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My Solution: Without loss of generality, let angle B be the largest angle. Let ABC be the patch to be turned over. Let D be the point of intersection between AC and its perpendicular through point B. Let E be the midpoint of BC. Let F be the midpoint of AB. The fur dresser cuts along DE and DF (which bisect BC and AB, respectively), and turns the 3 resulting parts over - the triangles around vertical axes and the quadrilateral around EF. When he sewed them together, patch ABC was reversed but preserved its shape.
rixar's solution: Construct the incenter of triangle ABC on the hide side. Call that point I. Connect I to AB, AC, and BC at points E, F, G, respectively, such that IE is perpendicular to AB, IF is perpendicular to AC, and IG is perpendicular to BC. The fur dresser cuts along IE, IF, and IG. He is left with three symmetrical quadrangles, which are the same shape when flipped on the axes created by lines constructing the incenter.
Kudos to Brian Smith to being the first to get my solution, and rixar for getting his own solution. There might be others :) |
