**Problem 3 of the Balkan Mathematical Olympiad 1988**

Let ABCD be a tetrahedron and let d be the sum of squares of its edges' lengths. Prove that the tetrahedron can be included in a region bounded by two parallel planes, the distances between the planes being at most 1/2 \sqrt{d/3} .

**Solution by Vo Duc Dien (dedicated to Mark Brockmann)**

Let E, F and M be the midpoints of AC, AB and DC, respectively; also let the edge’s length of the tetrahedron be l. From D and M draw the altitudes to plane ABC to meet it at H and I, respectively. We will prove that the tetrahedron fits into the parallel planes with DC and AB on either plane.

The sum of squares of six lengths is 6l² = d or l = \sqrt{d/6}

Consider equilateral triangle DAC, DE² = l² - l²/4 or DE = l \sqrt{3} /2. We also have BE = DE and since H is also the centroid of triangle ABC, HE = BE/3 = DE/3 = l \sqrt{3} /6.

Now consider right triangle DHE where DH² = DE² - HE² = 3l²/4 - 3l²/36 = 2l²/3 or DH = l \sqrt{2/3} = \sqrt{d} /3.

Now FM² = MI² + FI² = (DH/2)² + (2HE)² = 1/2 \sqrt{d/3} , but as we can see FM is orthogonal to AB (in triangle BMA) and it’s also orthogonal to DC (in triangle DFC), therefore, the plane containing AB and the plane containing DC that are both orthogonal to FM are parallel to each other. The tetrahedron, therefore, fits into the two planes being at most 1/2 \sqrt{d/3} apart.