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# How to Find 5 Triangles with Integer Sides

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Find five triangles with integers sides (without no commun divisors ) such that the median AD from angle A to side BC is m(a) = b + c - a where a, b , c are sides opposite to angles A, B, C respectively. The inner circle is tangent at point X to segment BD on side BC.
So find the 5 triangles of integers sides (with no commun divisor ) such that the length of median AD is m(a) = b + c - a.
For each of the five triangles, give separately the length of sides a, b, and c.

https://brainmass.com/math/triangles/find-triangles-integer-sides-493757

## SOLUTION This solution is FREE courtesy of BrainMass!

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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