Yes! Same as what I got. But with a much shrewder and concise derivation. ((( Because of the invariance of area ratios, and length ratios along any single line, I had used a 1, 1, sqrt(2) right triangle to find the intersection points of the r-, s-, and t-medians, and then used a determinant formula to get the area of the central triangle. Way hairier expressions involved. ))) The formula where r = s = t is g(s), where g(s) := f(s,s,s) = (1 - 2s)^2 / (s^2 - s + 1) And ask some questions whose answer I have no idea about so far: * Why is g(s) the ratio of two quadratic polynomials? * Why is f(r,s,t) the ratio of two sextic polynomials? * Is it even obvious a priori that the formulas should be rational functions? Also: It may be amusing in g(s) to take s = 1/N, N = 1,2,3,.... One gets an interesting sequence of fractions in lowest terms: N: 1 2 3 4 5 6 7 8 9 10 --------------------------------------------------------------------------------------- g(1/N): 1 0 1/7 4/13 3/7 16/31 25/43 12/19 49/73 64/91 N: 11 12 13 14 15 16 17 18 19 20 --------------------------------------------------------------------------------------- g(1/N): 27/37 100/133 121/157 48/61 169/211 196/241 75/91 256/307 289/343 108/127 . . . Seems to me the denominators are more often prime numbers than most fractions in lowest terms. Is this true as N -> oo ??? Which, free associating, makes me wonder: ----- PROBLEM: Find an asymptotic expression for the frequency of primes among denominators of fractions — say as we take all fractions K/L that are already in lowest terms, with sqrt(K^2 + L^2) <= R, as R -> oo. ----- ——Dan
On May 30, 2015, at 11:45 AM, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun