<PRE><B>Question 1109591</B>
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We know side BC is 5, because BM is 2 and MC is 3.&lt;br&gt;
Since AM bisects angle BAC, AB/AC = BM/MC = 2/3.&lt;br&gt;
Probably that is what you were missing, because that makes the problem relatively easy.&lt;br&gt;
You just need to find integers AB and AC for which AB/AC = 2/3 and for which AB, AC, and BC can form a triangle.&lt;br&gt;
For example, if AB=2 and AC=3, then the three sides would be 2, 3, and 5; but those lengths do not make a triangle.&lt;br&gt;
But if AB=4 and AC=6, then the three sides are 4, 5, and 6; and that DOES make a triangle.&lt;br&gt;
So there is one answer; but there are more (two more).  I leave it to you to find them.&lt;br&gt;
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Added in response to reader&#39;s question....&lt;br&gt;
In a triangle, an angle bisector divides the opposite side into two parts whose lengths are proportional to the lengths of the two sides that form the angle.&lt;br&gt;
For example, if AB=5 and BC=7, the bisector of angle B divides side AC into two parts whose lengths are in the ratio 5:7.</PRE>