448. Extensions or generalizations.
by Virgil Nicula, Sep 9, 2016, 4:27 PM
Lemma 1. Let an trapezoid 
and 
so that 
Prove that 
Proof. Let
with 
so that 
and 
i.e. 
Since
get 
Lemma 2 (Cristea). Let
with 
and 
for which denote 
Prove 
Proof 1. Let
so that 
Thus, 
Proof 2. Let
with 
Thus, 
![$\frac 1{AY}\cdot [DC(XD-BD)+DB(XD+DC)]=$](//latex.artofproblemsolving.com/a/c/f/acf7f275f814dfa3604da6e3be368a79280327b1.png)
i.e. relation 
Proof 3. Let
and apply Menelaus' theorem to 
i.e. 
Lemma 3. Let with incenter 
and 
so that 
and 
Prove that 
Proof. Let
Apply Menelaus" theorem to 
So ray 
is 
-bisector in 
and ray 
is the 
-bisector
in
So 
what is true.
Extension (Francisco Javier García Capitán) :
Lemma 4. Let with interior 
for which 
and 
for which 
Prove that 
Proof 1. Apply Menelaus' theorem to
See P1 from here and its extensions.
Proof 2. Let
and apply 
P1 (own). Let with the incenter for which denote 
and for which denote 
Prove that 
Proof.
i.e. 
and 
are isogonal
Extension. Let with interior for which 
and for which let 
Prove that 
Proof 1. If has barycentric coordinates 
w.r.t. 
where 
then prove 
Apply lemma 
Proof 2.
Apply the Menelaus' theorem to 
From 
and 
obtain that 
Particular case. Let and an interior which belongs to the -bisector 
, where 
Denote
and for which 
Prove that 
P2 (Baris Altay). Let convex be with 
so that 
and 
Prove that 
Proof. Let
Apply Menelaus' theorem to 
and
P3. Let and its interior with cevian 
where 
and 
Let 
Prove that 
and ![$\frac {[EAF]}{[BAC]}=\frac {[EPF]}{[BPC]}.$](//latex.artofproblemsolving.com/8/6/d/86d017d6c130d61ea45c4182d14bb2aa35f91571.png)
Proof. Let
and 
Is well known that division 
is harmonic, i.e. 
So
i.e. the true relation 
Apply Menelaus' theorem to
![$\frac {[EAF]}{[CAB]}=\frac {[EPF]}{[CPB]}.$](//latex.artofproblemsolving.com/9/6/a/96a1c5c5ceb803273dfc4edd548faa6fcbfdb6ed.png)
Otherwise. is harmonic 
![$\frac {[EAF]}{[EPF]}=\frac {LA}{LP}=\frac {DA}{DP}=\frac {[BAC]}{[BPC]}\implies$](//latex.artofproblemsolving.com/0/f/1/0f15ac4a15ed4b930cd3cf20b1392f6efa57e098.png)
![$\frac {[EAF]}{[EPF]}=\frac {[BAC]}{[BPC]}$](//latex.artofproblemsolving.com/c/e/6/ce6bfe87e9b9eb98b6935875c38929f3604cf46e.png)
P4. Let and an interior 
Let ![$\left\{\begin{array}{ccc}
x=[BPC]\ ; & y=[EPF]\ ; & z=[EAF]\\\\
u=[PBF]\ ; & v=[PCE]\ ; & S=[BAC]\end{array}\right\|,$](//latex.artofproblemsolving.com/e/b/c/ebc74f052264f8e1d024275835efbd76fc697eae.png)
where 
and
Prove that 
and 
Proof.![$\left\{\begin{array}{ccc}
\frac {FA}{FB}=\frac {[FEA]}{[FEB]} & = & \frac z{u+y}\\\\
\frac {FA}{FB}=\frac {[FCA]}{[FCB]} & = & \frac{v+y+z}{u+x}\end{array}\right\|$](//latex.artofproblemsolving.com/2/6/5/265330bcfd83d0254903b6a70df0cb9d0bb7aba3.png)
and ![$\left\{\begin{array}{c}
\frac {PB}{PE}=\frac {[PFB]}{[PFE]}=\frac uy\\\\
\frac {PB}{PE}=\frac {[PCB]}{[PCE]}=\frac xv\end{array}\right\|$](//latex.artofproblemsolving.com/a/4/5/a4515c30896c395c4f3a67f9427afa38d0b8a343.png)
.
Example. If
and 
then find 
Answer
P5.Given scalene
and 
where internal bisectors of angles 
meet 
respectively. Let 
meet
perpendicular bisector of
at 
Let 
meet perpendicular bisector of 
at 
Let meet perpendicular bisector of 
at 
Prove that 
Proof. Let
Thus, 
is isosceles. Thus, 
a.s.o.
Generalizare (Kostas Vitas). is given and let be an arbitrary point inwardly to it, with its cevian 
On 
we
define
as the harmonic conjugate of 
w.r.t. 
and similarly we define 
and 
Prove that midpoints of ![$[DD'],$](//latex.artofproblemsolving.com/9/3/7/937a7079b1579848699fbe90e56b965aa26abe36.png)
![$[EE']\,$](//latex.artofproblemsolving.com/9/0/b/90bf6714bd6a292ab3f30d2045442ffb6825257e.png)
![$[FF']$](//latex.artofproblemsolving.com/f/c/1/fc1330a4e0dd3cfb967354db364d4feadf4f964f.png)
are collinear.
Proof 1. It is easy to define 
as the intersections of the sidelines 
respectively, from the segment lines 
as well known.
These points are collinear ( well known and easy to prove ) and now, based on Gauss’s theorem, we conclude that the midpoints of the segments
as the
diagonals of the complete quadrilateral
are collinear and the proof is completed. This problem is coming from the past, as well known.
Proof 2. I"ll use a remarkable property of the harmonic division
where 
if 
is midpoint of ![$[BD],$](//latex.artofproblemsolving.com/7/0/d/70d4a23927687c9a81b0039f6882744169a38520.png)
then 
If denote
midpoints
of 
respectively, then 
P6 (nkht-tk14). Let
be the inner angled-bisectors of where and 
Denote the incircle 
the
circumcircle
and the -excircle 
Prove that 
the distance 
the inequality 
Proof 1 (TelvCohl). Let
be the excentral triangle of 
and let 
be the midpoint of 
Let 
be the projection of 
on 
respectively and
let
be the intersection of Clearly, 
lie on a circle with diameter 
(center ) and 
is the polar of 
w.r.t. 
so 
is
perpendicular to
respectively, hence notice 
are concyclic we conclude that 
P7 (M.O. Sanchez). Let
be a triangle with the incenter and the points 
so that 
and 
is cyclically. Prove that 
Proof. Denote
i.e. 
Apply the Cristea's relation 
Since
is cyclic obtain that
Therefore, 
and 
In conclusion, from the relations and 
obtain that
Extension. Let and its interior point Consider so that 
and is cyclically. Prove that 
Proof. Suppose w.l.o.g. that has the barycentric coordinates 
where 
and denote 
where 
and 
Apply the Cristea's relation
Since is cyclic obtain 
Thus, 
and 
In conclusion, from the relations and obtain that 
Remark.
where 
P8 (Poncelet's porism). Consider two circles
satisfying 
.
Then starting from any on 
, we can draw a triangle including which is inscribed in and circumscribes 
.
Proof 1. Invert in. Then inverts to 
, with radius 
. Choose a point 
on , and construct two circles through and with radius 
. Both these circles
touch at 
, and 
, so is the midpoint of 
. Let 
meet again at 
, and let 
. Note that the
circle on diameter
passes through . Easily 
, so 
has radius and is tangent to . Inverting back, the result follows.
P9. Let an-right with the midpoint of ![$[AB].$](//latex.artofproblemsolving.com/2/6/4/264577404140cd39a4b68d919fc0a13d5815fac5.png)
Prove that 
where 
Proof 1. Let
and 
where 
the symetric 
of w.r.t. 
so that 
and 
Thus,
and 
and 
Proof 2 (trigonometric). Denote
where 
Thus, 
Extension. Let with 
and the midpoint of Prove that 
Proof. The theorem of Sines to
Apply identity 
We have the equality if and only if 
Remark. Study the particular cases
P10. Prove that
![$\boxed{\boxed{\begin{array}{lllll}\\\
& \blacktriangleleft\text{\underline{Preliminary.}}\blacktriangleright\\\\
1\blacktriangleright & \prod\, \cos\frac {B-C}2=\frac {s^2+r^2+2Rr}{8R^2} \\ \\
\text{\underline{Proof.}} & \prod\, \cos\frac {B-C}2=\prod\, \frac {(b+c)}a\sin\frac A2=\frac {(a+b+c)(ab+bc+ca)-abc}{abc}\prod\, \sin\frac A2 \implies \\ \\
\ & \prod\, \cos\frac {B-C}2=\frac {2s\cdot (s^2+r^2+4Rr)-4Rrs}{4Rrs}\cdot\frac r{4R}=\frac {s^2+r^2+2Rr}{8R^2} \\ \\ \hline\hline \\\
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 \\ \\
\text{\underline{Proof.}} & \sum\, \frac 1{\cos^2\frac A2}=\frac {s^2+(4R+r)^2}{s^2}\ ;\ \sum\, \tan\frac A2=\frac {4R+r}s\ ;\ \sum\, \sin^2\frac A2=\frac {2R-r}{2R}\ (\ast) \\ \\
\ & \implies\left(\sum\, \cos\frac {B-C}2\right)^2=\sum\, \cos^2\frac {B-C}2+2\sum\, \cos\frac {A-B}2\cos\frac {A-C}2= \\ \\
\ & =\sum\, \left(\frac s{2R}\cdot\frac 1{\cos\frac A2}-\sin\frac A2\right)^2+2\prod\, \cos\frac {B-C}2\cdot\sum\, \frac 1{\cos\frac {B-C}2}\stackrel{(\ast)\wedge (1)}{=} \\ \\
\ & =\frac {s^2}{4R^2}\cdot\frac {s^2+(4R+r)^2}{s^2}-\frac sR\cdot\frac {4R+r}s+\frac {2R-r}{2R}+\frac {s^2+r^2+2Rr}{4R^2}\cdot\sum\, \frac 1{\cos\frac {B-C}2}= \\ \\
\ & =1+\frac {s^2+r^2+2Rr}{4R^2}+\frac {s^2+r^2+2Rr}{4R^2}\cdot\sum\, \frac 1{\cos\frac {B-C}2}\ \ldots\ \implies (2) \\\
\end{array}}\ \boxed{\begin{array}{cccc}
\\\
& \blacktriangleleft\text{\underline{Proof of the proposed problem.}}\blacktriangleright\\\\
\ & \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/b/1/bb17dcfbbf2b7669541c9c52a52d11fa0ae82ec8.png)
Remark.
Here is a nice form of this inequality: if the internal angle bisectors of 
intersect the circumcircle of again at 
respectively, then it holds that 
because 
a.s.o.
P11. Let with 
, the incenter , the centroid 
, 
and 
.
Suppose w.l.o.g.
. Prove that 
.
Proof. Suppose w.l.o.g.. Denote 
so that 
and 
where 
the midpoint of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Thus,
. Apply the Menelaus' theorem to the transversal 
.
Thus,
. In conclusion, 
P12. Let with the incircle 
which touches it at 
and 
Let 
Prove that 
Proof. Are wellknown
and 
and 
From the sum of the relations and get 
P13. Let be a triangle with 
and 
is its circumcircle. Construct the point 
so that 
and
the tangents
to , where 
Denote the midpoint of the side ![$[AB]\ .$](//latex.artofproblemsolving.com/f/4/3/f43654eb147de5ab7da28700a60e256a697a2e10.png)
Prove that 
Proof. Denote
and 
Observe that ![$[SC]$](//latex.artofproblemsolving.com/7/9/a/79a04a804a1df2e5dcb0b811b4d18eefb5adfdcc.png)
is a diameter of and 
Therefore,
Remark 
The line 
is the polar of w.r.t. the circle 
is the harmonic conjugate of w.r.t. ![$[BC]\implies$](//latex.artofproblemsolving.com/5/b/0/5b03e533335663bf8d03b77549a5236ed4125235.png)
In conclusion, 
and 
is the polar of w.r.t.
is the harmonic conjugate of w.r.t. ![$[AS]\implies$](//latex.artofproblemsolving.com/6/5/d/65d50e5c18e44ef2e276111eb90ef6667f4231b4.png)
In conclusion,
Apply Menelaus' theorem to 
over 
In conclusion, 
Remark. Apply Menelaus' theorem to
over 
P14 (M. O. Sanchez). Let with 
and the points 
so that 
and 
Prove that there is the relation 
Proof. Thus,![$S=[ABC]\ \stackrel{AB=1}{\implies}\ S=\frac {\sqrt 3}4$](//latex.artofproblemsolving.com/8/1/a/81a0cc20eef46a5264bd057c7a4ad20261156c46.png)
and 
where 
Apply the generalized Pythagoras' theorem to
Observe that 
I"ll use the identity 
for any 
Particular case
P15 (Carlos Hugo Olivera DIAZ). Let an acute with 
Denote the midpoint of and the projection
of on 
the excenters 
and 
Prove that 
Proof. Let
and 
Thus, 
Particular case
In this case 
and 
so that 
Prove that 
Proof. Let
with 
so that 
and 
i.e. 
Since
get 
Lemma 2 (Cristea). Let
with 
and 
for which denote 
Prove 
Proof 1. Let
so that 
Thus, 
Proof 2. Let
with 
Thus, 
![$\frac 1{AY}\cdot [DC(XD-BD)+DB(XD+DC)]=$](http://latex.artofproblemsolving.com/a/c/f/acf7f275f814dfa3604da6e3be368a79280327b1.png)
i.e. relation 
Proof 3. Let
and apply Menelaus' theorem to 
i.e. 
Lemma 3. Let
with incenter 
and 
so that 
and 
Prove that 
Proof. Let
Apply Menelaus" theorem to 
So ray 
is 
-bisector in 
and ray 
is the 
-bisectorin
So 
what is true.Extension (Francisco Javier García Capitán) :
Lemma 4. Let
with interior 
for which 
and 
for which 
Prove that 
Proof 1. Apply Menelaus' theorem to
See P1 from here and its extensions.Proof 2. Let
and apply 
P1 (own). Let
with the incenter for which denote 
and for which denote 
Prove that 
Proof.
i.e. 
and 
are isogonal Extension. Let
with interior for which 
and for which let 
Prove that 
Proof 1. If
has barycentric coordinates 
w.r.t. 
where 
then prove 
Apply lemma 
Proof 2.
Apply the Menelaus' theorem to 
From 
and 
obtain that 
Particular case. Let
and an interior which belongs to the -bisector 
, where 
Denote
and for which 
Prove that 
P2 (Baris Altay). Let convex
be with 
so that 
and 
Prove that 
Proof. Let
Apply Menelaus' theorem to 
and
P3. Let
and its interior with cevian 
where 
and 
Let 
Prove that 
and ![$\frac {[EAF]}{[BAC]}=\frac {[EPF]}{[BPC]}.$](http://latex.artofproblemsolving.com/8/6/d/86d017d6c130d61ea45c4182d14bb2aa35f91571.png)
Proof. Let
and 
Is well known that division 
is harmonic, i.e. 
So
i.e. the true relation 
Apply Menelaus' theorem to
![$\frac {[EAF]}{[CAB]}=\frac {[EPF]}{[CPB]}.$](http://latex.artofproblemsolving.com/9/6/a/96a1c5c5ceb803273dfc4edd548faa6fcbfdb6ed.png)
Otherwise. is harmonic 
![$\frac {[EAF]}{[EPF]}=\frac {LA}{LP}=\frac {DA}{DP}=\frac {[BAC]}{[BPC]}\implies$](http://latex.artofproblemsolving.com/0/f/1/0f15ac4a15ed4b930cd3cf20b1392f6efa57e098.png)
![$\frac {[EAF]}{[EPF]}=\frac {[BAC]}{[BPC]}$](http://latex.artofproblemsolving.com/c/e/6/ce6bfe87e9b9eb98b6935875c38929f3604cf46e.png)
P4. Let
and an interior 
Let ![$\left\{\begin{array}{ccc}
x=[BPC]\ ; & y=[EPF]\ ; & z=[EAF]\\\\
u=[PBF]\ ; & v=[PCE]\ ; & S=[BAC]\end{array}\right\|,$](http://latex.artofproblemsolving.com/e/b/c/ebc74f052264f8e1d024275835efbd76fc697eae.png)
where 
and
Prove that 
and 
Proof.
![$\left\{\begin{array}{ccc}
\frac {FA}{FB}=\frac {[FEA]}{[FEB]} & = & \frac z{u+y}\\\\
\frac {FA}{FB}=\frac {[FCA]}{[FCB]} & = & \frac{v+y+z}{u+x}\end{array}\right\|$](http://latex.artofproblemsolving.com/2/6/5/265330bcfd83d0254903b6a70df0cb9d0bb7aba3.png)
and ![$\left\{\begin{array}{c}
\frac {PB}{PE}=\frac {[PFB]}{[PFE]}=\frac uy\\\\
\frac {PB}{PE}=\frac {[PCB]}{[PCE]}=\frac xv\end{array}\right\|$](http://latex.artofproblemsolving.com/a/4/5/a4515c30896c395c4f3a67f9427afa38d0b8a343.png)
.Example. If
and 
then find 
Answer
P5.Given scalene
and 
where internal bisectors of angles 
meet 
respectively. Let 
meetperpendicular bisector of
at 
Let 
meet perpendicular bisector of 
at 
Let meet perpendicular bisector of 
at 
Prove that 
Proof. Let
Thus, 
is isosceles. Thus, 
a.s.o.Generalizare (Kostas Vitas).
is given and let be an arbitrary point inwardly to it, with its cevian 
On 
wedefine
as the harmonic conjugate of 
w.r.t. 
and similarly we define 
and 
Prove that midpoints of ![$[DD'],$](http://latex.artofproblemsolving.com/9/3/7/937a7079b1579848699fbe90e56b965aa26abe36.png)
![$[EE']\,$](http://latex.artofproblemsolving.com/9/0/b/90bf6714bd6a292ab3f30d2045442ffb6825257e.png)
![$[FF']$](http://latex.artofproblemsolving.com/f/c/1/fc1330a4e0dd3cfb967354db364d4feadf4f964f.png)
are collinear.Proof 1. It is easy to define
as the intersections of the sidelines 
respectively, from the segment lines 
as well known.These points are collinear ( well known and easy to prove ) and now, based on Gauss’s theorem, we conclude that the midpoints of the segments
as thediagonals of the complete quadrilateral
are collinear and the proof is completed. This problem is coming from the past, as well known.Proof 2. I"ll use a remarkable property of the harmonic division
where 
if 
is midpoint of ![$[BD],$](http://latex.artofproblemsolving.com/7/0/d/70d4a23927687c9a81b0039f6882744169a38520.png)
then 
If denotemidpoints
of 
respectively, then 
P6 (nkht-tk14). Let
be the inner angled-bisectors of where and 
Denote the incircle 
thecircumcircle
and the -excircle 
Prove that 
the distance 
the inequality 
Proof 1 (TelvCohl). Let
be the excentral triangle of 
and let 
be the midpoint of 
Let 
be the projection of 
on 
respectively andlet
be the intersection of Clearly, 
lie on a circle with diameter 
(center ) and 
is the polar of 
w.r.t. 
so 
isperpendicular to
respectively, hence notice 
are concyclic we conclude that 
P7 (M.O. Sanchez). Let
be a triangle with the incenter and the points 
so that 
and 
is cyclically. Prove that 
Proof. Denote
i.e. 
Apply the Cristea's relation 
Since is cyclic obtain that
Therefore, 
and 
In conclusion, from the relations and 
obtain that Extension. Let
and its interior point Consider so that 
and is cyclically. Prove that 
Proof. Suppose w.l.o.g. that
has the barycentric coordinates 
where 
and denote 
where 
and 
Apply the Cristea's relation
Since is cyclic obtain 
Thus, 
and 
In conclusion, from the relations and obtain that 
Remark.
where 
P8 (Poncelet's porism). Consider two circles
satisfying 
.Then starting from any
on 
, we can draw a triangle including which is inscribed in and circumscribes 
.Proof 1. Invert in
. Then inverts to 
, with radius 
. Choose a point 
on , and construct two circles through and with radius 
. Both these circlestouch
at 
, and 
, so is the midpoint of 
. Let 
meet again at 
, and let 
. Note that thecircle on diameter
passes through . Easily 
, so 
has radius and is tangent to . Inverting back, the result follows.P9. Let an
-right with the midpoint of ![$[AB].$](http://latex.artofproblemsolving.com/2/6/4/264577404140cd39a4b68d919fc0a13d5815fac5.png)
Prove that 
where 
Proof 1. Let
and 
where 
the symetric 
of w.r.t. 
so that 
and 
Thus,
and 
and 
Proof 2 (trigonometric). Denote
where 
Thus, 
Extension. Let
with 
and the midpoint of Prove that 
Proof. The theorem of Sines to
Apply identity 
We have the equality if and only if 
Remark. Study the particular cases
P10. Prove that
![$\boxed{\boxed{\begin{array}{lllll}\\\
& \blacktriangleleft\text{\underline{Preliminary.}}\blacktriangleright\\\\
1\blacktriangleright & \prod\, \cos\frac {B-C}2=\frac {s^2+r^2+2Rr}{8R^2} \\ \\
\text{\underline{Proof.}} & \prod\, \cos\frac {B-C}2=\prod\, \frac {(b+c)}a\sin\frac A2=\frac {(a+b+c)(ab+bc+ca)-abc}{abc}\prod\, \sin\frac A2 \implies \\ \\
\ & \prod\, \cos\frac {B-C}2=\frac {2s\cdot (s^2+r^2+4Rr)-4Rrs}{4Rrs}\cdot\frac r{4R}=\frac {s^2+r^2+2Rr}{8R^2} \\ \\ \hline\hline \\\
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 \\ \\
\text{\underline{Proof.}} & \sum\, \frac 1{\cos^2\frac A2}=\frac {s^2+(4R+r)^2}{s^2}\ ;\ \sum\, \tan\frac A2=\frac {4R+r}s\ ;\ \sum\, \sin^2\frac A2=\frac {2R-r}{2R}\ (\ast) \\ \\
\ & \implies\left(\sum\, \cos\frac {B-C}2\right)^2=\sum\, \cos^2\frac {B-C}2+2\sum\, \cos\frac {A-B}2\cos\frac {A-C}2= \\ \\
\ & =\sum\, \left(\frac s{2R}\cdot\frac 1{\cos\frac A2}-\sin\frac A2\right)^2+2\prod\, \cos\frac {B-C}2\cdot\sum\, \frac 1{\cos\frac {B-C}2}\stackrel{(\ast)\wedge (1)}{=} \\ \\
\ & =\frac {s^2}{4R^2}\cdot\frac {s^2+(4R+r)^2}{s^2}-\frac sR\cdot\frac {4R+r}s+\frac {2R-r}{2R}+\frac {s^2+r^2+2Rr}{4R^2}\cdot\sum\, \frac 1{\cos\frac {B-C}2}= \\ \\
\ & =1+\frac {s^2+r^2+2Rr}{4R^2}+\frac {s^2+r^2+2Rr}{4R^2}\cdot\sum\, \frac 1{\cos\frac {B-C}2}\ \ldots\ \implies (2) \\\
\end{array}}\ \boxed{\begin{array}{cccc}
\\\
& \blacktriangleleft\text{\underline{Proof of the proposed problem.}}\blacktriangleright\\\\
\ & \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/b/1/bb17dcfbbf2b7669541c9c52a52d11fa0ae82ec8.png)
Remark.
Here is a nice form of this inequality: if the internal angle bisectors of 
intersect the circumcircle of
again at 
respectively, then it holds that 
because 
a.s.o.P11. Let
with 
, the incenter , the centroid 
, 
and 
.Suppose w.l.o.g.
. Prove that 
.Proof. Suppose w.l.o.g.
. Denote 
so that 
and 
where 
the midpoint of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Thus,
. Apply the Menelaus' theorem to the transversal 
.Thus,
. In conclusion, 
P12. Let
with the incircle 
which touches it at 
and 
Let 
Prove that 
Proof. Are wellknown
and 
and 
From the sum of the relations
and get 
P13. Let
be a triangle with 
and 
is its circumcircle. Construct the point 
so that 
andthe tangents
to , where 
Denote the midpoint of the side ![$[AB]\ .$](http://latex.artofproblemsolving.com/f/4/3/f43654eb147de5ab7da28700a60e256a697a2e10.png)
Prove that 
Proof. Denote
and 
Observe that ![$[SC]$](http://latex.artofproblemsolving.com/7/9/a/79a04a804a1df2e5dcb0b811b4d18eefb5adfdcc.png)
is a diameter of and 
Therefore,
Remark 
The line 
is the polar of w.r.t. the circle 
is the harmonic conjugate of w.r.t. ![$[BC]\implies$](http://latex.artofproblemsolving.com/5/b/0/5b03e533335663bf8d03b77549a5236ed4125235.png)
In conclusion, 
and 
is the polar of w.r.t. 
is the harmonic conjugate of w.r.t. ![$[AS]\implies$](http://latex.artofproblemsolving.com/6/5/d/65d50e5c18e44ef2e276111eb90ef6667f4231b4.png)
In conclusion,
Apply Menelaus' theorem to 
over 
In conclusion, 
Remark. Apply Menelaus' theorem to
over 
P14 (M. O. Sanchez). Let
with 
and the points 
so that 
and 
Prove that there is the relation 
Proof. Thus,
![$S=[ABC]\ \stackrel{AB=1}{\implies}\ S=\frac {\sqrt 3}4$](http://latex.artofproblemsolving.com/8/1/a/81a0cc20eef46a5264bd057c7a4ad20261156c46.png)
and 
where 
Apply the generalized Pythagoras' theorem to 
Observe that 
I"ll use the identity 
for any 
Particular case
P15 (Carlos Hugo Olivera DIAZ). Let an acute
with 
Denote the midpoint of and the projection of on 
the excenters 
and 
Prove that 
Proof. Let
and 
Thus, 
Particular case
In this case 
This post has been edited 495 times. Last edited by Virgil Nicula, Mar 1, 2019, 1:45 PM
August 2018
January 2018