445. Mathematics "instant" for the high school.
by Virgil Nicula, Jun 8, 2016, 5:07 PM
P0. 
Proof. Prove easily that
Thus, 
i.e. the relation 
what is true. Thus, 
Geometrical interpretation 1. Let an
-right 
with the circumcircle 
Let 
the -altitude 
where 
so that 
doesn't separates
and
so that 
where 
is the tangent to 
at 
Prove easily that 
and 
Let 
and 
Relations
and 
are true and 
Prove easily that 
and 
Indeed,
Very nice !
Geometrical interpretation 2 (Rachid Moussaoui): sursa.
P1 (Sunken Rock). In
and 
are the bisectors of 
and 
where 
and 
respectively. Prove that the circumcenters of
and 
lie on 
Proof. The circumcenters of and lie on 
is cyclic 
the circumcenters of and lie on 
P2. Let
be the centroid of and 
, 
so that 
Prove that ![$[AMN]=[BMN]+[CMN]\ .$](//latex.artofproblemsolving.com/a/8/5/a8537e96ccd6444cc9bb8fb34581d42a778d4e85.png)
Denoted ![$[XYZ]$](//latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
- the area of 
Extension. Let
be an interior , so that 
and 
Prove that![$[AMN]\cdot\frac {PR}{PA}=[BMN]\cdot\frac {RC}{BC}+[CMN]\cdot\frac {RB}{BC}\ .$](//latex.artofproblemsolving.com/1/4/8/1489b533962aa042629ac25c01087d5eb50f0c1d.png)
Denoted - the area of
Proof. I"ll use an well known the Cristea's identity
Indeed, the relation
![$\frac {PR}{PA}\cdot BC=\frac {[BMN]}{[AMN]}\cdot RC+\frac {[CMN]}{[AMN]}\cdot RB\iff$](//latex.artofproblemsolving.com/0/0/5/005758902352aa16f6b45814d31cf666a6c93ba7.png)
Remark. An equivalent form of this identity is![$\boxed{[AMN]\cdot\frac {AR}{AP}=[ABN]\cdot\frac {CR}{CB}+[ACM]\cdot\frac {BR}{BC}}\ (**)\ .$](//latex.artofproblemsolving.com/4/0/6/4067204e2be4567869c6ea3e2c306df3ced16b9f.png)
See here the proof of the Cristea's identity (third relation from PP1).
Cazuri particulare.
- centrul de greutate ![$\Longrightarrow\ [AMN]=[BMN]+[CMN]\ .$](//latex.artofproblemsolving.com/7/c/d/7cd7aeb3cfa7a5b0ff2ececabcad642d959374df.png)
- centrul centrului inscris ![$\Longrightarrow\ a\cdot [AMN]=b\cdot [BMN]+c\cdot [CMN]\ .$](//latex.artofproblemsolving.com/e/4/2/e426a77f6e3a8f7783b6139b498cec37f33ec27d.png)
P3. Prove that
and 
Proof.
Let
Therefore, 
P4. Prove easily that
Show that 
Proof 1 (mudok). Suppose w.l.o.g. that
Prove easily that 
and 
Proof 2 (own).
I"ll use Gerretsen's 
Indeed, 
i.e. the relation 
is true. In conclusion, 
i.e. the relation 
P5. For are well known the inequalities 
Prove that 
adica
(Virgil Nicula & Cosmin Pohoata, Mathematical Reflections).
Proof. Exist
so that 
The first inequality is equivalent with 
what is the Schur's inequality under another form. The second inequality is equivalent with
![$\prod (s+x)+8xyz\leq 2s\cdot \left[s^2 +\sum (xy)\right]\iff$](//latex.artofproblemsolving.com/e/3/9/e39aa244007b4cc506cb78edef996d181a301557.png)
Otherwise. ![$\left\{\begin{array}{ccc}
x+y+z & \ge & 3\sqrt [3]{xyz}\\\\
xy+yz+zx & \ge & 3\sqrt [3]{x^2y^2z^2}\end{array}\right\|\ \bigodot$](//latex.artofproblemsolving.com/5/3/2/5323a43e03c66def6dec8836a4a5bea505c33e2e.png)
P6. Prove that
there are two equivalent inequalities 
Proof.
![$\frac 1{s^2}\cdot\left[\left(r_a+r_b+r_c\right)^2-2\left(r_ar_b+r_br_c+r_cr_a\right)\right]+\frac {2r}{R}\ge 2\iff$](//latex.artofproblemsolving.com/7/b/4/7b453ef82691ba31da4ab0260353492ff63cd506.png)
![$\frac 1{s^2}\cdot\left[(4R+r)^2-2s^2\right]+\frac {2r}{R}\ge 2\iff$](//latex.artofproblemsolving.com/b/1/4/b14e61ce06aba270de0525386e50340fd806c710.png)
Now I"ll give a short geometrical proof of the second inequality
If 
is the orthocenter and 
is the Gergonne's point, then is well known that
![$H\Gamma^2 =4R^2\cdot\left[1-\frac {2s^2(2R-r)}{R(4R+r)^2}\right]\ge 0\implies$](//latex.artofproblemsolving.com/d/9/5/d95d3f1e1fda511bb368da7b8e57f2b365a83c62.png)
i.e.
P7.Let
with incircle 
what is tangent to in 
the -excircle 
and midpoint 
of the -altitude ![$[AS]$](//latex.artofproblemsolving.com/8/1/f/81f842d87b22553e4b674bb2bf46a080da785009.png)
where 
Prove that 
Proof 1. Let
Thus, 
what is true.
Proof 2. Let
and 
Hence 
Consider the homothety with the center what transforms incircle
in-excircle, i.e. 
and 
are collinear. 
are homothetic and 
is the midpoint of ![$[D_1L_1]\ ,$](//latex.artofproblemsolving.com/6/3/4/6342f75b13f0e4df8a16756ee94c429de6fabeeb.png)
then 
pass by the midpoint of ![$[AS]\ .$](//latex.artofproblemsolving.com/1/c/7/1c7d89ce0de186aaed8a6383751ed9d1665d7d7d.png)
P8 (Sergey Yurin, Kvant). Let with 
and 
. Consider 
for which 
. Find 
.
• • • • • • • • • • • •
P9 (J. C. Miranda). Let with 
Denote 
so that 
and the midpoint of ![$[AM]\ .$](//latex.artofproblemsolving.com/e/6/4/e642a640cf5a59113546c863aa4d0f99da789766.png)
Ascertain the length of the segment ![$[BP]$](//latex.artofproblemsolving.com/c/d/a/cda0272ae887e314ff58215fbf55333fa558fefd.png)
(standard notations).
Proof. Observe that
is 
-isosceles, 
and 
Apply the theorem of median in 
where denote 
Therefore,![$4\cdot BP^2=2\cdot\left[BA^2+BM^2\right]-AM^2\ \stackrel{(*)}{\iff}\ 4x^2=$](//latex.artofproblemsolving.com/7/b/6/7b63be40913199fdcfad8dec1459176717447740.png)
![$2\left[c^2+(a-b)^2\right]-4b^2\sin^2\frac C2=$](//latex.artofproblemsolving.com/7/1/6/716f077ea9176b2a22de849c55663574095beb39.png)
![$4ax^2=2ac^2+2a(a-b)^2-b\left[c^2-(a-b)^2\right]\iff$](//latex.artofproblemsolving.com/7/5/b/75b8ef85c81953bcf1a0ecc33b3f3f94b9e90ae3.png)
Particular case
and 
P10 (Kadir Altintas - Turkey). Let
be a rectangle with the center 
Denote 
so that 
and 
Denote
the incircles
and 
of the triangles 
and 
respectively. Prove that 
Proof. Let the midpoint of ![$[CD]$](//latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
and 
Therefore, 
and 
On other hand, 
Observe that 
P11 (Conc.Saraghin 2014). Let
the feet 
of the 
-altitude and the -altitude 
the feet 
of the -bisector and the -bisector. Prove that 
Proof. the line 
is antiparallel direction to 
is cyclic 
P12 (Queensland Australia). Solve the following system
Proof 1. Let
Prove easily 
and 
i.e. 
Thus
P13. Let the functions
where 
and 
where 
Let the real roots
of the equation 
and the real roots 
of the equation 
Prove that 
Proof.
![$(m-p)\underline{[(m-r)-(p-n)]}+$](//latex.artofproblemsolving.com/1/4/d/14df002383e3cec1494d7aae41a49808dc008315.png)
![$(n-r)\underline{[(n-p)-(r-m)]}=$](//latex.artofproblemsolving.com/5/7/3/573dfebc40528796e98ac2a4f57f4f7fe2c30999.png)
![$[(m+n)-(p+r)][(m-p)+(n-r)]=$](//latex.artofproblemsolving.com/6/5/7/657e355bd2df0c03df955f518a83e697894c61e0.png)
![$[(m+n)-(p+r)]^2=(a-c)^2\ .$](//latex.artofproblemsolving.com/5/9/e/59e6b5a941b052cf7d70c7b2f8dbfbfddb45a53c.png)
In conclusion,
P14. Prove that for any
so that 
there is the inequality 
Proof.
The real function 
is convex and
Hence from the Jensen's inequality obtain that 
![$ f\left[\sum\limits_{cyc}\frac{a}{3}\cdot (a+2b) \right]=\sqrt{3}\ .$](//latex.artofproblemsolving.com/3/b/7/3b770c37d213329135754751496816b35353280b.png)
P15. Let an acute with the orthocenter 
Let 
be the bisectors of 
respectively , where 
Denote
the projections
of the point 
on 
respectively and the projections 
of the point 
on 
respectively. Prove that 
Proof. Let
and 
Define the projections 
of on respectively and the projections 
of on
respectively. Thus, 
and 
are rectangles. Hence 
In conclusion, 
and 
P16. Let
be an -right triangle with the altitude where 
Denote the incircles 
of 
respectively and 
Prove that 
Proof 1. Suppose w.l.o.g. that
Let 
and 
Thus, 
Thus, 
i.e. 
Proof 2.
is the orthocenter of 
Remark. Suppose w.l.o.g.
and denote 
and 
Prove easily that 
and 
In conclusion, 
and 
where Thus, 
P17. Let with the circumcircle 
and Let 
and the points 
so that 
Prove that 
Proof. If
are midpoints of ![$[AB]\ ,$](//latex.artofproblemsolving.com/b/a/3/ba35a1c584b9f3786aaed51acf1cbc01ce2968a4.png)
![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
respectively, then 
Hence 
Remark. I"ll prove the well-known relation
Indeed, 
P18. Let be an orthodiagonal and cyclical quadrilateral with the circumcircle 
Let 
the midpoints 
of ![$[BC]\ ,$](//latex.artofproblemsolving.com/6/5/6/6563ca77d31eb676ef646fa9c37ea181b812d864.png)
![$[CD]\ ,$](//latex.artofproblemsolving.com/5/f/7/5f745def914e8ed03d2015c00e2d98fd458adcab.png)
![$[DA]$](//latex.artofproblemsolving.com/7/b/e/7be1ee9448ef63ac6f1f6b8dd6982c30bad9bb31.png)
respectively and the projections 
of on respectively. Then 
is a rectangle and
the points
belong to the circle with the diameter ![$[MP]$](//latex.artofproblemsolving.com/4/8/2/4821ed42ce01e14fb61be739cc547c042e1a9005.png)
and its center in the midpoint of ![$[OE]$](//latex.artofproblemsolving.com/2/a/0/2a082802ed9a33ba2b0993de60b65d137ee48461.png)
because the quadrilaterals 
and 
are parallelograms.
P19. Prove that at least one from its medians has the length less than or equally to 
where 
Proof 1. Suppose w.l.o.g.
i.e. 
and 
i.e. 
I"ll use the well-known identities
Thus, 
Hence
![$3\left[2\left(b^2+c^2\right)-a^2\right]\le \left[a+(b+c)^2\right]\iff$](//latex.artofproblemsolving.com/2/3/c/23cf4cbd954f783589a3e59aa11a67239706c871.png)
Proof 2. Let
be the smallest and the greatest medians of 
Prove easily that 
Prove similarly that
Proof 3. Prove easily that
Hence
In conclusion, 
Prove similarly if 
, then
P20. Let with the orthocenter 
Prove the chain of the equivalencies 
is -right-angled triangle 
(standard notations).
P21 (Adil Abdullayev). Prove that
.
Proof. I"ll use only the identity
and the remarkable inequality 
Thus, 
Hence 
From
obtain that the inequality 
is true.
In conclusion,
P22 (Miguel Ochoa SANCHEZ). Click => here.
Proof.
where 
is the length of the -altitude. From the theorem of Cathetus 
obtain that 
Apply the Cristea's theorem
From the relations 
obtain the required identity, i.e. 
P23 (Kadir Altintas, Turkey). Let with incircle 
midpoint of -Gergonne's cevian 
where 
and the midpoint of ![$[AT]\ .$](//latex.artofproblemsolving.com/9/7/2/97274983169ba36eadaa26e420576baf0974199b.png)
Prove 
Proof (Emil Stoyanov). Let
and 
where 
is the Nagel's cevian. Thus, 
are the midpoints of segments 
Hence 
Nice!
Proof. Prove easily that
Thus, 
i.e. the relation 
what is true. Thus, 
Geometrical interpretation 1. Let an
-right 
with the circumcircle 
Let 
the -altitude 
where 
so that 
doesn't separates and
so that 
where 
is the tangent to 
at 
Prove easily that 
and 
Let 
and 
Relations
and 
are true and 
Prove easily that 
and 
Indeed,
Very nice !Geometrical interpretation 2 (Rachid Moussaoui): sursa.
P1 (Sunken Rock). In
and 
are the bisectors of 
and 
where 
and 
respectively. Prove that the circumcenters of
and 
lie on 
Proof. The circumcenters of
and lie on 
is cyclic 
the circumcenters of and lie on 
P2. Let
be the centroid of and 
, 
so that 
Prove that ![$[AMN]=[BMN]+[CMN]\ .$](http://latex.artofproblemsolving.com/a/8/5/a8537e96ccd6444cc9bb8fb34581d42a778d4e85.png)
Denoted ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
- the area of 
Extension. Let
be an interior , so that 
and 
Prove that
![$[AMN]\cdot\frac {PR}{PA}=[BMN]\cdot\frac {RC}{BC}+[CMN]\cdot\frac {RB}{BC}\ .$](http://latex.artofproblemsolving.com/1/4/8/1489b533962aa042629ac25c01087d5eb50f0c1d.png)
Denoted - the area of Proof. I"ll use an well known the Cristea's identity
Indeed, the relation
![$\frac {PR}{PA}\cdot BC=\frac {[BMN]}{[AMN]}\cdot RC+\frac {[CMN]}{[AMN]}\cdot RB\iff$](http://latex.artofproblemsolving.com/0/0/5/005758902352aa16f6b45814d31cf666a6c93ba7.png)
Remark. An equivalent form of this identity is
![$\boxed{[AMN]\cdot\frac {AR}{AP}=[ABN]\cdot\frac {CR}{CB}+[ACM]\cdot\frac {BR}{BC}}\ (**)\ .$](http://latex.artofproblemsolving.com/4/0/6/4067204e2be4567869c6ea3e2c306df3ced16b9f.png)
See here the proof of the Cristea's identity (third relation from PP1).
Cazuri particulare.
- centrul de greutate ![$\Longrightarrow\ [AMN]=[BMN]+[CMN]\ .$](http://latex.artofproblemsolving.com/7/c/d/7cd7aeb3cfa7a5b0ff2ececabcad642d959374df.png)
- centrul centrului inscris ![$\Longrightarrow\ a\cdot [AMN]=b\cdot [BMN]+c\cdot [CMN]\ .$](http://latex.artofproblemsolving.com/e/4/2/e426a77f6e3a8f7783b6139b498cec37f33ec27d.png)
P3. Prove that
and 
Proof.
Let
Therefore, 
P4. Prove easily that
Show that 
Proof 1 (mudok). Suppose w.l.o.g. that
Prove easily that 
and 
Proof 2 (own).
I"ll use Gerretsen's 
Indeed, 
i.e. the relation 
is true. In conclusion, 
i.e. the relation 
P5. For
are well known the inequalities 
Prove that 
adica
(Virgil Nicula & Cosmin Pohoata, Mathematical Reflections).Proof. Exist
so that 
The first inequality is equivalent with 
what is the Schur's inequality under another form. The second inequality is equivalent with![$\prod (s+x)+8xyz\leq 2s\cdot \left[s^2 +\sum (xy)\right]\iff$](http://latex.artofproblemsolving.com/e/3/9/e39aa244007b4cc506cb78edef996d181a301557.png)
Otherwise. ![$\left\{\begin{array}{ccc}
x+y+z & \ge & 3\sqrt [3]{xyz}\\\\
xy+yz+zx & \ge & 3\sqrt [3]{x^2y^2z^2}\end{array}\right\|\ \bigodot$](http://latex.artofproblemsolving.com/5/3/2/5323a43e03c66def6dec8836a4a5bea505c33e2e.png)
P6. Prove that
there are two equivalent inequalities 
Proof.
![$\frac 1{s^2}\cdot\left[\left(r_a+r_b+r_c\right)^2-2\left(r_ar_b+r_br_c+r_cr_a\right)\right]+\frac {2r}{R}\ge 2\iff$](http://latex.artofproblemsolving.com/7/b/4/7b453ef82691ba31da4ab0260353492ff63cd506.png)
![$\frac 1{s^2}\cdot\left[(4R+r)^2-2s^2\right]+\frac {2r}{R}\ge 2\iff$](http://latex.artofproblemsolving.com/b/1/4/b14e61ce06aba270de0525386e50340fd806c710.png)
Now I"ll give a short geometrical proof of the second inequality
If 
is the orthocenter and 
is the Gergonne's point, then is well known that![$H\Gamma^2 =4R^2\cdot\left[1-\frac {2s^2(2R-r)}{R(4R+r)^2}\right]\ge 0\implies$](http://latex.artofproblemsolving.com/d/9/5/d95d3f1e1fda511bb368da7b8e57f2b365a83c62.png)
i.e. P7.Let
with incircle 
what is tangent to in 
the -excircle 
and midpoint 
of the -altitude ![$[AS]$](http://latex.artofproblemsolving.com/8/1/f/81f842d87b22553e4b674bb2bf46a080da785009.png)
where 
Prove that 
Proof 1. Let
Thus, 
what is true.Proof 2. Let
and 
Hence 
Consider the homothety with the center what transforms incirclein
-excircle, i.e. 
and 
are collinear. 
are homothetic and 
is the midpoint of ![$[D_1L_1]\ ,$](http://latex.artofproblemsolving.com/6/3/4/6342f75b13f0e4df8a16756ee94c429de6fabeeb.png)
then 
pass by the midpoint of ![$[AS]\ .$](http://latex.artofproblemsolving.com/1/c/7/1c7d89ce0de186aaed8a6383751ed9d1665d7d7d.png)
P8 (Sergey Yurin, Kvant). Let
with 
and 
. Consider 
for which 
. Find 
.• • • • • • • • • • • •
. Therefore, 
.
so that the line 
is
obtain that 
. Therefore, 
.
so that 
and 
obtain that 
and 
obtain that the 
is 
, i.e. 
be an equilateral triangle so that 
separates 
from which results 
and
. Thus, 
. If denote 
, then 
, i.e. 
, where 
. Thus, 
and 
is equilateral triangle because 
and 
. Therefore, 
. Obtain that
is 
is 
, i.e. 
so that 
for which 
. Prove easily that 
is 
, 
and 
is a parallelogram because 
and 
. Therefore, 
is an 
, i.e. 
and 
.
. Observe : that 
is 
; 
is equilateral,
and 
; 
is 
-
; 
is 
. Thus, 
and 
is 
.
. Apply the theorem of Sinus 
![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Construct the point 
so that 
and the point 
so that
. Observe that 
from where obtain that 
and 
, i.e. 
.
. Apply the Sinus' theorem in the mentioned triangles:
P9 (J. C. Miranda). Let
with 
Denote 
so that 
and the midpoint of ![$[AM]\ .$](http://latex.artofproblemsolving.com/e/6/4/e642a640cf5a59113546c863aa4d0f99da789766.png)
Ascertain the length of the segment ![$[BP]$](http://latex.artofproblemsolving.com/c/d/a/cda0272ae887e314ff58215fbf55333fa558fefd.png)
(standard notations).Proof. Observe that
is 
-isosceles, 
and 
Apply the theorem of median in 
where denote 
Therefore,
![$4\cdot BP^2=2\cdot\left[BA^2+BM^2\right]-AM^2\ \stackrel{(*)}{\iff}\ 4x^2=$](http://latex.artofproblemsolving.com/7/b/6/7b63be40913199fdcfad8dec1459176717447740.png)
![$2\left[c^2+(a-b)^2\right]-4b^2\sin^2\frac C2=$](http://latex.artofproblemsolving.com/7/1/6/716f077ea9176b2a22de849c55663574095beb39.png)
![$4ax^2=2ac^2+2a(a-b)^2-b\left[c^2-(a-b)^2\right]\iff$](http://latex.artofproblemsolving.com/7/5/b/75b8ef85c81953bcf1a0ecc33b3f3f94b9e90ae3.png)
Particular case
and 
P10 (Kadir Altintas - Turkey). Let
be a rectangle with the center 
Denote 
so that 
and 
Denotethe incircles
and 
of the triangles 
and 
respectively. Prove that 
Proof. Let the midpoint
of ![$[CD]$](http://latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
and 
Therefore, 
and 
On other hand, 
Observe that 
P11 (Conc.Saraghin 2014). Let
the feet 
of the 
-altitude and the -altitude 
the feet 
of the -bisector and the -bisector. Prove that 
Proof.
the line 
is antiparallel direction to 
is cyclic 
P12 (Queensland Australia). Solve the following system
Proof 1. Let
Prove easily 
and 
i.e. 
Thus
P13. Let the functions
where 
and 
where 
Let the real roots
of the equation 
and the real roots 
of the equation 
Prove that 
Proof.
![$(m-p)\underline{[(m-r)-(p-n)]}+$](http://latex.artofproblemsolving.com/1/4/d/14df002383e3cec1494d7aae41a49808dc008315.png)
![$(n-r)\underline{[(n-p)-(r-m)]}=$](http://latex.artofproblemsolving.com/5/7/3/573dfebc40528796e98ac2a4f57f4f7fe2c30999.png)
![$[(m+n)-(p+r)][(m-p)+(n-r)]=$](http://latex.artofproblemsolving.com/6/5/7/657e355bd2df0c03df955f518a83e697894c61e0.png)
![$[(m+n)-(p+r)]^2=(a-c)^2\ .$](http://latex.artofproblemsolving.com/5/9/e/59e6b5a941b052cf7d70c7b2f8dbfbfddb45a53c.png)
In conclusion, P14. Prove that for any
so that 
there is the inequality 
Proof.
The real function 
is convex and
Hence from the Jensen's inequality obtain that 
![$ f\left[\sum\limits_{cyc}\frac{a}{3}\cdot (a+2b) \right]=\sqrt{3}\ .$](http://latex.artofproblemsolving.com/3/b/7/3b770c37d213329135754751496816b35353280b.png)
P15. Let an acute
with the orthocenter 
Let 
be the bisectors of 
respectively , where 
Denotethe projections
of the point 
on 
respectively and the projections 
of the point 
on 
respectively. Prove that 
Proof. Let
and 
Define the projections 
of on respectively and the projections 
of on respectively. Thus, 
and 
are rectangles. Hence 
In conclusion, 
and 
P16. Let
be an -right triangle with the altitude where 
Denote the incircles 
of 
respectively and 
Prove that 
Proof 1. Suppose w.l.o.g. that
Let 
and 
Thus, 
Thus, 
i.e. 
Proof 2.
is the orthocenter of 
Remark. Suppose w.l.o.g.
and denote 
and 
Prove easily that 
and 
In conclusion, 
and 
where Thus, 
P17. Let
with the circumcircle 
and Let 
and the points 
so that 
Prove that 
Proof. If
are midpoints of ![$[AB]\ ,$](http://latex.artofproblemsolving.com/b/a/3/ba35a1c584b9f3786aaed51acf1cbc01ce2968a4.png)
![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
respectively, then 
Hence 
Remark. I"ll prove the well-known relation
Indeed, 
P18. Let
be an orthodiagonal and cyclical quadrilateral with the circumcircle 
Let 
the midpoints 
of ![$[BC]\ ,$](http://latex.artofproblemsolving.com/6/5/6/6563ca77d31eb676ef646fa9c37ea181b812d864.png)
![$[CD]\ ,$](http://latex.artofproblemsolving.com/5/f/7/5f745def914e8ed03d2015c00e2d98fd458adcab.png)
![$[DA]$](http://latex.artofproblemsolving.com/7/b/e/7be1ee9448ef63ac6f1f6b8dd6982c30bad9bb31.png)
respectively and the projections 
of on respectively. Then 
is a rectangle andthe points
belong to the circle with the diameter ![$[MP]$](http://latex.artofproblemsolving.com/4/8/2/4821ed42ce01e14fb61be739cc547c042e1a9005.png)
and its center in the midpoint of ![$[OE]$](http://latex.artofproblemsolving.com/2/a/0/2a082802ed9a33ba2b0993de60b65d137ee48461.png)
because the quadrilaterals 
and 
are parallelograms.P19. Prove that
at least one from its medians has the length less than or equally to 
where 
Proof 1. Suppose w.l.o.g.
i.e. 
and 
i.e. 
I"ll use the well-known identities
Thus, 
Hence
![$3\left[2\left(b^2+c^2\right)-a^2\right]\le \left[a+(b+c)^2\right]\iff$](http://latex.artofproblemsolving.com/2/3/c/23cf4cbd954f783589a3e59aa11a67239706c871.png)
Proof 2. Let
be the smallest and the greatest medians of 
Prove easily that 
Prove similarly that
Proof 3. Prove easily that
Hence
In conclusion, 
Prove similarly if 
, then
P20. Let
with the orthocenter 
Prove the chain of the equivalencies 
is -right-angled triangle 
(standard notations).P21 (Adil Abdullayev). Prove that
.Proof. I"ll use only the identity
and the remarkable inequality 
Thus, 
Hence 
From
obtain that the inequality 
is true.In conclusion,
P22 (Miguel Ochoa SANCHEZ). Click => here.
Proof.
where 
is the length of the -altitude. From the theorem of Cathetus 
obtain that 
Apply the Cristea's theorem
From the relations 
obtain the required identity, i.e. 
P23 (Kadir Altintas, Turkey). Let
with incircle 
midpoint of -Gergonne's cevian 
where 
and the midpoint of ![$[AT]\ .$](http://latex.artofproblemsolving.com/9/7/2/97274983169ba36eadaa26e420576baf0974199b.png)
Prove 
Proof (Emil Stoyanov). Let
and 
where 
is the Nagel's cevian. Thus, 
are the midpoints of segments 
Hence 
Nice!This post has been edited 313 times. Last edited by Virgil Nicula, Apr 19, 2018, 8:33 AM
August 2018
January 2018