# Hexagons in a Square

In the interior of a square of side-length 3 there are several regular hexagons whose sum of perimeters is equal to 42 (the hexagons may overlap). Prove that there are two perpendicular lines such that each one of them intersects at least five of the hexagons.

(Math Magazine, VOL. 84, NO. 2, APRIL 2011, problem 1842. Proposed by Bianca-Teodora Iordache, student, National College "Carol I," Craiova, Romania. Solution by CMC 328, Carleton College, Northfield, MN.) In the interior of a square of side-length 3 there are several regular hexagons whose sum of perimeters is equal to 42 (the hexagons may overlap). Prove that there are two perpendicular lines such that each one of them intersects at least five of the hexagons.

### Solution

Firrst observe that when we project a regular hexagon of side length a onto a line its shortest possible projection is a3. To see that, note that the diameter of a circle inscribed into a regular hexagon of side a is a3. In addition, the projection of a hexagon is not shorter than that of its inscribed circle.

Project all the hexagons onto one of the sides of the square. Since the sum of all the hexagons' perimeters is 42, the sum of all of their side-lengths is 7. Hence, their projection length on one edge of the square is at least 73 ≈ 12.124. Since all of these projections are onto a segment of length 3, and 73/4 > 3, there must be some region in the segment covered by at least five of the projections. Pick a point in this region and draw a line through this point perpendicular to the edge; this line must intersect at least five hexagons. By carrying out this construction for two perpendicular edges of the square, we get the desired two perpendicular lines. • 200 points have been cosen on a circle, all with integer number of degrees. Prove that the points there are at least one pair of antipodes, i.e., the points 180° apart.
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