Track / Stage 6 / 65 of 400 #1065 of 1964

Problem 1065

National olympiad, first round

Geometry Difficulty 6.1 Prove it

8.3. In an equilateral triangle $ABC$ , through a random point inside it, three lines are drawn: parallel to $AB$ until intersecting with $BC$ and $CA$ ; parallel to $BC$ until intersecting with $AB$ and $CA$ ; parallel to $CA$ until intersecting with $BC$ and $AB$ . Prove that the sum of the three obtained segments is equal to twice the side of the triangle $ABC$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution: Let an arbitrary point $P$ be chosen inside the triangle. Draw the segments and label them as shown in the figure. It is obvious that triangles $D E P, P F G, P I H$ are equilateral, as all angles in them are 60 degrees.

Next, notice that $BFPE$ is a parallelogram, since the opposite sides in it are pairwise parallel. Therefore, $P F = B E$ . Similarly, $A D P I$ is a parallelogram, so $A D = P I$ .

It remains to notice that the sum of the three segments is $E H + F I + G D = E P$

!
$+ P H + F P + P I + G P + P D = 2 E D + 2 P F + 2 P I = 2 E D + 2 B E + 2 A D = 2 A B$ , which is what we needed to prove.

Criteria: Considering special cases of the position of point $P$ is worth nothing.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.