Construct any triangle ABC and its medians BD, AE, and CF.

Construct a second triangle (the "triangle of medians") with sides having the lengths of the medians. We will construct the triangle of medians, triangle HCF, using one of the actual medians CF as a side.

What are the relationships between the properties of the original triangle ABC and the triangle of medians HCF?

To get started, we will look at various special triangles and their triangles of medians.

Investigations show that triangle HCF is also equilateral.

Investigations show that triangle HCF is also isosceles.

In fact, we can prove our observation that an isosceles (or equilateral) triangle always produces an isosceles (or equilateral) triangle of medians.

Given: Triangle ABC

Segments AB and BC congruent

Midpoints M and N of AB and BC, respectively

Show: Medians CM and AN are congruent

Proof:

MB = 1/2 AB and NB = 1/2 CB by definition of midpoint. But AB = CB by hypothesis, so MB = NB. m<MBC = m<NBA since both angles are the same as <ABC. And BC = BA by hypothesis. So triangle MBC is congruent to triangle NBA by SAS. Therefore CM = AN by CPCTC.**

This lemma directly implies that the triangle of medians of an isosceles triangle is isosceles. Applying the lemma to two different pairs of congruent sides shows that the triangle of medians of an equilateral triangle is also equilateral.

This DOES NOT guarantee that HCF will be right. In fact, in the sketch below, ABC is an isosceles right triangle, but HCF is an isosceles triangle with angles measuring 72, 72, and 37 degrees.

This is a more difficult question. The sketch below shows HCF a right triangle, but ABC has no obvious special properties. ABC is acute and scalene, with its largest angle measuring 84 degrees.

Here is a promising clue. In this sketch, both triangles are right triangles and they are similar with angles measuring 90, 35, and 55 degrees.

Continuing my exploration, I have found two more situations where both triangles are right. In each case, both triangles had angles measuring 90,35, and 55 degrees.

Very suspicious, but this 90-35-55 condition seems far too specific. We need to experiment a bit more systematically.

Beginning by fixing BAC at 90 degrees and fixing the measure of BA at 1.0 inch (I was originally using the more sensible choice of 1.0, but converting to Mac from PC changed all my sketches.) Then systematically vary m<ABC from 1 through 89 degrees by moving vertex C along the side of right angle BAC.

These constraints do not cause us to lose any generality. If there were another case where a different angle of ABC were right, we could match up that angle to ABC in our experiment. If there were another case where neither leg of ABC had length 1.33, we could scale it back to a similar triangle with a leg of length 1.33. Similarity gives proportional medians, so the triangle of medians would also be similar to the one in our experiment, with its angle measure preserved at 90-35-55.

Here is the general setup:

When m<ABC is very close to 0 (and m<BCA is very close to 90), m<CFH is close to 180 and the other two angles m<HCF and m<CFH are the same and very close to 0.

As m<ABC increases from 0 to 35 (and m<BCA decreases from 90 to 55), m<CFH decreases steadily from near 180 to 90. Meanwhile, the other two angles increase steadily with m<FHC growing at a faster rate than m<HCF, until m<FHC reaches 55 and m<HCF reaches 35.

As m<ABC increases from 35 to 55 (and m<BCA decreases from 55 to 35), m<CFH decreases steadily from 90 to 55. Meanwhile, the m<FHC steadily increases from 55 to 90 (Their measures are equal at 72 when ABC is a 45-45-90). Also m<HCF remains fairly stable increasing slightly from 35 to 37 and returning to 35.

As m<ABC increases from 55 to 90 (and m<BCA decreases from 35 to near 0), m<CFH decreases steadily from 55 to 0. Meanwhile, the m<FHC increases from 90 to near 180. Meanwhile m<HCF decreases very slowly from 35 to near 0.

Because the variation of the angle measures in our experiment behaved very continuously, we observe the only time that HCF can be a right triangle is when ABC has angle measures 90, 35, and 55. And this occurs when HCF also has angle measures 90, 35, and 55.

Although our experiment is not proof, it is convincing enough for us to accept our conjecture.

Because a 90-35-55 triangle is not a particularly special case, it seems unlikely that we would be able to find a simple, purely geometrical proof.