 4 lines.

If the triangle is obtuse at $A$, then extending the sides $A B$ and $A C$ adds 2 steps to the number of steps. However, if we do not draw the Thales' circle over the largest side, only 1 extension is necessary.

Zaránd Rudolf (Bp. VIII., Piarista g. I. o. t.)

Remarks. 1. In the II. solution, we can save 1 step by using the ruler and compass simultaneously, which the problem statement does not mention but does not exclude. It is sufficient to set the perpendicular bisector of $B C$ with the ruler, and then, with the compass temporarily placed in the midpoint of $B C$, which has become visible, draw the Thales' circle as the 3rd visible step.

Szőkefalvi-Nagy Zoltán (Szeged, Radnóti M. g. I. o. t.)

2. Extending this idea, we can save all 4 arcs of the I. solution by first using two compasses, then one compass and the ruler simultaneously: with the compasses (possibly two-pointed, "piercing" compasses) opened to $c$ and $a$, we place one point of each in $A$ and $C$, respectively, and then bring the "free" points together on the paper to $B_{1}$. Then, replacing one compass with a ruler, we set and draw $B B_{1}$, and so on. However, this procedure cannot be considered a Euclidean construction in any case, as the mentioned alignment is only possible by trial and error (on the intersection circle of the spheres that can be drawn around $A$ and $C$ with radii $c$ and $a$). This is similar to the geometric experiment of angle trisection with a paper plane: we need to pay attention to 3 points: the two compass points and the paper.
3. We accepted solutions with more than 7 steps.