## Task 4 - 120714

Given a triangle $\triangle A B C$. A point $C_{1}$ should have the following properties:

(1) The triangle $\triangle A B C_{1}$ is equal in area to the triangle $\triangle A B C$,

(2) $A C = A C_{1}$.

(3) $C \neq C_{1}$.

a) Provide a construction method to obtain all points $C_{1}$ that have the properties (1), (2), (3)!

b) Investigate how the number of points $C_{1}$ with the properties (1), (2), (3) depends on the properties of the given triangle $\triangle A B C$! (Case distinction)
(I) Suppose a point $C_{1}$ has the properties (1), (2), (3). Then:

The triangles $\triangle A B C$ and $\triangle A B C_{1}$ share the side $A B$. Because of (1), they also agree in the height corresponding to this side. Therefore, the point $C_{1}$ lies on one of the parallels to $A B$ that are at the same distance from $A B$ as point $C$. Because of (2), $C_{1}$ also lies on the circle with radius $A C$ centered at $A$.

(II) From this, it follows that a point $C_{1}$ only has the properties (1), (2), (3) if it can be obtained by the following construction:

We draw the parallel $g$ to $A B$ through $C$ and the other parallel $g_{1}$ to $A B$ that is at the same distance from $A B$ as $g$. We draw the circle $k$ with radius $A C$ centered at $A$. If this circle intersects or touches $g$ or $g_{1}$ at a point different from $C$, then this point is called $C_{1}$.

(III) The proof that every such constructed point $C_{1}$ satisfies the conditions (1) to (3) follows from (II) and the fact that the triangles $\triangle A B C$ and $\triangle A B C_{1}$ agree in the height corresponding to their (common) side.
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0386.jpg?height=330&width=1032&top_left_y=1556&top_left_x=520)

(IV) Let $D$ be the foot of the perpendicular from $C$ to the line through $A$ and $B$. Now we distinguish the following two cases:

1. Case: If $\triangle A B C$ is not a right triangle at $A$, then $D \neq A$, so $\triangle A D C$ is a right triangle at $D$ and in it $A C$ is the longest side (hypotenuse), $D C$ is one of the other sides (cathetus), thus $D C < A C$. The distance that $g$ and $g_{1}$ have from $A B$ and consequently from $A$ is therefore less than the radius of $k$. Therefore, $k$ intersects each of the parallels $g, g_{1}$ at exactly two points. Of the four intersection points thus created, exactly one is $C$. The number of points $C_{1}$ that satisfy (1), (2), (3) is therefore 3.
2. Case: If $\triangle A B C$ is a right triangle at $A$, then $D = A$, so $D C = A C$. The distance that $g$ and $g_{1}$ have from $A B$ and consequently from $A$ is therefore equal to the radius of $k$. Therefore, $k$ touches each of the parallels $g, g_{1}$ at exactly one point. Of the two points of tangency thus created, exactly one is $C$.

The number of points $C_{1}$ that satisfy (1), (2), (3) is therefore 1.

Solutions of the First Round 1972 taken from $[5]$

### 4.14.2 Second Round 1972, Class 7