An acute isoceles triangle ABC has integer side lengths of a,b,c. The perpendicular feet are drawn from each angle and intersect at H. Then, each side is perpendicularly bisected to meet at O. Point N is drawn such that N is on segment \(\overline{OH}\). Point L is drawn at the midpoint of line \(\overline{AH}\) such that \(\overline{LN}=\frac{1}{2}\overline{OA}\). Triangle \(\triangle ABC\) has an area of 48. If the sum of the two smallest possible lengths of \(\overline{NH}\) can be expressed as \(\frac{x}{y}\) for positive co-prime \(integers\) x,y, find x+y.

Also, this is probably the worst problem that I've made cuz this is just stupid hard and uses practically unknown formulas.

HINT: try to see if you can do it without this information below

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One side length is 10 of both triangles.

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