Estonija, 2010

Geometry Difficulty 5.4 AIME, harder Prove it Estonia

Points $A'$ , $B'$ and $C'$ are chosen correspondingly on the sides $AB$ , $BC$ , and $CA$ of an equilateral triangle $ABC$ so that $\frac{|A'B|}{|AB|} = \frac{|B'C|}{|BC|} = \frac{|C'A|}{|CA|} = k$ . Find all positive real numbers $k$ for which the area of the triangle $A'B'C'$ is exactly half of the area of the triangle $ABC$ .

Solution

Let $\alpha$ be the angle at the vertex $A$ (Fig. 10).

The area of the triangle $AA'C'$ is $S_{AA'C'} = \frac{1}{2} \cdot |AA'| \cdot |AC'| \cdot \sin \alpha = \frac{1}{2} \cdot (1-k)|AB| \cdot k|AC| \cdot \sin \alpha = (1-k)k S_{ABC}$ .

Similarly $S_{BB'A'} = (1-k)k S_{ABC}$ and $S_{CC'B'} = (1-k)k S_{ABC}$ .

Hence the triangles $AA'C'$ , $BB'A'$ and $CC'B'$ are of equal area.

Fig. 10

the area of the triangle $A'B'C'$ is half of the area of the triangle $ABC$ iff the area of the triangle $AA'C'$ is one sixth of the area of the triangle $ABC$ , i.e. $(1-k)k = \frac{1}{6}$ .

The solutions of $k^2 - k + \frac{1}{6} = 0$ are $k_{1,2} = \frac{1}{2} \pm \frac{\sqrt{3}}{6}$ , both of them are positive.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.