1977 IMO Problems/Problem 1

Problem

In the interior of a square $ABCD$ we construct the equilateral triangles $ABK, BCL, CDM, DAN.$ Prove that the midpoints of the four segments $KL, LM, MN, NK$ and the midpoints of the eight segments $AK, BK, BL, CL, CM, DM, DN, AN$ are the 12 vertices of a regular dodecagon.

Solution

Just use complex numbers, with $a = 1$ , $b = i$ , $c = - 1$ and $d = - i$ . With some calculations, we have $k = \frac {\sqrt {3} - 1}{2}( - 1 - i)$ , $l = \frac {\sqrt {3} - 1}{2}(1 - i)$ , $m = \frac {\sqrt {3} - 1}{2}(1 + i)$ and $n = \frac {\sqrt {3} - 1}{2}( - 1 + i)$ . Now it's an easy job to calculate the twelve midpoints and to find out they are all of the form $\frac {\sqrt {3} - 1}{2}e^{\frac {k\pi}{6}i}$ , with $k\in\mathbb{N}: 0\le k\le 11$ , and the result follows.

The above solution was posted and copyrighted by Joao Pedro Santos. The original thread for this problem can be found here: [1]

1977 IMO (Problems) • Resources
Preceded by First Question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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1977 IMO Problems/Problem 1

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