Strange triangle perimeter problem - Russian Olympiad 2010 This problem came from Crux Mathematicorum - Canadian Mathematical Society.

The perimeter of triangle $\triangle ABC$ is equal to $4$. Points $X$ and $Y$ are marked on the rays $AB$ and $AC$ in such a way that $AX = AY = 1$. The segments $BC$ and $XY$ intersect at point $M$ in their interior. Prove that the perimeter of one of the triangles $\triangle ABM$ or $\triangle AMC$ is equal to $2$.

I created $MN$ parallel to $AB$, $N$ is on $AC$. Then I calculated the two pairs of similar triangles introduced by this parallel segment. I also used Stewart to calculate $AM$ It didn't go anywhere.
 A: 
What strange conditions have been given? We are told the lengths AX = AY = 1 and
the perimeter of ABC is 4, and effectively nothing else. The conclusion, which is an
either-or statement, is equally puzzling.
Let us reflect the point A over both X and Y to two points U and V so that AU = AV =
2. This seems slightly better, because AU = AV = 2 now, and the “two” in the perimeter
is now present. But what do we do? Recalling that s = 2 in the triangle, we find that U and
V is the tangency points of the excircle, call it a. Set IA the excenter, tangent to BC at TA. See Figure 2.7E.

Looking back, we have now encoded the AX = AY = 1 condition as follows: X and
Y are the midpoints of the tangents to the A-excircle. We need to show that one of ABM
or ACM has a perimeter equal to the length of the tangent.
Now the question is: how do we use this?
Let us look carefully again at the diagram. It would seem to suggest that in this case,
ABM is the one with perimeter two (and not ACM). What would have to be true in
order to obtain the relation AB + BM + MA = AU? Trying to bring the lengths closer to the triangle in question, we write AU = AB + BU = AB + BT. So we would need
BM + MA = BT , or MA = MT .
So it would appear that the points X, M, Y have the property that their distance to A
equals the length of their tangents to the A-excircle. This motivates the last addition to our
diagram: construct a circle of radius zero at A, say ω0. Then X and Y lie on the radical axis
of ω0 and Ta; hence so does M! Now we have MA = MT, as required.
Now how does the either-or condition come in? Now it is clear: it reflects whether T
lies on BM or CM. (It must lie in at least one, because we are told that M lies inside the
segment BC, and the tangency points of the A-excircle to BC always lie in this segment
as well.) This completes the solution, which we present concisely below.
Let IA be the center of the A-excircle, tangent to BC at T ,
and to the extensions of AB and AC at U and V. We see that AU = AV = s = 2. Then XY
is the radical axis of the A-excircle and the circle of radius zero at A. Therefore AM = MT.
Assume without loss of generality that T lies on MC, as opposed to MB. Then $ AB +
BM + MA = AB + BM + MT = AB + BT = AB + BU = AU = 2 $ as desired.
