452. Working page II.
by Virgil Nicula, Dec 3, 2016, 9:13 PM
P1. Let 
the midpoint 
of ![$[AB],$](//latex.artofproblemsolving.com/5/8/e/58e377d741d6612acac694994b20fa626c18e0ce.png)
the midpoint 
of ![$[CE]$](//latex.artofproblemsolving.com/e/a/8/ea8eadb930f32cceb220dee99277c24b202fb80c.png)
and the point 
so that 
Prove that 
and 
Proof 1. Denote the midpoint
of ![$[BD].$](//latex.artofproblemsolving.com/1/4/8/1487d6a49a34ef99e799a1a57b155de4717b3cb7.png)
Observe that ![$[EG]$](//latex.artofproblemsolving.com/6/b/9/6b9d8fd2dcaf49dc947a96aa6fa8bda74ac807d5.png)
is the middle line in 
Hence 
and 
Observe that
is the midpoint of ![$[CG],$](//latex.artofproblemsolving.com/f/3/e/f3eba4ff6ccddeaa0f2255421645303fd3ac9707.png)
i.e. ![$[DF]$](//latex.artofproblemsolving.com/4/8/7/487608ba746e637d846b20401f23cc2b80336338.png)
is the middle line in 
Hence 
and 
Since 
obtain that
i.e. 
From the product of the relations 
and 
obtain that 
i.e. 
i.e.
Proof 2. Apply the Menelaus' theorem to the
and the points 
Apply the Menelaus' theorem to the transversal
Proof 3 (slicing and without words). Let
be the point so that 
is the midpoint of ![$[AS].$](//latex.artofproblemsolving.com/7/3/e/73e820c3376a0ff5426931e6ff3bd1b036e6af21.png)
Prove easily or is evidently that is the centroid of 
and
P2. Indraznesc sa va propun si eu o problema de informatica. Se da un vector-memorie de intrare/iesire de (n+1) elemente si doua locatii de manevra si/sau indice. Sa se determine coeficientii polinomului monic (coeficient dominant egal cu 1) de grad n ale carui radacini sunt date (de intrare) si introduse pe n pozitii in vectorul dat (una ramane disponibila) si in urma prelucrarii tot aici sa se obtina si iesirea, adica cei (n+1) coeficienti ai polinomului (evident ca datele de intrare nu ma intereseaza in iesire). Sa nu incepeţi cu relaţii Viete ca ma apucă ameteala, desi sunt doctor in matematici de multa vreme ... Succes !
P3. Let 
so that 
and 
Prove that 
Proof 1.
Proof 2.
P4 (clasa a IX - a). Let two exterior circles
where 
and its common exterior tangents are 
so that 
Let two points 
and 
so that the lines 
and 
separate the points 
Prove that 
is cyclic 
is cyclic.
https://cdn.artofproblemsolving.com/images/...5a6844c24eb.jpg
P5 (clasa a IX - a). Let
with the incircle 
Denote: 
the intersections 
of 
with the angled-bisectors from
so that 
the midpoint 
of ![$[UV]$](//latex.artofproblemsolving.com/a/1/3/a136f90fcd9d8d96d81566cab1c035ac43940789.png)
and the intersection 
Prove that 
Proof 1. Suppose w.l.o.g.
Thus, 
(is well-known!) and 
Apply the Menelaus' theorem to 
The relation 
is well-known. Hence 
i.e.
Proof 2. Suppose w.l.o.g. Thus, (is well-known!) and 
Apply Cristea's identity: 
i.e. 
P6. Let the secant circles
for which denote 
A common tangent line touches 
at 
and 
at 
, the line
being closer to
than to 
The line 
meets again at the point 
Prove that the line 
is a bisector of the angle 
Proof.
P7. Let with the altitude 
and the median 
where the points 
lie in this order on the line 
Suppose
that the incenter
of 
lie on the line and the incenter 
of 
lies on the line 
Find angles of 
Proof 1. Observe that
Therefore, 
In conclusion, 
Proof 2.
P8 (Julio Orihuela). Let
be a triangle and the interior point for which 
and 
Prove that 
Proof 1. Observe that
i.e. 
and prove easily that 
Apply the trigonometric form
of the Ceva's theorem
Proof 2. Let
Thus, 
the ray 
is the bisector of 
Thus,
Proof 3 (synthetic). Let
be the symmetrical point of w.r.t. the line 
Prove easily that 
is equilateral, i.e. 
and
Thus, 
P9 . Let
be a rhombus with 
and its incircle 
what touches at 
and 
Let 
and 
so that 
is tangent to 
at 
Prove that 
Proof 1. Denote
and 
Hence 
Let 
so that 
Thus, 
Apply the theorem of Cosinus to 
![$(y+x)\left[y^2-y(u+v)-uv\right]=(y-x)\left[y^2-y(u+v)+uv\right]\iff$](//latex.artofproblemsolving.com/3/7/5/3754cdbb0981c0447aad189bf136a830f2b41e34.png)
Let 
Hence 
![$(LD+DF)\cdot QP=LA\cdot QF\ \stackrel{(2)}{\iff}\ \left[x+\frac {(x+v)(y-u)}{y-v}\right]\cdot \cancel{(u+v)}=y\cdot \frac {\cancel{(v+u)}(v+x)}{y-v}\iff$](//latex.artofproblemsolving.com/e/2/1/e210d025bbaa00b69076fbc6b6ee7dac69d9c6df.png)
![$\left[x(y-v)+(x+v)(y-u)\right]=y(v+x)\iff$](//latex.artofproblemsolving.com/2/2/6/2269408c67a4a952dac130d73df098f86dbd1fc5.png)
i.e. the relation what is true. In conclusion, 
Remark. Denote
Therefore, 
i.e. the relation
Proof 2 (Kostas Theodoros Rekoumis). I"ll use same notations from upper proof. Apply the Brianchon's theorem to the particular case of the degenerated
hexagon
there is 
Therefore, 
Very nice idea!
P10 (Miguel Ochoa Sanchez). Let with the incircle and 
Prove that the area ![$[PAQ]=\frac 14\cdot [BAC]\cdot\csc^2\frac A2.$](//latex.artofproblemsolving.com/d/1/7/d17791ce9c12dda8397fddfcaae3d8ece62cf8ef.png)
Proof. Denote the area
of 
and 
Apply the particular case of the Cristea's theorem for the incircle 
![$\frac {[PAQ]}S=\frac {AP}{AB}\cdot \frac {AQ}{AC}=$](//latex.artofproblemsolving.com/d/9/c/d9c0555ff0911cbbd338751646f1ed78ea041893.png)
![$\frac {[PAQ]}S=\frac 14\cdot\csc^2\frac A2\ .$](//latex.artofproblemsolving.com/3/4/a/34ad5dedcdbc726fff262b7f06f8890e915e8d0c.png)
Remark.![$[PAQ]=[BPQC]\iff$](//latex.artofproblemsolving.com/7/2/3/723c74e838007bf79e794fd97e32bf05bae2bf93.png)
![$2\cdot [PAQ]=S\iff$](//latex.artofproblemsolving.com/2/3/f/23fb69ea1134e87aabc41034553680a0902e65ac.png)
![$\frac {[PAQ]}S=\frac 12\iff$](//latex.artofproblemsolving.com/9/1/3/9136307ebda292e9d6242d4bef754e06c861d6b2.png)
P11 (Kadir Altintas). Let with 
the centroid and the Lemoine's point 
Prove that 
Proof 1. Denote
![$\left[2\left(b^2+c^2\right)-a^2\right]+\left[2\left(a^2+c^2\right)-b^2\right]=9c^2\iff$](//latex.artofproblemsolving.com/9/b/0/9b06a68537cb5a6bebfe7bc28bfdd9eb495372ae.png)
Prove easily that the length 
of the -symmedian, where 
is given by the relation 
where 
is the length of the-median and 
With the van Aubel's relation obtain 
Therefore,
![$\frac 19\cdot \left[2\left(b^2+c^2\right)-a^2-2\left(a^2+c^2\right)+b^2\right]=$](//latex.artofproblemsolving.com/a/a/7/aa7f4da351f11b3bbc3305473cba1592ac23c1ec.png)
Observe that 
![$\frac {c^2}{\left(a^2+b^2+c^2\right)^2}\cdot \left\{b^2\left[2\left(b^2+c^2\right)-a^2\right]-a^2\left[2\left(a^2+c^2\right)-b^2\right]\right\}=$](//latex.artofproblemsolving.com/a/3/d/a3d021eb3ccb4a60014bbd03d802c48e39deb6f7.png)
![$\frac {c^2}{\left(a^2+b^2+c^2\right)^2}\cdot \left[2\left(b^4-a^4\right)+2c^2\left(b^2-a^2\right)\right]=$](//latex.artofproblemsolving.com/2/b/2/2b21b188e7caa6cbf3502cf58f1093a4986dfcf0.png)
Hence 
Remark. Let
be the incenter of 
Is well-known 
a.s.o. Hence 
Therefore,
In conclusion, 
Proof 2.
P12 (Kadir Altintas). Let with the incircle and the centriod 
Prove that 
Proof. Observe that
where is the midpoint of ![$[BC]\ ;$](//latex.artofproblemsolving.com/2/c/e/2cee46ce04305480ad31681565572a638aaf48b5.png)
is wellknown (or prove easily) that 
Apply the Stewart's relation
to the ray
where 
i.e. 
![$\left|\frac 13\cdot(bc-4Rr)+\frac 16\cdot \left[4r^2+(b-c)^2\right]=r^2+\frac 29\cdot m_a^2\right|\ \odot (18)\ \iff$](//latex.artofproblemsolving.com/4/e/d/4ed5297bc38c04a50ec107747bf1fb70c44d964c.png)
![$6\left(bc-4Rr\right)+3\left[4r^2+(b-c)^2\right]=18r^2+2\left(b^2+c^2\right)-a^2\iff$](//latex.artofproblemsolving.com/d/f/d/dfd58eaed965d280a364ca6d70bf62ae6f5c519d.png)
Is well-known that
The relation becomes 
Particular case. If is -isosceles, i.e. 
then the relation becomes 
For the substitution 
i.e.
Since 
obtain only the solution 
P13 (Kadir Altintas). Let with the incircle and the centroid Prove that ![$IG=2r\iff 8\left(a^3+b^3+c^3\right)+9abc=7[ab(a+b)+bc(b+c)+ca(c+a)]\ .$](//latex.artofproblemsolving.com/b/5/9/b5982365bfc3637f0ee21d44e21062a7af2996c8.png)
Proof. Observe that where is the midpoint of is wellknown (or prove easily) that Apply the Stewart's relation to 
where 
![$\left|\frac 13\cdot(bc-4Rr)+\frac 16\cdot \left[4r^2+(b-c)^2\right]=4r^2+\frac 29\cdot m_a^2\right|\ \odot (18)\ \iff$](//latex.artofproblemsolving.com/e/7/e/e7ec6d714f882148fd67029890edcc5949d0751f.png)
![$6\left(bc-4Rr\right)+3\left[4r^2+(b-c)^2\right]=72r^2+2\left(b^2+c^2\right)-a^2\iff$](//latex.artofproblemsolving.com/6/0/c/60cfc883e8b2c980271f3b73ced5a78619889508.png)
In conclusion, 
Denote 
Is well-known that 
![$\sum bc(b+c)=\sum bc\left[(a+b+c)-a\right]=(a+b+c)(ab+bc+ca)-3abc=$](//latex.artofproblemsolving.com/f/4/9/f4948a26abea5f8d41d41b95d2efcbc157ae27b2.png)
Therefore, 
From the upper equivalencies and obtain the conclusion of our problem, more exactly 
P14 (Kadir Altintas). Let an-isoscelrs with the incircle and the circumcircle 
Prove that 
Proof. From Euler's relation
get 
Let the midpoint of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
Thus,
where 
- the golden ratio.
P15 (Thanos Kalogerakis). Let-isosceles with incircle 
and Nagel's point 
Prove that 
Proof. Let:
Apply
Menelaus' theorem to transversal
From 
get 
An easy extension. Let with the incircle and the Nagel's point Prove that 
P16 (Adil Abdullayev - Azerbaiyan). Let with the incircle where 
Prove that 
Proof. The following identities are well-known
a.s.o. 
In conclusion, 
I used the standard notations.
P17 (Kadir Altintas). The interior of has barycentric coordinates 
and 
Proof. Let
and from the Aubel's relations 
obtain 
Let the projections 
and of 
and on the opposite
sides respectively. Thus,
a.s.o. Hence 
Particular cases.
(Nagel's point). 
![$\frac s{\sum\left[\frac {s-a}s\cdot (s-b)(s-c)\right]}=\frac {s^2}{sr^2}=\frac s{r^2}\implies$](//latex.artofproblemsolving.com/6/e/2/6e2ecc8ebcfd7471a2412258227cad383cc0b6a0.png)
I used well-known identity 
(the centroid). 
I used the well-known inequality 
(the Lemoine's point). 
I used the simple inequality![$4(s-b)(s-c)\le [(s-b)+(s-c)]^2=a^2\implies$](//latex.artofproblemsolving.com/0/5/3/053c6a3ddb283fc32feae1dadc296300661ec6de.png)
P18 (Martin Lukarevski - Skopje). Prove that
there is the trigonometrical inequality 
Thus, 
i.e.
what is true. Otherwise. 
P19 (Daniel SITARU). Prove that there is the inequality 
Proof. We shall apply the wellknown identity
Indeed, 
i.e.
Hence the our inequality becomes 
Since 
and
what is true. Therefore, 
i.e. the required inequality is true.
P20 (Daniel SITARU). Prove that there is the inequality 
Proof 1.
i.e. the required inequality is true.
Proof 2. I"will use the well-known identity
a.s.o. Thus, 
![$s\cdot\left[\sum s\left(r_a-r\right)-2sr\right]=$](//latex.artofproblemsolving.com/8/b/6/8b62720b3611b30f6e9d1b6c3bde1f7f702d1c3a.png)
![$s^2\cdot\left[\sum\left(r_a-r\right)-2r\right]=$](//latex.artofproblemsolving.com/e/c/d/ecd0e93c8ba1547cf6a4b87717da1fb6d3782422.png)
i.e. 
We shall apply
now the remarkably inequalities
and 
i.e. 
Remark. You can try
where 
a.s.o.
Lemma.Prove that
there is the inequality 
Proof.
Is well-known that 
In conclusion, 
P21 (George APOSTOLOPOULOS). Prove that there is the inequality 
where 
is the inradii of the triangles 
.
Proof. I"ll use the upper lemma. The area![$[BIC]=\frac {ar}2=\frac {x(a+BI+CI)}2\ ,$](//latex.artofproblemsolving.com/b/5/2/b52b258a369faba3c1629c3801ebab3dfd07ae97.png)
i.e. 
a.s.o. Hence 
![$\frac 1{2Rr}\cdot \left[\sum a+\sum (BI+CI)\right]=$](//latex.artofproblemsolving.com/8/c/4/8c4936578135ae6fc428ad56dd13f3ad609f9478.png)
![$\boxed{\sum\frac {\sin A}x=\frac 1{Rr}\cdot \left(s+\sum IA\right)}\stackrel{(1)}{\le}\ \frac 1{Rr}\cdot \left[\frac {3R\sqrt 3}2+2(R+r)\right]=\frac {3\sqrt 3}{2r}+\frac 2r+\frac 2R\implies$](//latex.artofproblemsolving.com/f/7/5/f75c9926d091d1a21aa580455523b459227a9d22.png)
P22 (Seyran IBRAHIMOV). Prove that there is the inequality 
Proof. Apply the inequalities Chebyshev (1) , Gerretsen (2) and Euler (3)
![$4\cdot 3r\sqrt 3\cdot \left[\left(16Rr-5r^2\right)-r^2-4Rr\right]=$](//latex.artofproblemsolving.com/2/f/6/2f657e79d1518eb4879fe5c05a582a9127e80562.png)
P23 (Daniel SITARU). Prove that there is the inequality 
(standard notations).
Proof. Apply the simple inequality
Therefore, ![$4\cdot\sum\frac {\left(m_a^2\right)^2}{h_bh_c}\ge 4\cdot \sum\frac {[s(s-a)]^2\cdot 4R^2}{ac\cdot ab}=$](//latex.artofproblemsolving.com/9/a/8/9a87fd89912167ca4b7a2992fe701d267f3808fa.png)
![$ \frac {4Rs}r\cdot \frac {[\sum (s-a)]^2}{\sum a}=$](//latex.artofproblemsolving.com/8/1/6/81625bc53a58759c9db5c371dce65f89c44a81a2.png)
Remain to prove 
i.e. 
Since 
is sufficiently to prove
i.e. 
what is true.
P24 (IMO 2009). Let an isosceles and bisectors of 
meet 
at 
respectively. Let incenter of and 
. Find all possible values of .
Proof. Denote
. Thus, 
.
An easy extension. Let
with the incentre 
. Denote 
, 
and the incentre 
of 
. Prove that 
.
Proof.
.
the midpoint 
of ![$[AB],$](http://latex.artofproblemsolving.com/5/8/e/58e377d741d6612acac694994b20fa626c18e0ce.png)
the midpoint 
of ![$[CE]$](http://latex.artofproblemsolving.com/e/a/8/ea8eadb930f32cceb220dee99277c24b202fb80c.png)
and the point 
so that 
Prove that 
and 
Proof 1. Denote the midpoint
of ![$[BD].$](http://latex.artofproblemsolving.com/1/4/8/1487d6a49a34ef99e799a1a57b155de4717b3cb7.png)
Observe that ![$[EG]$](http://latex.artofproblemsolving.com/6/b/9/6b9d8fd2dcaf49dc947a96aa6fa8bda74ac807d5.png)
is the middle line in 
Hence 
and 
Observe that
is the midpoint of ![$[CG],$](http://latex.artofproblemsolving.com/f/3/e/f3eba4ff6ccddeaa0f2255421645303fd3ac9707.png)
i.e. ![$[DF]$](http://latex.artofproblemsolving.com/4/8/7/487608ba746e637d846b20401f23cc2b80336338.png)
is the middle line in 
Hence 
and 
Since 
obtain that
i.e. 
From the product of the relations 
and 
obtain that 
i.e. 
i.e. Proof 2. Apply the Menelaus' theorem to the
and the points 
Apply the Menelaus' theorem to the transversal
Proof 3 (slicing and without words). Let
be the point so that 
is the midpoint of ![$[AS].$](http://latex.artofproblemsolving.com/7/3/e/73e820c3376a0ff5426931e6ff3bd1b036e6af21.png)
Prove easily or is evidently that is the centroid of 
and P2. Indraznesc sa va propun si eu o problema de informatica. Se da un vector-memorie de intrare/iesire de (n+1) elemente si doua locatii de manevra si/sau indice. Sa se determine coeficientii polinomului monic (coeficient dominant egal cu 1) de grad n ale carui radacini sunt date (de intrare) si introduse pe n pozitii in vectorul dat (una ramane disponibila) si in urma prelucrarii tot aici sa se obtina si iesirea, adica cei (n+1) coeficienti ai polinomului (evident ca datele de intrare nu ma intereseaza in iesire). Sa nu incepeţi cu relaţii Viete ca ma apucă ameteala, desi sunt doctor in matematici de multa vreme ... Succes !
P3. Let
so that 
and 
Prove that 
Proof 1.
Proof 2.
P4 (clasa a IX - a). Let two exterior circles
where 
and its common exterior tangents are 
so that 
Let two points 
and 
so that the lines 
and 
separate the points 
Prove that 
is cyclic 
is cyclic.https://cdn.artofproblemsolving.com/images/...5a6844c24eb.jpg
P5 (clasa a IX - a). Let
with the incircle 
Denote: 
the intersections 
of 
with the angled-bisectors from
so that 
the midpoint 
of ![$[UV]$](http://latex.artofproblemsolving.com/a/1/3/a136f90fcd9d8d96d81566cab1c035ac43940789.png)
and the intersection 
Prove that 
Proof 1. Suppose w.l.o.g.
Thus, 
(is well-known!) and 
Apply the Menelaus' theorem to 
The relation 
is well-known. Hence 
i.e. Proof 2. Suppose w.l.o.g.
Thus, (is well-known!) and 
Apply Cristea's identity: 
i.e. 
P6. Let the secant circles
for which denote 
A common tangent line touches 
at 
and 
at 
, the linebeing closer to
than to 
The line 
meets again at the point 
Prove that the line 
is a bisector of the angle 
Proof.
P7. Let
with the altitude 
and the median 
where the points 
lie in this order on the line 
Supposethat the incenter
of 
lie on the line and the incenter 
of 
lies on the line 
Find angles of 
Proof 1. Observe that
Therefore, 
In conclusion, 
Proof 2.
P8 (Julio Orihuela). Let
be a triangle and the interior point for which 
and 
Prove that 
Proof 1. Observe that
i.e. 
and prove easily that 
Apply the trigonometric formof the Ceva's theorem
Proof 2. Let
Thus, 
the ray 
is the bisector of 
Thus,
Proof 3 (synthetic). Let
be the symmetrical point of w.r.t. the line 
Prove easily that 
is equilateral, i.e. 
and
Thus, 
P9 . Let
be a rhombus with 
and its incircle 
what touches at 
and 
Let 
and 
so that 
is tangent to 
at 
Prove that 
Proof 1. Denote
and 
Hence 
Let 
so that 
Thus, 
Apply the theorem of Cosinus to 
![$(y+x)\left[y^2-y(u+v)-uv\right]=(y-x)\left[y^2-y(u+v)+uv\right]\iff$](http://latex.artofproblemsolving.com/3/7/5/3754cdbb0981c0447aad189bf136a830f2b41e34.png)
Let 
Hence 
![$(LD+DF)\cdot QP=LA\cdot QF\ \stackrel{(2)}{\iff}\ \left[x+\frac {(x+v)(y-u)}{y-v}\right]\cdot \cancel{(u+v)}=y\cdot \frac {\cancel{(v+u)}(v+x)}{y-v}\iff$](http://latex.artofproblemsolving.com/e/2/1/e210d025bbaa00b69076fbc6b6ee7dac69d9c6df.png)
![$\left[x(y-v)+(x+v)(y-u)\right]=y(v+x)\iff$](http://latex.artofproblemsolving.com/2/2/6/2269408c67a4a952dac130d73df098f86dbd1fc5.png)
i.e. the relation what is true. In conclusion, 
Remark. Denote
Therefore, 
i.e. the relation Proof 2 (Kostas Theodoros Rekoumis). I"ll use same notations from upper proof. Apply the Brianchon's theorem to the particular case of the degenerated
hexagon
there is 
Therefore, 
Very nice idea!P10 (Miguel Ochoa Sanchez). Let
with the incircle and 
Prove that the area ![$[PAQ]=\frac 14\cdot [BAC]\cdot\csc^2\frac A2.$](http://latex.artofproblemsolving.com/d/1/7/d17791ce9c12dda8397fddfcaae3d8ece62cf8ef.png)
Proof. Denote the area
of 
and 
Apply the particular case of the Cristea's theorem for the incircle 
![$\frac {[PAQ]}S=\frac {AP}{AB}\cdot \frac {AQ}{AC}=$](http://latex.artofproblemsolving.com/d/9/c/d9c0555ff0911cbbd338751646f1ed78ea041893.png)
![$\frac {[PAQ]}S=\frac 14\cdot\csc^2\frac A2\ .$](http://latex.artofproblemsolving.com/3/4/a/34ad5dedcdbc726fff262b7f06f8890e915e8d0c.png)
Remark.
![$[PAQ]=[BPQC]\iff$](http://latex.artofproblemsolving.com/7/2/3/723c74e838007bf79e794fd97e32bf05bae2bf93.png)
![$2\cdot [PAQ]=S\iff$](http://latex.artofproblemsolving.com/2/3/f/23fb69ea1134e87aabc41034553680a0902e65ac.png)
![$\frac {[PAQ]}S=\frac 12\iff$](http://latex.artofproblemsolving.com/9/1/3/9136307ebda292e9d6242d4bef754e06c861d6b2.png)
P11 (Kadir Altintas). Let
with 
the centroid and the Lemoine's point 
Prove that 
Proof 1. Denote
![$\left[2\left(b^2+c^2\right)-a^2\right]+\left[2\left(a^2+c^2\right)-b^2\right]=9c^2\iff$](http://latex.artofproblemsolving.com/9/b/0/9b06a68537cb5a6bebfe7bc28bfdd9eb495372ae.png)
Prove easily that the length 
of the -symmedian, where 
is given by the relation 
where 
is the length of the
-median and 
With the van Aubel's relation obtain 
Therefore,
![$\frac 19\cdot \left[2\left(b^2+c^2\right)-a^2-2\left(a^2+c^2\right)+b^2\right]=$](http://latex.artofproblemsolving.com/a/a/7/aa7f4da351f11b3bbc3305473cba1592ac23c1ec.png)
Observe that 
![$\frac {c^2}{\left(a^2+b^2+c^2\right)^2}\cdot \left\{b^2\left[2\left(b^2+c^2\right)-a^2\right]-a^2\left[2\left(a^2+c^2\right)-b^2\right]\right\}=$](http://latex.artofproblemsolving.com/a/3/d/a3d021eb3ccb4a60014bbd03d802c48e39deb6f7.png)
![$\frac {c^2}{\left(a^2+b^2+c^2\right)^2}\cdot \left[2\left(b^4-a^4\right)+2c^2\left(b^2-a^2\right)\right]=$](http://latex.artofproblemsolving.com/2/b/2/2b21b188e7caa6cbf3502cf58f1093a4986dfcf0.png)
Hence 
Remark. Let
be the incenter of 
Is well-known 
a.s.o. Hence 
Therefore,
In conclusion, 
Proof 2.
P12 (Kadir Altintas). Let
with the incircle and the centriod 
Prove that 
Proof. Observe that
where is the midpoint of ![$[BC]\ ;$](http://latex.artofproblemsolving.com/2/c/e/2cee46ce04305480ad31681565572a638aaf48b5.png)
is wellknown (or prove easily) that 
Apply the Stewart's relationto the ray
where 
i.e. 
![$\left|\frac 13\cdot(bc-4Rr)+\frac 16\cdot \left[4r^2+(b-c)^2\right]=r^2+\frac 29\cdot m_a^2\right|\ \odot (18)\ \iff$](http://latex.artofproblemsolving.com/4/e/d/4ed5297bc38c04a50ec107747bf1fb70c44d964c.png)
![$6\left(bc-4Rr\right)+3\left[4r^2+(b-c)^2\right]=18r^2+2\left(b^2+c^2\right)-a^2\iff$](http://latex.artofproblemsolving.com/d/f/d/dfd58eaed965d280a364ca6d70bf62ae6f5c519d.png)
Is well-known that
The relation becomes 
Particular case. If
is -isosceles, i.e. 
then the relation becomes 
For the substitution 
i.e.
Since 
obtain only the solution 
P13 (Kadir Altintas). Let
with the incircle and the centroid Prove that ![$IG=2r\iff 8\left(a^3+b^3+c^3\right)+9abc=7[ab(a+b)+bc(b+c)+ca(c+a)]\ .$](http://latex.artofproblemsolving.com/b/5/9/b5982365bfc3637f0ee21d44e21062a7af2996c8.png)
Proof. Observe that
where is the midpoint of is wellknown (or prove easily) that Apply the Stewart's relation to 
where 
![$\left|\frac 13\cdot(bc-4Rr)+\frac 16\cdot \left[4r^2+(b-c)^2\right]=4r^2+\frac 29\cdot m_a^2\right|\ \odot (18)\ \iff$](http://latex.artofproblemsolving.com/e/7/e/e7ec6d714f882148fd67029890edcc5949d0751f.png)
![$6\left(bc-4Rr\right)+3\left[4r^2+(b-c)^2\right]=72r^2+2\left(b^2+c^2\right)-a^2\iff$](http://latex.artofproblemsolving.com/6/0/c/60cfc883e8b2c980271f3b73ced5a78619889508.png)
In conclusion, 
Denote 
Is well-known that 
![$\sum bc(b+c)=\sum bc\left[(a+b+c)-a\right]=(a+b+c)(ab+bc+ca)-3abc=$](http://latex.artofproblemsolving.com/f/4/9/f4948a26abea5f8d41d41b95d2efcbc157ae27b2.png)
Therefore, 
From the upper equivalencies
and obtain the conclusion of our problem, more exactly 
P14 (Kadir Altintas). Let an
-isoscelrs with the incircle and the circumcircle 
Prove that 
Proof. From Euler's relation
get 
Let the midpoint of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
Thus,
where 
- the golden ratio.P15 (Thanos Kalogerakis). Let
-isosceles with incircle 
and Nagel's point 
Prove that 
Proof. Let:
ApplyMenelaus' theorem to transversal
From 
get 
An easy extension. Let
with the incircle and the Nagel's point Prove that 
P16 (Adil Abdullayev - Azerbaiyan). Let
with the incircle where 
Prove that 
Proof. The following identities are well-known
a.s.o. 
In conclusion, 
I used the standard notations. P17 (Kadir Altintas). The interior
of has barycentric coordinates 
and 
Proof. Let
and from the Aubel's relations 
obtain 
Let the projections 
and of 
and on the oppositesides respectively. Thus,
a.s.o. Hence 
Particular cases.
(Nagel's point). 
![$\frac s{\sum\left[\frac {s-a}s\cdot (s-b)(s-c)\right]}=\frac {s^2}{sr^2}=\frac s{r^2}\implies$](http://latex.artofproblemsolving.com/6/e/2/6e2ecc8ebcfd7471a2412258227cad383cc0b6a0.png)
I used well-known identity 
(the centroid). 
I used the well-known inequality 
(the Lemoine's point). 
I used the simple inequality
![$4(s-b)(s-c)\le [(s-b)+(s-c)]^2=a^2\implies$](http://latex.artofproblemsolving.com/0/5/3/053c6a3ddb283fc32feae1dadc296300661ec6de.png)
P18 (Martin Lukarevski - Skopje). Prove that
there is the trigonometrical inequality 
Thus, 
i.e.
what is true. Otherwise. 
P19 (Daniel SITARU). Prove that
there is the inequality 
Proof. We shall apply the wellknown identity
Indeed, 
i.e.
Hence the our inequality becomes 
Since 
and
what is true. Therefore, 
i.e. the required inequality is true. P20 (Daniel SITARU). Prove that
there is the inequality 
Proof 1.
i.e. the required inequality is true.Proof 2. I"will use the well-known identity
a.s.o. Thus, 
![$s\cdot\left[\sum s\left(r_a-r\right)-2sr\right]=$](http://latex.artofproblemsolving.com/8/b/6/8b62720b3611b30f6e9d1b6c3bde1f7f702d1c3a.png)
![$s^2\cdot\left[\sum\left(r_a-r\right)-2r\right]=$](http://latex.artofproblemsolving.com/e/c/d/ecd0e93c8ba1547cf6a4b87717da1fb6d3782422.png)
i.e. 
We shall applynow the remarkably inequalities
and 
i.e. 
Remark. You can try
where 
a.s.o.Lemma.Prove that
there is the inequality 
Proof.
Is well-known that 
In conclusion, 
P21 (George APOSTOLOPOULOS). Prove that
there is the inequality 
where 
is the inradii of the triangles 
.Proof. I"ll use the upper lemma. The area
![$[BIC]=\frac {ar}2=\frac {x(a+BI+CI)}2\ ,$](http://latex.artofproblemsolving.com/b/5/2/b52b258a369faba3c1629c3801ebab3dfd07ae97.png)
i.e. 
a.s.o. Hence 
![$\frac 1{2Rr}\cdot \left[\sum a+\sum (BI+CI)\right]=$](http://latex.artofproblemsolving.com/8/c/4/8c4936578135ae6fc428ad56dd13f3ad609f9478.png)
![$\boxed{\sum\frac {\sin A}x=\frac 1{Rr}\cdot \left(s+\sum IA\right)}\stackrel{(1)}{\le}\ \frac 1{Rr}\cdot \left[\frac {3R\sqrt 3}2+2(R+r)\right]=\frac {3\sqrt 3}{2r}+\frac 2r+\frac 2R\implies$](http://latex.artofproblemsolving.com/f/7/5/f75c9926d091d1a21aa580455523b459227a9d22.png)
P22 (Seyran IBRAHIMOV). Prove that
there is the inequality 
Proof. Apply the inequalities Chebyshev (1) , Gerretsen (2) and Euler (3)
![$4\cdot 3r\sqrt 3\cdot \left[\left(16Rr-5r^2\right)-r^2-4Rr\right]=$](http://latex.artofproblemsolving.com/2/f/6/2f657e79d1518eb4879fe5c05a582a9127e80562.png)
P23 (Daniel SITARU). Prove that
there is the inequality 
(standard notations).Proof. Apply the simple inequality
Therefore, ![$4\cdot\sum\frac {\left(m_a^2\right)^2}{h_bh_c}\ge 4\cdot \sum\frac {[s(s-a)]^2\cdot 4R^2}{ac\cdot ab}=$](http://latex.artofproblemsolving.com/9/a/8/9a87fd89912167ca4b7a2992fe701d267f3808fa.png)
![$ \frac {4Rs}r\cdot \frac {[\sum (s-a)]^2}{\sum a}=$](http://latex.artofproblemsolving.com/8/1/6/81625bc53a58759c9db5c371dce65f89c44a81a2.png)
Remain to prove 
i.e. 
Since 
is sufficiently to prove
i.e. 
what is true.
P24 (IMO 2009). Let an
isosceles and bisectors of 
meet 
at 
respectively. Let incenter of and 
. Find all possible values of .Proof. Denote
. Thus, 
.An easy extension. Let
with the incentre 
. Denote 
, 
and the incentre 
of 
. Prove that 
.Proof.
.This post has been edited 477 times. Last edited by Virgil Nicula, Feb 24, 2019, 2:45 PM
August 2018
January 2018