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• Dear friends. The following question is stated in the current Monthly: prove that cycsum(a^2)-4sqrt(3)S is not less than 2 cycsum( a^2(x^2-yz)/x)/(x+y+z) for
Message 1 of 4 , Dec 2 12:37 AM
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Dear friends.
The following question is stated in the current Monthly: prove that
cycsum(a^2)-4sqrt(3)S is not less than
2 cycsum( a^2(x^2-yz)/x)/(x+y+z) for all positive x,y,z where a,b,c are the sides of a triangle and S its area
(1) a more handy form is : define
f(x,y,z) = cycsum( a^2/x*(-x^2+xz+xy+2yz)) /(x+y+z) and show that
f(x,y,z) is not less than 4sqrt(3)S
(2) f is homogeneous and thus define a function acting over (barycentric) points in the triangle plane
(3) using standard numeric values, see that over triangle centers in ETC (lying inside ABC), min f is obtained for X(13)
(4) use formal barycentrics of X(13) and prove that, if true, property is optimal since f(X(13))= 4sqrt(3)S
(5) read the doc : X(13) minimizes a sum of lengths (that would introduce horrific square roots), but also satisfies angle equalities.
(6) therefore compute g(X) = cycsum(1/tan(XB,XC)) and check that
g = -f /4/S
(7) in the right domain, extremum is obtained when angles are equal to 120°, leading to 3*(1/sqrt(3)) as needed.

Best regards,
Pierre.
• Dear Pierre If for a point X inside ABC with homogenous barycentrics (x : y : z) p = angle (XB, XC) q = angle(XC, XA) r = angle(XA, XB) with cot(p) = P, cot(q)
Message 2 of 4 , Dec 3 7:16 AM
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Dear Pierre

If for a point X inside ABC
with homogenous barycentrics (x : y : z)
p = angle (XB, XC)
q = angle(XC, XA)
r = angle(XA, XB)
with cot(p) = P, cot(q) = Q, cot(r) = R
where p + q + r = 2.pi (i)
gives
P.Q + Q.R + R.P = 1 (ii)
and S the area of ABC
from FG200921 we get that
x = k/(cotA - P)
y = k/(cotB - Q)
z = k/(cotC - R)
where
cotA = (bb + cc - aa)/4S . . .
and from (ii)
k = (aayz + bbzx + ccxy)/[2S(x+y+z)] .
Substituting
x, y, z and k we get
P + Q + R = -f/(4S) (iii)
where f = f(x,y,z) = cycsum( a^2/x*(-x^2+xz+xy+2yz)) /(x+y+z)
and since the function cot(x) is convex we have
by Jensen's inequality
P + Q + R = cot(p) + cot(q) + cot(r) >= 3.cot[(p+q+r)/3] or
P + Q + R >= -sqrt(3) or from (iii)
f <= 4S.sqrt(3)

Best regards

> Dear friends.
> The following question is stated in the current Monthly:
> prove that
> cycsum(a^2)-4sqrt(3)S is not less than
> 2 cycsum( a^2(x^2-yz)/x)/(x+y+z) for all positive x,y,z
> where a,b,c are the sides of a triangle and S its area
> (1) a more handy form is : define
> f(x,y,z) = cycsum( a^2/x*(-x^2+xz+xy+2yz)) /(x+y+z) and
> show that
> f(x,y,z) is not less than 4sqrt(3)S
> (2) f is homogeneous and thus define a function acting over
> (barycentric) points in the triangle plane
> (3) using standard numeric values, see that over triangle
> centers in ETC (lying inside ABC), min f is obtained for
> X(13)
> (4) use formal barycentrics of X(13) and prove that, if
> true, property is optimal since f(X(13))= 4sqrt(3)S
> (5) read the doc : X(13) minimizes a sum of lengths (that
> would introduce horrific square roots), but also satisfies
> angle equalities.
> (6) therefore compute g(X) = cycsum(1/tan(XB,XC)) and check
> that
> g = -f /4/S
> (7) in the right domain, extremum is obtained when angles
> are equal to 120°, leading to 3*(1/sqrt(3)) as needed.
>
> Best regards,
> Pierre.
>
>
>

___________________________________________________________
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• Dear Pierre, Sorry I said nonsense that the function cot(x) is convex and that P + Q + R = -sqrt(3). So my proof is not correct. My proof is correct in the
Message 3 of 4 , Dec 3 11:49 AM
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Dear Pierre,
Sorry I said nonsense that the
function cot(x) is convex
and that P + Q + R >= -sqrt(3).
So my proof is not correct.
My proof is correct in the direction
P + Q + R <= -sqrt(3)
and hence f >= 4S.sqrt(3)
if p, q, r > pi/2 since the function
cot(x) is concave in the interval
(pi/2, pi).

Best regards

> Dear Pierre
>
> If for a point X inside ABC
> with homogenous barycentrics (x : y : z)
> p = angle (XB, XC)
> q = angle(XC, XA)
> r = angle(XA, XB)
> with cot(p) = P, cot(q) = Q, cot(r) = R
> where p + q + r = 2.pi  (i)
> gives
> P.Q + Q.R + R.P = 1  (ii)
> and S the area of ABC
> from FG200921 we get that
> x = k/(cotA - P)
> y = k/(cotB - Q)
> z = k/(cotC - R)
> where
> cotA = (bb + cc - aa)/4S  . . .
> and from (ii)
> k = (aayz + bbzx + ccxy)/[2S(x+y+z)] .
> Substituting
> x, y, z and k  we get
> P + Q + R = -f/(4S) (iii)
> where f = f(x,y,z) = cycsum( a^2/x*(-x^2+xz+xy+2yz))
> /(x+y+z)
> and since the function cot(x) is convex we have
> by Jensen's inequality
> P + Q + R = cot(p) + cot(q) + cot(r) >= 3.cot[(p+q+r)/3]
> or
> P + Q + R >= -sqrt(3) or from (iii)
> f <= 4S.sqrt(3)
>
> Best regards
>
> > Dear friends.
> > The following question is stated in the current
> Monthly:
> > prove that
> > cycsum(a^2)-4sqrt(3)S is not less than
> > 2 cycsum( a^2(x^2-yz)/x)/(x+y+z) for all positive
> x,y,z
> > where a,b,c are the sides of a triangle and S its
> area
> > (1) a more handy form is : define
> > f(x,y,z) = cycsum( a^2/x*(-x^2+xz+xy+2yz)) /(x+y+z)
> and
> > show that
> > f(x,y,z) is not less than 4sqrt(3)S
> > (2) f is homogeneous and thus define a function acting
> over
> > (barycentric) points in the triangle plane
> > (3) using standard numeric values, see that over
> triangle
> > centers in ETC (lying inside ABC), min f is obtained
> for
> > X(13)
> > (4) use formal barycentrics of X(13) and prove that,
> if
> > true, property is optimal since f(X(13))= 4sqrt(3)S
> > (5) read the doc : X(13) minimizes a sum of lengths
> (that
> > would introduce horrific square roots), but also
> satisfies
> > angle equalities.
> > (6) therefore compute g(X) = cycsum(1/tan(XB,XC)) and
> check
> > that
> > g = -f /4/S
> > (7) in the right domain, extremum is obtained when
> angles
> > are equal to 120°, leading to 3*(1/sqrt(3)) as
> needed.
> >
> > Best regards,
> > Pierre.
> >
> >
> >
>
>
>
>
> ___________________________________________________________
>
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>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>

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• Dear Pierre, now I found a proof without restriction for p, q, r. P + Q = cot(p) + cot(q) = sin(p+q)/sin(p).sin(q) = -sin(r)/sin(p).sin(q). Since 0
Message 4 of 4 , Dec 3 12:32 PM
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Dear Pierre,
now I found a proof without
restriction for p, q, r.
P + Q = cot(p) + cot(q) = sin(p+q)/sin(p).sin(q)
= -sin(r)/sin(p).sin(q).
Since 0 < sin(p).sin(q) = (1/2)[cos(p-q)-cos(p+q)] <
< (1/2)[1-cos(p+q)] = (1/2)[1-cos(r)] = sin^2(r/2)
we have P + Q <= -sin(r)/sin^2(r/2) = -2.cot(r/2)
Similarly working we get
P + Q + R <= -(cot(p/2) + cot(q/2) + cot(r/2))
<= -3cot[(p+q+r)/6] = -sqr(3)
since the function cot(x) is convex in the
interval (0, pi/2).

Best regards

> Dear Pierre,
> Sorry I said nonsense that the
> function cot(x) is convex
> and that P + Q + R >= -sqrt(3).
> So my proof is not correct.
> My proof is correct in the direction
> P + Q + R <= -sqrt(3)
> and hence f >= 4S.sqrt(3)
> if p, q, r > pi/2 since the function
> cot(x) is concave in the interval
> (pi/2, pi).
>
> Best regards

>

___________________________________________________________
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