Re: [math-fun] A New Year's Puzzle
Thanks, but this is only putting geometrically what I put algebraically, and doesn't really show what's going on, because your eventual equation is one between four-dimensional volumes. There ought to be a solution which stays in at most two dimensions. R.
there ought?? that the final relation
[BPQ]^2 = [ ... ] = ([ABQ] + [BCP] + [PDQ])^2 - 4 [ABQ] [BCP] .
is quadratic in areas suggests that there won't be such a solution. my solution had the mild benefit that being quadratic in areas arose from ratios of edge lengths. but yes, after that it's all algebraic. by the way, we went down a similar path some time ago when someone asked a related question about hero(n)'s formula: is there a reasonable 4-dimensional interpretation for s(s - a)(s - b)(s - c) ? i don't think anyone posted one. mike
B----F---------C /|\_ / / / / /_ / / | / \_ / / / / \_ / / | / \/ E----X------_--P / |/ __-- / / /_-- / A----Q---------D
On Saturday 15 December 2007, Michael Reid wrote:
by the way, we went down a similar path some time ago when someone asked a related question about hero(n)'s formula: is there a reasonable 4-dimensional interpretation for s(s - a)(s - b)(s - c) ? i don't think anyone posted one.
Having failed to come up with a sensible way to make A^2 = s(s-a)(s-b)(s-c), I wonder occasionally: what about Ar = (s-a)(s-b)(s-c)? Three dimensions might be easier than four :-). -- g
On 12/15/07, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Saturday 15 December 2007, Michael Reid wrote:
by the way, we went down a similar path some time ago when someone asked a related question about hero(n)'s formula: is there a reasonable 4-dimensional interpretation for s(s - a)(s - b)(s - c) ? i don't think anyone posted one.
Having failed to come up with a sensible way to make A^2 = s(s-a)(s-b)(s-c), I wonder occasionally: what about Ar = (s-a)(s-b)(s-c)? Three dimensions might be easier than four :-).
I must have missed the last discussion, so I'll seize the opportunity put in my two-pennyworth this time round. There is a school of thought (very much in the minority) that the involvement of square roots and trigonometric functions in elementary geometry is an unnecessary complication: in particular, that instead of considering distance, area, volume, ..., one should deal only with their squares. In terms of squares, Heron's formula becomes both rational and quadratic: (4A)^2 = 2( (a^2)(b^2) + (b^2)(c^2) + (c^2)(a^2) ) - ( (a^2)^2 + (b^2)^2 + (c^2)^2 ) . As I've pointed out on a previous occasion, the analogous Cayley-Menger determinant for a tetrahedron or higher-dimensional simplex does not factorise; so once out of the plane, we are to some extent obliged to adopt this "sqrts out" attitude. Fred Lunnon
participants (3)
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Fred lunnon -
Gareth McCaughan -
Michael Reid