Re: [EMHL] From Atul Dixit: Kenmotu configuration

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• ... Probably we have already discussed this, but let s try it anyway: C* A B* / / / A / Ab
Message 1 of 1 , Jun 1, 2001
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>If we consider the Kenmotu configuration. Draw lines through the
>vertices of the three squares, nearer to the respective vertices of
>the triangle, parallel to the respective diagonals of the three
>squares. Do these three lines form a triangle perspective to triangle
>ABC?

Probably we have already discussed this, but let's try it anyway:

C* A B*
/\
/ \
/ A' \
/ \
Ab Ac
/ \
/ \
Ba J Ca
/ \
/ \
/ B' C' \
/ \
B--------Bc---Cb---------C

A*

Let JAbA'Ac, JBcB'Ba, JCaC'Cb be the three congruent squares.

Let A*B*C* be the vertices of the triangle formed by the parallels.
(B*C* // AbAc, etc)

By some angle computations, which I hope they were correct, I found that
the triangles A*BC, B*CA, C*AB are isosceles: (A*BC) = (A*CB) = A, etc.

Now (by Kiepert triangles theory) we have:

Triangles: A*B*C*, ABC perspective <==>

cot(A*BC) + cotB cot(B*CA) + cotC cot(C*AB) + cotA
---------------- x ----------------- x ---------------- = 1
cot(A*CB) + cotC cot(B*AC) + cotA cot(C*BA) + cotB

or

cotA + cotB cotB + cotC cotC + cotA
----------- x ------------ x ----------- = 1
cotA + cotC cotB + cotA cotC + cotB

True.

Perspector: The perspector is the point:

((cotB + cotC) ::) = ((sinA / sinBsinC) ::) = (sin^2A ::) = K : Lemoine P.
in Barycentrics.

And a problem (generalization), by the way:

Construct the three Kenmotu squares but so that their sides
be proportional to given p, q, r. Special case: p:q:r = a:b:c
(In the standard Kenmotu problem we have that p:q:r = 1:1:1)

Antreas
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