New year's triangle challenge. Can you solve it?

Good work Track P (where did your answer go?) and torquest. But what are the SIDES of the triangle?

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There were two answers, but now none. Oh well!

Perhaps it is time for a small hint: my solution has one edge of the triangle with length less than 100, with the other two edges MUCH longer, but nearly equal. Feel free to use a computer to help you find it. If no one gets it I will eventually post the answer here.

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Congratulations, Philip J/ψ, I just checked your answer and it is correct! In the process I checked my hint, and it is WRONG! SORRY EVERYONE. (It was for a simpler problem where the product of the tangents of the half angles was 2^2 * 5^2 * 7^2 / (3^8 * 11^4) ). Unfortunately my notes from 15 years ago are mostly missing and I mixed it up with the other problem.

Philip J/ψ, I would be interested to see a sketch of what you did, but don't feel a huge obligation. This problem is hardly worth so much time.

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CONCLUDING REMARKS:

Philip J/ψ, I have read your pdf file. Wonderful! You deserve more than 10 points!

The "very original" triangle was in fact the one posed here, not the one of my hint. See this link:

http://groups.google.com/group/rec.puzzles/browse_...

A little about the background of this problem: back around 1993 I became interested in "square Heron triangles", triangles with integer sides and square-integer area. The simplest example is [a,b,c] = [9,10,17] area 6^2. In this example the tangents of the half-angles [t1,t2,t3] = [2/9, 1/4, 2], and the product t1*t2*t3 = 1/9, a rational square. Finding triangles with t1*t2*t3 = rational square, turns out to be an equivalent problem.

In the present question t1*t2*t3 is the given rational square, and the problem is to find a triangle. The solution triangle (found by Philip J/ψ) has area = 4084080^2 = (2^4 * 3 * 5 * 7 * 11 * 13 * 17)^2 and so is a square-heron triangle.

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CONCLUDING REMARKS (part 2):

The interesting part is this: associated with each half-angle (say t1) of any particular square-Heron triangle is an elliptic curve of positive rank whose rational points correspond with all the square-Heron triangles having one angle t1, but t2 and t3 different from the starting triangle. Thus infinitely many pairwise dissimilar square-Heron triangles share half-angle t1.

In this way, each square-Heron triangle gives rise to three positive rank elliptic curves, and each rational point on each of these curves gives rise to two more positive rank elliptic curves. In short there is an "elliptic surface" each rational point of which corresponds to a different and dissimilar square-Heron triangle!

However, given a rational number t1, there may be no square-Heron triangles with half-angle t1. New problem: for which rational t1 does there exist a square-Heron triangle with one half-angle equal to t1? The answer is not known.

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CONCLUDING REMARKS (part 3--last):

So I considered the less ambitious question: for which positive integers, e, does there exist a square-Heron triangle with cotangent of one of the half angles equal to e? For example, in the triangle [9,10,17], t2=1/4, so e=4 possesses square-Heron triangles.

It was when I considered e=20 that I discovered the New Year's Challenge triangle of this question. It has one half-angle with cotangent equal to 20 (equivalently t1=1/20).

The 'next' square-heron triangle with t1=1/20 has t1*t2*t3 = (47119182670560 / 224045250732161)^2

There are many interesting unanswered questions is this venue.

Thanks to all who expressed interest, and worked on this 'new year's challenge', and thanks especially to Philip J/ψ for a very fine solution.

--JB

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EDIT: above where I refer to t1, t2, t3 as angles I should have made clear they are tangents of the triangle's half angles, not the angles themselves.