Problem:
A paper equilateral triangle ABC has side length 12. The paper triangle is folded so that vertex A touches a point on side BC a distance 9 from point B. The length of the line segment along which the triangle is folded can be written as nmpββ, where m,n, and p are positive integers, m and n are relatively prime, and p is not divisible by the square of any prime. Find m+n+p.
Solution:
Let the fold line intersect sides AB and AC at distances x and y from point A, respectively, as shown. Let the fold line have length z. Then the Law of Cosines gives 92+(12βx)2β2β 9β (12βx)β cosB=x2, which simplifies to x=539β. Similarly, 32+(12βy)2β2β 3β (12βy)β cosC=y2, which simplifies to y=739β.
Finally, z2=(539β)2+(739β)2β2β 539ββ 739ββ cosA, which simplifies to z=353939ββ. The requested sum is 39+39+35=113β.