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Selected Problems from the Final Round of National Olympiad

Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Estonia

The size of the angle $ABC$ , expressed in degrees, in a right triangle $ABC$ is an integer. It is known that for some positive integer $n$ , one can choose points $K_0 = A$ , $K_2, \dots, K_{2n}$ on the hypotenuse $AB$ and points $K_1 = C$ , $K_3, \dots, K_{2n+1} = B$ on the leg $CB$ in such a way that each triangle $K_{i-1}K_iK_{i+1}$ with $i = 1, \dots, 2n$ is isosceles with base $K_{i-1}K_{i+1}$ . Find all possible values of the size of angle $ABC$ .

Solution

Let $\angle ABC = \alpha$ (Fig. 12). Then the base angle of the last isosceles triangle $K_{2n-1}K_{2n}K_{2n+1}$ is $\alpha$ . The base angle of the second last isosceles triangle $K_{2n-2}K_{2n-1}K_{2n}$ has the size $180^\circ - (180^\circ - 2\alpha) = 2\alpha$ . The base angle of the next triangle before it, $K_{2n-3}K_{2n-2}K_{2n-1}$ , has the size

$180^\circ - (180^\circ - 4\alpha) - \alpha = 3\alpha$ .

Generally, the size of the base angle of triangle $K_{2n-i}K_{2n-i+1}K_{2n-i+2}$ is $180^\circ - (180^\circ - 2\cdot(i - 1)\alpha) - (i - 2)\alpha = i\alpha$ ( $i = 3, \ldots, 2n$ ). Thus the base angle of triangle $ACK_2 = K_0K_1K_2$ has the size $2n\alpha$ .

Now in the triangle $ABC$ we get $90^\circ = \angle BAC + \angle ABC = 2n\alpha + \alpha$ , whence

$2n + 1$ is an odd divisor of $90$ and is greater than $1$ (as a triangle cannot have two angles of the size $90^\circ$ ). Such divisors are $3$ , $5$ , $9$ , $15$ , and $45$ that give the solutions $30^\circ$ , $18^\circ$ , $10^\circ$ , $6^\circ$ , and $2^\circ$ , respectively.

Fig. 12

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.