text: math problem challenge? ok here is problem i want to put out there lets see how many of you can slove this.  there is a right triangle with side x side y and hypotenuse being 12.  there is a box 4 x 4 inside the triagle so it looks like this:  here is a link for the pic it is not drawen to scale.  http://i17.photobucket.com/albums/b51/paki_munda/d361re2.jpg i assume you want the values of x and y?  top triangle hypotenuse: sqrt( (y-4)^2 + 16 )   top triangle vertical leg: y-4  full triangle hypotenuse: 12  full triangle vertical leg: y  by similar triangles sqrt( (y-4)^2 + 16 ) / (y-4) = 12 / y  multiply by (y-4) and y: y sqrt( (y-4)^2 + 16 ) = 12 (y-4)  simplify: y sqrt( y^2 - 8y -16 + 16 ) = 12y - 48 y sqrt( y^2 - 8y ) = 12y - 48 y sqrt( y (y-8) ) = 12y - 48  square both sides: y^2 * y(y-8) = (12y - 48)^2 y^4 - 8y^3 = 144y^2 - 8*144y - 48^2 y^4 - 8y^3 - 144y^2 + 8*144y + 48^2 = 0  now just solve for y, and from there it is easy to solve for x...  after a bit of playing with an excel spreadsheet, i'm getting the following pairs:  x = 6.680915, y = 9.968218 x = 9.968218, y = 6.680915  notice how these are symetric answer depending on whether x is a long leg or y is a long leg.  double checking the answers we get:  x^2 + y^2 = 12 sqrt((x-4)^2 + 4^2) + sqrt((y-4)^2 + 4^2) = 12  the two solutions are therefore: x = 6.680915, y = 9.968218 (y is the long leg) x = 9.968218, y = 6.680915 (x is the long leg)  if you assume y > x, then the first solution is the only solution.