Corners are cut off from an equilateral triangle T to produce a regular…

Concept: An equilateral triangle can be partitioned, by trisecting each side, into 9 congruent equilateral triangles. Removing the three corner triangles (one at each vertex) leaves 6 congruent triangles that combine into a regular hexagon — so whenever this particular corner-cut is the one that produces a genuinely REGULAR hexagon H, area(H) : area(T) = 6 : 9.

Application

1. Let the side of triangle T be 3a, and mark the two trisection points on each of its three sides.

2. Join the trisection points nearest each vertex — this cuts off a small equilateral triangle of side a from each of the 3 corners, removing 3 congruent triangles of side a in all.

3. The trisection lines split T into 9 congruent equilateral triangles of side a. Removing the 3 corner ones leaves the other 6, which fit together into hexagon H (regular, since the construction is symmetric about all three vertices).

4. So area(H) accounts for 6 of these unit triangles and area(T) accounts for all 9, giving area(H) : area(T) = 6 : 9 = 2 : 3.

Cross-check: Take the side of T as s = 3a, so area(T) = (√3/4)·s2 = 9·(√3/4)·a2. A regular hexagon of side a is exactly six equilateral triangles of side a, so its area = 6·(√3/4)·a2. Hence area(H) / area(T) = 6/9 = 2/3 — confirming the ratio 2 : 3 independently of the corner-cutting argument.