460. Working page VI.
by Virgil Nicula, Oct 27, 2017, 7:17 PM
P1 (Vasile MASGRAS). Let 
be a square with the circumcircle 
and the point 
which belongs to the little arc 
Prove that 
(See here).
Proof 1 (metric). Supppose w.l.o.g.
i.e. 
and apply Ptolemy's theorem to cyclic quadrilaterals 
Very nice problem !
Proof 2 (trigonometric). Denote
where 
Thus, 
Extension. Let be a rectangle with the circumcircle 
and 
Prove that 
where 
so that 
Generalizare (Vasile MASGRAS). Prove that in any cyclic pentagon
there is the identity 
P2 (British Mathematical Olympiad). Prove that
there is the following equivalence 
(standard notations).
Proof 1 (trig).
![$2\cos A[\cos A-\cos (B-C)]=0\iff$](//latex.artofproblemsolving.com/3/4/9/3492d2c266ec527753dd2c0b9d7c043af17e6782.png)
![$\cos A[\cos (B+C)+\cos (B-C)]=0\iff$](//latex.artofproblemsolving.com/5/d/8/5d8afde1a62ca8f11a0f2d8ded2f56140ed8f851.png)
Remark. Prove easily that
Hence 
Proof 2 (metric).
a.s.o. Thus,
Remark. Prove easily that
Thus, 
In conclusion,
i.e. the equivalence 
P3 (C. V. Durrel). Prove that
there is the equivalence 
(standard notations).
Proof 1 (metric). I"ll use the well-known identities
Therefore, 
Remark. Prove that 
with equality 
Proof 2 (trigonometric). I"ll use the well-known identity
Thus, 
Remark. Prove that 
Remark. Prove that
there are the following equivalencies: 
, where 
-centroid and 
-incenter of 
P4 (Vadym MITROFANOV). Prove that for any
there is the inequality 
(standard notations).
Proof. I"ll use identity
theorem of Sines 
and remarkable inequalities 
Hence 
![$\frac s{2R}\cdot \left[(4R+r)-3r\right]=$](//latex.artofproblemsolving.com/f/d/3/fd3af3fbc4d33fc98945d7b9f547c7544ea1be53.png)
Hence 
In conclusion, 
P5 (Van Khea). Prove that for any there is the identity 
(standard notations).
Proof. Are well-known or prove easily the identities
Therefore, 
![$4R^2+2R\left[\left(r_a+r_b+r_c\right)-r\right]=4R^2+2R\left[\left(4R+\cancel r\right)-\cancel r\right]=4R^2+8R^2=12R^2\ ,$](//latex.artofproblemsolving.com/f/5/8/f583d728b19274e1be85740f4bf56f0257448a62.png)
i.e. 
P6 (Kadir ALTINTAS). Prove that there are the equivalencies 
Proof.
I"ll use well-known identities 
i.e. 
are directly proportional to 
Thus,
Hence 
I"ll use the identities 
i.e. 
are directly proportional to 
Thus, 
Hence 
P7 (Adil Abdullayev). Prove that there is the identity 
(standard notations).
Proof 1. I"ll use the well known identities
and 
Indeed, 
Hence 
In conclusion, 
Remark. 
Proof 2. I"ll use the well known identities
and 
i.e. 
Hence 
i.e. 
In conclusion, 
P8 (Hung Nguyen Viet). Prove that
an acute 
there is the inequality 
(standard notations).
Proof. I"ll use the well known identities![$\boxed{\cos A+\cos B+\cos C=1+\frac rR\ }\ (1)\ ,\ \boxed{\ a^2+b^2+c^2=2\left[s^2-r(4R+r)\right]\ }\ (2)$](//latex.artofproblemsolving.com/4/7/3/473e340ba8c8e0bd1f063ee3d65410495bbf6b74.png)
and the inequalities 
and 
Thus, 
and 
what is true. Remain to prove
i.e. 
But 
because is true.
P9 (Dũng Nguyễn Tiến, Viet). Prove that there is the equivalence 
where
are the projections of the centroid on 
respectively (standard notations).
Proof.![$[BG]\ ,$](//latex.artofproblemsolving.com/9/0/6/9067a26a1b520d41aae9874a77ee7b28e31d4ed5.png)
![$[CG]$](//latex.artofproblemsolving.com/a/4/c/a4c498243ffacd1cd9abbf11f51cc2b074d05b93.png)
are the diameters of the cyclical quadrilaterals 
respectively 
Therefore,
The sum of the relations 
and 
P10 (Boris COLAKOVIC, Serbia). Prove that there is the relation 
(standard notations).
Proof. Observe that
and 
Thus, ![$\frac {b+c}a\in \left(1,\sqrt 2\right]\iff$](//latex.artofproblemsolving.com/1/b/8/1b8adf60674cc7cd3a674f736590b53fa868cfce.png)
![$1+\frac {b+c}a\in \left(2,1+\sqrt 2\right]\ \stackrel{(*)}{\iff}\ \frac {h_a}r\in \left(2,1+\sqrt 2\right]\iff$](//latex.artofproblemsolving.com/e/4/8/e484dd2bb40bd0e816d159586ab74300d9490106.png)
P11. Prove that there is the following relations identity 
the chain of the inequalities
(standard notations).
Proof.
I"ll use the well known identity
and the remarkable bilateral inequality 
P12 (own). Prove that there is the inequality 
(standard notations).
Proof.
Prove easily
the inequality
a.s.o. and the identity 
Thus, 
a.s.o. Therefore, 
In conclusion,
P13. Let with 
and 
Prove that 
(standard notations).
Proof 1. Remark that
i.e. 
and 
Apply the theorem of Sines 
Denote 
The relation 
becomes
P14. Let with centroid 
midpoint of ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and 
Prove that 
Proof. Denote
and 
Observe that 
and 
i.e. 
Apply the Menelaus' theorem to
the transversal
From the relations 
and 
obtain easily the required relation.
P15. In denote 
Prove that 
is cyclically (standard notations).
Proof. Denote
. Observe that 
, 
and 
. Apply the theorem of Sines in 
. Apply the theorem of Cosines
in the triangles
and 
. Observe that 
and
. In conclusion, 
. Let 
. Since 
and 
are cyclically obtain that 
is cyclically.
P16 and its proof (Đỗ Hữu Đức Thịnh). We can try to use the well known Ptolemy's inequality in a convex quadrilateral
For example,
a.s.o. 
P17 (Nguyen Viet Hung - Vietnam).
Proof.
Prove easily that 
(Euler's inequality), what is true. Hence and the inequality is also true. From the
inequality and the Gerretsen's inequality 
obtain that 
i.e. 
Prove easily that 
what is true. In conclusion, and inequality 
is true.
what is the Blundon's inequality. In conclusion, and the inequality 
is true.
![$\frac 1{sr}\cdot \left[\cancel{2s^2}-2\left(\cancel{s^2}-r^2-4Rr\right)\right]=$](//latex.artofproblemsolving.com/d/9/1/d91aadd0b00cd951bd73d3e0534567a3eb098301.png)
Prove easily that 
what is true.
Hence and the inequality
is true. From the well known inequality 
obtain that 
i.e. 
Remarks.
Observe that 
Hence can apply the Chebyshev's inequality 
![$\blacktriangleright\ \frac 13\cdot\sum\frac a{r_a}\ge\sqrt [3]{\prod\frac a{r_a}}=\sqrt [3]{\frac {abc\cdot r}{r_ar_br_c\cdot r}}=\sqrt [3]{\frac {4R\cancel Sr}{S^{\cancel 2}}}=\sqrt [3]{\frac {4R}s}\ \stackrel{2s\le 3R\sqrt 3}{\ge}\ \sqrt [3]{\frac{4\cdot 2}{3\sqrt 3}}=\sqrt[3]{\left(\frac 2{\sqrt 3}\right)^3}=\frac 2{\sqrt 3}\implies\sum\frac {a}{r_a}\ge 2\sqrt{3}\ .$](//latex.artofproblemsolving.com/9/a/c/9ac9461f2ed8db135d7ef2f0d13b389aabf245f7.png)
P18 (own). Prove that there is the relations 
(standard notations).
Proof. I"ll use the identities
and the inequalities
and ![$\sum (s-b)(s-c)\le \left[\sum (s-a)\right]^2\iff\boxed{\ 3r(4R+r)\le s^2\ }\ (2)\ .$](//latex.artofproblemsolving.com/1/c/a/1ca0a5b2c01abb426a54cb155ab038159d27192c.png)
Therefore
![$\sum \left[s+\frac {(s-b)(s-c)}{s-a}\right]=$](//latex.artofproblemsolving.com/b/e/b/beb692c70a858dd2eac2f4c9fd9e458618500ccc.png)
![$[3r+(4R+r)]\sqrt 3=$](//latex.artofproblemsolving.com/9/0/6/9060ba107ad32abb4ecfc13f89c18a3531927b33.png)
On other hand 
P19 (own). Prove that there is the inequality 
(standard notations).
Proof. Prove easily that
Hence 
and 
P20 (Nguyễn Đăng Khoa, Vietnam). Let with the incircle 
and the midpoint of the side ![$[AB]\ .$](//latex.artofproblemsolving.com/f/4/3/f43654eb147de5ab7da28700a60e256a697a2e10.png)
Suppose that 
Prove that 
Proof.
Therefore, 
Since 
and 
obtain that 
P21 (Carlos Olivera Diaz). Let
and two mobile points 
and 
so that 
Ascertain the minimum value of the segment ![$[MN]\ .$](//latex.artofproblemsolving.com/8/f/3/8f33a629708367aee0e2e495cd6641b16ceacdae.png)
Proof. Denote![$AM=x\in (0,c]\ ,$](//latex.artofproblemsolving.com/e/5/b/e5b6f3021cbe3212c46b1b69ac9ed4ab5a9b77ea.png)
i.e. 
and ![$AN=y\in (0,b]\ ,$](//latex.artofproblemsolving.com/8/1/0/810e4e25cfc1aea6398883d70a195c0a6a9280ed.png)
i.e. 
Hence 
Apply the
C.B.S. inequality
and 
where 
P22 (Dung Nguyen Tien, Vietnam). Let with the excircles 
and the incircle 
Denote 
and
Prove that the implication 
is cyclic 
where 
(golden ratio).
Proof. Observe that
and 
Denote 
where is the circumcircle of 
I"ll apply the power of the points 
w.r.t.
Thus, 
![$2(s-a)(b+c)=a^2\implies (b+c)[(b+c)-a]=a^2$](//latex.artofproblemsolving.com/6/8/5/6850385d3cdd52ee2673f7cdddd9e672e4cc6383.png)
In conclusion, 
where 
P23 (Ercole SUPPA, Italy).
Proof. Suppose w.l.o.g.
i.e. 
Let the midpoint 
of ![$[AD]\ ,$](//latex.artofproblemsolving.com/2/6/7/26748bb1d79775245b0f0bbbd36d03d1342c4737.png)
i.e. 
the incircle 
of 
and the incircle of 
Apply the theorem of the
-bisector in 
Apply the well known identity 
in the
-right 
i.e. 
In conclusion, 
P24 (Problem 733, SANGAKU).
Proof.
P25 (Ercole SUPPA, Italy).
Proof.
P26 (Valentin VORNICU). It is known that
is the smallest angle in the triangle 
. The points 
and 
divide the circumcircle
of the triangle into two arcs. Let
be an interior point of the arc between and which does not contain 
. The perpendicular bisectors
of
and 
meet the line 
at 
and 
, respectively. The lines 
and 
meet at 
. Show that 
.
Alternative formulation: Four different points
are chosen on a circle 
such that the triangle 
is not right-angled. Prove that:
(a) The perpendicular bisectors of and meet 
at certain and 
respectively, and that 
and 
meet at a certain 
(b) The length of one of the line segments
and 
is the sum of the lengths of the other two.
Proof.
and 
In conclusion,![$TB+TC=\frac{R}{\cos A}\cdot [\sin (B-x)+\sin (C-y)]=$](//latex.artofproblemsolving.com/3/3/8/338215a42c3ac53520634bcd874623b038c39b57.png)
be a square with the circumcircle 
and the point 
which belongs to the little arc 
Prove that 
(See here).Proof 1 (metric). Supppose w.l.o.g.
i.e. 
and apply Ptolemy's theorem to cyclic quadrilaterals 
Very nice problem !Proof 2 (trigonometric). Denote
where 
Thus, 
Extension. Let
be a rectangle with the circumcircle 
and 
Prove that 
where 
so that 
Generalizare (Vasile MASGRAS). Prove that in any cyclic pentagon
there is the identity 
P2 (British Mathematical Olympiad). Prove that
there is the following equivalence 
(standard notations).Proof 1 (trig).
![$2\cos A[\cos A-\cos (B-C)]=0\iff$](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)
Remark. Prove easily that
Hence 
Proof 2 (metric).
a.s.o. Thus, 
Remark. Prove easily that
Thus, 
In conclusion,
i.e. the equivalence 
P3 (C. V. Durrel). Prove that
there is the equivalence 
(standard notations).Proof 1 (metric). I"ll use the well-known identities
Therefore, 
Remark. Prove that 
with equality 
Proof 2 (trigonometric). I"ll use the well-known identity
Thus, 
Remark. Prove that 
Remark. Prove that
there are the following equivalencies: 
, where 
-centroid and 
-incenter of 
P4 (Vadym MITROFANOV). Prove that for any
there is the inequality 
(standard notations).Proof. I"ll use identity
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, 
P5 (Van Khea). Prove that for any
there is the identity 
(standard notations).Proof. Are well-known or prove easily the identities
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. 
P6 (Kadir ALTINTAS). Prove that
there are the equivalencies 
Proof.
I"ll use well-known identities 
i.e. 
are directly proportional to 
Thus,
Hence 
I"ll use the identities 
i.e. 
are directly proportional to 
Thus, 
Hence 
P7 (Adil Abdullayev). Prove that
there is the identity 
(standard notations).Proof 1. I"ll use the well known identities
and 
Indeed, 
Hence 
In conclusion, 
Remark. 
Proof 2. I"ll use the well known identities
and 
i.e. 
Hence 
i.e. 
In conclusion, 
P8 (Hung Nguyen Viet). Prove that
an acute 
there is the inequality 
(standard notations).Proof. I"ll use the well known identities
![$\boxed{\cos A+\cos B+\cos C=1+\frac rR\ }\ (1)\ ,\ \boxed{\ a^2+b^2+c^2=2\left[s^2-r(4R+r)\right]\ }\ (2)$](http://latex.artofproblemsolving.com/4/7/3/473e340ba8c8e0bd1f063ee3d65410495bbf6b74.png)
and the inequalities 
and 
Thus, 
and 
what is true. Remain to prove
i.e. 
But 
because is true.P9 (Dũng Nguyễn Tiến, Viet). Prove that
there is the equivalence 
where
are the projections of the centroid on 
respectively (standard notations).Proof.
![$[BG]\ ,$](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 
P10 (Boris COLAKOVIC, Serbia). Prove that
there is the relation 
(standard notations).Proof. Observe that
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)
P11. Prove that
there is the following relations identity 
the chain of the inequalities
(standard notations).Proof.
I"ll use the well known identity
and the remarkable bilateral inequality 
P12 (own). Prove that
there is the inequality 
(standard notations).Proof.
Prove easilythe inequality
a.s.o. and the identity 
Thus, 
a.s.o. Therefore, 
In conclusion,
P13. Let
with 
and 
Prove that 
(standard notations).Proof 1. Remark that
i.e. 
and 
Apply the theorem of Sines 
Denote 
The relation 
becomes
P14. Let
with centroid 
midpoint of ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and 
Prove that 
Proof. Denote
and 
Observe that 
and 
i.e. 
Apply the Menelaus' theorem tothe transversal
From the relations 
and 
obtain easily the required relation. P15. In
denote 
Prove that 
is cyclically (standard notations).Proof. Denote
. Observe that 
, 
and 
. Apply the theorem of Sines in 
. Apply the theorem of Cosinesin the triangles
and 
. Observe that 
and
. In conclusion, 
. Let 
. Since 
and 
are cyclically obtain that 
is cyclically. P16 and its proof (Đỗ Hữu Đức Thịnh). We can try to use the well known Ptolemy's inequality in a convex quadrilateral
For example,
a.s.o. 
P17 (Nguyen Viet Hung - Vietnam).
Proof.
Prove easily that 
(Euler's inequality), what is true. Hence and the inequality is also true. From theinequality
and the Gerretsen's inequality 
obtain that 
i.e. 
Prove easily that 
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.Hence and the inequality
is true. From the well known inequality 
obtain that 
i.e. 
Remarks.
Observe that 
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)
P18 (own). Prove that
there is the relations 
(standard notations).Proof. I"ll use the identities
and the inequalities
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 
P19 (own). Prove that
there is the inequality 
(standard notations).Proof. Prove easily that
Hence 
and 
P20 (Nguyễn Đăng Khoa, Vietnam). Let
with the incircle 
and the midpoint of the side ![$[AB]\ .$](http://latex.artofproblemsolving.com/f/4/3/f43654eb147de5ab7da28700a60e256a697a2e10.png)
Suppose that 
Prove that 
Proof.
Therefore, 
Since 
and 
obtain that 
P21 (Carlos Olivera Diaz). Let
and two mobile points 
and 
so that 
Ascertain the minimum value of the segment ![$[MN]\ .$](http://latex.artofproblemsolving.com/8/f/3/8f33a629708367aee0e2e495cd6641b16ceacdae.png)
Proof. Denote
![$AM=x\in (0,c]\ ,$](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 theC.B.S. inequality
and 
where 
P22 (Dung Nguyen Tien, Vietnam). Let
with the excircles 
and the incircle 
Denote 
and
Prove that the implication 
is cyclic 
where 
(golden ratio).Proof. Observe that
and 
Denote 
where is the circumcircle of 
I"ll apply the power of the points 
w.r.t.
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 
P23 (Ercole SUPPA, Italy).
Proof. Suppose w.l.o.g.
i.e. 
Let the midpoint 
of ![$[AD]\ ,$](http://latex.artofproblemsolving.com/2/6/7/26748bb1d79775245b0f0bbbd36d03d1342c4737.png)
i.e. 
the incircle 
of 
and the incircle of 
Apply the theorem of the
-bisector in 
Apply the well known identity 
in the
-right 
i.e. 
In conclusion, 
P24 (Problem 733, SANGAKU).
Proof.
P25 (Ercole SUPPA, Italy).
Proof.
P26 (Valentin VORNICU). It is known that
is the smallest angle in the triangle 
. The points 
and 
divide the circumcircleof the triangle into two arcs. Let
be an interior point of the arc between and which does not contain 
. The perpendicular bisectorsof
and 
meet the line 
at 
and 
, respectively. The lines 
and 
meet at 
. Show that 
.Alternative formulation: Four different points
are chosen on a circle 
such that the triangle 
is not right-angled. Prove that:(a) The perpendicular bisectors of
and meet 
at certain and 
respectively, and that 
and 
meet at a certain 
(b) The length of one of the line segments
and 
is the sum of the lengths of the other two.Proof.
and 
In conclusion,
![$TB+TC=\frac{R}{\cos A}\cdot [\sin (B-x)+\sin (C-y)]=$](http://latex.artofproblemsolving.com/3/3/8/338215a42c3ac53520634bcd874623b038c39b57.png)
This post has been edited 495 times. Last edited by Virgil Nicula, May 23, 2018, 9:03 AM
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