VI OM - II - Task 3

What should be the angle at the vertex of an isosceles triangle so that a triangle can be constructed with sides equal to the height, base, and one of the remaining sides of this isosceles triangle?
We will adopt the notations indicated in Fig. 9. A triangle with sides equal to $a$, $c$, $h$ can be constructed if and only if the following inequalities are satisfied:






Since in triangle $ADC$ we have $a > h$, $\frac{c}{2} + h > a$, the first two of the above inequalities always hold, so the necessary and sufficient condition for the existence of a triangle with sides $a$, $c$, $h$ is the inequality






From triangle $ADC$ we have $h = a \cos \frac{x}{2}$, $\frac{c}{2} = a \sin \frac{x}{2}$; substituting into inequality (1) gives






or






and since $\frac{x}{4} < 90^\circ$, the required condition takes the form






or






Approximately, $4 \arctan -\frac{1}{2} \approx 106^\circ$ (with a slight deficit).