## Task 4 - 130724

Let $\triangle A B C$ be a triangle in which the size $\gamma$ of the interior angle $B C A$ is smaller than the sizes of the other two interior angles.

Construct all points $P$ on the sides $A C$ and $B C$ such that $\angle B P A=2 \gamma$!

Describe and justify your construction; determine the number of points $P$ with the required property!
(I) Suppose a point $P$ on $A C$ has the desired property. By the exterior angle theorem for $\triangle B C P$, it follows that $\angle C B P = \angle B P A - \angle B C P = \gamma$.

(II) Therefore, a point $P$ on $A C$ meets the conditions of the problem if and only if it can be obtained by the following construction:

(1) At $B$, measure an angle of size $\gamma$ along $B C$ on the side of the line through $B$ and $C$ where $A$ lies.

(2) If the free arm of this angle intersects the side $A C$, let $P$ be the intersection point.

(III) Proof that every point $P$ constructed in this way meets the conditions of the problem: By construction (2), $P$ lies on $A C$. Furthermore, by the exterior angle theorem and construction (1), $\angle B P A = \angle B C P + \angle C B P = 2 \gamma$.

(IV) Construction step (1) is always uniquely executable. Since $\gamma < \beta$, the free arm of the angle constructed in (1) intersects the interior of triangle $A B C$ and cuts the side $A C$ between $A$ and $C$;

Thus, construction step (2) yields exactly one point $P$ on $A C$ that has the desired property. If we swap $A$ and $B$ in the considerations (I) to (IV), we obtain: There is exactly one (additional) point $P'$ on $B C$ that has the desired property. Therefore, there are always exactly 2 such points.

Solutions of the 2nd Round 1973 adopted from [5]

### 4.15.3 3rd Round 1973, Class 7