Solution:

Write $b = AC,$ $c = AB,$ $a = BC,$ and $k = \cos A.$ Right triangles $APC$ and $AQB$ give $AP = bk$ and $AQ = ck,$ so triangle $APQ$ is similar to triangle $ACB$ with ratio $k,$ whence $PQ = ak = 25.$ The points on the line occur in the order $X, P, Q, Y,$ so the power of $P$ gives $XP \cdot PY = 10 \cdot 40 = 400 = AP \cdot PB,$ and the power of $Q$ gives $YQ \cdot QX = 15 \cdot 35 = 525 = AQ \cdot QC.$ With $u = bk$ and $v = ck$ these read $$u(c - u) = 400, \qquad v(b - v) = 525,$$ that is, $w - u^2 = 400$ and $w - v^2 = 525,$ where $w = \frac{uv}{k} = bck.$

By the law of cosines, $a^2 = b^2 + c^2 - 2bck,$ so $a^2k^2 = u^2 + v^2 - 2uvk = 625.$ Substituting $u^2 = w - 400$ and $v^2 = w - 525,$ and $uv = wk,$ gives $2w - 925 - 2wk^2 = 625,$ so $wk^2 = w - 775.$ Then $$(uv)^2 = w^2k^2 = w(w - 775) = (w - 400)(w - 525),$$ which simplifies to $150w = 210000,$ so $w = 1400$ and $k^2 = \frac{1400 - 775}{1400} = \frac{25}{56}.$

Thus $k = \frac{5}{2\sqrt{14}}$ and $$bc = \frac{w}{k} = 1400 \cdot \frac{2\sqrt{14}}{5} = 560\sqrt{14},$$ so $m + n = 560 + 14 = 574.$