Equal Circles, Medial Triangle and Orthocenter

Solution

Any three equal circles having for centers the vertices of a given triangle cut the respective sides of the medial triangle in six points equidistant from the orthocenter of the given triangle.

Note: The problem is equivalent to saying that the six points in question are concyclic; thus it claims the existence of a circle. One can easily surmise that the circle exists even if only four of the six points are available.

Let ΔABC be given, with M_a, M_b, M_c the midpoints and the orthocenter H.

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Assume circle centered at A meets M_bM_c in U and U'. Let B be the intersection of the altitude from A and M_bM_c. The two right triangles APU and APU' are equal. Applying the Pythagorean identity gives

PU² = PU'² = AU² - PU² = HU² - HP².

Hence,

implying

HU² = AU² - AH·DH.