Problem 732

Consider the task as solved. In our diagram, which also shows the projection, $AC > CB$, and the segment $B_{2}C_{2}$ of the line $e$ that falls within the triangle is denoted by $e$.

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Our task is solved if we can construct the segment $AC_{2} = x$. According to the problem,

$$\frac{cm}{2} = xe$$

If we denote the projection of the side $AC$ onto $AC_{1}$ by $d$, then it is clearly

$$\frac{d}{m} = \frac{x}{e}$$

Multiplying (1) and (2) (and swapping the two sides)

$$x^{2} = \frac{c}{2} d$$

Thus, $x$ is the geometric mean between $\frac{c}{2} = AC_{3}$ and $d = AC_{1}$.

Csekö Sarolta (Bp., I., Szilágyi E. lg. II. o. t.)

Remark: Our task can also be stated as follows: Transform the triangle $AC_{3}C$ - whose area is half of the given triangle - into a right-angled triangle $AC_{2}B_{2}$ of equal area, such that the angle $\angle A$ is common in both triangles. In this formulation, our task is a special case of the problem discussed on pages 72-73 of the textbook for the second year of high school (Tankönyvkiadó, 1955).