10. Walker's inequality and its extension.
by Virgil Nicula, Apr 20, 2010, 12:58 AM
Walker's inequality. Prove that in any acute or right triangle exists the inequality 
.
Method 1 (own).
. I used the well-known identities 
and 
.
Remark.
and 
(Euler's relation in an acute triangle) 
. I denote 
- the distance of the circumcenter 
of 
to the sideline 
.
Method 2. Denote the projections
, 
and 
of the orthocenter 
on the sidelines 
, 
and 
respectively. Thus, 
![$3-\sum\frac {[BHC]}{[ABC]}=3-1=2$](//latex.artofproblemsolving.com/8/1/d/81dee2e8caa6e2bf4af9b94ca3890af98fd89def.png)
.
I used well-known relations
a.s.o. and 
a.s.o. See here the Cezar Lupu's article.
Remark. In an acute triangle
there are the inequalities 
.
Can obtain the second inequality with the well-known metrical relation
.
Lemma. In any triangle
with incenter 
the following inequality holds : 
.
Proof. Show easily that
a.s.o. Thus, we have :
The last inequality is due to Gerretsen i.e.
. Therefore, we have shown that : .[/quote]
Method 3 (M.C.). For an acute triangle denote its orthocenter and let 
, 
, 
be its altitudes . Thus, is the incenter of the orthic triangle 
.
Applying the upper lemma for
we have : 
, where 
and 
are the circumradius and inradius of 
.
Taking into account that
and 
, the last inequality becomes : 
Walker's inequality .
Remark. In the above proof I used the well-known relations
and 
(in an acute triangle) .
Extension of the Walker's inequality (see here).
Proof. For the point
denote 
, 
, 
. Since 
a.s.o. obtain that
![$ \left[\sum (y + z)\cdot AD\right]^2\le \sum (y + z)\cdot\sum (y + z)\cdot AD^2 = 2\cdot\sum (y + z)\cdot AD^2 =$](//latex.artofproblemsolving.com/6/1/a/61a68346b194f8c8d1cbc154cc7357e9655204fe.png)
. In conclusion 
with equality iff 
.
Particular cases.
.
.
.
.Method 1 (own).
. I used the well-known identities 
and 
.Remark.
and 
(Euler's relation in an acute triangle) 
. I denote 
- the distance of the circumcenter 
of 
to the sideline 
.Method 2. Denote the projections
, 
and 
of the orthocenter 
on the sidelines 
, 
and 
respectively. Thus, 
![$3-\sum\frac {[BHC]}{[ABC]}=3-1=2$](http://latex.artofproblemsolving.com/8/1/d/81dee2e8caa6e2bf4af9b94ca3890af98fd89def.png)
.I used well-known relations
a.s.o. and 
a.s.o. See here the Cezar Lupu's article.Remark. In an acute triangle
there are the inequalities 
.Can obtain the second inequality with the well-known metrical relation
.Lemma. In any triangle
with incenter 
the following inequality holds : 
.Proof. Show easily that
a.s.o. Thus, we have :
The last inequality is due to Gerretsen i.e.
. Therefore, we have shown that : .[/quote]Method 3 (M.C.). For an acute triangle
denote its orthocenter and let 
, 
, 
be its altitudes . Thus, is the incenter of the orthic triangle 
.Applying the upper lemma for
we have : 
, where 
and 
are the circumradius and inradius of 
.Taking into account that
and 
, the last inequality becomes : 
Walker's inequality .Remark. In the above proof I used the well-known relations
and 
(in an acute triangle) .Extension of the Walker's inequality (see here).
Quote:
Let 
be an interior point of the triangle with the barycentrical coordinates 
w.r.t.
. Then there is an inequality 
.
be an interior point of the triangle with the barycentrical coordinates 
w.r.t. 
. Then there is an inequality 
.Proof. For the point
denote 
, 
, 
. Since 
a.s.o. obtain that
![$ \left[\sum (y + z)\cdot AD\right]^2\le \sum (y + z)\cdot\sum (y + z)\cdot AD^2 = 2\cdot\sum (y + z)\cdot AD^2 =$](http://latex.artofproblemsolving.com/6/1/a/61a68346b194f8c8d1cbc154cc7357e9655204fe.png)
. In conclusion 
with equality iff 
.Particular cases.
.
.
.This post has been edited 13 times. Last edited by Virgil Nicula, Nov 22, 2015, 10:01 PM
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