164. Some metrical problems (5) in a circle.
by Virgil Nicula, Oct 21, 2010, 3:13 PM
PP1. Prove that in any
with circumcircle
exists the equivalence
(M. Chirita). Generalization.
Proof. Since
obtain

.
Particular cases for
. Denote
for which
and
for which
.
.
Proof.
. From
obtain


.
PP2. Let
be a triangle with circum radius
, inradius
and semiperimeter
and let
be a real number.
Prove that exists the relation
.
Solve the equation
, where
(Virgil Nicula).
Proof. Denote
. Observe that 


because
and
. For the second point of the conclusion observe that for
the above proved identity becomes

. Thus, our equation
, where 
is equivalently with the equation
, i.e.
.
PP3. For the cyclical convex quadrilateral with the circumcircle
denote
and
. Prove that :
.
.
, where the powers of
,
w.r.t.
is
,
respectively.
Proof. Prove easily all !
PP4. Let
be a triangle with the incircle
. Denote
. Consider two points
,
. Prove that :
is tangent to
.
is tangent to
, where
and
is the length of the circumradius (Toshio Seimyia).
Proof. Denote
. From
and
obtain
.
On other hand


, i.e. the relation
.
Otherwise. From
obtain


, i.e. the relation
.
![$2\cdot [MDN]=(bc-xy)\cdot\sin A-(s-b)(c-x)\cdot\sin B-(s-c)(b-y)\cdot\sin C=$](//latex.artofproblemsolving.com/d/d/2/dd2a539dd126e380da0cdc89e8a79ffe62e040ad.png)

. I used the relation
. Therefore,

.
In conclusion,
, i.e. the ratio
is constant.
Remarks.
The relation
can write equivalently
. Observe that
is maximum

is maximum
is maximum
the sum
is maximum from the relation
he sum
is
minimum. From the relation
results that the product
is constant. Thus, the area
is maximum
, i.e.
.
If the triangle
is equilateral, then the relation
becomes
. For
,
obtain
(from the Van Aubel's relation)
, i.e.
what means that the point
belongs to the
-middle line.
PP5. Let
be a triangle
with circumcircle
and incircle
. Consider an inner point
of given triangle and three circles with same radius 
which pass through
and each from them is tangent only to two sides of
. Prove that
and
. (Aram Yagubyants, Rostov, Rusia).
Proof. Denote : the circle
which is tangent to
,
, the circle
which is tangent to
,
and the circle
which is
tangent to
,
. Since
obtain that
. Observe that

. Thus,
, i.e.
. Since
is incenter of
obtain
, i.e.
.



Proof. Since









Particular cases for






Quote:
Generalization. Prove that in any
with circumcircle
and incircle 
exists the equivalence
.
Remarks.
with left equality
.
with left equality
.



exists the equivalence

Remarks.








Proof.











PP2. Let










Proof. Denote



![$R\left[\cos\left(A-\frac x2\right)+\cos\left(B-\frac x2\right)+\cos\left(C-\frac x2\right)-\cos\frac {3x}{2}\right]=$](http://latex.artofproblemsolving.com/a/d/7/ad7d43d75fd0b270b8667451cfbf68f5219ff894.png)
![$R\left[(\cos A+\cos B+\cos C)\cdot \cos\frac x2+(\sin A+\sin B+\sin C)\cdot \sin \frac x2-\cos\frac {3x}{2}\right]=$](http://latex.artofproblemsolving.com/a/b/2/ab2736e770f5e778025398a5c6233e16c73266b0.png)











is equivalently with the equation


PP3. For the cyclical convex quadrilateral with the circumcircle















Proof. Prove easily all !
PP4. Let












![$4R\cdot [MDN]=(s-a)\cdot BM\cdot CN$](http://latex.artofproblemsolving.com/6/7/d/67d9abed7b96c8f69a2d66162a3af91394f39c10.png)


Proof. Denote


![$[MAN]+[AEIF]=$](http://latex.artofproblemsolving.com/c/7/d/c7d7f8eae8d5a26d22df5cbe7f025274eea7582e.png)
![$2\left([AIM]+[AIN]\right)$](http://latex.artofproblemsolving.com/8/1/7/817c960c7ed03bb7a8e7684284714f5b63c15c50.png)





On other hand













Otherwise. From











![$\blacktriangleright\ \ [MDN]=S-[MAN]-[BDM]-CDN]$](http://latex.artofproblemsolving.com/d/7/6/d766c062240cdee83142b8207dd1608c88920d97.png)

![$2\cdot [MDN]=(bc-xy)\cdot\sin A-(s-b)(c-x)\cdot\sin B-(s-c)(b-y)\cdot\sin C=$](http://latex.artofproblemsolving.com/d/d/2/dd2a539dd126e380da0cdc89e8a79ffe62e040ad.png)






![$4R\cdot[MDN]=xb(s-b)+yc(s-c)-xya=$](http://latex.artofproblemsolving.com/4/6/2/462f6981b26d15436d14e57392d876b7bb2e266f.png)
![$xb(s-b)+yc(s-c)-xy[(s-b)+(s-c)]=$](http://latex.artofproblemsolving.com/f/0/2/f023686cb1a7be2c91e3ef5dfa2f492525368052.png)

![$(c-x)(b-y)\left[(s-b)\cdot\frac {x}{c-x}+(s-c)\cdot\frac {y}{b-y}\right]=$](http://latex.artofproblemsolving.com/6/1/5/6152c293cd9dbc5ccec4f3ac82fe75644abb97d7.png)


In conclusion,
![$4R\cdot [MDN]=BM\cdot (s-a)$](http://latex.artofproblemsolving.com/2/3/5/235f91726e428f1aae2388bc943590442d7d85fd.png)
![$\frac {[MDN]}{BM\cdot CN}=\frac {s-a}{4R}$](http://latex.artofproblemsolving.com/e/d/b/edb93e2fcb9d46caaed3b8ec12e43a7d72f0f315.png)
Remarks.




![$[MAN]$](http://latex.artofproblemsolving.com/9/d/4/9d4025159fbb1d28f2be8dead001aedc17d9e422.png)


is maximum








minimum. From the relation


![$[MDN]$](http://latex.artofproblemsolving.com/a/3/0/a300a6a46e8633012165240dd40f5a006a6ef541.png)









(from the Van Aubel's relation)




PP5. Let






which pass through




Proof. Denote : the circle







tangent to

















This post has been edited 82 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:38 PM