Given that angle ACD and angle BDC are less than 90, what is the smallest possible area of trapezoid ABCD* in which the height each side is a different integer length?

Explanation:

A good way to get the answer is to play around with Pythagorean triples. 3:4:5 is the most obvious one. Yet, if we make BD = 5 (the hypotenuse of a 3:4:5) and the height either 4 or 3, there is no Pythagorean triple that can be formed with 3 or 4 besides a 3:4:5 (remember each side has to be a different integer length). The next triple that leaps to mind is a 5:12:13. We can make 12 the height. That means AC can be 15 and if we draw a straight line down from the vertex at A to CD (let’s call this point E), then CE = 9. We will want to make AB has small as possible, while still ensuring it is an integer. That gives us AB = 1. When we add this all up, we get CD = 15. That is problematic because each side has to be a different integer length. An easy way to correct this is to make AB = 2, thereby making CD = 16.

(16 + 2)/2 x 12 = 108.

Yeah, the answer!

Well, be careful. Just because we got a value doesn’t mean we’ve got the small possible value. Is there any other Pythagorean triple that is also small that we have failed to consider? Well, there is the much forgotten 8:15:17. This doesn’t seem like an answer because 17 is pretty big, but if you notice that 8 also turns up in a 6:8:10 triangle, which is smaller than the 9:12:15 triangle, then it is worth calculating the numbers.

The height is 8 and BD equals 10 and BF = 6.

AC = 17 and CE equals 15.

We can make AB 1 without any redundancy in the length of side, the way we had with 5:12:13.

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