Let the triangle be the intersection of the plane x+y+z=1 with the positive octant, so that its vertices are X=(1,0,0), Y=(0,1,0), Z=(0,0,1). The r-median joins X to R=(0,r,1-r). The s-median joins Y to S=(1-s,0,s). The t-median joins Z to T=(t,1-t,0). The r-median consists of the convex linear combination XR = (1-u)X + uR. The s-median consists of the convex linear combination YS = (1-v)Y + vS. Their intersection C = XR.YS can be found with a little algebra. C = XR.YS = ((1-r)(1-s), rs, (1-r)s) / (1-r+rs). By cyclically permuting both (r,s,t) and the coordinates (x,y,z) we also have A = YS.ZT = ((1-s)t, (1-s)(1-t), st) / (1-s+st), B = ZT.XR = (tr, (1-t)r, (1-t)(1-r)) / (1-t+tr). The 3-space linear transformation L that takes X to A, Y to B, and Z to C is the matrix whose columns are the coordinates of A, B, and C The problem asks for Area(ABC)/Area(XYZ). Since the volume of a pyramid is (1/3)base*height, this equals Vol(OABC)/Vol(OXYZ). But this latter quantity equals det(L). Cranking out the determinant, we get for the area ratio a fraction with numerator ((1-r)(1-s)(1-t) - rst)^2, and denominator (1-r+rs)(1-s+st)(1-t+tr). -- Gene From: Eugene Salamin via math-fun <math-fun@mailman.xmission.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, May 29, 2015 11:06 AM Subject: Re: [math-fun] Triangle puzzle Ratios of lengths and ratios of areas are invariant under affine transformations, so the answer should hold for arbitrary triangles, not just equilateral ones. -- Gene From: Dan Asimov <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, May 29, 2015 9:10 AM Subject: [math-fun] Triangle puzzle For s in [0,1], define an s-median of a triangle (ABC in counterclockwise order) from vertex A to be the line segment connecting A to the point P of the opposite side BC such that BP : PC = s : (1-s) . (E.g., an ordinary median is a (1/2)-median.) A / \ / \ / \ / \ / \ B------P---------C Old math puzzle: Suppose we draw all three s-medians of an equilateral triangle ABC, where s = 1/3. Let delta be the triangle bounded by segments of the three s-medians. Find the ratio area(delta) / area(ABC). New math puzzle: Suppose we draw the r-, s- and t-medians of an equilateral triangle ABC for r, s, t in [0,1], again assuming ABC go counterclockwise around the perimeter. Let delta be the triangle bounded by segments of the three "medians". Find the ratio area(delta) / area(ABC) . ——Dan