398. Geometry problems (II).
by Virgil Nicula, Sep 9, 2014, 7:11 PM
PP1 (Miguel Ochoa Sanchez). Let an equilateral 
with the circumcircle 
. Consider two circles 
exterior tangent one to other and which are interior tangent to and
together are tangent to
. Denote the length 
of the exterior common tangent for 
and 
and the lengths 
of the tangents from 
to respectively. Prove that 
Proof. Denote
. Apply the Casey's theorem to: 
. Thus,
An easy extension. Let with the circumcircle . Consider two circles exterior tangent one to other and which are interior tangent to and together are tangent to .
Denote the length of the exterior common tangent for and and the lengths of the tangents from to respectively. Prove that exist the relation 
PP2 (J. Orihuela). Let
with diameter ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, its chord ![$[CD]$](//latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
so that 
doesn't separate 
, 
and the midpoint 
of and 
so 
. Find 
Proof 1. Denote
. Since 
is cyclically obtain that 
. Since 
is cyclically obtain that 
, i.e. 
.
Proof 2. Prove easily that
. Therfore, 
, i.e. 
is 
-right-angled. In conclusion, .
PP3 (Ruben Dario Auqui). Let
be a rectangle with 
. Denote the midpoint of . Prove that the circles 
and 
are tangent iff 
.
and in this case calculate the lengths
and 
of the radii for the circles what belong to inside of and are tangent to the given circles and to , 
respectively.
Proof. The circles and are tangent iff ![$BC^2=\left(1+\frac 12\right)^2-\left(\frac 12\right)^2=\left[\left(1+\frac 12\right)+\frac 12\right]\cdot\left[\left(1+\frac 12\right)-\frac 12\right]\iff$](//latex.artofproblemsolving.com/f/4/4/f44ba3c712f1b094c667b188c90f484e8019cd2b.png)
.
Otherwise. The given circles are tangent
is common tangent to given circles 
.
.
.
PP4 (Ruben Dario Auqui). Let with the incircle 
what touches , 
at , respectively. Denote the incircle 
,
the projection
of 
on 
and the point 
such that 
. Prove that 
.
Proof. Denote the midpoint
of ![$[EF]$](//latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
. Since 
is a rectangle obtain that 
and 
. Remark. 
.
PP5. Let an interior
of so that 
and the incenters 
, 
, 
of 
, 
, 
. Prove that 
.
Proof.
so that 
. Let 
so that 
. Thus, 
, where 
is the incentre of . Let 
and apply the Menelaus' teorem to 
, what is symetrically w.r.t. 
. So 
(Sharygin G.O. - 2013).
PP6 (MIguel Ochoa Sanchez). Let with the excentre 
. Denote the midpoint of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the foot 
of the 
-bisector 
,
the midpoint
of ![$[BV]$](//latex.artofproblemsolving.com/4/a/c/4accff37106f4e62208903f62838ea9f7ae7d6fb.png)
and the intersections 
, 
. Prove that 
(an easy extension).
Proof. Apply the Menelaus' theorem to
.
Let
. Thus, the division 
is harmonically, i.e. 
. So 
. In conclusion, 
.
PP7 (Carnot). Let with the circumcircle and the midpoints 
of , ![$[CA]$](//latex.artofproblemsolving.com/4/5/c/45c1acd47628de406680d04c09fe6314c3847acf.png)
, respectively. Prove that 
.
Proof. Apply the Ptolemy's theorem to the cyclical quadrilaterals
, 
and 
. Appear two cases, when is acute or nonacute:
![$\blacktriangleright\ \triangle ABC\ \mathrm{is\ acute}\ :\ \left\{\begin{array}{cccc}
OPAN\ : & R\cdot \frac a2=OP\cdot\frac b2+ON\cdot \frac c2 & \implies & b\cdot OP+c\cdot ON=aR\\\\
OMBP\ : & R\cdot\frac b2=OM\cdot \frac c2+OP\cdot\frac a2 & \implies & c\cdot OM+a\cdot OP=bR\\\\
ONCM\ : & R\cdot\frac c2=ON\cdot \frac a2+OM\cdot\frac b2 & \implies & a\cdot ON+b\cdot OM=bR\\\\
& [OBC]+[OCA]+[OAB]=2S & \implies & a\cdot OM+b\cdot ON+c\cdot OP=2sr\end{array}\right\|$](//latex.artofproblemsolving.com/7/5/b/75bbd513db8259b460efa595a843d0d6ab4a4c89.png)
.
![$\blacktriangleright\ A\ge 90^{\circ}\ :\ \left\{\begin{array}{cccc}
OPAN\ : & R\cdot \frac a2=OP\cdot\frac b2+ON\cdot \frac c2 & \implies & b\cdot OP+c\cdot ON=aR\\\\
OMBP\ : & R\cdot\frac b2=-OM\cdot \frac c2+OP\cdot\frac a2 & \implies & -c\cdot OM+a\cdot OP=bR\\\\
ONCM\ : & R\cdot\frac c2=ON\cdot \frac a2-OM\cdot\frac b2 & \implies & a\cdot ON-b\cdot OM=cR\\\\
& -[OBC]+[OCA]+[OAB]=2S & \implies & -a\cdot OM+b\cdot ON+c\cdot OP=2sr\end{array}\right\|$](//latex.artofproblemsolving.com/9/5/4/9547fbc6e7c4c3a306006e2e4b8bfb454de4bce2.png)
.
PP8 (Ruben Dario). Let an-isosceles for which denote 
and the bisector 
of
, where 
. Prove that 
.
Proof. Denote
, 
and 
so that 
. Apply the theorem of the -bisector in 
. Observe that 
and 
. Therefore,
is cyclical deltoid 
. Otherwise,
apply the Pythagoras' theorem in the
-right 
, where 
. Therefore, 
. In conclusion,
. Particular case. 
.
PP9 (Ruben Dario). Let an-right with 
so that 
. Prove that 
.
Proof 1. Prove easily that
and 
. The area ![$[NMP]=\frac {1+\frac zy\cdot \frac xz\cdot \frac yx}{\left(1+\frac zy\right)\left(1+\frac xz\right)\left(1+\frac yx\right)}\cdot [ABC]=$](//latex.artofproblemsolving.com/8/8/d/88d2424047fd98818e49dfa339f857f457e45a81.png)
![$\boxed{[NMP]=\frac{xyz}{y+z}}$](//latex.artofproblemsolving.com/d/2/3/d23839db1c23a17b5bb4031d52dfd273fe9d2a53.png)
. Apply the identity for any 
. Thus, 
![$4\cdot [NMP]=$](//latex.artofproblemsolving.com/7/9/e/79e6af14da37c37a31c468b3ca56e788f0a61b35.png)
, what is truly.
Proof 2. Denote
, where 
and 
. Observe that 
. Therefore,
.
.
![$\frac {y^2(z-x)(y+x)+z^2(z+x)(y-x)}{yz[(z+x)(y+x)-(z-x)(y-x)]}=$](//latex.artofproblemsolving.com/2/f/0/2f08e71d6209aae0ea13ce51481016b9ce4b3152.png)
![$\frac {y^2[(yz+xz)-x(x+y)]+z^2[(yz+xy)-x(x+z)]}{2xyz(y+z)}\ \stackrel{(1\wedge 2)}{=}$](//latex.artofproblemsolving.com/f/e/d/fedd424026c41632cf8b5f0f86cfd2dfb15ade54.png)
![$\frac {y^2[(yz+xz)-z(y-x)]+z^2[(yz+xy)-y(z-x)]}{2xyz(y+z)}=$](//latex.artofproblemsolving.com/d/d/8/dd82b8172a0e625e3b3b86fa3d07117c94e0dc66.png)
An easy extension (own). Let with so that and 
. Prove that 
.
Lemma.
, where 
. Prove that ![$\left\{\begin{array}{ccc}
[DEF] & = & f(x,y,z)\cdot [ABC]\ \ \mathrm{where}\\\\
f(x,y,z) & = & (1-x)(1-y)(1-z)+xyz\end{array}\right\|\ .$](//latex.artofproblemsolving.com/1/a/d/1ad4788ecf5793d02f900e20b5daac3daae15c17.png)
Proof.
and ![$\frac {[DEF]}{[ABC]}=\frac {[ABC]-[AEF]-[BFD]-[CDE]}{[ABC]}=1-\frac {[AEF]}{[ACB]}-\frac {[BFD]}{BAC]}-\frac {[CDE]}{[CBA]}=$](//latex.artofproblemsolving.com/6/9/5/695389b45f434aa317adca454eb5c335de4268f4.png)
![$\boxed{\frac {[DEF]}{[ABC]}=(1-x)(1-y)(1-z)+xyz}\ (*)$](//latex.artofproblemsolving.com/d/c/0/dc02bb939c7a4f1be6a79f14848fb11ab900d089.png)
.
Application. Let and 
, where . Prove that ![$[DEF]=[MNP]$](//latex.artofproblemsolving.com/d/6/e/d6eb353f9cf20ea6f454e835f0ba95b109e16356.png)
.
Proof. Apply the relation
of the upper lemma using the evident remark 
for any 
.
Lema. Let with 
and the orthocentre . For a line 
, 
denote
and 
so that 
. Prove that 
.
Proof. Denote the midpoint of ![$[BC]\ ,\ BC=a$](//latex.artofproblemsolving.com/6/e/c/6ec98a4eba03f0780d01c123f5ee2a05551a32cf.png)
and I"ll apply an well-known relation 
Denote 
, 
and the previous
relation becomes
i.e. 
PP10. Let with 
and the orthocentre . For 
let and so that 
. Prove that 
Proof. Denote w.l.o.g.
separate , and 
Thus, 
and
. Therefore, 
In conclusion,
, what is truly from the previous lemma.
PP11. Let be a quadrilateral with 
Denote the points 
Prove that 
Proof 1. Denote
Thus, 
Proof 2. Denote
Thus, ![$\left\{\begin{array}{ccccccc}
BE\parallel AD & \implies & [ABD]=[AED] & \implies & [DIE]=[AIB] & \implies & ID\cdot IE=IA\cdot IB\\\\
AF\parallel BC & \implies & [BAC]=[BFC] & \implies & [CIF]=[AIB] & \implies & IC\cdot IF=IA
\cdot IB\end{array}\right\|$](//latex.artofproblemsolving.com/0/4/7/0478ca0ec2a66ace2dbc3f7c21b94138a6ef4a7b.png)
I used the notation ![$[XYZ]$](//latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
- the area of the triangle 
.[/color]
PP12. Let with 
, the incircle 
and 
. Prove that 
.
Proof 1. Denote
. Thus, 
the ray 
is the bisector of the angle 
. Prove analogously that the ray 
is the bisector of the angle 
.
Remark. Since the division
is harmonically and using an well-known property obtain that 
. Can easily prove the relation . Let
so that 
and 
. Observe that 
is an equilaterally, 
and 
.
Proof 2 (Mihai Miculiţa). Observe that the ray
is the exterior -bisector and the ray 
is the interior -bisector of 
. Thus, the point is the -excenter of
and belongs to the-exterior bisector of , i.e. the ray is the bisector of 
. Prove analogously that the ray is the bisector of 
.
Remarca. Superba solutie, dle profesor ! Deseori prof. N.N. Mihaileanu (l-am avut profesor la facultate) spunea ca "geometria
este o "fantana cu apa vie " si daca vrei sa mai descoperi ceva in matematica intoarce-te la ea pentru a-ti potoli setea".
PP13 (GMB1 - 2015). Let and the cevian 
of the interior point , where , 
, 
.
Denote: the midpoint of the side ![$[BC]\ ;\ S\in FM\cap AC\ ;$](//latex.artofproblemsolving.com/7/8/6/7864a7834fc5c773981803d4b11aae09410985e8.png)
the point 
so that 
. Prove that 
.
Proof. Apply the Menelaus' theorem to the transversal
. Observe that 
. Therefore, 
because is the cevian triangle of . In conclusion, 
, i.e. .
PP14. Let a square with and the equilateral so that 
and doesn't
separate and . Prove that 
so that 
.
Proof 1. Denote
, 
, the midpoint 
of ![$[DF]$](//latex.artofproblemsolving.com/4/8/7/487608ba746e637d846b20401f23cc2b80336338.png)
and 
. Thus, 
denote 
so that 
.
Proof 2. Denote and . Thus, 
. Hence 
.
PP15 (M. O. Sanchez). Let
with the centroid 
the circumcircle and 
Prove that 
Proof. Observe that
PP16. Let with the incenter 
the circumcircle and 
Prove that 
Proof. Observe that
Extension. Let with the circumcircle and for an interior point with the barycentrical coordinates 
w.r.t. denote 
Prove that 
Proof.
Apply the Stewart's relation to the cevian in and obtain that 
Therefore, 
Particular cases
with the circumcircle 
. Consider two circles 
exterior tangent one to other and which are interior tangent to andtogether are tangent to
. Denote the length 
of the exterior common tangent for 
and 
and the lengths 
of the tangents from 
to respectively. Prove that 
Proof. Denote
. Apply the Casey's theorem to: 
. Thus, An easy extension. Let
with the circumcircle . Consider two circles exterior tangent one to other and which are interior tangent to and together are tangent to .Denote the length
of the exterior common tangent for and and the lengths of the tangents from to respectively. Prove that exist the relation 
PP2 (J. Orihuela). Let
with diameter ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, its chord ![$[CD]$](http://latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
so that 
doesn't separate 
, 
and the midpoint 
of and 
so 
. Find 
Proof 1. Denote
. Since 
is cyclically obtain that 
. Since 
is cyclically obtain that 
, i.e. 
.Proof 2. Prove easily that
. Therfore, 
, i.e. 
is 
-right-angled. In conclusion, .PP3 (Ruben Dario Auqui). Let
be a rectangle with 
. Denote the midpoint of . Prove that the circles 
and 
are tangent iff 
.and in this case calculate the lengths
and 
of the radii for the circles what belong to inside of and are tangent to the given circles and to , 
respectively.Proof. The circles
and are tangent iff ![$BC^2=\left(1+\frac 12\right)^2-\left(\frac 12\right)^2=\left[\left(1+\frac 12\right)+\frac 12\right]\cdot\left[\left(1+\frac 12\right)-\frac 12\right]\iff$](http://latex.artofproblemsolving.com/f/4/4/f44ba3c712f1b094c667b188c90f484e8019cd2b.png)
.Otherwise. The given circles are tangent
is common tangent to given circles 
.
.
.PP4 (Ruben Dario Auqui). Let
with the incircle 
what touches , 
at , respectively. Denote the incircle 
,the projection
of 
on 
and the point 
such that 
. Prove that 
.Proof. Denote the midpoint
of ![$[EF]$](http://latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
. Since 
is a rectangle obtain that 
and 
. Remark. 
.PP5. Let an interior
of so that 
and the incenters 
, 
, 
of 
, 
, 
. Prove that 
.Proof.
so that 
. Let 
so that 
. Thus, 
, where 
is the incentre of . Let 
and apply the Menelaus' teorem to 
, what is symetrically w.r.t. 
. So 
(Sharygin G.O. - 2013). PP6 (MIguel Ochoa Sanchez). Let
with the excentre 
. Denote the midpoint of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the foot 
of the 
-bisector 
,the midpoint
of ![$[BV]$](http://latex.artofproblemsolving.com/4/a/c/4accff37106f4e62208903f62838ea9f7ae7d6fb.png)
and the intersections 
, 
. Prove that 
(an easy extension).Proof. Apply the Menelaus' theorem to
.Let
. Thus, the division 
is harmonically, i.e. 
. So 
. In conclusion, 
.PP7 (Carnot). Let
with the circumcircle and the midpoints 
of , ![$[CA]$](http://latex.artofproblemsolving.com/4/5/c/45c1acd47628de406680d04c09fe6314c3847acf.png)
, respectively. Prove that 
.Proof. Apply the Ptolemy's theorem to the cyclical quadrilaterals
, 
and 
. Appear two cases, when is acute or nonacute:![$\blacktriangleright\ \triangle ABC\ \mathrm{is\ acute}\ :\ \left\{\begin{array}{cccc}
OPAN\ : & R\cdot \frac a2=OP\cdot\frac b2+ON\cdot \frac c2 & \implies & b\cdot OP+c\cdot ON=aR\\\\
OMBP\ : & R\cdot\frac b2=OM\cdot \frac c2+OP\cdot\frac a2 & \implies & c\cdot OM+a\cdot OP=bR\\\\
ONCM\ : & R\cdot\frac c2=ON\cdot \frac a2+OM\cdot\frac b2 & \implies & a\cdot ON+b\cdot OM=bR\\\\
& [OBC]+[OCA]+[OAB]=2S & \implies & a\cdot OM+b\cdot ON+c\cdot OP=2sr\end{array}\right\|$](http://latex.artofproblemsolving.com/7/5/b/75bbd513db8259b460efa595a843d0d6ab4a4c89.png)
.![$\blacktriangleright\ A\ge 90^{\circ}\ :\ \left\{\begin{array}{cccc}
OPAN\ : & R\cdot \frac a2=OP\cdot\frac b2+ON\cdot \frac c2 & \implies & b\cdot OP+c\cdot ON=aR\\\\
OMBP\ : & R\cdot\frac b2=-OM\cdot \frac c2+OP\cdot\frac a2 & \implies & -c\cdot OM+a\cdot OP=bR\\\\
ONCM\ : & R\cdot\frac c2=ON\cdot \frac a2-OM\cdot\frac b2 & \implies & a\cdot ON-b\cdot OM=cR\\\\
& -[OBC]+[OCA]+[OAB]=2S & \implies & -a\cdot OM+b\cdot ON+c\cdot OP=2sr\end{array}\right\|$](http://latex.artofproblemsolving.com/9/5/4/9547fbc6e7c4c3a306006e2e4b8bfb454de4bce2.png)
.PP8 (Ruben Dario). Let an
-isosceles for which denote 
and the bisector 
of
, where 
. Prove that 
.Proof. Denote
, 
and 
so that 
. Apply the theorem of the -bisector in 
. Observe that 
and 
. Therefore,
is cyclical deltoid 
. Otherwise,apply the Pythagoras' theorem in the
-right 
, where 
. Therefore, 
. In conclusion,
. Particular case. 
. PP9 (Ruben Dario). Let an
-right with 
so that 
. Prove that 
.Proof 1. Prove easily that
and 
. The area ![$[NMP]=\frac {1+\frac zy\cdot \frac xz\cdot \frac yx}{\left(1+\frac zy\right)\left(1+\frac xz\right)\left(1+\frac yx\right)}\cdot [ABC]=$](http://latex.artofproblemsolving.com/8/8/d/88d2424047fd98818e49dfa339f857f457e45a81.png)
![$\boxed{[NMP]=\frac{xyz}{y+z}}$](http://latex.artofproblemsolving.com/d/2/3/d23839db1c23a17b5bb4031d52dfd273fe9d2a53.png)
. Apply the identity for any 
. Thus, 
![$4\cdot [NMP]=$](http://latex.artofproblemsolving.com/7/9/e/79e6af14da37c37a31c468b3ca56e788f0a61b35.png)
, what is truly.Proof 2. Denote
, where 
and 
. Observe that 
. Therefore,
.
.
![$\frac {y^2(z-x)(y+x)+z^2(z+x)(y-x)}{yz[(z+x)(y+x)-(z-x)(y-x)]}=$](http://latex.artofproblemsolving.com/2/f/0/2f08e71d6209aae0ea13ce51481016b9ce4b3152.png)
![$\frac {y^2[(yz+xz)-x(x+y)]+z^2[(yz+xy)-x(x+z)]}{2xyz(y+z)}\ \stackrel{(1\wedge 2)}{=}$](http://latex.artofproblemsolving.com/f/e/d/fedd424026c41632cf8b5f0f86cfd2dfb15ade54.png)
![$\frac {y^2[(yz+xz)-z(y-x)]+z^2[(yz+xy)-y(z-x)]}{2xyz(y+z)}=$](http://latex.artofproblemsolving.com/d/d/8/dd82b8172a0e625e3b3b86fa3d07117c94e0dc66.png)
An easy extension (own). Let
with so that and 
. Prove that 
.Lemma.
, where 
. Prove that ![$\left\{\begin{array}{ccc}
[DEF] & = & f(x,y,z)\cdot [ABC]\ \ \mathrm{where}\\\\
f(x,y,z) & = & (1-x)(1-y)(1-z)+xyz\end{array}\right\|\ .$](http://latex.artofproblemsolving.com/1/a/d/1ad4788ecf5793d02f900e20b5daac3daae15c17.png)
Proof.
and ![$\frac {[DEF]}{[ABC]}=\frac {[ABC]-[AEF]-[BFD]-[CDE]}{[ABC]}=1-\frac {[AEF]}{[ACB]}-\frac {[BFD]}{BAC]}-\frac {[CDE]}{[CBA]}=$](http://latex.artofproblemsolving.com/6/9/5/695389b45f434aa317adca454eb5c335de4268f4.png)
![$\boxed{\frac {[DEF]}{[ABC]}=(1-x)(1-y)(1-z)+xyz}\ (*)$](http://latex.artofproblemsolving.com/d/c/0/dc02bb939c7a4f1be6a79f14848fb11ab900d089.png)
.Application. Let
and 
, where . Prove that ![$[DEF]=[MNP]$](http://latex.artofproblemsolving.com/d/6/e/d6eb353f9cf20ea6f454e835f0ba95b109e16356.png)
.Proof. Apply the relation
of the upper lemma using the evident remark 
for any 
.Lema. Let
with 
and the orthocentre . For a line 
, 
denote
and 
so that 
. Prove that 
.Proof. Denote the midpoint
of ![$[BC]\ ,\ BC=a$](http://latex.artofproblemsolving.com/6/e/c/6ec98a4eba03f0780d01c123f5ee2a05551a32cf.png)
and I"ll apply an well-known relation 
Denote 
, 
and the previousrelation becomes
i.e. 
PP10. Let
with 
and the orthocentre . For 
let and so that 
. Prove that 
Proof. Denote w.l.o.g.
separate , and 
Thus, 
and
. Therefore, 
In conclusion,
, what is truly from the previous lemma.PP11. Let
be a quadrilateral with 
Denote the points 
Prove that 
Proof 1. Denote
Thus, 
Proof 2. Denote
Thus, ![$\left\{\begin{array}{ccccccc}
BE\parallel AD & \implies & [ABD]=[AED] & \implies & [DIE]=[AIB] & \implies & ID\cdot IE=IA\cdot IB\\\\
AF\parallel BC & \implies & [BAC]=[BFC] & \implies & [CIF]=[AIB] & \implies & IC\cdot IF=IA
\cdot IB\end{array}\right\|$](http://latex.artofproblemsolving.com/0/4/7/0478ca0ec2a66ace2dbc3f7c21b94138a6ef4a7b.png)
I used the notation ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
- the area of the triangle 
.[/color]PP12. Let
with 
, the incircle 
and 
. Prove that 
.Proof 1. Denote
. Thus, 
the ray 
is the bisector of the angle 
. Prove analogously that the ray 
is the bisector of the angle 
.Remark. Since the division
is harmonically and using an well-known property obtain that 
. Can easily prove the relation . Let
so that 
and 
. Observe that 
is an equilaterally, 
and 
.Proof 2 (Mihai Miculiţa). Observe that the ray
is the exterior -bisector and the ray 
is the interior -bisector of 
. Thus, the point is the -excenter of and belongs to the
-exterior bisector of , i.e. the ray is the bisector of 
. Prove analogously that the ray is the bisector of 
.Remarca. Superba solutie, dle profesor ! Deseori prof. N.N. Mihaileanu (l-am avut profesor la facultate) spunea ca "geometria
este o "fantana cu apa vie " si daca vrei sa mai descoperi ceva in matematica intoarce-te la ea pentru a-ti potoli setea".
PP13 (GMB1 - 2015). Let
and the cevian 
of the interior point , where , 
, 
.Denote: the midpoint
of the side ![$[BC]\ ;\ S\in FM\cap AC\ ;$](http://latex.artofproblemsolving.com/7/8/6/7864a7834fc5c773981803d4b11aae09410985e8.png)
the point 
so that 
. Prove that 
. Proof. Apply the Menelaus' theorem to the transversal
. Observe that 
. Therefore, 
because is the cevian triangle of . In conclusion, 
, i.e. .PP14. Let a square
with and the equilateral so that 
and doesn'tseparate
and . Prove that 
so that 
.Proof 1. Denote
, 
, the midpoint 
of ![$[DF]$](http://latex.artofproblemsolving.com/4/8/7/487608ba746e637d846b20401f23cc2b80336338.png)
and 
. Thus, 
denote 
so that 
.Proof 2. Denote
and . Thus, 
. Hence 
.PP15 (M. O. Sanchez). Let
with the centroid 
the circumcircle and 
Prove that 
Proof. Observe that
PP16. Let
with the incenter 
the circumcircle and 
Prove that 
Proof. Observe that
Extension. Let
with the circumcircle and for an interior point with the barycentrical coordinates 
w.r.t.
denote 
Prove that 
Proof.
Apply the Stewart's relation to the cevian in and obtain that 
Therefore, 
Particular cases
This post has been edited 428 times. Last edited by Virgil Nicula, Feb 11, 2018, 3:05 PM
August 2018
January 2018