How to Find 5 Triangles with Integer Sides

Call the side lengths a, b, and c, and the median m. Note that the sentence "The inner circle is tangent at point X to segment BD on side BC" has no bearing on the problem, since X is not mentioned again.

We will use Apollonius' Theorem for the length of a median (http://en.wikipedia.org/wiki/Apollonius%27_theorem):

c^2 + b^2 = 2[m^2 + (a/2)^2]

Substituting m = b + c - a, as given in the problem, we get

c^2 + b^2 = 2[(b+c-a)^2 + (a/2)^2]
c^2 + b^2 = 2[(a^2 + b^2 + c^2 - 2ab - 2ac + 2bc + a^2/4]
c^2 + b^2 = 2a^2 + 2b^2 + 2c^2 - 4ab - 4ac + 4bc + a^2/2
[Equation 1] 0 = (5/2)a^2 + b^2 + c^2 - 4ab - 4ac + 4bc

Equation 1 is the Diophantine equation we must solve (meaning we're looking for a, b, c which are integers). Because a, b, and c form a triangle, we have the additional conditions

a, b, c > 0
a + b > c
b + c > a
c + a > b

We also observe that Equation 1 implies that a is even, because an odd a would result in the right side of the equation being an integer plus 1/2 while the left side is 0. Also, Equation 1 is symmetric in b, and c, so when we try possible triples we can skip many duplicates. For example, if we check (7, 6, 5), we do not also need to check (7, 5, 6).

Here are two triples which satisfy all the conditions, and therefore form triangles which solve the problem:

(6, 5, 5)
(22, 15, 23)

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