Problem:
In △ABC,BC=23,CA=27, and AB=30. Points V and W are on AC with V on AW, points X and Y are on BC with X on CY, and points Z and U are on AB with Z on BU. In addition, the points are positioned so that UV∥BC,WX∥AB, and YZ∥CA. Right angle folds are then made along UV,WX, and YZ. The resulting figure is placed on a level floor to make a table with triangular legs. Let h be the maximum possible height of a table constructed from △ABC whose top is parallel to the floor. Then h can be written in the form nkm​​, where k and n are relatively prime positive integers and m is a positive integer that is not divisible by the square of any prime. Find k+m+n.
Solution:
Call a table functional if its top is parallel to the floor. Let BC=a, CA=b, and AB=c, with a≤b≤c. The height of a functional table is the common height of triangles AUV,CXW, and BYZ, where each height is measured to the fold. For a functional table to be of maximum height, two of the folds must intersect on a side of the triangle. If this were not the case, then each fold could be shifted by the same small amount to obtain a functional table with greater height. Thus Z=U,V=W, or X=Y. Assume that Z=U. Let the common height of △AUV and △BUY be hc​, and let AU=x, so that BU=BZ=c−x. Because △AUV is similar to △ABC, it follows that UV=a⋅cx​ and [AUV]=(cx​)2[ABC].
Thus
hc​=UV2[AUV]​=ac2x​⋅[ABC].
Similarly, by using △BYZ, it follows that hc​=YZ2[BYZ]​=bc2(c−x)​⋅[ABC]. Equating these two expressions and solving for x yields x=a+bac​, and hence hc​=a+b2[ABC]​. By similar arguments, the heights ha​ and hb​ that result from X=Y and V=W, respectively, are
ha​=b+c2[ABC]​ and hb​=c+a2[ABC]​.
Thus the maximum height h of a functional table is the minimum of ha​,hb​, and hc​, which is ha​=b+c2[ABC]​, because if the height were larger than this, some of the folds would intersect.
For the given triangle, a=23,b=27, and c=30, and by Heron's Formula, [ABC]=20221​. Because the two longer sides have lengths 27 and 30 , the formula yields
