## Task 3 - 340723

In the triangles $A B C$, let the measures of the interior angles at $A, B$, and $C$ be denoted by $\alpha, \beta$, and $\gamma$ respectively. Furthermore, assume that the perpendicular bisector of side $A B$ and the angle bisector passing through $A$ intersect at a point $D$ that lies on side $B C$.

a) Derive a formula that, in every triangle that meets this condition, expresses $\gamma$ in terms of $\beta$! For which values of $\beta$ does there exist a triangle that meets the given conditions; for which values of $\beta$ does no such triangle exist?

b) Determine all values of $\beta$ for which a triangle that meets the given conditions is a right triangle!

c) Determine all values of $\beta$ for which a triangle that meets the given conditions is isosceles!
a) (see figure) Since $D$ lies on the perpendicular bisector of $AB$, it follows that $AD = BD$. By the base angle theorem for triangle $ABD$, this implies $\angle BAD = \angle ABD = \beta$.

Since $AD$ is the angle bisector, it follows that $a = 2\beta$ (1) and thus, by the angle sum property, the desired formula is $\gamma = 180^\circ - 3\beta$. (2)

Furthermore,

I. For all $\beta \geq 60^\circ$, there is no triangle of the required type; for if there were such a triangle, then, as shown above, (2) would hold. This is not possible, as it would result in a value $\gamma \leq 0$.

II. For all positive $\beta < 60^\circ$, there is a triangle of the required type.

For if one chooses the values of $\alpha$ and $\gamma$ from (1) and (2) for such a $\beta$, one has three positive angle measures $\alpha, \beta, \gamma$ with $\alpha + \beta + \gamma = 180^\circ$.

Therefore, there is a triangle $ABC$ with these interior angles. If $AD$ is the angle bisector in this triangle, then triangle $ABD$ is isosceles by the converse of the base angle theorem, so $D$ also lies on the perpendicular bisector of $AB$.

b) With $\beta = 90^\circ$, a triangle that meets the conditions stated in the problem cannot be a right triangle; for with this value of $\beta$, there is no such triangle according to a) I.

On the other hand, there is such a triangle that is right-angled with $\alpha = 90^\circ$; for according to (1), one can choose $\beta = 45^\circ$ and proceed as in a) II. Similarly, $\gamma = 90^\circ$ is achievable with $\beta = 30^\circ$ according to (2).

The values sought in (b) are thus exactly $\beta = 30^\circ$ and $\beta = 45^\circ$.

c) With $\alpha = \beta$, a triangle that meets the conditions stated in the problem cannot be isosceles; for according to (1), it would follow that $2\beta = \beta$, thus $\beta = 0$ and no triangle would exist.

On the other hand, $\beta = \gamma$ is achievable according to (2) with $\beta = 180^\circ - 3\beta$, thus $\beta = 45^\circ$, and similarly, $\alpha = \gamma$ is achievable according to (1) and (2) with $2\beta = 180^\circ - 3\beta$, thus $\beta = 36^\circ$. The values sought in (c) are therefore exactly $\beta = 45^\circ$ and $\beta = 36^\circ$.