286. ABC is right triangle iff s=2R+r and other similar pp.
by Virgil Nicula, Jun 14, 2011, 2:38 AM
Proposed problem. Prove that in any 
there is the equivalence 
.
Proof 1. Denote
. Thus, ![$4P=2\cos A\left[\cos (B+C)+\cos (B-C)\right]=$](//latex.artofproblemsolving.com/a/c/d/acd65e5420ed735bda3f1fb003a57782ee2b6336.png)
.
Since
![$2\left[s^2-(2R+r)^2\right]$](//latex.artofproblemsolving.com/f/2/a/f2aff73b51ede25de3ebc20bdb5b8815adbafe80.png)
obtain that the required identity 
from where obtain
is right-angled.
Proof 2. Denote
. Thus, 
![$\frac {\prod\left[\left(a^2+b^2+c^2\right)-2a^2\right]}{8(abc)^2}=$](//latex.artofproblemsolving.com/2/3/5/235d7eb62f55a5258fa458fb1901137fbdf1dcf7.png)
![$\frac {\left(a^2+b^2+c^2\right)\left[2\sum b^2c^2-\sum a^4\right]-8(abc)^2}{8(abc)^2}=$](//latex.artofproblemsolving.com/0/c/0/0c03f5abbdb7c049265777ba2100feedb5626799.png)
, i.e. 
a.s.o.
Proof 3.
or 
.
Observe that
, i.e. 
.
In conclusion, from upper relations
and 
obtain that 
is right-angled.
Proof 4. I"ll show
. Prove easily that 
.
Therefore,
.
Proof 5. 
.
Remark. Observe that
. Thus, 
.
PP1. Prove that in any triangle
there is the inequality 
.
Proof. Using Mollweide's formula i.e.
, the proposed inequality can be written as : 
. Observe that for 
we have the chains 
.
Therefore, by Chebyshev's inequality it follows that :
, as desired.
PP2. Prove that in any triangle there is the inequality 
.
Proof. Using the well known relations in a triangle :
the given inequality
is equivalent to :
, which is well-known.
PP3. Prove that in with the circumcenter 
and orthocenter 
exists the relation 
.
Proof (Mateescu Constantin). The following sharper inequality is also true :
.
Taking into account the well-known distances in a triangle :
, the inequality 
is equivalent to:
, which is weaker than the well-known Gerretsen i.e. 
.
Now let's draw our attention to
, where 
represents Nagel's point. Since : 
we obtain that :
.
Similarly we can obtain the analogous inequalities and thus the boxed inequality is proved.
PP4. Prove that
Proof.
Remark.
.
PP5. Prove that in any exists the inequality 
.
Proof.![$\left(\sqrt{\frac {s-a}{s-b}}+\sqrt{\frac {s-b}{s-c}}+\sqrt{\frac {s-c}{s-a}}\right)^2\ \stackrel{\text{(C.B.S.)}}{\le}\ \left[(s-a)+(s-b)+(s-c)\right]$](//latex.artofproblemsolving.com/c/9/2/c92da5ec7ff54547179df21793a20fa460820740.png)
![$\cdot\left[\frac 1{s-b}+\frac 1{s-c}+\frac 1{s-a}\right]=$](//latex.artofproblemsolving.com/a/d/c/adc5b33c609f056eb16a94162f1ce68252894dd3.png)
. I used the well-known identity 
.
PP6. Let
be a triangle. Prove that 
.
Proof 1. Prove easily that
![$s\cdot \prod\left(1-\sqrt 3\cdot\tan\frac A2\right)=4\left[s-(R+r)\sqrt 3\right]$](//latex.artofproblemsolving.com/b/b/9/bb97fcfef30ebb6bc76cfd50dbd6863349456b33.png)
.
Proof 2. Prove easily that
. Therefore, 
. For 
, where
apply the identity 
. In conclusion, 
.
Remarks.
The equivalence 
is false because the implication
is true and 
is false.
This equivalence 
is true.
PP7. Prove that
Proof (Mateescu Constantin). We assume that
, 
, 
are the angles of a triangle labelled in such a way that and are "on the same side" of 
, i.e.
or 
. In either case, we have :
. On the other hand,
since
we can weaken the above inequality in order to obtain :
. Thus, it's enough to show
that :
.
Taking into account the identity :
, the inequality is equivalent to :
Now,
,
so in either case the inequality
holds, and the proof is complete.
Remark. The proposed inequality is equivalent to the following nice form :
.
PP8. Prove that
.
Preliminary.
.
Proof.
.
![$2\blacktriangleright\ \sum\frac 1{\cos\frac {B-C}2}=\frac {4R^2}{s^2+r^2+2Rr}\cdot\left[\left(\sum\, \cos\frac {B-C}2\right)^2-1\right]-1$](//latex.artofproblemsolving.com/d/0/5/d055413225007192e570e3bbc34f3c57094dfce1.png)
.
Proof.
.
Proof of the proposed problem (Mateescu Constantin).
![$\begin{array}{cccc}
\\\
\ & \left(\cos\frac {B-C}2-1\right)\left(\cos\frac {C-A}2-1\right)\left(\cos\frac {A-B}2-1\right)\, \le\, 0 \\ \\
\iff & 1+\sum\, \cos\frac {A-B}2\cos\frac {A-C}2\, \ge\, \sum\, \cos\frac {B-C}2+\prod\, \cos\frac {B-C}2 \\ \\
\iff & 1+\prod\, \cos\frac {B-C}2\cdot\sum\, \frac 1{\cos\frac {B-C}2}\, \ge\, \sum\, \cos\frac {B-C}2+\prod\, \cos\frac {B-C}2 \\ \\
\stackrel{(1)}{\iff} & 1+\frac {s^2+r^2+2Rr}{8R^2}\cdot\left(\sum\, \frac 1{\cos\frac {B-C}2}-1\right)\, \ge\, \sum\, \cos\frac {B-C}2 \\ \\
\stackrel{(2)}{\iff} & 1+\frac {s^2+r^2+2Rr}{8R^2}\cdot\left\{\frac {4R^2}{s^2+r^2+2Rr}\cdot\left[\left(\sum\, \cos\frac {B-C}2\right)^2-1\right]-2\right\}\, \ge\, \sum\, \cos\frac {B-C}2 \\ \\
\iff & 1+\frac 12\cdot\left[\left(\sum\, \cos\frac {B-C}2\right)^2-1\right]-\sum\, \cos\frac {B-C}2\, \ge\, \frac {s^2+r^2+2Rr}{4R^2} \\ \\
\iff & 2+\left(\sum\, \cos\frac {B-C}2\right)^2-1-2\sum\, \cos\frac {B-C}2\, \ge\, \frac {s^2+r^2+2Rr}{2R^2} \\ \\
\iff & \left(\sum\, \cos\frac {B-C}2-1\right)^2\, \ge\, \frac {s^2+r^2+2Rr}{2R^2} \\ \\
\iff & \sum\, \cos\frac {B-C}2\, \ge\, 1+\sqrt {\frac {s^2+r^2+2Rr}{2R^2}} \\ \\
\iff & \boxed{\ \frac {h_a}{l_a}+\frac {h_b}{l_b}+\frac {h_c}{l_c}\ \ge\ 1+\sqrt {\frac {s^2+r^2+2Rr}{2R^2}}\ } \\\
\end{array}$](//latex.artofproblemsolving.com/b/2/a/b2a802e8ce293a353cf978912bfb7187a1f42e2d.png)
PP9
Proof (Mateescu Constantin). We will transform the given inequality (in any triangle) into one restricted to an acute angled triangle by using the following substitutions :
, 
and thus we are left to prove that: 
. For convenience,
we will return to the notation of a triangle in the last inequality and denote by 
, 
and 
its usual elements (circumradius, inradius and
semiperimeter respectively). Hence, the inequality
becomes: 
Using the well-known identities:
the last inequality is now equivalent to:
, which is actually true in any
triangle since :
, where
I made use of the known inequalities:
(Gerretsen) , 
and 
(Euler).
there is the equivalence 
.Proof 1. Denote
. Thus, ![$4P=2\cos A\left[\cos (B+C)+\cos (B-C)\right]=$](http://latex.artofproblemsolving.com/a/c/d/acd65e5420ed735bda3f1fb003a57782ee2b6336.png)
.Since
![$2\left[s^2-(2R+r)^2\right]$](http://latex.artofproblemsolving.com/f/2/a/f2aff73b51ede25de3ebc20bdb5b8815adbafe80.png)
obtain that the required identity 
from where obtain
is right-angled.Proof 2. Denote
. Thus, 
![$\frac {\prod\left[\left(a^2+b^2+c^2\right)-2a^2\right]}{8(abc)^2}=$](http://latex.artofproblemsolving.com/2/3/5/235d7eb62f55a5258fa458fb1901137fbdf1dcf7.png)
![$\frac {\left(a^2+b^2+c^2\right)\left[2\sum b^2c^2-\sum a^4\right]-8(abc)^2}{8(abc)^2}=$](http://latex.artofproblemsolving.com/0/c/0/0c03f5abbdb7c049265777ba2100feedb5626799.png)
, i.e. 
a.s.o.Proof 3.
or 
.Observe that
, i.e. 
.In conclusion, from upper relations
and 
obtain that 
is right-angled.Proof 4. I"ll show
. Prove easily that 
.Therefore,
.Proof 5.
.Remark. Observe that
. Thus, 
.PP1. Prove that in any triangle
there is the inequality 
.Proof. Using Mollweide's formula i.e.
, the proposed inequality can be written as : 
. Observe that for 
we have the chains 
.Therefore, by Chebyshev's inequality it follows that :
, as desired.PP2. Prove that in any triangle
there is the inequality 
.Proof. Using the well known relations in a triangle :
the given inequalityis equivalent to :
, which is well-known.PP3. Prove that in
with the circumcenter 
and orthocenter 
exists the relation 
.Proof (Mateescu Constantin). The following sharper inequality is also true :
.Taking into account the well-known distances in a triangle :
, the inequality 
is equivalent to:
, which is weaker than the well-known Gerretsen i.e. 
.Now let's draw our attention to
, where 
represents Nagel's point. Since : 
we obtain that :
.Similarly we can obtain the analogous inequalities and thus the boxed inequality is proved.
PP4. Prove that
Proof.
Remark.
. PP5. Prove that in any
exists the inequality 
.Proof.
![$\left(\sqrt{\frac {s-a}{s-b}}+\sqrt{\frac {s-b}{s-c}}+\sqrt{\frac {s-c}{s-a}}\right)^2\ \stackrel{\text{(C.B.S.)}}{\le}\ \left[(s-a)+(s-b)+(s-c)\right]$](http://latex.artofproblemsolving.com/c/9/2/c92da5ec7ff54547179df21793a20fa460820740.png)
![$\cdot\left[\frac 1{s-b}+\frac 1{s-c}+\frac 1{s-a}\right]=$](http://latex.artofproblemsolving.com/a/d/c/adc5b33c609f056eb16a94162f1ce68252894dd3.png)
. I used the well-known identity 
.PP6. Let
be a triangle. Prove that 
.Proof 1. Prove easily that
![$s\cdot \prod\left(1-\sqrt 3\cdot\tan\frac A2\right)=4\left[s-(R+r)\sqrt 3\right]$](http://latex.artofproblemsolving.com/b/b/9/bb97fcfef30ebb6bc76cfd50dbd6863349456b33.png)
.Proof 2. Prove easily that
. Therefore, 
. For 
, where
apply the identity 
. In conclusion, 
.Remarks.
The equivalence 
is false because the implication
is true and 
is false. This equivalence 
is true.PP7. Prove that
Proof (Mateescu Constantin). We assume that
, 
, 
are the angles of a triangle labelled in such a way that and are "on the same side" of 
, i.e.
or 
. In either case, we have :
. On the other hand,since
we can weaken the above inequality in order to obtain :
. Thus, it's enough to showthat :
.Taking into account the identity :
, the inequality is equivalent to :
Now,
,so in either case the inequality
holds, and the proof is complete.Remark. The proposed inequality is equivalent to the following nice form :
.PP8. Prove that
.Preliminary.
.Proof.
.![$2\blacktriangleright\ \sum\frac 1{\cos\frac {B-C}2}=\frac {4R^2}{s^2+r^2+2Rr}\cdot\left[\left(\sum\, \cos\frac {B-C}2\right)^2-1\right]-1$](http://latex.artofproblemsolving.com/d/0/5/d055413225007192e570e3bbc34f3c57094dfce1.png)
.Proof.
.Proof of the proposed problem (Mateescu Constantin).
![$\begin{array}{cccc}
\\\
\ & \left(\cos\frac {B-C}2-1\right)\left(\cos\frac {C-A}2-1\right)\left(\cos\frac {A-B}2-1\right)\, \le\, 0 \\ \\
\iff & 1+\sum\, \cos\frac {A-B}2\cos\frac {A-C}2\, \ge\, \sum\, \cos\frac {B-C}2+\prod\, \cos\frac {B-C}2 \\ \\
\iff & 1+\prod\, \cos\frac {B-C}2\cdot\sum\, \frac 1{\cos\frac {B-C}2}\, \ge\, \sum\, \cos\frac {B-C}2+\prod\, \cos\frac {B-C}2 \\ \\
\stackrel{(1)}{\iff} & 1+\frac {s^2+r^2+2Rr}{8R^2}\cdot\left(\sum\, \frac 1{\cos\frac {B-C}2}-1\right)\, \ge\, \sum\, \cos\frac {B-C}2 \\ \\
\stackrel{(2)}{\iff} & 1+\frac {s^2+r^2+2Rr}{8R^2}\cdot\left\{\frac {4R^2}{s^2+r^2+2Rr}\cdot\left[\left(\sum\, \cos\frac {B-C}2\right)^2-1\right]-2\right\}\, \ge\, \sum\, \cos\frac {B-C}2 \\ \\
\iff & 1+\frac 12\cdot\left[\left(\sum\, \cos\frac {B-C}2\right)^2-1\right]-\sum\, \cos\frac {B-C}2\, \ge\, \frac {s^2+r^2+2Rr}{4R^2} \\ \\
\iff & 2+\left(\sum\, \cos\frac {B-C}2\right)^2-1-2\sum\, \cos\frac {B-C}2\, \ge\, \frac {s^2+r^2+2Rr}{2R^2} \\ \\
\iff & \left(\sum\, \cos\frac {B-C}2-1\right)^2\, \ge\, \frac {s^2+r^2+2Rr}{2R^2} \\ \\
\iff & \sum\, \cos\frac {B-C}2\, \ge\, 1+\sqrt {\frac {s^2+r^2+2Rr}{2R^2}} \\ \\
\iff & \boxed{\ \frac {h_a}{l_a}+\frac {h_b}{l_b}+\frac {h_c}{l_c}\ \ge\ 1+\sqrt {\frac {s^2+r^2+2Rr}{2R^2}}\ } \\\
\end{array}$](http://latex.artofproblemsolving.com/b/2/a/b2a802e8ce293a353cf978912bfb7187a1f42e2d.png)
PP9
Proof (Mateescu Constantin). We will transform the given inequality (in any triangle) into one restricted to an acute angled triangle by using the following substitutions :
, 
and thus we are left to prove that: 
. For convenience,we will return to the notation of a triangle
in the last inequality and denote by 
, 
and 
its usual elements (circumradius, inradius andsemiperimeter respectively). Hence, the inequality
becomes: 
Using the well-known identities:
the last inequality is now equivalent to:
, which is actually true in anytriangle since :
, whereI made use of the known inequalities:
(Gerretsen) , 
and 
(Euler).This post has been edited 76 times. Last edited by Virgil Nicula, Nov 21, 2015, 3:56 PM
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