## Re: [EMHL] de Longchamps point

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• A computation error (not typo!): ... ^^^^^^^^^^^^^^^^^^^ The trilinears of de L point are: cos(B-C) + cos(B+C) (cosA - cosBcosC ::) = (cosA -
Message 1 of 6 , Jun 30, 2000
A computation error (not typo!):

I wrote:

>Let ABC be a triangle. Draw a cevian AA' so that:
>if A'b, A'c are the orth. projections of A' on AC, AB, resp., then
>the lines AA', BA'b, CA'c are concurrent.
>
>
>Similarly, draw the cevians BB', CC'.
>
>The three cevians AA', BB', CC' concur at de Longchamps p. (cosA - cos(B-C)::)
^^^^^^^^^^^^^^^^^^^

The trilinears of de L point are:

cos(B-C) + cos(B+C)
(cosA - cosBcosC ::) = (cosA - ------------------ ::) =
2

cos(B-C) - cosA
(cosA - --------------- ::) = (3cosA - cos(B-C) ::)
2

Also, (8),(9),(10), in the solution I posted, read:

sinA_1 cosC - cosAcosB 3cosC - cos(A-B)
------ = --------------- = --------------- (8)
sinA_2 cosB - cosAcosC 3cosB - cos(A-C)

sinB_1 cosA - cosBcosC 3cosA - cos(B-C)
------ = --------------- = --------------- (9)
sinB_2 cosC - cosBcosA 3cosC - cos(B-A)

sinC_1 cosB - cosCcosA 3cosB - cos(C-A)
------ = --------------- = --------------- (10)
sinC_2 cosA - cosCcosB 3cosA - cos(C-B)

Sorry!

Antreas
• Dear friends, ... The triangle A A bA c is the pedal triangle of A and is in perpective with ABC therefore A belongs to the Darboux cubic (isogonal cubic
Message 2 of 6 , Jul 1, 2000
Dear friends,

>
> Let ABC be a triangle. Draw a cevian AA' so that:
> if A'b, A'c are the orth. projections of A' on AC, AB, resp., then
> the lines AA', BA'b, CA'c are concurrent.
>
>
> A
> /\
> / \
> / \
> / \
> / \
> / \
> / \
> A'c A'b
> / \
> / \
> B---------A'---------C
>
> Similarly, draw the cevians BB', CC'.
>
> The three cevians AA', BB', CC' concur at de Longchamps p.

The triangle A'A'bA'c is the pedal triangle of A' and is in perpective with
ABC therefore A' belongs to the Darboux cubic (isogonal cubic with pivot L =
de Longchamps) and is the trace of L on BC. CQFD.

Best wishes

Bernard
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