- Dear friends,

Let M be a given point in the triangle ABC. Extend the lines AM, BM, and CM to intersect the sides of the triangle ABC at A0, B0, and C0, respectively. Next construct the parallel to A0C0 through M, which intersects BA and BC at C1 and A2, respectively. Analogously, draw the parallel through M to B0A0 (and to B0C0) to find A1 and B2 (and B1 and C2). It is easy to prove that A1B2||A2B1, B1C2||B2C1 and C1A2||C2A1.

Areas will be denoted by square brackets. Denote T1=[MC1B2], T2=[MA1C2], T3=[MB1A2], S1=[MA1A2], S2=[MB1B2], S3=[MC1C2], P1=[AB2C1], P2=[BC2A1], P3=[CA2B1].

I find that the inequality P1+P2+P3+7(S1+S2+S3)>=8(T1+T2+T3) holds true and equality occurs if and only if the central point M is a midpoint of medians or centroid of triangle ABC. I find also that 7 is the best constant fot this inequality.

I ask all the friends to inform me if the midpoints of medians are in the Encyclopedia of Kimberling and are there any other similar interesting properties of these points.

Yakub - Dear Yakub,

Let M=p:q:r (barycentrics) and

s=p+q+r, t= pq+qr+rp, u= pqr then

T1+T2+T3=4*u*(-t+s^2)/(s^2*(s*t-u)),

S1+S2+S3=2*u*(t+s^2)/(s^2*(s*t-u)),

P1+P2+P3=(s^3*t-7*s^2*u+2*u*t)/(s^2*(s*t-u))

(times the area of triangle ABC).

control : T1+T2+T3+S1+S2+S3+P1+P2+P3=1

When M=G, relation P+(k-1)S = k T hold for every k.

Otherwise, positive factor s*t-u that appears in all denominators

can be replaced by u (same global degree) and formula

P+(k-1)S >= k T follows with k=8 (your result).

The centroid is not involved in k=8, since, locally around G,

k=9 is sufficient.

Best regards.

--- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@...> wrote:

>

> Dear friends,

> Let M be a given point in the triangle ABC. Extend the lines AM, BM, and CM to intersect the sides of the triangle ABC at A0, B0, and C0, respectively. Next construct the parallel to A0C0 through M, which intersects BA and BC at C1 and A2, respectively. Analogously, draw the parallel through M to B0A0 (and to B0C0) to find A1 and B2 (and B1 and C2). It is easy to prove that A1B2||A2B1, B1C2||B2C1 and C1A2||C2A1.

> Areas will be denoted by square brackets. Denote T1=[MC1B2], T2=[MA1C2], T3=[MB1A2], S1=[MA1A2], S2=[MB1B2], S3=[MC1C2], P1=[AB2C1], P2=[BC2A1], P3=[CA2B1].

> I find that the inequality P1+P2+P3+7(S1+S2+S3)>=8(T1+T2+T3) holds true and equality occurs if and only if the central point M is a midpoint of medians or centroid of triangle ABC. I find also that 7 is the best constant fot this inequality.

> I ask all the friends to inform me if the midpoints of medians are in the Encyclopedia of Kimberling and are there any other similar interesting properties of these points.

> Yakub

> - Dear friend you wrote:
>

What means this part

> Let M=p:q:r (barycentrics) and

> s=p+q+r, t= pq+qr+rp, u= pqr then

>

> T1+T2+T3=4*u*(-t+s^2)/(s^2*(s*t-u)),

> S1+S2+S3=2*u*(t+s^2)/(s^2*(s*t-u)),

> P1+P2+P3=(s^3*t-7*s^2*u+2*u*t)/(s^2*(s*t-u))

> (times the area of triangle ABC).

>

> control : T1+T2+T3+S1+S2+S3+P1+P2+P3=1

>

> When M=G, relation P+(k-1)S = k T hold for every k.

> Otherwise, positive factor s*t-u that appears in all denominators

> can be replaced by u (same global degree) and formula

> P+(k-1)S >= k T follows with k=8 (your result).

>

> The centroid is not involved in k=8, since, locally around G,

> k=9 is sufficient.

>

...positive factor s*t-u that appears in all denominators

can be replaced by u (same global degree)...?

Thank you for your interest.

Yakub. - Dear Yakub,

In the proof of Thm 1.1 of your :

http://f1.grp.yahoofs.com/...files_of_Hyacinthos.../20092516nev.pdf

you wrote "after simplifications"... but you don't give the details.

The question was to prove relation 0 <= P+(k-1)S - k T (1.1)

where P,S,T are as given in p,q,r (or more efficiently in s,t,u)

by transforming this relation into your formula (1.2) that holds

for all p,q,r positive. This formula is given in a multiplicative form, in order to allow the next step (1.2) ==> (1.3). But to prove (1.1) ==> (1.2), we better restate (1.2) into

0 <= t*s/u + 48*t/s^2 - 25 (1.2b)

and see that this (1.2b) is quite the same as (1.1) for the choice k=8 .... except from denominators s^2*u versus s^2(s*t-u).

Since s*t-u is is clearly positive inside triangle ABC, the proof is obtained.

If we only discard the denominators in (1.1), the relation (1.2b) will not be obtained by a simple transposition of terms (denominator u is required, as well as global degree 0).

Therefore proving (1.1) ==> (1.2b) is replacing s*t-u by u. Afterwards, one can remark that not only the sign is preserved by this substitution, but also the global degree (in p,q,r).

Best regards,

Pierre.

-----------------------------------

By the way : in the 'Files' Directory, files are sorted by name. What is the trick to sort by date ?

-----------------------------------

--- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@...> wrote:

>

> Dear friend you wrote:

> >

> > Let M=p:q:r (barycentrics) and

> > s=p+q+r, t= pq+qr+rp, u= pqr then

> >

> > T1+T2+T3=4*u*(-t+s^2)/(s^2*(s*t-u)),

> > S1+S2+S3=2*u*(t+s^2)/(s^2*(s*t-u)),

> > P1+P2+P3=(s^3*t-7*s^2*u+2*u*t)/(s^2*(s*t-u))

> > (times the area of triangle ABC).

> >

> > control : T1+T2+T3+S1+S2+S3+P1+P2+P3=1

> >

> > When M=G, relation P+(k-1)S = k T hold for every k.

> > Otherwise, positive factor s*t-u that appears in all denominators

> > can be replaced by u (same global degree) and formula

> > P+(k-1)S >= k T follows with k=8 (your result).

> >

> > The centroid is not involved in k=8, since, locally around G,

> > k=9 is sufficient.

> >

>

> What means this part

>

> ...positive factor s*t-u that appears in all denominators

> can be replaced by u (same global degree)...?

>

> Thank you for your interest.

> Yakub.

> - Dear Pierre and other friends, I find that the inequality P+7S>=8T

(see Theorem 1.1 in

http://f1.grp.yahoofs.com/v1/0HLqStqb7vBjiOMMgKGBq5FMeB9xbBkbPqsNl4-HXSh0fzx6jtEPPjBWv7v41t65knwkD1RSxI75QBmw6_wjSRdruKvGhJRVyA/20092516nev.pdf

)

can be reduced to a special quintic triangle inequality. Namely, our inequality is equivalent to

9*S{a^5}-15*S{(ab^4}+6*S{a^3b^2}+25*S{abc^3}-16*S{ab^2c^2}>=0,

where "S{.}" means "Symmetric Sum". For example, S(a)=a+b+c.

It is a special case of the inequality with parameter u:

(u+1)^2*SS{a^5}-(u+3)(u+1)*SS{(ab^4}+2(u+1)*SS{a^3b^2}+(u+3)^2*SS{abc^3}-4(u+2)*SS{ab^2c^2}>=0,

appeared in a paper by William R. Harris:

Real even symmetric ternary forms, Journal of Algebra, 222, 204-245 (1999).

The above inequality and therefore our inequality P+7S>=8T correspond to the case u=2. The author (Harris) at the end of his paper give some information:

"There is disscussion in the literature on quartic and sextic triangle inequalities, but we have found no previous work on quintic triangle inequalties"

Therefore, after 10 years, I ask you all to inform me if there is any other example of discussion of quintic triangle inequality in recent literature. Note also that I was unfamiliar with the paper of Harris when I published my results. For this reason and unfortunately he did not appeared in my references.

Yakub.

--- In Hyacinthos@yahoogroups.com, "pldx1" <pldx1@...> wrote:

>

> Dear Yakub,

>

> In the proof of Thm 1.1 of your :

> http://f1.grp.yahoofs.com/...files_of_Hyacinthos.../20092516nev.pdf

> you wrote "after simplifications"... but you don't give the details.

>

> The question was to prove relation 0 <= P+(k-1)S - k T (1.1)

> where P,S,T are as given in p,q,r (or more efficiently in s,t,u)

> by transforming this relation into your formula (1.2) that holds

> for all p,q,r positive. This formula is given in a multiplicative form, in order to allow the next step (1.2) ==> (1.3). But to prove (1.1) ==> (1.2), we better restate (1.2) into

>

> 0 <= t*s/u + 48*t/s^2 - 25 (1.2b)

>

> and see that this (1.2b) is quite the same as (1.1) for the choice k=8 .... except from denominators s^2*u versus s^2(s*t-u).

>

> Since s*t-u is is clearly positive inside triangle ABC, the proof is obtained.

>

> If we only discard the denominators in (1.1), the relation (1.2b) will not be obtained by a simple transposition of terms (denominator u is required, as well as global degree 0).

>

> Therefore proving (1.1) ==> (1.2b) is replacing s*t-u by u. Afterwards, one can remark that not only the sign is preserved by this substitution, but also the global degree (in p,q,r).

>

> Best regards,

> Pierre.

>

> -----------------------------------

> By the way : in the 'Files' Directory, files are sorted by name. What is the trick to sort by date ?

> -----------------------------------

>

>

>

> --- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@> wrote:

> >

> > Dear friend you wrote:

> > >

> > > Let M=p:q:r (barycentrics) and

> > > s=p+q+r, t= pq+qr+rp, u= pqr then

> > >

> > > T1+T2+T3=4*u*(-t+s^2)/(s^2*(s*t-u)),

> > > S1+S2+S3=2*u*(t+s^2)/(s^2*(s*t-u)),

> > > P1+P2+P3=(s^3*t-7*s^2*u+2*u*t)/(s^2*(s*t-u))

> > > (times the area of triangle ABC).

> > >

> > > control : T1+T2+T3+S1+S2+S3+P1+P2+P3=1

> > >

> > > When M=G, relation P+(k-1)S = k T hold for every k.

> > > Otherwise, positive factor s*t-u that appears in all denominators

> > > can be replaced by u (same global degree) and formula

> > > P+(k-1)S >= k T follows with k=8 (your result).

> > >

> > > The centroid is not involved in k=8, since, locally around G,

> > > k=9 is sufficient.

> > >

> >

> > What means this part

> >

> > ...positive factor s*t-u that appears in all denominators

> > can be replaced by u (same global degree)...?

> >

> > Thank you for your interest.

> > Yakub.

> >

>