Re: equilateral tringles
- From: "Philippe 92" <nospam@xxxxxxxxxxxx>
- Date: Thu, 01 Dec 2005 01:12:56 +0100
Dave Rusin wrote :
> In article <3950172.1133377985893.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
> eugene <jane1806@xxxxxxx> wrote:
>
>> Let ABC be an equilateral triangle. Points A_1,B_1,C_1 are chosen
> inside the triangle in such a way that A_1 \in CC_1, B_1 \in AA_1,
> C_1 \in BB_1 and AB_1=B_1A_1, BC_1=C_1B_1, CA_1=C_1A_1. Prove that
> the triangle A_1B_1C_1 is also equilateral.
>
> Nice.
>
> The word "chosen" is misleading here: the points A1, B1, C1 are
> uniquely specified by the condition that each of them is a midpoint
> of one of these three segments (and an endpoint of another).
> But then, that means that a one-third turn of the figure preserves
> all the interior line segments,
Why that ? Yes if AA_1 = BB_1 = CC_1, or if angles are equal...
that is if A_1B_1C_1 equilateral, you proove that it is equilateral !
> If thus proving A_1 B_1 is congruent
> to B_1 C_1, and that one to C_1 A_1.
Only once you have prooved it is equilateral then :
> To your drawing, add the three other line segments AC_1, BA_1, CB_1.
> This partitions the original triangle into seven. Computing area
> as a product (1/2)(base)(height), it is easy to argue that various
> pairs of triangles have equal area because (relative to an appropriate
> line containing an edge from each) the triangles have equal bases
> and equal heights. Thus all seven triangles have equal area, each
> 1/7 of the original total area.
fine relation !
Regards.
--
philippe
mail : chephip at free dot fr
site : http://chephip.free.fr/
.
- Follow-Ups:
- Re: equilateral tringles
- From: Dave Rusin
- Re: equilateral tringles
- References:
- equilateral tringles
- From: eugene
- Re: equilateral tringles
- From: Dave Rusin
- equilateral tringles
- Prev by Date: Bound on standard deviation ?
- Next by Date: Re: Well Ordering the Reals
- Previous by thread: Re: equilateral tringles
- Next by thread: Re: equilateral tringles
- Index(es):
Relevant Pages
|
