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My Three Triangle Puzzle

© Copyright 2001, Jim Loy

What do these three triangles have in common, besides a side of seven? You might want to think about it before you go on to the next paragraph.

Comments: Unless you recognize the second and third triangles, the above question seems a little difficult to answer. The areas are all different, and the perimeters are all different. Most of the angles are different. Could the upper angle in all three triangles be the same (60 degrees)? It looks like that might be the answer, but how do we prove that? The Pythagorean Theorem comes to mind (let's ignore the law of cosines for now). How could we possibly apply the Pythagorean theorem?

If those angles are 60 degrees, then making a copy of the middle triangle, next to the rightmost triangle, would produce an equilateral triangle. See the drawing on the right. It looks like a triangle. Can we prove that the right side is a straight line? This is the same puzzle as the above, but from a different perspective. Does that help us any?

Solution: I guess the easiest way of solving our problem is to look at the equilateral triangle on the left, and show that the two appropriate triangles are congruent with those in the previous diagram. By the Pythagorean Theorem, the altitude (h) of the equilateral triangle is 4sqr(3) (4 times the square root of 3). A further application of the Pythagorean Theorem shows that the hypotenuse (x) of the skinny right triangle (with sides h, 1, and x) is indeed 7. And so the two appropriate triangles are congruent in the two diagrams (by SSS (see Congruence Of Triangles, Part I)). The appropriate supplementary angles are also congruent. And so the rightmost line in the upper right diagram is indeed a straight line. And incidentally, the angles that we thought might be 60 degrees are indeed 60 degrees, which answers our original question.

Above, I mentioned the law of cosines, which is that in any triangle ABC, c^2=a^2+b^2-2ab(cos(C)) (with c^2 meaning c squared). The cosine of 60 degrees is 1/2. We could have applied this formula to the original triangles, and shown that the cosine of the upper angles was 1/2 in all three triangles. That would have been too easy.

One further diagram, I've redrawn all three triangles on the same base. And we see that all five vertices are on a circle. This is the result of one of Euclid's theorems: An angle inscribed in a circle is half the central angle of the inscribed arc. So all inscribed angles with the same arc are congruent angles.

On the left, we see some more triangles that have integer-length sides. The leftmost angle is 60 degrees (there is an equilateral triangle in the diagram).

Let's call the following Jim's Theorem (sort of like the Pythagorean Theorem, but Jim is 30 degrees short of a right angle): In a triangle in which C=60 degrees, c^2=a^2+b^2-ab. The proof follows directly from the law of cosines. Then we have pairs of Jim triples (see Pythagorean Triples), ignoring those that are multiples of smaller Jim triples:

Each pair fits together to form an equilateral triangle.

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