- Let ABC be a triangle. Construct a line intersecting CA, CB at Ca, Cb, resp,
and another line intersecting BA, BC at Ba, Bc, resp., such that:
(ACa = BCb = ABa = CBc) AND (CaCb = BaBc)
A trig. solution leads to a quadratic equation:
Let ACa = BCb = ABa = CBc = x
Triangle (CaCbC): (CaCb)^2 = (b-x)^2 + (a-x)^2 - 2(b-x)(a-x)cosC (1)
Triangle (BaBcB): (BaBc)^2 = (c-x)^2 + (a-x)^2 - 2(c-x)(a-x)cosB (2)
CaCb = BaBc, and (1) and (2) ==>
(b-x)^2 - (c-x)^2 = 2(b-x)(a-x)cosC - 2(c-x)(a-x)cosB
and from this we get the x.
PS: Probably the point of intersection A' of CaCb, BaBc,
(if there is only one such real point !), and the similarly defined
ones B', C', are of some interest.
(for example: Are the lines AA', BB', CC' concurrent?)
- Dear Hyacinthians,
let me give you the following construction problem.
Construct the triangle ABC if given in position, the
line on which the base BC is located,and the feet of
the Cevians of O to the two lateral sides.
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- Dear all
Given the first Brocard triangle of triangle ABC,how to construct
the triangle ABC ?
My solution has become unnecessarily long.Will anybody tell me a precise
solution of it?
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- Johnson, page 280:
The solution depends onthe fact that any triangle is inversively similar to
its first Brocard triangle.
We locate the Brocard points of the given triangle, also its first Brocard
triangle; then by similarity we locate on the circumcircle of the given
triangle the Brocard points of the required triangle.
The vertices of the latter can then be found at once.
Is this what you did?
::From: Atul Dixit [mailto:atul_dixie@...]
::Sent: Saturday, August 31, 2002 6:42 PM
::Subject: [EMHL] Construction problem
:: Given the first Brocard triangle of triangle ABC,how to construct
::the triangle ABC ?
::My solution has become unnecessarily long.Will anybody tell me a precise
::solution of it?
::Send and receive Hotmail on your mobile device: http://mobile.msn.com
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