462. Diverse probleme de geometrie.
by Virgil Nicula, Aug 3, 2018, 8:20 AM
P1. Let 
with the circumcircle 
and with the incircle 
. Define the points 
. Prove that 
.
Proof. Denote the power
of 
w.r.t. the circle 
. Thus, 
.
The relations
a.s.o. are well-known. Apply the Stewart's theorem to the rays 
in 
respectively :
. Hence 
, i.e. .
P2 (IMO 2010). Let an interior
of . Lines 
meet again the circumcircle 
at 
respectively. The tangent at 
to meets 
at 
. Show that 
.
Proof. Consider the circle which is tangent to in and to 
in . Denote 
, where 
. From the well-known property obtain 
,
i.e.
. Since 
obtain 
(the line is tangent to the circumcircle of 
) , i.e. 
.
In conclusion,
, 
, i.e. 
. Nice problem !
P3 (Balcan geo 2003). Let and the tangent to the circumcircle of at 
meet 
at 
. The perpendicular to at 
meets the perpendicular
bisector of
at 
. The perpendicular to at 
meets the perpendicular bisector of 
at 
. Prove that the points 
. (Valentin Vornicu).
Proof (Virgil NICULA).
.
P4 (OME, 2009). Let
be an acute triangle with the incircle 
and the circumcircle 
.
Denote
for which 
and 
. Prove that 
.
Proof. From the identity
for 
obtain 
.
Therefore,
. In conclusion, 
, i.e. 
.
Extension (Virgil NICULA). Let an acute with the incircle 
and the circumcircle . Let
so that , 
and 
, 
. Prove that 
.
Particular cases.
what are evidently. Thus we
verified the upper identity ! Also we can find easily the points
for which 
, i.e. 
.
Proof. Let
for which 
. Apply the generalized Pythagoras' in 
: 
![$ \left[sx - (s - a)h_a\right]\left(x - h_a\right) + h_a(s - a)\left(2R - h_a\right) =$](//latex.artofproblemsolving.com/7/9/e/79eda64de53c040d2adf3e07dc1a231e7f8e83ea.png)
. Since 
obtain 
.
P5 (USA TST 2006). Let be a triangle. Triangles 
and 
are constructed outside of triangle such that 
and 
. Segments 
and 
meet at 
. Let 
be the circumcenter of triangle 
. Prove that 
Proof. I"ll show that
, i.e. 
. Let 
, the circumcircle 
of 
and 
. Thus, 
the quadrilaterals 
, 
are cyclically with the radical axis 
. Observe that 
the
ray
is the bisector of the angle 
. Prove easily that 
. Thus, 
and 
. Apply the generalized Pythagoras'
to
, 
.
I used the following simple identities
and 
.
P6. For and 
construct the parallelogram 
. Let 
.
Prove that the circumcenter of
belongs to the circumcircle of 
.
Proof.
P7 (France TST 2007). Let such that 
. Let ![$D\in [AC]$](//latex.artofproblemsolving.com/c/5/a/c5acd77baad8d15d98170c2ac60f572eeeed5507.png)
such that 
. The incircle 
of
touches its sides in![$K\in [AB]$](//latex.artofproblemsolving.com/f/7/c/f7c55ea4f0642b6ee4395cbe8c000c6fec53e3ef.png)
and ![$L\in [AC]$](//latex.artofproblemsolving.com/c/4/5/c45e50a5a8063a6607d6afd834b15cea829eb243.png)
. Let 
be the incircle of 
. Denote 
. Prove that 
.
Proof.. The incircle
touches in 
, 
and 
. Observe that 
. Thus,
. Since 
obtain that 
. Denote 
. The property 
is well-known (see the lower remark). Thus,
is cyclically. Analogously 
, i.e. 
. Observe that 
. Apply the Menelaus' theorem to transversal 
in 
: 
.
So 
the quad. 
is cyclically 
or 
P8. Let
be a triangle with 
. Let 
bisect 
and let 
bisect 
, where 
and 
. If 
, then what are the angles of the triangle ?
Proof I (metric). Let
. Thus, 
. Thus, 
, 
because .
Proof II (trigonometric) Denote
. I"ll apply the Sinus' theorem in the triangles 
, 
: 
![$ \sin (60+x)\left[\sin (30+2x)+\frac 12\right]=\sin (30+2x)(\sin x+\sin 60)$](//latex.artofproblemsolving.com/f/9/0/f9014e97ba223e408eef58d3d00ae7e6856a69a3.png)
![$ [\cos (x-30)-\cos (90-2x)]+$](//latex.artofproblemsolving.com/6/c/f/6cfc307118123ab15122148835ecb7deb5cf7715.png)
![$[\sin 3x+\sin (60-3x)]=$](//latex.artofproblemsolving.com/9/4/f/94fe11441f683123ef54d217847bfd74ccc07ca4.png)
![$ [\cos (x+30)-\cos (x-30)]+\cos (2x-30)$](//latex.artofproblemsolving.com/b/0/5/b05b68ab2f09bc2aa220f250038619fa84e1d0e7.png)
![$2\sin\left(60-\frac {3x}{2}\right)\left[\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)\right]$](//latex.artofproblemsolving.com/d/6/7/d6787a30786e7e73494cf66e267856c792fca8d8.png)
.
Thus,
or 
.
Extension (Virgil NICULA). Let an acute with the incircle and the circumcircle . Let
so that , and , . Prove that .
with the circumcircle 
and with the incircle 
. Define the points 
. Prove that 
.Proof. Denote the power
of 
w.r.t. the circle 
. Thus, 
.The relations
a.s.o. are well-known. Apply the Stewart's theorem to the rays 
in 
respectively :
. Hence 
, i.e. .P2 (IMO 2010). Let an interior
of . Lines 
meet again the circumcircle 
at 
respectively. The tangent at 
to meets 
at 
. Show that 
.Proof. Consider the circle which is tangent to
in and to 
in . Denote 
, where 
. From the well-known property obtain 
,i.e.
. Since 
obtain 
(the line is tangent to the circumcircle of 
) , i.e. 
.In conclusion,
, 
, i.e. 
. Nice problem !P3 (Balcan geo 2003). Let
and the tangent to the circumcircle of at 
meet 
at 
. The perpendicular to at 
meets the perpendicularbisector of
at 
. The perpendicular to at 
meets the perpendicular bisector of 
at 
. Prove that the points 
. (Valentin Vornicu).Proof (Virgil NICULA).
.P4 (OME, 2009). Let
be an acute triangle with the incircle 
and the circumcircle 
.Denote
for which 
and 
. Prove that 
.Proof. From the identity
for 
obtain 
.Therefore,
. In conclusion, 
, i.e. 
.Extension (Virgil NICULA). Let an acute
with the incircle 
and the circumcircle . Let so that
, 
and 
, 
. Prove that 
.Particular cases.
what are evidently. Thus weverified the upper identity ! Also we can find easily the points
for which 
, i.e. 
.Proof. Let
for which 
. Apply the generalized Pythagoras' in 
: 
![$ \left[sx - (s - a)h_a\right]\left(x - h_a\right) + h_a(s - a)\left(2R - h_a\right) =$](http://latex.artofproblemsolving.com/7/9/e/79eda64de53c040d2adf3e07dc1a231e7f8e83ea.png)
. Since 
obtain 
.P5 (USA TST 2006). Let
be a triangle. Triangles 
and 
are constructed outside of triangle such that 
and 
. Segments 
and 
meet at 
. Let 
be the circumcenter of triangle 
. Prove that 
Proof. I"ll show that
, i.e. 
. Let 
, the circumcircle 
of 
and 
. Thus, 
the quadrilaterals 
, 
are cyclically with the radical axis 
. Observe that 
theray
is the bisector of the angle 
. Prove easily that 
. Thus, 
and 
. Apply the generalized Pythagoras'to
, 
.I used the following simple identities
and 
.P6. For
and 
construct the parallelogram 
. Let 
.Prove that the circumcenter of
belongs to the circumcircle of 
.Proof.
P7 (France TST 2007). Let
such that 
. Let ![$D\in [AC]$](http://latex.artofproblemsolving.com/c/5/a/c5acd77baad8d15d98170c2ac60f572eeeed5507.png)
such that 
. The incircle 
of touches its sides in
![$K\in [AB]$](http://latex.artofproblemsolving.com/f/7/c/f7c55ea4f0642b6ee4395cbe8c000c6fec53e3ef.png)
and ![$L\in [AC]$](http://latex.artofproblemsolving.com/c/4/5/c45e50a5a8063a6607d6afd834b15cea829eb243.png)
. Let 
be the incircle of 
. Denote 
. Prove that 
.Proof.. The incircle
touches in 
, 
and 
. Observe that 
. Thus,
. Since 
obtain that 
. Denote 
. The property 
is well-known (see the lower remark). Thus,
is cyclically. Analogously 
, i.e. 
. Observe that 
. Apply the Menelaus' theorem to transversal 
in 
: 
.So
the quad. 
is cyclically 
or 
P8. Let
be a triangle with 
. Let 
bisect 
and let 
bisect 
, where 
and 
. If 
, then what are the angles of the triangle ?Proof I (metric). Let
. Thus, 
. Thus, 
, 
because .Proof II (trigonometric) Denote
. I"ll apply the Sinus' theorem in the triangles 
, 
: 
![$ \sin (60+x)\left[\sin (30+2x)+\frac 12\right]=\sin (30+2x)(\sin x+\sin 60)$](http://latex.artofproblemsolving.com/f/9/0/f9014e97ba223e408eef58d3d00ae7e6856a69a3.png)
![$ [\cos (x-30)-\cos (90-2x)]+$](http://latex.artofproblemsolving.com/6/c/f/6cfc307118123ab15122148835ecb7deb5cf7715.png)
![$[\sin 3x+\sin (60-3x)]=$](http://latex.artofproblemsolving.com/9/4/f/94fe11441f683123ef54d217847bfd74ccc07ca4.png)
![$ [\cos (x+30)-\cos (x-30)]+\cos (2x-30)$](http://latex.artofproblemsolving.com/b/0/5/b05b68ab2f09bc2aa220f250038619fa84e1d0e7.png)
![$2\sin\left(60-\frac {3x}{2}\right)\left[\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)\right]$](http://latex.artofproblemsolving.com/d/6/7/d6787a30786e7e73494cf66e267856c792fca8d8.png)
.Thus,
or 
.Extension (Virgil NICULA). Let an acute
with the incircle and the circumcircle . Let so that
, and , . Prove that .This post has been edited 134 times. Last edited by Virgil Nicula, Mar 28, 2019, 1:43 PM
![$\boxed{\ F=\frac 1{\sqrt[3]4+\sqrt [3]{10}+\sqrt[3]{25}}\ }$](http://latex.artofproblemsolving.com/c/2/6/c26fe3f2df13221941ec084d95c7e4d971393aa2.png)
(clasa a VII - a).![$\sqrt[3]2=a$](http://latex.artofproblemsolving.com/e/e/0/ee027c0cafaeda75a6c476843aefd5be0282dc4c.png)
si ![$\sqrt [3]5=b\ ,$](http://latex.artofproblemsolving.com/7/b/4/7b4d7e346587b74adc48c2859ad915ec879ed76a.png)
atunci fractia 
devine 
adica 
![$\frac {b-a}{b^3-a^3}=\frac{\sqrt [3]5-\sqrt [3]2}{5-2}\implies$](http://latex.artofproblemsolving.com/d/6/a/d6a6b215bcae70aa641f5662b506a55d4a50fe12.png)
![$\boxed{\ F=\frac{\sqrt [3]5-\sqrt [3]2}3\ }\ .$](http://latex.artofproblemsolving.com/2/a/2/2a29f7e07a605fc730d87fea4bcad46a63a55f17.png)
(standard notations).
i.e. 
and 
i.e. 
Observe that 
From 
obtain that 
Remain to prove 
Indeed, from the relation 
obtain that 
i.e. the true relation 
![$4\cos A[\cos (B-C)-\cos A]=1\iff$](http://latex.artofproblemsolving.com/b/c/e/bcea818b361410c7a961034941cd3955509e9aca.png)
![$[2\cos A-\cos (B-C)]^2+\sin^2(B-C)=0\iff$](http://latex.artofproblemsolving.com/a/5/c/a5ced1ab893e0290d435a64f927dea84271dec56.png)
and 
i.e. 
and 
and I"ll use the well known identities 
Thus, 
Hence 
is equivalent with the inequality 
![$\sum a^2(s - a)^2 = 2r^2\left[\left(4R + r\right)^2 - s^2\right]$](http://latex.artofproblemsolving.com/e/b/e/ebe8cb16e10d2978d1b538cc640432589c9aa758.png)
starting from the development of ![$\left[\sum a(s-a)\right]^2 = \ldots$](http://latex.artofproblemsolving.com/6/3/6/6365dc8e28d3a4fac36ff0ee1f8c56f6436bfb7b.png)
The inequality becomes ![$2r^2\left[(4R+r)^2-s^2\right]\le 2Rs^2r\iff$](http://latex.artofproblemsolving.com/7/5/5/755b6fd0176637590612904d0cfa469e7a2f0601.png)
![$r\left[(4R+r)^2-s^2\right]\le Rs^2\iff$](http://latex.artofproblemsolving.com/9/b/e/9be35f474a516a395fbad4528b907d944384c58b.png)
Is well known 
and observe that 
Hence 
In conclusion, our inequality 
is true. Very nice!
Thus, 
Appear two cases 
Observe that
or 
i.e. 
obtain 
i.e. 
for 
obtain 
The lines 
and 
meet again 
in 
and 
![$\frac Rr\cdot\left[3-\left(1+\frac rR\right)\right]=$](http://latex.artofproblemsolving.com/d/e/8/de82e2264b4a3bf0fdac9af3994552c4f297ffcb.png)
because 
what is true. In conclusion, 
I used the well known relations 
and the power of 
w.r.t. 
meet again 
Apply the Stewart's relation to the cevian 
where 
and 
![$\sum\left[b^2(s-b)+c^2(s-c)\right]\ \stackrel{(1\wedge 2)}{\iff}\sum a\cdot AD^2+2sr(2R-r)=$](http://latex.artofproblemsolving.com/6/5/e/65e91e3885dbd2efb2e3ea65a3d4dd1050733eae.png)
![$\sum a\cdot AD^2=2sr\left[4(R+r)-(2R-r)\right]=2sr(2R+5r)\implies$](http://latex.artofproblemsolving.com/d/8/0/d80cb7d130960d1c7c30f864b2d80daa5818fc1a.png)
Apply the well known inequalities 
and 
In conclusion, 
![$1.\blacktriangleright\ \sum a(s-b)(s-c)=\sum[s(s-b)(s-c)-(s-a)(s-b)(s-c)]=sr(4R+r)-3sr^2=$](http://latex.artofproblemsolving.com/b/0/a/b0a56ae071d42c4887535458d775624b7acd3b74.png)
![$sr[(4R+r)-3r]=sr(4R-2r)=\boxed{\ 2sr(2R-r)\ }\ .$](http://latex.artofproblemsolving.com/2/a/6/2a6ce536e921305d169f828804ae3790a9cca33b.png)
![$\sum \left[x\left(y^2+z^2\right)+2xyz\right]=$](http://latex.artofproblemsolving.com/1/a/5/1a5ebeabcd1a2eca9fa4a993f77cdda4ac707006.png)
![$\sum [yz(x+y+z)-xyz+2xyz]=$](http://latex.artofproblemsolving.com/e/3/4/e34a94bc05e0aaaf07acbc2addcca9174a971bb4.png)
and 
Hence 
and 
![$2s\left(s^2-r^2-4Rr\right)\left[(a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc\right]=$](http://latex.artofproblemsolving.com/5/9/9/599bc2a78a8b83ce993469ca1c8a748563c9b1e7.png)
I used the notations and the identities 
and 
there is the followung inequality 
Therefore, the required inequality 
what is true.
(standard notations).
i.e. 
Therefore, 
what is true.
and 
Hence 
Remark. 
of its sides ![$([BC],[CA],[AB])$](http://latex.artofproblemsolving.com/f/2/5/f25000944ae05d06f52ffa17d5fad2fb6a8a6329.png)
respectively. Prove that 
is orthodiagonal 
Hence 
so that 
and 
Prove that
Therefore, 
is orthodiagonal 
![$\frac{c^2\left(n-1\right)}{(n+1)}+\frac{b^2\left(m-1\right)}{(m+1)}+b^2+c^2=2bc\cdot\cos A\cdot\left[1+\frac{mn}{(m+1)(n+1)}\right]\implies$](http://latex.artofproblemsolving.com/3/b/d/3bd04cb1b235e3a4a54832e56f76e793c9144981.png)
In the particular case 
obtain that 
i.e. the proposed problem P11.
there is the following inequality ![$:\ \boxed{\ \left[I_aBC\right]+\left[I_bCA\right]+\left[I_cAB\right]\ge 3[ABC]\ }\ .$](http://latex.artofproblemsolving.com/2/b/1/2b15c0ced9f7875f61311f0feee323b7c61e4b39.png)
Therefore, ![$\sum\left[I_aBC\right]=\frac 12\cdot\sum ar_a\ \stackrel{(*)}{=}\ \frac 12\cdot \sum s\left(r_a-r\right)=$](http://latex.artofproblemsolving.com/f/6/3/f63187d5d5c9abfdcfe703c1544c9c56a86a8673.png)
![$\boxed{\ \sum\left[I_aBC\right]=s(2R-r)\ }\ (1)\ .$](http://latex.artofproblemsolving.com/d/a/d/dadadd87779e099c4464368567c52d89c86026ba.png)
Observe that ![$\sum\left[I_aBC\right]\ge 3S\ \stackrel{(1)}{\iff}\ \cancel s(2R-r)\ge 3\cancel sr\iff$](http://latex.artofproblemsolving.com/e/4/8/e4846a8894b4c062b3eb3bae9e807f8785a79bc8.png)
an acute 
the following relationship holds 
(standard notations).
of ![$[BC]\ ,$](http://latex.artofproblemsolving.com/6/5/6/6563ca77d31eb676ef646fa9c37ea181b812d864.png)
the circumcenter 
Observe that 
Thus, 
Hence 
![$\boxed{\ 2\sqrt[3]{2x-1}=x^3+1\ }\ (*)\ .$](http://latex.artofproblemsolving.com/c/c/7/cc74fff7f3af27592235b14bd15b11ba35d1d958.png)
Se observa ca prin diferenta celor doua relatii se obtine 
(standard notations).
Therefore, 
Let ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and the circumcircle 
Thus,
Otherwise. 
(standard notations).
and the identity 
Hence 
In conclusion, 
Find the solutions (zeroes) of the following equations ![$\ :\ \left[\frac {x+2}3\right]=\frac {x-1}4\ ;\ \left[\frac {x+1}3\right]=\frac {x-1}2\ ;\ \left[\frac {4x+1}{5x+3}\right]=\frac {x}{x+8}\ .$](http://latex.artofproblemsolving.com/9/1/3/9136a0123e9c48648a1d40850892c0afa32b7272.png)
Let 
so that 
and 
Prove that 
Find 
where 
are the roots of the equation 
![$1.1\blacktriangleright=\frac {x-1}4=\left[\frac {x+2}3\right]=z\iff$](http://latex.artofproblemsolving.com/8/d/1/8d1debc595fa94843cfceb09642353f051a6ab42.png)
and 
![$1.2\blacktriangleright\ \frac {x-1}2=\left[\frac {x+1}3\right]=z\in\mathbb Z\implies$](http://latex.artofproblemsolving.com/9/9/5/9952e44b4e527f77f9bd04614409f9d42334d9d2.png)
and 
![$z\in (-1,2]\cap \mathbb Z\iff$](http://latex.artofproblemsolving.com/2/3/8/2383fb0c15c4d8408b24f12e2bcc8114f389f777.png)
![$1.3\blacktriangleright\ \frac x{x+8}=\left[\frac {4x+1}{5x+3}\right]=z\in \mathbb Z\iff$](http://latex.artofproblemsolving.com/c/9/e/c9ef90f8fdcaf118e4ab7941b68eaea7117ff9c4.png)
and 
![$\left(z-\frac {31z+1}{37z+3}\right)\left[(z+1)-\frac {31z+1}{37z+3}\right]\le 0$](http://latex.artofproblemsolving.com/d/2/0/d2079c76fb2d1cfb8fb93667de0967004a7bffff.png)
and 
and 
and 
there is the following inequality 
Thus, 
![$\boxed{\ \sum a^2=2\left[s^2-r^2(4R+r)^2\right]\ }\ ;$](http://latex.artofproblemsolving.com/1/1/7/11757e7e0a5c7f7104272455d1f0890ed0dc32cd.png)
Hence, 
![$\cancel 4s^2\left(\cancel {s^2}-3r^2\right)\ge \cancel 4\left[\cancel {s^4}-r^2(4R+r)^2\right]\iff$](http://latex.artofproblemsolving.com/2/6/1/261dd5bceff878c50188044cec4573de1b9cfc00.png)
what is true. Indeed, 
![$\frac16\sum \left[(a-c)^2+b(a+c-2b)\right]\ge 0\iff$](http://latex.artofproblemsolving.com/1/e/0/1e07968709e0d5f50f4dc16ea74381cc43dff8f0.png)
which is fourth degree Schur inequality and holds for all reals.
and the midpoints 
of the
respectively. Prove that the area of the convex hexagon 
is equally to 
where 
![$\left(\ [BC]\ ,\ [CA]\ ,\ [AB]\ \right)$](http://latex.artofproblemsolving.com/7/c/d/7cd9ff27a312b766ff4c9b4e0b209e4e29f27d26.png)
respectively. Prove easily that 
a.s.o. and 
are the exradii of 
Thus, ![$[APBMCN]=S+\sum [BMC]=S+\frac 12\cdot \sum (BC\cdot XM)\ \stackrel{(1)}{=}\ sr+\sum \frac {a\left(r_a-r\right)}4\ \stackrel{(*)}{=}$](http://latex.artofproblemsolving.com/7/3/b/73b19c2d498b2505f602219340a50850bccb0195.png)
![$sr+\frac 14\cdot\left[\sum s\left(r_a-r\right)-2sr\right]=$](http://latex.artofproblemsolving.com/7/4/9/7495341befe9b7d577e3f8ac246ee2f196abcb74.png)
![$sr+\frac s4\cdot\left[\sum \left(r_a-r\right)-2r\right]=$](http://latex.artofproblemsolving.com/a/0/8/a08407dbc3f3315900e5398208a5015dc34485ab.png)
![$\boxed{\ [APBMCN]=sR\ }\ .$](http://latex.artofproblemsolving.com/9/a/1/9a1bba20fbf8dec65825f7cf7b28cbca07bdc60f.png)
and apply the Menelaus' theorem for the transversal 
to the triangles 
and 
respectively: 
From the relations 
and 
There is generally the identity 
.
be the foot of the altitude from 
and 
the circumcircle
be the circle tangent to 
,
and 
be the circle tangent to 
and 
be the interior common
if and only if 
.
be a circle which is interior tangent in the point 
,
denote the intersections
between the circle 
is the bisector of the angle 
.
of the triangle 
are interior tangent to the circle 
.
be a square with the circumcircle 
Prove that 
i.e. 
and apply Ptolemy's theorem to cyclic quadrilaterals 
where 
Thus, 
and 
Prove that 
where 
so that 
there is the identity 
there is the following equivalence 
(standard notations).
![$2\cos A[\cos A-\cos (B-C)]=0\iff$](http://latex.artofproblemsolving.com/3/4/9/3492d2c266ec527753dd2c0b9d7c043af17e6782.png)
![$\cos A[\cos (B+C)+\cos (B-C)]=0\iff$](http://latex.artofproblemsolving.com/5/d/8/5d8afde1a62ca8f11a0f2d8ded2f56140ed8f851.png)
Hence 
a.s.o. Thus, 
Thus, 
i.e. the equivalence 
there is the equivalence 
(standard notations).
Therefore, 
Remark. Prove that 
with equality 
Thus, 
Remark. Prove that 
, where 
(standard notations).
theorem of Sines 
and remarkable inequalities 
Hence 
![$\frac s{2R}\cdot \left[(4R+r)-3r\right]=$](http://latex.artofproblemsolving.com/f/d/3/fd3af3fbc4d33fc98945d7b9f547c7544ea1be53.png)
Hence 
In conclusion, 
(standard notations).
Therefore, 
![$4R^2+2R\left[\left(r_a+r_b+r_c\right)-r\right]=4R^2+2R\left[\left(4R+\cancel r\right)-\cancel r\right]=4R^2+8R^2=12R^2\ ,$](http://latex.artofproblemsolving.com/f/5/8/f583d728b19274e1be85740f4bf56f0257448a62.png)
i.e. 
i.e. 
are directly proportional to 
Hence 
i.e. 
are directly proportional to 
Thus, 
Hence 
(standard notations).
and 
Indeed, 
Hence 
In conclusion, 
Remark. 
and 
i.e. 
Hence 
i.e. 
In conclusion, 
there is the inequality 
(standard notations).![$\boxed{\cos A+\cos B+\cos C=1+\frac rR\ }\ (1)\ ,\ \boxed{\ a^2+b^2+c^2=2\left[s^2-r(4R+r)\right]\ }\ (2)$](http://latex.artofproblemsolving.com/4/7/3/473e340ba8c8e0bd1f063ee3d65410495bbf6b74.png)
and the inequalities 
and 
Thus, 
and 
i.e. 
But 
because is true.
respectively (standard notations).![$[BG]\ ,$](http://latex.artofproblemsolving.com/9/0/6/9067a26a1b520d41aae9874a77ee7b28e31d4ed5.png)
![$[CG]$](http://latex.artofproblemsolving.com/a/4/c/a4c498243ffacd1cd9abbf11f51cc2b074d05b93.png)
are the diameters of the cyclical quadrilaterals 
respectively 
Therefore,
The sum of the relations 
and 
(standard notations).
and 
Thus, ![$\frac {b+c}a\in \left(1,\sqrt 2\right]\iff$](http://latex.artofproblemsolving.com/1/b/8/1b8adf60674cc7cd3a674f736590b53fa868cfce.png)
![$1+\frac {b+c}a\in \left(2,1+\sqrt 2\right]\ \stackrel{(*)}{\iff}\ \frac {h_a}r\in \left(2,1+\sqrt 2\right]\iff$](http://latex.artofproblemsolving.com/e/4/8/e484dd2bb40bd0e816d159586ab74300d9490106.png)
(standard notations).
and the remarkable bilateral inequality 
(standard notations).
Prove easily
a.s.o. and the identity 
Thus, 
a.s.o. Therefore, 
and 
Prove that 
(standard notations).
i.e. 
and 
Apply the theorem of Sines 
Denote 
The relation 
midpoint ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and 
Prove that 
and 
Observe that 
and 
i.e. 
Apply the Menelaus' theorem to
From the relations 
and 
obtain easily the required relation. 
Prove that 
is cyclically (standard notations).
. Observe that 
, 
and 
. Apply the theorem of Sines in 
. Apply the theorem of Cosines
and 
. Observe that 
. In conclusion, 
. Let 
. Since 
and 
are cyclically obtain that 
a.s.o. 
Prove easily that 
(Euler's inequality), what is true. Hence and the inequality 
obtain that 
i.e. 
what is true. In conclusion, and inequality 
is true.
what is the Blundon's inequality. In conclusion, and the inequality 
is true.
![$\frac 1{sr}\cdot \left[\cancel{2s^2}-2\left(\cancel{s^2}-r^2-4Rr\right)\right]=$](http://latex.artofproblemsolving.com/d/9/1/d91aadd0b00cd951bd73d3e0534567a3eb098301.png)
Prove easily that 
what is true.
is true. From the well known inequality 
i.e. 
Hence can apply the Chebyshev's inequality 
![$\blacktriangleright\ \frac 13\cdot\sum\frac a{r_a}\ge\sqrt [3]{\prod\frac a{r_a}}=\sqrt [3]{\frac {abc\cdot r}{r_ar_br_c\cdot r}}=\sqrt [3]{\frac {4R\cancel Sr}{S^{\cancel 2}}}=\sqrt [3]{\frac {4R}s}\ \stackrel{2s\le 3R\sqrt 3}{\ge}\ \sqrt [3]{\frac{4\cdot 2}{3\sqrt 3}}=\sqrt[3]{\left(\frac 2{\sqrt 3}\right)^3}=\frac 2{\sqrt 3}\implies\sum\frac {a}{r_a}\ge 2\sqrt{3}\ .$](http://latex.artofproblemsolving.com/9/a/c/9ac9461f2ed8db135d7ef2f0d13b389aabf245f7.png)
(standard notations).
and ![$\sum (s-b)(s-c)\le \left[\sum (s-a)\right]^2\iff\boxed{\ 3r(4R+r)\le s^2\ }\ (2)\ .$](http://latex.artofproblemsolving.com/1/c/a/1ca0a5b2c01abb426a54cb155ab038159d27192c.png)
Therefore 
![$\sum \left[s+\frac {(s-b)(s-c)}{s-a}\right]=$](http://latex.artofproblemsolving.com/b/e/b/beb692c70a858dd2eac2f4c9fd9e458618500ccc.png)
![$[3r+(4R+r)]\sqrt 3=$](http://latex.artofproblemsolving.com/9/0/6/9060ba107ad32abb4ecfc13f89c18a3531927b33.png)
On other hand 
(standard notations).
Hence 
and 
and the midpoint ![$[AB]\ .$](http://latex.artofproblemsolving.com/f/4/3/f43654eb147de5ab7da28700a60e256a697a2e10.png)
Suppose that 
Prove that 
Therefore, 
Since 
and 
obtain that 
and two mobile points 
so that 
Ascertain the minimum value of the segment ![$[MN]\ .$](http://latex.artofproblemsolving.com/8/f/3/8f33a629708367aee0e2e495cd6641b16ceacdae.png)
![$AM=x\in (0,c]\ ,$](http://latex.artofproblemsolving.com/e/5/b/e5b6f3021cbe3212c46b1b69ac9ed4ab5a9b77ea.png)
i.e. 
and ![$AN=y\in (0,b]\ ,$](http://latex.artofproblemsolving.com/8/1/0/810e4e25cfc1aea6398883d70a195c0a6a9280ed.png)
i.e. 
Hence 
Apply the
and 
where 
and the incircle 
Denote 
Prove that the implication 
is cyclic 
where 
(golden ratio).
and 
Denote 
where 
I"ll apply the power of the points 
Thus, 
![$2(s-a)(b+c)=a^2\implies (b+c)[(b+c)-a]=a^2$](http://latex.artofproblemsolving.com/6/8/5/6850385d3cdd52ee2673f7cdddd9e672e4cc6383.png)
In conclusion, 
where 
i.e. 
Let 
of ![$[AD]\ ,$](http://latex.artofproblemsolving.com/2/6/7/26748bb1d79775245b0f0bbbd36d03d1342c4737.png)
i.e. 
the incircle 
of 
and the incircle 
Apply the well known identity 
i.e. 
In conclusion, 
is the smallest angle in the triangle 
be an interior point of the arc between 
at 
and 
,
and 
meet at 
.
.
are chosen on a circle 
such that the triangle 
is not right-angled. Prove that:
at certain 
respectively, and that 
and 
meet at a certain 
and 
is the sum of the lengths of the other two.
and 
![$TB+TC=\frac{R}{\cos A}\cdot [\sin (B-x)+\sin (C-y)]=$](http://latex.artofproblemsolving.com/3/3/8/338215a42c3ac53520634bcd874623b038c39b57.png)
class.
Remark. 
a.s.o.
where denote the midpoint ![$[AB]\ ,$](http://latex.artofproblemsolving.com/b/a/3/ba35a1c584b9f3786aaed51acf1cbc01ce2968a4.png)
the midpoint ![$[CM]$](http://latex.artofproblemsolving.com/f/e/2/fe2250ed20133e86505c3901649c4007cf77e6e7.png)
and 
where 
Find the ratio 
-
Construct an 
so that 
and 
Prove that 
and 
and 
Observe that 
and 
Apply the Stewart's
![$[CE]/\triangle ABC\ :\ CE^2=$](http://latex.artofproblemsolving.com/9/b/d/9bd28af92173b8c28084940d346296996d4cf5fe.png)
From the relations 
obtain that 
Remark. 
so that 
and 
Prove that 
i.e. 
there is the inequality 
Apply the Chebyshev's inequality 
i.e. 
and its incircle 
and
Prove that 
find the value of the angle 
if we know that 
Applly Stewart's relation to the cevians 
Therefore, 
In conclusion, 
and 
Prove easily that
Remark. 
i.e. 
is cyclic a.s.o. Is well-known that 
i.e. 
Can use 
if 
so that 
i.e. ![$\log (abc)=\log a+\log b+\log c=k[(b-c)+(c-a)+(a-b)]=0$](http://latex.artofproblemsolving.com/b/f/e/bfe880c4cd9ef6519a60cf8be543de702f92f57e.png)
so that 
Prove 
and 
are parallelograms, where 
is the orthocenter of 
![$\frac {[BDC]}{[BAC]}=$](http://latex.artofproblemsolving.com/0/3/b/03bc1123af5725f1c7a01c4c213819e67366348e.png)
![$\frac {[BHC]}{[BAC]}=$](http://latex.artofproblemsolving.com/2/7/2/2723edebb87068ce0d8baba103da7f54ff6da14f.png)
![$\frac {[BHC]}S=$](http://latex.artofproblemsolving.com/f/8/0/f80f20e810d666d5fe150b3d4696f08815039bbd.png)
![$\frac 1S\cdot [BHC]\implies$](http://latex.artofproblemsolving.com/a/b/4/ab41ea0c9ec3f8b2ab360c95f42191a388958e35.png)
![$\frac 1S\cdot\sum [BHC]=$](http://latex.artofproblemsolving.com/7/3/6/736970a53831282e5ebc49a9c56b49596f674c07.png)
Remark. Can apply the well-known property from
and 
Prove that 
what is the orthocenter of 
Observe that 
and apply the theorem of SINES in the triangles 
and 
so that 
and 
Prove easily 
is 
![$[UV]$](http://latex.artofproblemsolving.com/a/1/3/a136f90fcd9d8d96d81566cab1c035ac43940789.png)
in 
In conclusion, 
a.s.o.
(standard notations).
and 
Denote 
and the equivalence 
and 
Hence 
(standard notation). Remark. 
![$\frac 1{sr}\cdot \left[2s\left(s^2-r^2-4Rr\right)-2s\left(s^2-3r^2-6Rr\right)\right]=$](http://latex.artofproblemsolving.com/6/c/2/6c24cda6c55b88ec40a967a80570af98f864c084.png)
![$\frac 2r\cdot\left[\left(\cancel{s^2}-r^2-4Rr\right)-\left(\cancel{s^2}-3r^2-6Rr\right)\right]=$](http://latex.artofproblemsolving.com/8/c/4/8c4145ecc8002d431e5ece1e330870cf3be752df.png)
Remark. Observe that ![$\sum\frac {a^2}{6r_a}=\frac{2(R+r)}3\in [2r,R]\ .$](http://latex.artofproblemsolving.com/a/e/c/aec5afba6e8ee6d6c8271be52300e1ab057eb58b.png)
![$6xyz+\sum yz[(x+y+z)-x]=$](http://latex.artofproblemsolving.com/8/8/6/8861ee8285ffbcb0155820775961447f39b474b1.png)
and 
Prove that 
where 
![$bc\left(r^2+rr_a\right)+r\left[b^2(s-c)+c^2(s-b)\right]=$](http://latex.artofproblemsolving.com/5/d/d/5ddf7ea2507d57e435fa63b7fd8f772fd37f577a.png)
![$r\left[bc\left(r+r_a\right)+b^2(\underline s-c)+c^2(\underline s-b)\right]=$](http://latex.artofproblemsolving.com/7/9/7/797d58794018d5bce14e554ba62305d64f9b2abe.png)
![$r\left[\cancel{bc\left(b+c\right)}+sa^2-\cancel{bc(b+c)}\right]=$](http://latex.artofproblemsolving.com/2/e/9/2e958c4d02394915aa88d95b68fea3e427faf84e.png)
its circumcircle 
Prove that 
and 
Thus, 
a.s.o. In conclusion, 
i.e. 
Thus, 
and the remarkable property PP1 from
and 
Thus, 
![$3+\sum\frac {[BHC]}{[BAC]}=4\implies$](http://latex.artofproblemsolving.com/1/3/a/13a9b599821a490f0cf446f1efacdbd818d07004.png)
(standard notations).
I"ll use an well-known identities 
where 
Therefore,
i.e. 
where 
is the circumcircle of 
Thus, 
![$\frac 1{Rr}\cdot\left[\cancel{2s^2}-2\left(\cancel{s^2}-r^2-4Rr\right)\right]=$](http://latex.artofproblemsolving.com/9/2/c/92c561bfac0a1c109cf2433d60950a884384da83.png)
Indeed, 
On other hand 
In conclusion, 
Hence 
![$2\cos A\left[\cos (B+C)+\cos (B-C)\right]=$](http://latex.artofproblemsolving.com/5/e/c/5ec9443af5fc54beec4b93b407742a6b1b5d3047.png)
![$2\cos A[\cos (B-C)-\cos A]=$](http://latex.artofproblemsolving.com/3/a/e/3ae4fd1663000723a4e0895535e1a91401ba04ef.png)
![$-2\cos^2A-2\cos (B+C)\cos (B-C)]=$](http://latex.artofproblemsolving.com/5/d/9/5d9557dce473fb92d5f64d91f8b289324583275a.png)
where 
I"ll apply the distancies between two remarkable points from here: 
![$2\cdot 9\cdot\left[R^2-\frac {2\left(s^2-r^2-4Rr\right)}9\right]\iff$](http://latex.artofproblemsolving.com/8/d/3/8d39e1146bbfd2c6f3662cd45f8062b31c5c9129.png)
![$[BI_aC]=\frac 12\cdot BC\cdot I_aE=\frac {ar_a}2\iff$](http://latex.artofproblemsolving.com/8/1/a/81ab4f374c39de3de0bdb994bf1f0251534a1ae3.png)
![$\boxed{\ [BI_aC]=\frac {ar_a}2\ }\ (1)\ .$](http://latex.artofproblemsolving.com/d/c/4/dc44fc449ea076417b5b8cd819a4076294147533.png)
Find analogously ![$[CI_bA]=\frac {br_b}2$](http://latex.artofproblemsolving.com/c/e/b/cebe6224856795d064ae0f777950f310f08e720f.png)
and ![$[AI_cB]=\frac {cr_c}2\ .$](http://latex.artofproblemsolving.com/4/d/9/4d94923dbfce8e6ff6dfa7e02edd5fee55ba84fb.png)
On other hand ![$[DEF]=[DI_aE]+[FI_aE]-[DI_aF]=$](http://latex.artofproblemsolving.com/f/6/a/f6a63be56b4aaefd02be272f9bd48b15eb42819b.png)
![$\boxed{\ [DEF]=\frac {r_a^2(s-a)}{2R}\ }\ (2)\ .$](http://latex.artofproblemsolving.com/8/5/9/85910502ae6c7a45f3810def8e9454c0384c8c9f.png)
Find analogously ![$[LKM]=\frac {r_b^2(s-b)}{2R}$](http://latex.artofproblemsolving.com/e/7/d/e7dbceab8fc2fee2bc696d212a03a443e1059f4b.png)
and ![$[PNR]=\frac {r_c^2(s-c)}{2R}\ .$](http://latex.artofproblemsolving.com/7/d/f/7dfb6df725cd213f58bccd9912b67bb2f7428207.png)
Therefore,![$\frac {[BI_aC]}{[DEF]}\ \stackrel{(1\wedge 2)}{=}\ \frac {a\cancel{r_a}}{\cancel 2}\cdot\frac {\cancel 2R}{r_a^{\cancel 2}(s-a)}=\frac {aR}{r_a(s-a)}=\frac {aR}S\implies$](http://latex.artofproblemsolving.com/c/9/b/c9bccce5852ec0b05e1c23afc1d7950ec6ec3a25.png)
![$\boxed{\ \frac {[BI_aC]}{[DEF]}=\frac {aR}S\ }\ (3)\ .$](http://latex.artofproblemsolving.com/6/6/1/661cd91830a3b6be729c72caaa2a920289be16e9.png)
In conclusion, ![$\sum\frac {[BI_aC]}{[DEF]}\ \stackrel{(3)}{=}\ \sum\frac {aR}S=\frac RS\cdot \sum a=\frac R{sr}\cdot 2s=\frac {2R}r\implies$](http://latex.artofproblemsolving.com/b/4/7/b4736cab77f6855b648955bcc08dc3be2498fdd1.png)
![$\sum\frac {[BI_aC]}{[DEF]}=\frac {2R}r\ .$](http://latex.artofproblemsolving.com/2/c/4/2c4d796974f8b984f62e33283433807587c00196.png)
Observe that 
for any 
the points 
in this order and the equilateral triangles 
and 
so that the line
doesn't separate the points 
and 
Denote 
and 
Prove that 
and the quadrilateral 
is circumscriptible, i.e. 
where 
and apply theorem of Sines in 
i.e. 
is equilateral and the line 
Observe that 
and 
Thus, 
![$\left(b^2+a^2+2ba\right)+\left(\cancel{b^2}+c^2+2bc\right)-\left[\cancel{b^2}+b(a+c)+ac\right]=$](http://latex.artofproblemsolving.com/4/1/1/411b01870d205afae9b6ab8b901da72c5f83ba15.png)
Thus, 
what is true. Very nice problem!
and 
so that 
and 
Prove that 
The theorem of Sines in 
and 
Hence 
Define 
(Dũng Nguyễn Tiến's proof).![$[CA]\ ,$](http://latex.artofproblemsolving.com/7/9/0/79053f4e90a202762296b3a7020733fab1b27fe1.png)
![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
at 
Prove that 
(standard notations).
I used the well known inequality 
Thus,
a.s.o. 
Prove easily 
using only the well known inequality 
![$[DE]\ .$](http://latex.artofproblemsolving.com/1/d/a/1da2706ed712f1a6100a01df004875c2de6a8679.png)
Observe that 
Apply generalized Phytagoras' theorem in 
a.s.o. and the well known inequality 
a.s.o. Thus, 
and 
Remain to prove 
Indeed, 
![$\underline{\underline{s^2\le \left[2R+\left(3\sqrt 3-4\right)r\right]^2}}\ .$](http://latex.artofproblemsolving.com/4/7/d/47d7da1a4b7f80505437625df8b203faff3f2ed7.png)
I"ll apply now the well known inequality 
Remain to prove only![$4R^2+4Rr+3r^2\le \left[2R+\left(3\sqrt 3-4\right)r\right]^2\ ,$](http://latex.artofproblemsolving.com/3/c/a/3ca62e969160f6235594561b6cfd34967ab55f3e.png)
i.e. 
what is true (Gabi Cuc Cucoanes' proof).
and the required relation 
is equivalent with the relation
The generalized Phytagoras' theorem in 
Using the Ptolemy's theorem to the quadrilateral 
obtain that 
Hence the relation 
i.e. 
what is the true relation 
.
a.s.o. Here is 
and 
where 
Prove that 
Prove that 
exist the following
and 
a.s.o.![$\blacktriangleright\ \left\{\begin{array}{ccccc}
\sum\limits_{\mathrm{cyc}}a^2(b+c) & = & \sum\limits_{\mathrm{cyc}}a^2[(a+b+c)-a] & = & s_1S_2-S_3\\\\
\sum\limits_{\mathrm{cyc}}bc(b+c) & = & \sum\limits_{\mathrm{cyc}}bc[(a+b+c)-a] & = & s_1s_2-3s_3\\\\
\sum\limits_{\mathrm{cyc}}a\left(b^2+c^2\right) & = & \sum\limits_{\mathrm{cyc}}a\left[\left(b^2+c^2+a^2\right)-a^2\right] & = & s_1S_2-S_3\end{array}\right\|\
\mathrm{\underline{{and}}}\ \left\{\begin{array}{ccccc}
\sum\limits_{\mathrm{cyc}}a^3(b+c) & = & \sum\limits_{\mathrm{cyc}}a^3[(a+b+c)-a] & = & s_1S_3-S_4\\\\
\sum\limits_{\mathrm{cyc}}a\left(b^3+c^3\right) & = & \sum\limits_{\mathrm{cyc}}a[(a^3+b^3+c^3)-a^3] & = & s_1S_3-S_4\\\\
\sum\limits_{\mathrm{cyc}}bc\left(b^2+c^2\right) & = & \sum\limits_{\mathrm{cyc}}bc\left[\left(b^2+c^2+a^2\right)-a^2\right] & = & s_2S_2-s_1s_3\end{array}\right\|\ .$](http://latex.artofproblemsolving.com/2/c/f/2cfcbe95cd152352126b9ff1c85f40c46624933b.png)
![$\sum\limits_{\mathrm{cyc}}a^2[(ab+ac+bc)-bc]=$](http://latex.artofproblemsolving.com/3/f/b/3fb227763602b68b1b1dbd3c315f51c2f01421a9.png)
![$\sum\limits_{\mathrm{cyc}}a[(ab+ac+bc)-bc]=$](http://latex.artofproblemsolving.com/e/3/e/e3e00f39178369eb29ddd901695842da555c4602.png)
![$\sum\limits_{\mathrm{cyc}}a\left(b^4+c^4\right)=s_1S_4-S_5\
;\ \sum\limits_{\mathrm{cyc}}a^4(b+c)=\sum\limits_{\mathrm{cyc}}a^3[(ab+ac+bc)-bc]=$](http://latex.artofproblemsolving.com/a/2/5/a259e2ba40fda4f22177c3052d26633aa35979ec.png)
-
-
Altfel: 
etc.
If 
prove that 
is constant for a given triangle.
and the projection 
and 
are cyclic with the diameters ![$[BC]\ .$](http://latex.artofproblemsolving.com/5/f/a/5fad78e281930919485d791e012363fda8c76507.png)
Hence 
i.e. 
is equilateral 
Apply generalized Pythagoras' theorem in 
with 
(constant).
and denote 
where 
Thus, 
and
what touches given triangle at 
and 
Denote 
Prove that 
and denote 
and 
Apply the Menelaus' theorem to the collinear points 
Observe that 
Prove easily that 
Therefore, 
In conclusion, 
and the diameter ![$[AS]$](http://latex.artofproblemsolving.com/8/1/f/81f842d87b22553e4b674bb2bf46a080da785009.png)
of the circumcircle 
Observe that 
and 
so that 
Let 
Prove that ![$\frac 1{[FBE]}+\frac 1{[FCE]}=\frac 2{[FDE]}\ .$](http://latex.artofproblemsolving.com/d/3/0/d30f249ebf72ed3eac4dff437d1f279db6f1441c.png)
of 
on 
and denote 
and 
The relation 
is well known in the trapezoid 
Since 
are proportional with 
then the relation 
i.e. 
But 
are proportional with ![$\left([FBE],[FCE],[FDE]\right)$](http://latex.artofproblemsolving.com/a/0/1/a01cca34845e1a882e9d34cd9b44fadef75f6791.png)
and let 
so that 
Thus, 
and 
so that 
Prove that 
Observe that ![$B+D=\phi +\left[180^{\circ}-\left(\alpha +\beta\right)\right]=$](http://latex.artofproblemsolving.com/6/c/0/6c0bcb0eac3d939489e4829439867c2b094c0a07.png)
and 
Apply the
i.e. 
Thus, 
are proportional with 
Nice problem!
and 
are well-known. Hence ![$\sum a^2\le 2\left[\left(4R^2+\cancel{4Rr}+3r^2\right)-r^2-\cancel{4Rr}\right]=$](http://latex.artofproblemsolving.com/c/c/2/cc2739c1a6b19ff27cb0b565fb1cccfe9c6c8dfe.png)
what is true. In conclusion and the required inequality 
what is true only on the first quadrant because the function 
is convex: 
and the circle 
Prove that 
there is the relation 
and 
and 
Observe that 
and 
Apply the theorem of median in the triangles 
i.e. the required relation 
Particular cases 
Hence 
![$\sqrt 3-\left[\sin(60^{\circ}-2x)+\sin (60^{\circ}-2y)\right]=$](http://latex.artofproblemsolving.com/b/4/3/b43f0279f8f30137f2ea2c2e4948a1462e4d2d86.png)
![$\sqrt 3-2\sin\left[60^{\circ}-(x+y)\right]\cos(x-y)\ \stackrel{x+y=60^{\circ}}{=}\ \underline{\underline{\sqrt 3}}\ .$](http://latex.artofproblemsolving.com/8/e/f/8ef279f1b26bcc843651be7406cdd01db8c0063d.png)
where 
Thus, 
Hence in this case 
![$\sum [(1+x)(1-y)(1-z)]=\prod (1+x)\iff$](http://latex.artofproblemsolving.com/3/9/2/3927d4eef93da7e9dd448779e7b284a5b489b3fc.png)
![$\sum[1-y-\cancel z+\cancel {yz}+x(\cancel 1-\cancel y-z+yz)]=\prod (1+x)\iff$](http://latex.artofproblemsolving.com/f/5/9/f59cecf85a264fee2e101e4b96c29bc1bb4692ee.png)
![$\sum[1-y+x(-z+yz)]=\prod (1+x)\iff$](http://latex.artofproblemsolving.com/c/e/e/cee336088438bace1eb813c61dc562858cd6be0c.png)
Hence 
what is the true relation 
de forma 
stiind ca 
Frumoasa problema pentru clasa a IV - a !
Cifra 
ca prima cifra de la stanga a lui 
Apar cazurile 
si 
fiind de doua cifre trebuie ca 
Asadar, solutia 
de numere naturale 
s
mai putin cele 
numere care au cel putin o cifra egala cu zero. In concluzie, 
Suppose against all reason 
i.e. 
what is false.
be the quadratic equation for what 
Find the value of 
Therefore, 
where 
and 
Find the image 
i.e. 
of 
Observe that 
Therefore, 
In conclusion, 
and 
and 
so that 
so that 
is the 
Prove ![$[AEDF]=[BDH]$](http://latex.artofproblemsolving.com/9/2/c/92c129e50ea4ce55c20966e33f02258293549a3f.png)
![$S=[ABC]$](http://latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
and apply the Menelaus' theorem to the transversal 
![$\frac {[AEB]}{[ACB]}=\frac {FE}{FC}=\frac {c^2}{a(a+b)}=\frac {a-b}a\implies$](http://latex.artofproblemsolving.com/4/b/c/4bc568ba161850108e561b8a2be686d3f92ca4ce.png)
![$\boxed{\ [AEB]=\frac {a-b}a\cdot S\ }\ (1)\ .$](http://latex.artofproblemsolving.com/f/f/d/ffd463de4016650cf4a03edd93b09caf8289eae4.png)
Observe that ![$\frac {[BFH]}{[ABC]}=\frac {BF}{BA}\cdot \frac {BH}{BC}=\frac {\cancel a}{a+b}\cdot \frac {c^2}{a\cancel{^2}}=\frac {c^2}{a(a+b)}=\frac {a-b}a\implies$](http://latex.artofproblemsolving.com/2/9/b/29bcd35003718c6284b2431b7d3df89268456317.png)
![$\boxed{\ [BFH]=\frac {a-b}a\cdot S\ }\ (2)\ .$](http://latex.artofproblemsolving.com/3/6/5/365e27022af59364c662ec27346ac8df98e7a594.png)
![$[AEB]=[BFH]\iff$](http://latex.artofproblemsolving.com/c/9/f/c9f478fadcab9df36b12acb1872589d90158c9ab.png)
![$[AEB]-[BDF]=[BFH]-[BDF]\iff$](http://latex.artofproblemsolving.com/8/3/4/83473f1537f461a0db09d5e3fc50cf2e292fa92b.png)
![$[AEDF]=[BDH]\ .$](http://latex.artofproblemsolving.com/3/d/f/3df8f2c23aec085716457566dab0870d36b67993.png)
the circumcenter 
and 
Hence 
![$\left[r^2+(s-a)^2\right]-\left[r^2+\frac {(b-c)^2}4\right]=$](http://latex.artofproblemsolving.com/1/2/e/12ee454a4e48899ce3c6a0fbe7cc4ece4798fcad.png)
Very nice problem !
with the incircle 
the Nagel's point 
(standard notations).
and apply the Van Aubel's relation for the point 
Apply the
i.e. 
Prove easily that 
Indeed 
and a point 
are the barycentrical coordinates of 
Hence 
with the barycentrical coordinates w.r.t.
where 
Prove that there is the following identity 
Prove that 
![$R^2-\left[\frac {a^2}4-\frac {(b-c)^2}4\right]=$](http://latex.artofproblemsolving.com/c/9/6/c96e598d1a126b9785f399d57e6ff8090397774b.png)
![$(bc-4Rr)-r^2=\cancel{R^2}-\left[\cancel{R^2}-(s-b)(s-c)\right]\iff$](http://latex.artofproblemsolving.com/a/4/9/a495e595bc43d47020cb10b62185c824ab81b063.png)
In conclusion, 
and its interior point 
w.r.t. 
where 
of the sides ![$[AC],[AB]$](http://latex.artofproblemsolving.com/b/c/1/bc1910f00c43994a0b4711c712cab791d59d05bd.png)
respectively. Prove that 
(standard notations).![$\boxed{\ \mathrm{Prove\ that}\ \forall\ \triangle ABC\ \mathrm{there\ is\ the\ identity}\ :\ ar_a+br_b+cr_c=2\left(\left[I_aI_bI_c\right]-\left[ABC\right]\right)\ \mathrm{(standard\ notations)}\ .}$](http://latex.artofproblemsolving.com/c/3/1/c31fe95e7d1abab58f5a346a9f121e48f89106da.png)
![$[ABC]=sr=(s-a)r_a\implies$](http://latex.artofproblemsolving.com/e/4/4/e44600a7eb26a382818dc66d73de817f3dde1987.png)
Hence ![$\left[I_aI_bI_c\right]=\sum\left([BIC]+\left[BI_aC\right]\right)=[ABC]+\frac 12\cdot \sum ar_a\implies$](http://latex.artofproblemsolving.com/f/5/1/f511da419cc86ad690ccc423ab3bcfc77ae17017.png)
![$2\left(\left[I_aI_bI_c\right]-[ABC]\right)\ \stackrel{(*)}{=}\ \sum ar_a\ .$](http://latex.artofproblemsolving.com/7/9/f/79f8b13142f3cd7ff52df0c097f02eea047dddd3.png)
![$\frac {a(s-b)(s-c)}{S[(s-c)+(s-b)]}=$](http://latex.artofproblemsolving.com/0/1/b/01b96566cd99976cfb0a90ea571cccc99104aa5a.png)
and 
where 
, 
the points on 
, 
and denote by 
Circle 
at two points, the closer of which to the vertex 
. Therefore, 
. Apply Menelaus' theorem to 
. From the relations 
.
Let 
.Prove easily that 
and 
. Hence 
, i.e. 
.
(standard notations).
Apply the theorem of median in the triangles
I"ll prove that the required bilateral inequality is equivalently with the Gerretsen's inequality:
Indeed: 
the bilateral inequality 
a.s.o. Thus 
i.e. the relation ![$[AN]$](http://latex.artofproblemsolving.com/b/0/6/b065e2d64ee016911f4b23fe8c308311c71bfa54.png)
of the circumcircle 
is a parallelogram, i.e. 
and 
a.s.o. In conclusion, 
i.e. the relation 
for the centroid 
i.e. the relation 
, i.e. 
(standard notations).
. Thus, the relation 
. I used the well-known identity 
. Since 
get easily the relation 
. Indeed, with remarkable relations 
get 
. In conclusion, 
.
, where 
. Apply the CBS inequality 
. Observe that 
.
.
. Thus, 
.
.
, where 
. Thus, our inequality becomes 
, what is truly.
![$\sum\frac {a^2}{a(b+c-a)}\ \stackrel{(CBS)}{\ge}\ \frac {(a+b+c)^2}{\sum [a(b+c-a)]}=$](http://latex.artofproblemsolving.com/f/c/0/fc0203cdaca6931f848a9515f0af1f8d47f4ce4a.png)
because 
is well-known.
because 
.
. In conclusion, we obtained the stronger inequality 
. 
.![$\sum\frac {(b+c)^2}{-a+b+c}\ \stackrel{(CBS)}{\ge}\ \frac {\left[\sum (b+c)\right]^2}{\sum (-a+b+c)}=$](http://latex.artofproblemsolving.com/0/9/6/09646d1fd1b0fcd0fa1e9487e0627c3e3397a6a1.png)
. Denote 
.
![$\sum\left[2x+2(y+z)+\frac {(y+z)^2}{2x}\right]\implies$](http://latex.artofproblemsolving.com/8/c/8/8c88c52d2021cea45cc0678f9d137748e609601e.png)
, where 
a.s.o.
because 
.
.
. Our inequality is equivalently with
.
. From the Schur's inequality for 
obtain that
.
what results from 
and 
.
. See
.
obtain that 
with the equality 
.
![$ \left[\prod\left(b^2+c^2\right)\right]^2 \ge \left[abc\prod (b+c)\right]^2\iff$](http://latex.artofproblemsolving.com/0/8/d/08df06d35201489d8513fa24ea06235af6ee1f64.png)
.
.
.
.
and 
.
.
.![$\blacktriangleright\ \sum \left[a(\cos B+\cos C)\right]\ge\frac {2r}R\cdot\sum\sqrt {\frac {a^3}{b+c-a}}\iff$](http://latex.artofproblemsolving.com/9/d/c/9dc559b6b321b4737525a674a43e9d621d49e1bb.png)
.
![$[(\underline{b+c-a})+\underline a]^2\ge 4\underline a(\underline {b+c-a})$](http://latex.artofproblemsolving.com/3/b/e/3be16d2c271b60db84ab31a81e4c1004f252a9da.png)
,
,
.
. Prove that 
.
which results from the well-known inequality 
.
so that the equation 
has at least a zero in 
.
. So, 
. Also, 
.
(Its also obvious from the original equation itself). Also, we get solutions (easy to check) at both the boundary values of 
. So, ![$m\in\left[0,\frac{1}{\sqrt{2}}\right]$](http://latex.artofproblemsolving.com/9/7/f/97f20b06005f43a632d3d3e18b307344d7bea217.png)
.
.
![$ \frac {1}{16S^2}\cdot\frac 13\cdot\left[\sum\left(b^2 + c^2 - a^2\right)\right]^2 = \left(\frac {a^2 + b^2 + c^2}{4S\sqrt 3}\right)^2\ge 1$](http://latex.artofproblemsolving.com/d/3/2/d32aa98a7fa97a9655d8fc6ceb57f1ab24f43baf.png)
.
and 
![$\left[\frac {(a+b)+(c+d)}2\right]^2\cdot \frac {\sum a}4=$](http://latex.artofproblemsolving.com/4/7/c/47c5ecca8cece6ac8cb403db74a0ec82f554fe6d.png)
Remark. 
Adunam 
stanga/dreapta si folosim 
Inmultim cu 
si 
Insa 
Inmultim cu 
si folosim 
Simplificam prin 
si stim ca 
![$\left(b^2+c^2\right)\left[a^2-(b-c)^2\right]\le 2a^2bc$](http://latex.artofproblemsolving.com/7/b/7/7b713bbdcc01db7cee4ef488dcc06ad5caa4a469.png)
![$a^2\left[\left(b^2+c^2\right)-2bc\right]\le\left(b^2+c^2\right)(b-c)^2$](http://latex.artofproblemsolving.com/6/1/f/61feb45586797476f75d95fdea44e8dddd9b366a.png)
deoarece 
si 
Procedam asemanator si pentru 
Folosim 
Asadar, 
Inlocuind pe 
si ridicand la patrat, inegalitatea de demonstrat devine 
ceea ce este adevarat. Egalitatea are loc daca si numai daca 
Asadar, 
Altfel. 
Astfel, 
Insa 
Asadar 
insa mai slaba decat cea propusa deoarece 
Indicatie "tare" 
![$2[p^2-R(R-2r)]\ \le\boxed{\sum\frac {a(a^2+2bc)}{b+c}\ \le\ \frac {4(R+r)^2(2R-r)}{R}\ \le\ 9R(2R-r)}\ .$](http://latex.artofproblemsolving.com/8/e/8/8e8ae8b9f11edcbf2126678eb6e73da3ed1c27e6.png)
![$\sum\ \frac{a[(b+c)^2-2bc\cos A]}{b+c}=$](http://latex.artofproblemsolving.com/6/7/5/675d081ccf27cf3a2af1180b16291aefad0fd2b1.png)
Asadar, 
Intr-adevar, aplicand inegalitatea C.B.S. si folosind faptul ca
si 
obtinem 
Revenind in 
In continuare aplicam inegalitatea Gerretsen 
(notatii standard).![$\boxed{H\Gamma^2=4R^2\left[1-\frac{2s^2(2R-r)}{R(4R+r)^2}\right]\ge 0}\ (**)\ ,$](http://latex.artofproblemsolving.com/d/2/d/d2d4b67eb833aad72ea15c638a9da3112ffa1411.png)
unde 
Asadar, 
![$4Rr\cdot \left[\frac {r(4R+r)}{sr^2}+\frac {s^2+r(4R+r)}{4Rrs}\right]=$](http://latex.artofproblemsolving.com/c/b/3/cb3bf87fc4035103f5fced0c7345a07dcd37ce4c.png)
In concluzie, inegalitatea propusa este echivalenta cu inegalitatea 
^2\iff$](http://latex.artofproblemsolving.com/4/7/d/47d549273abe981032368e5a4d136eb5761667d6.png)
, adica ineg. adevarata 
, care este cunoscuta a fi adevarata.
folosind identitatea de tip Leibniz 
unde 
in raport cu 
Se stie ca 
Se arata usor relatia
Asadar, 
Din relatia
![$H\Gamma^2=4R^2-\frac{4s^2}{(4R+r)^2}\cdot[(4R+r)(R-r)+r(R+r)]=$](http://latex.artofproblemsolving.com/e/2/7/e275948075ae34e03b1db91fa91c1a957b0f19a5.png)
![$H\Gamma^2=4R^2\left[1-\frac{2s^2(2R-r)}{R(4R+r)^2}\right]\ .$](http://latex.artofproblemsolving.com/f/5/6/f56935bcbe7dcd3ae4caea7287053ebaa7f73b14.png)
and the circumcircle 
Prove that 
with equality iff 
with equality iff 
Apply the Stewart's relation to the cevian 
cu egalitate iff 
But 
The last inequality is equivalent with 
so that 
Prove that 
and find the maximum of the sum 
![$\boxed{\ \{a,b,c,d\}\subset [1,4]\ }\ .$](http://latex.artofproblemsolving.com/5/f/e/5feea9c9590b47eca825f78d29277c23baefc0ce.png)
![$\boxed{x_k\in [a,b]\ ,\ k\in\overline {1,n}\implies n^2\le \sum x_k\cdot\sum\frac {1}{x_k}\le \frac {(a+b)^2}{4ab}\cdot n^2}$](http://latex.artofproblemsolving.com/5/b/1/5b1953acfb36fd59adc3ee41357a236461833de5.png)
and obtain that ![$\sum \frac 1a\in \left[\frac 85,\frac 52 \right]\ .$](http://latex.artofproblemsolving.com/2/b/0/2b0e1dd745da79e50c50ae538fc03fe8bd4d4bed.png)
si 
(notatii standard).
what is true.
so that 
Prove that 
so that 
and 
where 
Observe
Equality occurs for 
then 
![$\{a,b\}\subset\left[0,\frac{\pi}4\right]\iff$](http://latex.artofproblemsolving.com/3/b/7/3b733ec202547c7e1ca1cf9b5712d150f1ba5e0b.png)
if 
or 
then 
or 
or 
Prove that 
Thus, 
what is true.
and denote 
Thus, 
and 
Denote 
Observe that 
i.e. 
for any 
so that 
Prove that ![$[MNP]\le \frac S4\ ,$](http://latex.artofproblemsolving.com/e/f/1/ef16ce173346bc54d44009227a7e17c5eacc0cd7.png)
where ![$S=[ABC]\ .$](http://latex.artofproblemsolving.com/4/4/2/4425fc906cee3e28fbf07d7426f8e85fef08bc4c.png)
i.e. 
where 
Thus, 
From the previous PP27 (the second implication) obtain 
Prove easily that ![$\frac {[MNP]}{[ABC]} =$](http://latex.artofproblemsolving.com/f/0/1/f010545e08eaae43818d829593645f9b2d2bfc46.png)
![$1-\frac {[PAN]}{[BAC]}-\frac {[MBP]}{[CBA]}-\frac {[NCM]}{[ACB]}=$](http://latex.artofproblemsolving.com/1/1/8/1183df5b5ca8436075a704957fa105db9e039618.png)
i.e. ![$[MNP]\le\frac S4\ .$](http://latex.artofproblemsolving.com/1/1/6/116f74eafc62d37c102fd16cb9d4e545fe970f99.png)
). I don't think there
whose normalised barycentric
and 
respectively, is given by the following relation ![$:\ \boxed{ {PP^{\prime}}^2 = - \left[(\beta - \beta^{\prime}) (\gamma - \gamma^{\prime}) a^2 + (\gamma - \gamma^{\prime}) (\alpha - \alpha^{\prime}) b^2 + (\alpha - \alpha^{\prime}) (\beta - \beta^{\prime}) c^2\right] }\ .$](http://latex.artofproblemsolving.com/9/5/5/9557e21c31c5c31fdfa3e36b8b0072331cd6f883.png)
and 
w.r.t. 
Now let 
and 
Using the above corollary we succesively obtain that
![$\frac {a^3 + b^3 + c^3}{ab + bc + ca} \le \sum \frac {-\left[\beta\gamma a^2 + \gamma\left(\alpha - 1\right) b^2 + \left(\alpha - 1\right)\beta c^2\right]}a\iff $](http://latex.artofproblemsolving.com/f/8/7/f8718495e9783e6686af93f0453459b8d0ef2fcf.png)
is the isotomic of the incenter 
![$\frac {[GIK]}{[ABC]}=\left|\begin{array}{ccc}
\frac 13 & \frac 13 & \frac 13\\\\
\frac {a}{\sum a} & \frac {b}{\sum a} & \frac {c}{\sum a}\\\\
\frac {a^2}{\sum a^2} & \frac {b^2}{\sum a^2} & \frac {c^2}{\sum a^2}\end{array}\right|=$](http://latex.artofproblemsolving.com/c/2/8/c281c8bd86d3f1cf1c89afa0c5f7cafbdd06177a.png)
![$\boxed{\ \frac {[GIK]}{[ABC]}=\frac {|(c-a)(c-b)(b-a)|}{3(a+b+c)\left(a^2+b^2+c^2\right)}\ }\ .$](http://latex.artofproblemsolving.com/a/b/b/abbd533aecc62a541b1703e6e2d3e517d1a4689d.png)
is a polynomial with the real coefficients and 
has only real roots 
Find 
depending on 
and 
obtain 
![$\left[\left(m^2-2n\right)^3+3\left(m^3-3mn+3p\right)^2+m^2\left(m^4-4m^2n+4mp+2n^2\right)\right]=$](http://latex.artofproblemsolving.com/b/3/e/b3e00b4e3255d74403e455df50d9c058b1c2f889.png)
Observe that 
where 
and 
Find the polynominal ![$P\in \mathbb R[X]\ ,\ \mathrm{gr}(P)=2$](http://latex.artofproblemsolving.com/8/2/d/82dfce7fd6012b946f54c5d8c7d77f63a1553f7e.png)
such that 
where 
![$P(X)=(m+n+p)X^2-[m(b+c)+n(c+a)+p(a+b)]X+mbc+nac+pab\implies$](http://latex.artofproblemsolving.com/f/6/5/f6577b41e3491881baa0583cbb096032b14cf648.png)
See
and from upper problem P1 get 
Denote 
Therefore 
and 
satisfying the relation 
Find the value 
where 
and 
where 
Hence, 
In conclusion, 
where 
and three collinear points 
the tangent 
to graph 
at the point 
meet again 
Prove that the points 
are collinearly.
are collinearly 
![$\left(x_2-x_1\right)\left(x_3-x_1\right)\left(x_2-x_3\right)\left[a_0\left(x_1+x_2+x_3\right)+a_1\right]=0\iff$](http://latex.artofproblemsolving.com/3/0/a/30a3050c952c35c9c11f22f073b6992469b715c2.png)
Let 
Thus, 
are collinearly 
the points 
are collinearly.
the following system of the equations 
.
and 
Our system becomes 
Thus, ![$D\left(S^2-P\right)=74\ \stackrel{1\wedge 2}{\iff}\ D\left[\left(52-D^2\right)-\frac {26-D^2}2\right]=74\iff$](http://latex.artofproblemsolving.com/9/c/9/9c979a20a989faec61341a49f4235f35d6f3b471.png)
a.s.o.
i.e. 
Observe that 
From the system of the relations 
with the substitution 
obtain that 
and the equation 
where 
where 
and 
where 
Thus, 
In conclusion, 
and 
and 
where 
standard notations for 
Hence 
Therefore, 
![$\sum \left[2xyz+x\left(y^2+z^2\right)\right]=$](http://latex.artofproblemsolving.com/e/8/3/e83ecfb8ba20f12bb40966d2af5d7b388bd67dcf.png)
![$6xyz+\sum yz[(x+y+z)-x]=6xyz+\sum yz\cdot \sum x-\sum (xyz)=$](http://latex.artofproblemsolving.com/c/2/3/c23df91762163b42f3aa2f842030d3e036f06fc1.png)
![$3(s-a)(s-b)(s-c)+s\cdot \sum [bc-s(s-a)]=$](http://latex.artofproblemsolving.com/4/6/0/46077c8c017270d14d6429ea38a0bd912ff1423d.png)
We have equality iff 
i.e. 
I used the identities 
the midpoint of ![$[AB]\ ;\ D\ -$](http://latex.artofproblemsolving.com/0/f/6/0f6ae9475386cfe0fb972445db1067bbeae1410d.png)
the projection of 
Prove that the equivalence 
and 
Thus, 
Thus, 
Thus, 
and 
the incenter 
the orthocenter 
the Nagel's point ![$[IH].$](http://latex.artofproblemsolving.com/6/c/2/6c29e278279eaafabad7aa9ec99e32a0819f06e8.png)
Prove that 
From the Aubel's relation obtain that 
Apply the Stewart's relation to the ray 
in 
Apply the theorem of median to the following triangles 
Product of the substruction between the previous relations becomes 
![$2\cdot\left[\frac {ac(s-b)}s-\frac {ab(s-c)}s\right]+2\cdot 4R^2\left(\cos^2B-\cos^2C\right)=2\cdot [(ac-4Rr)-(ab-4Rr)]+2\cdot 4R^2\left(\sin^2C-\sin^2B\right)=$](http://latex.artofproblemsolving.com/4/9/d/49d27c91add8785dee07db9dfa18a04873cdd716.png)
Apply the Stewart's relation to the ray 
![$\boxed{s^2\cdot NB^2+a(s-a)\cdot AD^2=s\cdot\left[c^2(s-a)+a(s-c)^2\right]}\ .$](http://latex.artofproblemsolving.com/d/8/0/d8038270e4c115b700d7342f73da7fb1cc195a6b.png)
Get analogously the similar relation![$\boxed{s^2\cdot NC^2+a(s-a)\cdot AD^2=s\cdot \left[b^2(s-a)+a(s-b)^2\right]}\ .$](http://latex.artofproblemsolving.com/c/9/3/c93ca01a9904348b391edde9961c21832cf605c9.png)
From the substruction between the previous relations get 
![$(c-b)\left[(s-a)(b+c)-a^2\right]=(c-b)[s(b+c)-a(a+b+c)]=s[(c-b)[(b+c)-2a]\iff$](http://latex.artofproblemsolving.com/a/9/d/a9dae866453f213786012cbca7c5348bd3740531.png)
In conclusion, 
From PP3 we have 
Therefore, 
![$(\cancel{c-b})[(b+c)-2a]=\frac {(\cancel{c-b})(c+b)}2\iff$](http://latex.artofproblemsolving.com/9/0/9/909429a8f540f25a581080ab8f947877b54e5143.png)
In conclusion, 
the incenter of 
i.e. 
From the previous remark obtain that
i.e. 
the Euler's nine-point center 
Prove that 
and 
where 
is the length of the 
is the length of the 
and
Prove easily that 
Therefore, 
![$\frac {a^2}{\left(a^2+b^2+c^2\right)^2}\cdot\left[c^2\left(2a^2+2c^2-\cancel{b^2}\right)-b^2\left(2a^2+2b^2-\cancel{c^2}\right)\right]=$](http://latex.artofproblemsolving.com/d/b/1/db15e2fee9ac2fec35789e534c53ccba5ed9e5cf.png)
![$\frac {a^2\left[2a^2\left(c^2-b^2\right)+2\left(c^4-b^4\right)\right]}{\left(a^2+b^2+c^2\right)^2}=$](http://latex.artofproblemsolving.com/6/d/3/6d372506a5e15d7677e68888395401e990766e5d.png)
Prove easily that 
Hence 
In conclusion, 
In the previous remark 2 proved that 
Hence 
In conclusion, 
From the previous
Hence 
In conclusion, 
be the point with barrycentrical coordinates w.r.t. 
Prove that 
(standard notations).
where 
then ![$\boxed{M_1M_2^2=-\left[\left(y_1-y_2\right)\left(z_1-z_2\right)\cdot a^2+\left(z_1-z_2\right)\left(x_1-x_2\right)\cdot b^2+\left(x_1-x_2\right)\left(y_1-y_2\right)\cdot c^2\right]}\ .$](http://latex.artofproblemsolving.com/3/9/b/39b30805163bbf2cc82395882c930ac96fe0c140.png)
For example, if 
then:![$MA^2=-\left[yz\cdot a^2+z(x-1)\cdot b^2+(x-1)y\cdot c^2\right]\implies\odot$](http://latex.artofproblemsolving.com/b/b/2/bb28628f0f468c7206c708c5209e358d74358590.png)
![$\begin{array}{ccc}
\nearrow & MB^2=-\left[zx\cdot b^2+x(y-1)\cdot c^2+z(y-1)\cdot a^2\right] & \searrow\\\\
\searrow & MC^2=-\left[xy\cdot c^2+y(z-1)\cdot a^2+x(z-1)\cdot b^2\right] & \nearrow\end{array}\odot\implies$](http://latex.artofproblemsolving.com/e/0/e/e0ec0f3e77db2133d28d3241a3c3e5f0a1df40b6.png)
unde 
sunt varfurile triunghiului 
- mijlocul lui ![$[BC]\ ;\ A_1\left(-1,1,1\right)$](http://latex.artofproblemsolving.com/4/6/2/462e169d3cde1ea3f13a53deb9464c0016607b06.png)
- simetricul lui 
- simetricul lui 
- centru de greutate al triunghiului 
- centrul cercului inscris 
- centrul cercului circumscris 
- ortocentrul
- centrul cercului exinscris 
- punctul lui Nagel
- punctul lui Gergonne
- punctul lui Lemoine (centrul simedian). Daca 
lungimea 
lungimea 
si 
![$\boxed{an_a^2+a(s-b)(s-c)=c^2(s-b)+b^2(s-c)\implies \odot\begin{array}{ccccc}
\nearrow & s^2\cdot NB^2+a(s-a)n_a^2 & = & s\cdot \left[c^2(s-a)+a^2(s-c)^2\right] & \searrow\\\\
\searrow & s^2\cdot NC^2+a(s-a)n_a^2 & = & s\cdot \left[b^2(s-a)+a^2(s-b)^2\right] & \nearrow \end{array} \odot \implies NB^2 - NC^2 = (c-b)(b+c-2a)}$](http://latex.artofproblemsolving.com/2/b/6/2b68044f0d93ea81f89be9216de2150115f62017.png)
where 
I"ll use the substitution 
where 
and 
where 
and 
Prove easily that 
i.e. 
is increasing 
on ![$\left[-\sqrt 2,\sqrt 7-4\right]$](http://latex.artofproblemsolving.com/c/4/5/c45be20f11e99870d3a8384b91dc01e07119b36c.png)
and decreasing
on ![$\left[\sqrt 7-4,\sqrt 2\right]\ .$](http://latex.artofproblemsolving.com/1/3/c/13c4d1100264822ff9ffc5a6c6a0b181752e2db5.png)
Thus, 
For 
or 
![$\boxed{\begin{array}{ccccccc}
t & \| & -\sqrt 2 & & \sqrt 7-4 & & \sqrt 2\\\
== & == & === & == & === & == & ===\\\
g^{\prime}(t) & \| & + & + & 0 & - & -\\\
== & == & === & == & === & == & ===\\\
g(t) & \| & \frac {2\left(4+\sqrt 2\right)}7 & \nearrow & \frac {2+\sqrt 7}3 & \searrow & \frac {2\left(4-\sqrt 2\right)}7\end{array}}\iff\left\|\begin{array}{c}
\min\limits_{t\in\left[-\sqrt 2,\sqrt 2\right]}g(t)=\min\limits_{x\in\mathbb R}f\left(\theta\right)=\frac {2\left(4-\sqrt 2\right)}7\\\\
\max\limits_{t\in \left[-\sqrt 2,\sqrt 2\right]}g(t)=\max\limits_{x\in\mathbb R}f\left(\theta\right)=\frac {2+\sqrt 7}3\end{array}\right\|\iff$](http://latex.artofproblemsolving.com/2/0/6/2067bdd83d28df22f27548fe90c27b8562d13984.png)
![$f\left(I\right)=\left[\frac {2\left(4-\sqrt 2\right)}7,\frac {2+\sqrt 7}3\right],$](http://latex.artofproblemsolving.com/9/5/f/95fd934f42b42e6c3427153de36af5261035b5af.png)
where ![$I=\left[-\sqrt 2,\sqrt 2\right]$](http://latex.artofproblemsolving.com/9/d/8/9d87fe6e212cf87ca6277728f5e0763f7d381493.png)
with the excircle 
for what denote 
Prove that 
has 
In conclusion,
and the required perimeter 
for which denote 
Prove that 
where 
Thus, 
are isosceles trapezoids and 
is the orthocenter of 
a.s.o. Remark. If 
then 
there is the inequality 
what is truly. Hence 
(standard notations).
so that 
and 
I"ll use the substitution 
where ![$t\in\left(1,\sqrt 2\right]$](http://latex.artofproblemsolving.com/7/3/0/730188fa3fd4143fa848773d480938130806d18a.png)
and 
The our inequality
![$t\in\left[-1-2\sqrt 2,\sqrt 2\right]\cap \left(1,\sqrt 2\right]\iff$](http://latex.artofproblemsolving.com/f/5/8/f58cd2c91d5b689d38c919f28b5ec797adbdaa21.png)
![$t\in \left(1,\sqrt 2\right]$](http://latex.artofproblemsolving.com/b/8/1/b818469b7826fc5b6bd1f816354ebc2a491b2f62.png)
what is truly.
obtain that 
I"ll use the previous lemma in the particular case 
From the sum of the relations 
where 
and ![$I=\left[0,\frac {\pi}2\right]\ .$](http://latex.artofproblemsolving.com/2/1/7/217fe3ab50978741732e4f853405ce418d95a618.png)
Prove easily that 
is strict
on 
Let 
Therefore,
such that 
Find the value of 
For example 
and 
Observe that
In conclusion, from the sum of 
Very nice problem!
with 
and the circle 
For a mobile point 
so that 
Prove that the area 
is 
![$[PQ]$](http://latex.artofproblemsolving.com/2/1/c/21ca08816cf8b23ddf756ce9ae098ad327f2443d.png)
and 
Suppose w.l.o.g. that 
i.e. 
Observe that 
and 
Therefore, 
i.e. 
and 
Hence 
Denote 
where 
In conclusion, 
Thus, 
is 
with
(constant) 
Prove that ![$2\cdot [ABCP]=AB^2\ .$](http://latex.artofproblemsolving.com/8/a/6/8a614b24f872f46c4f3cc46f32e15ecd44abff54.png)
and 
where 
and 
Hence 
![$2\cdot [ABCP]=PB\cdot (AC\sin\phi )\ \stackrel{(*)}{=}\ PB\cdot PB=PB^2\implies$](http://latex.artofproblemsolving.com/a/9/2/a924ffd17e8681ce0a9513117acb6b4192040974.png)
![$ [ABCP]=\frac {PB^2}2\ .$](http://latex.artofproblemsolving.com/3/f/1/3f181a339897124969a3c999d73702bc157099b0.png)
are isosceles: 
![$2\cdot [ABP]+2\cdot [CBP]=$](http://latex.artofproblemsolving.com/f/5/a/f5a70b0dbde5fee050bda2c74b73f4e057b4372c.png)
![$\underline{2\cdot [ABCP]}$](http://latex.artofproblemsolving.com/d/5/3/d530d853c90d10e3347e95c2610c8f120c079b91.png)
(S. Fulger).
Prove that 
In conclusion,
Prove analogously that 
a.s.o.
with the circumcircle 
and the trigonometrical identity 
where 
(standard notations). 
because is wellknown the inequality 
for any 
and denote 
Thus, our inequality becomes 
i.e. 
Denote 
In conclusion, the inequality becomes ![$[x+(y+z)]^2\le$](http://latex.artofproblemsolving.com/7/e/d/7ed7f632a954db2fe84d8dc4687f7d2f9a9e2044.png)
![$\cancel {x^2}+2Sx+\cancel{S^2}\le \cancel{x^2}+1+\left[S^2+(P-1)^2\right]\cdot x^2+$](http://latex.artofproblemsolving.com/8/d/3/8d3205ef7f2eefa1d904e1414555df238532d7e3.png)
![$\left[S^2+(P-1)^2\right]\cdot \underline x^2-2S\cdot \underline x+$](http://latex.artofproblemsolving.com/0/e/8/0e8ca67e6e1227482b6cb12443bc02213dfae028.png)
![$\left[(P-1)^2+1\right]\ge 0$](http://latex.artofproblemsolving.com/e/5/a/e5ab94b453d311d3d555488ca52c32523e7437c3.png)
what is true because 
![$S^2-\left[S^2+(P-1)^2\right]\cdot\left[(P-1)^2+1\right]=-(P-1)^2\left[\left(S^2+1\right)+(P-1)^2\right]\le 0\ .$](http://latex.artofproblemsolving.com/3/4/1/3414202e815b06aef3942f44c3e617260fd9656a.png)
We have equality iff 
and touches 
the semicircle is 
touches 
at 
where 
Denote 
and apply the Pythagoras' theorem
Observe
In conclusion, 
Nice problem!
Prove that 
(standard notations). 
is cyclic, i.e. 
I"ll use the well-known identities 
and 
and 
From the Pythagoras' theorem obtain that 
Prove that 
Observe that 
i.e. 
Apply the well-known Cristea's theorem to the points 
Done. Nice problem !
w.r.t. 
Prove that 
(standard notations).
is a parallelogram 
and 
a.s.o. 
Median 
![$2\left(HB^2+HC^2\right)-BC^2\ \stackrel{(1)}{=}\ 2\left[\left(4R^2-b^2\right)+\left(4R^2-c^2\right)\right]-a^2$](http://latex.artofproblemsolving.com/0/8/a/08a055d35cfc0b4ebd32dd22f9cecdca2fedc9e8.png)
Median 
![$2\left[\left(4R^2-a^2\right)+\left(16R^2-2b^2-2c^2-a^2\right)\right]-4R^2=$](http://latex.artofproblemsolving.com/5/a/0/5a068e09ee417da38378d4703f7666c04fe4b21f.png)
Hence can apply the generalized Pythagoras' theorem to 
![$5R^2-a^2+2R^2\left[\left(1-2\sin^2B\right)+\left(1-2\sin^2C\right)\right]=$](http://latex.artofproblemsolving.com/b/7/f/b7fdd3d651fa8e18edd606ed8d46d4e108b6a130.png)
I used the identities 
and 
![$-1-2\cos A[\cos (B-C)-\cos A]=-1-2\cos A[\cos (B-C)+\cos (B+C)]=-1-4\cos A\cos B\cos C\implies$](http://latex.artofproblemsolving.com/1/a/2/1a2ba8cbc23cc09a4dd7a0a1eaadd44d7fddbc86.png)
The power of 
i.e. 
.
and the well-known inequality 
Indeed, 
Prove easily that 
I"ll prove the inequality 
![$\left\{\begin{array}{ccccc}
3s^2=3\sum r_br_c\le \left(r_a+r_b+r_c\right)^2=(4R+r)^2 & \iff & \boxed{s\sqrt 3\le 4R+r} & \iff & 9s\sqrt 3\le 9(4R+r)\\\\
3\cdot\sqrt[3]{\prod (s-a)}\le\sum (s-a)=s\iff 27sr^2\le s^3 & \iff & \boxed{3r\sqrt 3\le s} & \implies & 9r\le s\sqrt 3\end{array}\right\|$](http://latex.artofproblemsolving.com/f/9/f/f9ff07dab53ef08bc244e92d7e6222c26948a255.png)
where ![$f:\left(0,\frac {\pi}2\right]\rightarrow \mathbb R\ ,\ f(x)=\frac {\sin x-x\cos x}{x-\sin x}$](http://latex.artofproblemsolving.com/b/0/f/b0fc5111c33d1c5cce113cf1ae56634089a91f95.png)
and 
If we'll prove that the function continuous 
and 
then from the Darboux's property get 
Indeed, 
![$\sin x\cdot\left[x-\frac {1-\cos x}{2\sin x}\cdot\left(1-\sqrt{4\cos x+5}\right)\right]\cdot\left[x-\frac {1-\cos x}{2\sin x}\cdot \left(1+\sqrt{4\cos x+5}\right)\right]\implies$](http://latex.artofproblemsolving.com/1/2/7/127c85401e1fee1efc888f4e97ed4291ea2ff499.png)
But ![$g'(x)\ \mathrm{.a.s.}\ (1-\cos x)\left[(1+2\cos x\sqrt{4\cos x+5}-\left(2\cos^2x+4\cos x+3\right)\right]\ \mathrm{.a.s.}\ (1+2\cos x)^2(4\cos x+5)$](http://latex.artofproblemsolving.com/1/c/c/1ccb5c5314b468c703d21ceda6f9f704bf5f8645.png)
for any 
the function 
i.e. ![$\boxed{(\forall ) x\in\left(0,\frac {\pi}2\right]\ ,\ \frac 2{\pi -2}<\frac {\sin x-x\cos x}{x-\sin x}<2}\ (*)\ .$](http://latex.artofproblemsolving.com/2/1/5/2156dc25e5cb79f5d2fd60936cbca0dafda0b495.png)
The inequality 
to the left side of the axis 
where 
so that for any point 
where 
the length 
of the arc ![$l(\overarc{AM})<[AT]\ ,$](http://latex.artofproblemsolving.com/a/b/f/abf34fa37e2a9ed139cea1928a616c7bb0e19533.png)
where 
and 
For 
obtain 
for any 
Let 
i.e. 
and 
From 
obtain 
Therefore, 
for any 
i.e. 
With other words, the most far point 
to the left side of the axis 
for which 
is 
where 
August 2018
January 2018