Tuesday, November 1, 2016

Solution to Halloween Challenge

Triangle ABC has area 780 square units. Side c is 78 units long. Angles A and B satisfy


cot(A) − cot(B) = 6.

Find the lengths of sides a and b and the measures of all three angles.

I tried to devise a scary-sounding problem, but the problem isn't as bad as it sounds. Given that the area is 780 square units and side c is 78 units, the altitude of the triangle is easily found: 20 units. Now suppose first that the altitude meets side c and separates it into two parts x and y, so that x + y = 78. From the given information, we would have x − y = 120. But if two positive numbers sum to 78, then they can't have a difference of 120...so in fact, the altitude doesn't meet side c. Our triangle "overhangs" side c (out past vertex B, say).

Suppose the triangle overhangs vertex B by distance x. Then cot(A) = (78 + x)/20 and cot(B) = − x/20, so from the given information (78 + x)/20 − (−x/20) = 6. Solve for x = 21.

Knowing the altitude and the overhang, we can use the Pythagorean theorem to find one side of our triangle as 29 units. Then a second application of the Pythagorean theorem (to the largest right triangle) gives the last side as 101 units.


With all three sides of the triangle known, finding the angles is straightforward. The answers are A = sin−1(20/101) ≈ 11.4°, B = π − sin−1(20/29) ≈ 136.4°, and C = π − AB ≈ 32.2°.

A devious way to solve the problem would be to play a hunch: namely, that all three side lengths are going to be whole numbers. Then because the area is also a whole number, the triangle will be what's known as a Heronian triangle, after Heron of Alexandria (AD c. 10– c. 70). A little googling then turns up this table of Heronian triangles, and CTRL-F finds the solution without much trouble.

In fact, I created this triangle by looking for Pythagorean triples that share a number in common. The triples that I chose were (20, 21, 29) and (20, 99, 101). There is no reason why the side lengths of the triangle had to be integers, but I wanted the difference of cotangents to look dramatic.

I'll end with one more method for solving the problem: first derive an area formula, ½(cot + cot B)−1c2, applicable to any triangle. (This is the ASA analogue of ½ a b sin C for the SAS case.) With this we can write down the value of sum of cotangents, which in conjunction with the difference of the cotangents determines the two cotangents (angles) individually. Then with all three angles and one side known, the law of sines gives the other two sides. 

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