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Topic: Triangle Problem
Replies: 18   Last Post: May 10, 2006 6:17 AM

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Michael Lambrou

Posts: 70
Registered: 12/3/04
RE: Triangle Problem
Posted: May 4, 2006 3:23 AM
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> -----Original Message-----
> From:
> [] On Behalf Of Philippe 92
> Sent: Wednesday, May 03, 2006 11:54 PM
> To:; geometry-
> Subject: Re: Triangle Problem

> Avni Pllana wrote :
> > Let ABC be a triangle such that CA < 2*BC and BC < 2*CA .
> > Construct points D and E on the sides BC and CA respectively, such that
> > BD = DE = EA .

and Philippe replied:

> Just a hint, as I won't solve this for you.
> Consider to construct a triangle A'B'C' similar to ABC, from any
> arbitrary choosen distance d = B'D' = D'E' = E'A',
> with D' on B'C' and E' on A'C'.
> This is done by constructing d on AC, d on BC, and move BC to B'C' so
> that D'E' becomes = d, and AB'C' remains similar to ABC.

No! This is a false, non-ruler and compass construction. The main error
is the step "and move BC to B'C' so that D'E' becomes = d".

Here is a way to solve the problem:
Let x = BD = DE = EA (an unknown quantity),
so that BC = a-x, EC = b-x.
By the cosine rule on DCE we have
DE^2 = CD^2 + CE^2 - 2.CD.CE. cosC
= CD^2 + CE^2 - 2.CD.CE.(a^2 + b^2 - c^2)/(2ab)
so that

x^2 =
(a-x)^2 + (b-x)^2 -2(a-x)(b-x).(a^2 + b^2 - c^2)/(2ab).

This is a quadratic in x and, as is well known, its roots are
ruler and compass constructible. That is all.

One more remark: Only one of the roots is positive.
If you interpret negative roots as going in the
opposite direction, you have a second solution
to the problem, but this time externally to the
given triangle.

All the best,
Michael Lambrou.

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