## Re: Cevians and wqual circles

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• Hello Vladimir, ... two ... concurrence, but ... I have done a similar experiment and find that although close, I do not think they are perspective. Call A
Message 1 of 3 , Feb 18, 2005

> I've been asked the following question:
>
> do the three cevians each of which divides a given triangle into
two
> triangles with equal inradii concur?
>
> A quick GSP experiment shows that they are very close to
concurrence, but
> there's a lot of room for doubt.
>
> Does anybody know the answer?

I have done a similar experiment and find that although close, I do
not think they are perspective.

Call A' the meet of the A cevian with BC. Then

A' = {0, c (b + Sqrt[s^2 - a s]), b (c + Sqrt[s^2 - a s])}
B' = {c (a + Sqrt[s^2 - b s]), 0, a (c + Sqrt[s^2 - b s])}
C' = {b (a + Sqrt[s^2 - c s]), a (b + Sqrt[s^2 - c s]), 0}

from which A'B'C' is not perspective.

The line I of BAA' to I of BCC'
The line I of CBB' to I of CAA'
The line I of ACC' to I of ABB'
are of course concurrent in I of ABC.

The radii of the incircles are rA' = r (s - Sqrt[s^2 - a s]) / a

Best regards,
Peter.
• Dear Vladimir ... two ... concurrence, but ... If A is the foot of the A-cevian, we have cos
Message 2 of 3 , Feb 19, 2005
> I've been asked the following question:
>
> do the three cevians each of which divides a given triangle into
two
> triangles with equal inradii concur?
>
> A quick GSP experiment shows that they are very close to
concurrence, but
> there's a lot of room for doubt.
>
> Does anybody know the answer?

If A' is the foot of the A-cevian, we have cos <BA'A = (b-c)/a and
it follows that the three cevian lines don't concur.
Friendly. Jean-Pierre
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