Selected Problems from the Final Round of National Olympiad

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 $180^{\circ }-(180^{\circ }-2\cdot (i-1)\alpha )-(i-2)\alpha =i\alpha$ ($i=3,\dots ,2n$). Thus the base angle of triangle $AC{K}_2=K_0{K}_1{K}_2$ 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