Estonian Mathematical Olympiad

A square $ABCD$ and its centre $E$ determine 8 distinct right triangles: $ABC$, $BCD$, $CDA$, $DAB$, $AEB$, $BEC$, $CED$, $DEA$.

Fig. 48

We show that more than 8 right triangles is impossible. Firstly, note that among any 5 points, one can choose 4 points that are vertices of a rectangle in at most one way. Indeed, suppose that this can be done in two different ways. Then these two quadruples of points have 3 points in common. But 3 vertices of a rectangle (even parallelogram) uniquely determine the last vertex.

Secondly, note that a quadrangle whose every three vertices form a right triangle is a rectangle. Indeed, these four triangles must have right angle at distinct vertices, otherwise three points would lie on a line. Now if $A$, $B$, $C$ are the vertices of a right triangle with right angle at $B$ then the remaining vertex $D$ must be located in such a way that some triangle could have right triangle at $A$ and some triangle could have right angle at $C$. Hence $D$ must lie on a line passing through $A$ and perpendicular to either $AB$ or $AC$, and also on a line passing through $C$ and perpendicular to either $AC$ or $BC$ (in Fig. 49, these lines are drawn using dashes of distinct colours).

Fig. 49

As lines perpendicular to $AC$ do not meet, $D$ must lie on the line passing through $A$ perpendicular to $AB$ or on the line passing through $C$ perpendicular to $BC$. If $D$ lies on both these lines, $ABCD$ is a rectangle. Hence consider the case with $D$ lying on exactly one of these lines. W.l.o.g., let $D$ lie on the line passing through $A$ perpendicular to $AB$ and on the line passing through $C$ perpendicular to $AC$. But then $BCD$ is not a right triangle. Hence the only possibility for $D$ is such that $ABCD$ is a rectangle.

Let now 5 points be chosen on a plane. Let us count right triangles with vertices in the chosen points by quadrangles, among the vertices of which the vertices of the triangle are. As there are 5 possibilities to extract 4 points out of the given 5 points and, by the above, at most one of the obtained quadruples enable 4 right triangles and the others consequently enable at most 3 right triangles, we can have at most $4+4\cdot 3=16$ right triangles. But each of these right triangles is counted twice as the fourth vertex can be either of the two remaining points. Hence there are at most 8 distinct right triangles.

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Alternative solution.

As in Solution 1, we show that it is possible to find 8 right triangles.

Now we prove that more than 8 right triangles is impossible. Suppose the opposite, i.e., there can be 9 right triangles. As there are $C_5^3=10$ triples of chosen points in total, at most one triple does not determine a right triangle. Let $A$ and $B$ be chosen points at maximal distance from each other (if there are several pairs of points with this property, take any of them). By choice, $AB$ can only be the hypotenuse of a right triangle. Hence at least two of the remaining three chosen points must lie on the circle with diameter $AB$; let these points be $C$ and $D$. At least one of $ACD$ and $BCD$ must be a right triangle. As all vertices of this triangle lie on the circle with diameter $AB$, two vertices of this triangle must be the endpoints of another diameter of this circle. These points can only be $C$ and $D$. Therefore also $CD$ is a line segment of maximal length and can only be the hypotenuse of a right triangle.

If the last chosen point $E$ also lies on the circle with diameter $AB$ then $ACE$ and $BCE$ are not right triangles because no pair of vertices can be the endpoints of a diameter of this circle. But if $E$ does not lie on this circle then $AEB$ and $CED$ are not right angles, implying that $ABE$ and $CDE$ cannot be right triangles. The contradiction shows that there cannot be 9 right triangles.