Re: [math-fun] Triangle 1/7th of generating ABC triangle
ERic wroter: << Draw a triangle ABC (clockwise labels) with each side devided into 3 equal parts (labels A,A1,A2,B,B1,B2,C,C1,C2): AA1=A1A2=A2B, BB1=B1B2=B2C, CC1=C1C2=C2A. Now draw AB1, BC1, CA1 -- those lines shape an "inside" triangle "I" having no common point with ABC. Question: Does an ABC triangle exist such that the surface of the inner "I" triangle is exactly 1/7th of ABC's surface?
This looks to me like putting the answer before the question! A slightly more general problem, and fun to solve, is: Suppose 0 <= p <= 1. Given a triangle ABC, mark the 3 points that are the fraction p of the way from A to B; from B to C; and from C to A. Now draw the segment from each of A,B,C to the marked point on the opposite side. Express the area of the interior triangle thus created, as a fraction F(p) of the area of ABC ? (As a check, F(1/3) = 1/7.) --Dan
On Saturday 09 June 2007 16:33, Dan Asimov wrote:
Suppose 0 <= p <= 1. Given a triangle ABC, mark the 3 points that are the fraction p of the way from A to B; from B to C; and from C to A. Now draw the segment from each of A,B,C to the marked point on the opposite side.
Express the area of the interior triangle thus created, as a fraction F(p) of the area of ABC ? (As a check, F(1/3) = 1/7.)
Working in coordinates (taking ABC to be, say, the canonical isosceles right triangle) yields the answer pretty quickly, especially if you notice at the outset that it obviously has to be a function of pq where q=1-p. But it's rather pedestrian and doesn't particularly give me the feeling of understanding *why* the answer is what it is. Is there a better solution? (Or you can appeal to something in the Ceva-Menelaus-etc. complex, but that's kinda cheating.) I suspect there's a rabbit-out-of-hat proof that finds an obvious reason why the degrees of the numerator and denominator (as polynomials in pq) have to be what they are, and then just checks a couple of obvious values -- e.g. when pq=1/4 we have to get 0, and when pq=0 we have to get 1. But I don't currently see what that obvious reason might be. -- g
Gareth McCaughan wrote:
...Is there a better solution?
If you draw ABC properly within a tessellation of unit equilateral triangles, it is clear from the Pythagorean theorem that its edge length is sqrt[7]. See figure 5 of: http://www.georgehart.com/radiolaria/radiolaria.html George http://www.georgehart.com
On Sunday 10 June 2007 02:11, George W. Hart wrote:
Gareth McCaughan wrote:
...Is there a better solution?
If you draw ABC properly within a tessellation of unit equilateral triangles, it is clear from the Pythagorean theorem that its edge length is sqrt[7]. See figure 5 of: http://www.georgehart.com/radiolaria/radiolaria.html
Very nice (and, completely irrelevantly, I like your ray-trace-olarian) but that only answers the question for p=1/3. ... No, wait. If you do the same construction but at a different angle (the picture you want to end up with is an equiangular hexagon with sides a,b,a,b,a,b with 3+3 of its diagonals drawn), you rapidly get that for a,b integers F(a/(a+b)) = (a-b)^2 / (a^2 + ab + b^2) or F(p) = (1-4pq) / (1-pq) which by continuity means that the latter is true for all p. In fact, forgetting about the underlying triangular grid, we can just draw the aforementioned equiangular hexagon and get the result for arbitrary p straight off. Very nice, and it does seem to me to give some indication of why the thing is true. Thanks! -- g
On Sunday 10 June 2007 10:45, I wrote:
On Sunday 10 June 2007 02:11, George W. Hart wrote: ...
If you draw ABC properly within a tessellation of unit equilateral triangles, it is clear from the Pythagorean theorem that its edge length is sqrt[7]. See figure 5 of: http://www.georgehart.com/radiolaria/radiolaria.html
Very nice (and, completely irrelevantly, I like your ray-trace-olarian) but that only answers the question for p=1/3.
and then went on to observe that the construction generalizes. If I'd looked at the discussion in the vicinity of figure 5 and not merely at the pretty picture I'd have noticed that George does consider the general case (though not in the context of areas or side division), so he probably already knew how to generalize the proof. :-) -- g
It is well known that a convenient mapping of a spherical surface to a flat surface involves projecting from the center onto the circumscribed cube, then unfolding the cube. As with any mapping of a spherical surface to a flat one, there is some amount of distortion. Of course, one could simply project onto an icosohedron instead and reduce the distortion, but using a cube is particularly convenient because one can use square arrays of pixels to represent the faces. (Of course there are hundreds of years of other kinds of projections.) So my question is, keeping these conveniently square faces, how much less distortion can you get by altering the mapping onto a face? For my purposes, what I care about is how distorted the square pixels become when mapped back to the sphere, because this determines the resolution of the flat representation. Is this an already solved problem? My thoughts follow... Using x,y,z coordinates on a unit sphere (x^2+y^2+z^2=1), and centering the cube's top face at <0,0,1>, the unadulterated projection to said top face is <x,y,z> -> <x/z,y/z> (where |x| and |y| are <= z). For this mapping, the linear density (Cartesian distance between two closely spaced points on the face divided by angular distance between the corresponding points on the sphere) varies between 1 and 3 over the face (although only by a factor of at most sqrt(3) as you vary direction at any given point); while the areal [is that a word?] density varies between 1 and 3sqrt(3) over the face. If we adulterate the projection so <x,y,z> -> <f(x/z),f(y/z)>, where f is a monotonic function, which may as well be antisymmetric about 0, and with f(0) = 0 and f(1) = 1, we can do better (and is that the only way to do better while keeping the square a square?). If we define g(s) = df(s)/ds, the areal density at <x,y,z> is g(x/z)g(y/z)/|z^3|, while the linear density is sqrt((g^2(x/z)((1-y^2)c+xys)^2 + g^2(y/z)((1-x^2)s+xyc)^2)/(1-(xs-yc)^2))/z^2 where c and s are the cos and sin of the direction (I'm pretty sure I computed this correctly). One obvious idea is to make the linear density constant along the face diagonals, giving f(s) = atan(sqrt(2)s)/atan(sqrt(2)), or constant along the horizontal and vertical lines through the center, giving f(s) = atan(s)/atan(1), or, more generally, f(s) = atan(ks)/atan(k) for some k. (The limit as k->0 gives the unadulterated projection.) For k = 1, the linear density ratio (max/min over the face) is sqrt(3), and the areal density ratio (max/min over the face) is sqrt(2). Perhaps a better idea is to make the areal density constant along the face diagonals, giving f(s) = integral((1+2s^2)^-(3/4),0,s)/integral((1+2s^2)^-(3/4),0,1) which I believe gives the least possible areal density ratio ((2/sqrt(3))^(3/2), a bit less than 5/4), although its linear density ratio (sqrt(2sqrt(3)), a bit less than 15/8) is not the least possible ratio between max and min linear densities (which would be sqrt(3), found along vs across the diagonal at the corners), but once you determine g(0) and g(1) the linear and areal densities at <0,0>,<1,0>, and <1,1> are all determined. BTW, this f(s) is suprisingly close to atan(ks)/atan(k) with k = 1.1488. --ms
On 6/9/07, Dan Asimov <dasimov@earthlink.net> wrote:
A slightly more general problem, and fun to solve, is:
Suppose 0 <= p <= 1. Given a triangle ABC, mark the 3 points that are the fraction p of the way from A to B; from B to C; and from C to A. Now draw the segment from each of A,B,C to the marked point on the opposite side.
Express the area of the interior triangle thus created, as a fraction F(p) of the area of ABC ? (As a check, F(1/3) = 1/7.)
And a still more general problem: draw three lines which cut sides AB, BC, CA in (directed) ratios p,q,r; q,r,p; r,p,q respectively. Then the ratio of the areas of secondary to primary triangle depends only on p,q,r [where by Menelaus' theorem, p q r + (1 - p)(1 - q)(1 - r) = 0]. Dan Asimov mentions the special case q = 0, r = 1, where the lines meet the vertices A,B,C; deVilliers also mentions the case q = p, r = (1-p)^2/(1-2p+2p^2), where the secondary triangle touches the sides of the primary; also a pair of analogues for parallelograms. But he doesn't actually mention the general case --- I don't know whether he was aware of it, nor an expression for the general ratio! Fred Lunnon
Many years ago I solved the even more general problem, and discovered that when stated slightly differently the formula comes out beautifully. Given triangle ABC, A' divides side BC in the ratio of m:1, B' divides CA in the ratio n:1, and C' divides AB in the ratio p:1. Now draw cevians AA', BB', and CC'. The ratio of the area of the center triangle to the area of the original triangle is: f(m,n,p) = (mnp-1)^2/((mn+n+1)(mp+m+1)(np+p+1)) Checking, well known cases: f(2,2,2) = 1/7, and f(p,p,p) = (p-1)^3/(p^3-1) (equivalent to but nicer than the formula in the Feynman's Triangle reference). Also the case where mnp=1 (area=0) is precisely Ceva's Theorem. I'm curious if anyone has seen this or similar result before. Nick Fred lunnon wrote:
On 6/9/07, Dan Asimov <dasimov@earthlink.net> wrote:
A slightly more general problem, and fun to solve, is:
Suppose 0 <= p <= 1. Given a triangle ABC, mark the 3 points that are the fraction p of the way from A to B; from B to C; and from C to A. Now draw the segment from each of A,B,C to the marked point on the opposite side.
Express the area of the interior triangle thus created, as a fraction F(p) of the area of ABC ? (As a check, F(1/3) = 1/7.)
And a still more general problem: draw three lines which cut sides AB, BC, CA in (directed) ratios p,q,r; q,r,p; r,p,q respectively. Then the ratio of the areas of secondary to primary triangle depends only on p,q,r [where by Menelaus' theorem, p q r + (1 - p)(1 - q)(1 - r) = 0].
Dan Asimov mentions the special case q = 0, r = 1, where the lines meet the vertices A,B,C; deVilliers also mentions the case q = p, r = (1-p)^2/(1-2p+2p^2), where the secondary triangle touches the sides of the primary; also a pair of analogues for parallelograms.
But he doesn't actually mention the general case --- I don't know whether he was aware of it, nor an expression for the general ratio!
Fred Lunnon
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Given triangle ABC, A' divides side BC in the ratio of m:1, B' divides CA in the ratio n:1, and C' divides AB in the ratio p:1. Now draw cevians AA', BB', and CC'. The ratio of the area of the center triangle to the area of the original triangle is: f(m,n,p) = (mnp-1)^2/((mn+n+1)(mp+m+1)(np+p+1))
Lovely. Is there an equally lovely proof? -- g
On 6/16/07, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
Given triangle ABC, A' divides side BC in the ratio of m:1, B' divides CA in the ratio n:1, and C' divides AB in the ratio p:1. Now draw cevians AA', BB', and CC'. The ratio of the area of the center triangle to the area of the original triangle is: f(m,n,p) = (mnp-1)^2/((mn+n+1)(mp+m+1)(np+p+1))
Lovely. Is there an equally lovely proof?
I was wondering what symmetry this expression needs to have. I think I see why f(m,n,p) = f(1/n, 1/m, 1/p). Are there other symmetries I'm missing? --Joshua
On Saturday 16 June 2007 14:31, Joshua Zucker wrote:
On 6/16/07, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
Given triangle ABC, A' divides side BC in the ratio of m:1, B' divides CA in the ratio n:1, and C' divides AB in the ratio p:1. Now draw cevians AA', BB', and CC'. The ratio of the area of the center triangle to the area of the original triangle is: f(m,n,p) = (mnp-1)^2/((mn+n+1)(mp+m+1)(np+p+1))
Lovely. Is there an equally lovely proof?
I was wondering what symmetry this expression needs to have. I think I see why f(m,n,p) = f(1/n, 1/m, 1/p). Are there other symmetries I'm missing?
Cyclic permutation of m,n,p. -- g
On 6/16/07, Nick Baxter <nickb@baxterweb.com> wrote:
Many years ago I solved the even more general problem, and discovered that when stated slightly differently the formula comes out beautifully.
Given triangle ABC, A' divides side BC in the ratio of m:1, B' divides CA in the ratio n:1, and C' divides AB in the ratio p:1. Now draw cevians AA', BB', and CC'. The ratio of the area of the center triangle to the area of the original triangle is: f(m,n,p) = (mnp-1)^2/((mn+n+1)(mp+m+1)(np+p+1))
Checking, well known cases: f(2,2,2) = 1/7, and f(p,p,p) = (p-1)^3/(p^3-1) (equivalent to but nicer than the formula in the Feynman's Triangle reference). Also the case where mnp=1 (area=0) is precisely Ceva's Theorem. I'm curious if anyone has seen this or similar result before.
All of which prompts the thought that the "real" theorem simply expresses the ratio of the areas of two general triangles A_1 A_2 A_3, B_1 B_2 B_3, as a function of the linear ratios p_ij in which side i of A cuts side j of B; or alternatively q_ij in which B cuts A. Re-scaling as Nick suggests might improve the aesthetics. As before, setting area ratio to 0 gives a (probably not terribly interesting) generalisation of Ceva. And while we're at it, what is the relation between the p_ij and q_ij? This is all starting to resemble barycentric coordinates ... WFL
participants (7)
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Dan Asimov -
Fred lunnon -
Gareth McCaughan -
George W. Hart -
Joshua Zucker -
Mike Speciner -
Nick Baxter