## A Geometric Inequality

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• Hi, all. My conjecture is the following : Let P be in the interior of triangle ABC, and let the lines AP, BP, CP intersect the sides BC, CA, AB in L, M, N,
Message 1 of 4 , Jan 26, 2001
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Hi, all.

My conjecture is the following :

Let P be in the interior of triangle ABC, and let the lines AP, BP,
CP intersect the sides BC, CA, AB in L, M, N, respectively.
Then, 3/2 >= MN/BC + NL/CA + LM/AB ??

Is this true?

Yours sincerely,
Hojoo Lee
• Dear Hojoo, ... This is obviously false : take (BC,CA,AB)=(13,9,6) and P on the median AA . Let f=MN/BC + NL/CA + LM/AB When P- A, f- 5sqrt(65)/36 = 1.1...
Message 2 of 4 , Jan 27, 2001
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Dear Hojoo,

you wrote :

> My conjecture is the following :
>
> Let P be in the interior of triangle ABC, and let the lines AP, BP,
> CP intersect the sides BC, CA, AB in L, M, N, respectively.
> Then, 3/2 >= MN/BC + NL/CA + LM/AB ??
>
> Is this already known result?
> Is this true?
>
This is obviously false :
take (BC,CA,AB)=(13,9,6) and P on the median AA'.

Let f=MN/BC + NL/CA + LM/AB

When P->A, f->5sqrt(65)/36 = 1.1...
When P->A', f->101/36 = 2.8...

Now, my question : since f is bounded inside the triangle, what are its
limit sup and limit inf ?

Best regards

Bernard
• ... If the triangle ABC is acute-angled and P is its orthocenter, then the inequality is true: MN/BC = cosA, NL/CA=cosB,LM/AB=cosC and cosA + cosB + cosC
Message 3 of 4 , Jan 27, 2001
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--- In Hyacinthos@y..., insight_love@h... wrote:
> Hi, all.
>
> My conjecture is the following :
>
> Let P be in the interior of triangle ABC, and let the lines AP, BP,
> CP intersect the sides BC, CA, AB in L, M, N, respectively.
> Then, 3/2 >= MN/BC + NL/CA + LM/AB ??
>
> Is this already known result?
> Is this true?
>
> Yours sincerely,
> Hojoo Lee

If the triangle ABC is acute-angled and P is its orthocenter, then
the inequality is true:
MN/BC = cosA, NL/CA=cosB,LM/AB=cosC and
cosA + cosB + cosC <=3/2 (a well-known inequality)

Achilleas Sinefakopoulos
• If we take A=(0,0); B=(0,1) C=(n,1) L (in segment BC) close enough to B. M=(m2,m2) (in segment AC) with m1=1 It is easy to see( taking n big enough) that the
Message 4 of 4 , Feb 3, 2001
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If we take A=(0,0); B=(0,1) C=(n,1)
L (in segment BC) close enough to B.
M=(m2,m2) (in segment AC) with m1=1
It is easy to see( taking n big enough) that the function
MN/BC + NL/CA + LM/AB
take values as small as we want.

manuel

> > Let P be in the interior of triangle ABC, and let the lines AP,
BP,
> > CP intersect the sides BC, CA, AB in L, M, N, respectively.
> > Then, 3/2 >= MN/BC + NL/CA + LM/AB ??
> >
> > Is this already known result?
> > Is this true?
> >
> This is obviously false :
> take (BC,CA,AB)=(13,9,6) and P on the median AA'.
>
> Let f=MN/BC + NL/CA + LM/AB
>
> When P->A, f->5sqrt(65)/36 = 1.1...
> When P->A', f->101/36 = 2.8...
>
> Now, my question : since f is bounded inside the triangle, what are
its
> limit sup and limit inf ?
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