Enigmatics |
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Posted on: Saturday 1/12, 2007; 5:33 PM The domain in which the integer-valued solution can lie is extremely restricted, so I use a straightforward brute force approach. Use the Law of Cosines to define a function for computing all the possible triangles (as lists of rules for the lengths of the sides) with integer length sides having a specified angle between one pair of the sides. For the purposes of this Enigma none of the sides can be greater than 24 units in length because the total triangle perimeter is less than 50 units. For triangles APC and PBC generate lists in the form of replacement rules defining all possible pairs of right angled triangles in Harry's triangle.
Construct a list of all possible pairs from the above two lists, keeping only the cases where (1) the common side PC has the same length for both right angled triangles, and (2) the two triangles are non-identical. I assume that this second condition means that the triangles are not reflections of each other about their common side PC.
Repeat these steps for the triangles APC and PBC in Tom's triangle. I use that APC is not an equilateral triangle.
Construct a list of all possible pairs from triangles1 and triangles2, keeping only the cases where the perimeters (1) differ by one unit of length, and (2) are both less than 50 units in length. There are two solutions which are mirror images of each other, as expected.
Compute the perimeters of Harry's and Tom's triangles (both mirror image solutions are shown).
Compute the length of PC in Harry's and Tom's triangles (both mirror image solutions are shown).
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