200. Some nice applications of the inversion.
by Virgil Nicula, Dec 28, 2010, 4:30 PM
Proposed problem 1. Let 
be a 
-right triangle with 
. Denote the midpoint 
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and for a mobile point 
denote the othogonal projections
, 
of the point 
on 
, 
respectively. Is given a line 
so that 
and 
. Denote 
and the projection 
of the point on 
.
1) Prove that
are collinearly.
2) Find the place of the point for which the area of 
is maximum.
3) Prove that for any point
the line 
pass through a fixed point.
Proof.
Observe that 
is cyclically (the circle 
with diameter 
). Hence 
and 
.
Since
(by symmetry) obtain 
. Since 
is cyclically obtain
. From three upper relations results 
. But 
.
Remark. Denote
. The point is the orthocenter of the triangle 
and 
. The line 
is image
by the inversion of the circle with the pole 
and the power 
because 
.
Since 
belongs to the circle with the diameter ![$ [AB]$](//latex.artofproblemsolving.com/d/7/a/d7a8027c238eec9cf67de0f7ec6cb1df4df49a61.png)
, then the area ![$ [ABH]$](//latex.artofproblemsolving.com/1/b/f/1bf71adb3e4982c95be2c9acb77aa2aed9721d5a.png)
is maximum iff
the distance of the point to the diameter is maximum, i.e. 
what means 
.
The point belongs to the circumcircle (fixed) of the triangle 
. Since 
obtain that the ray
is the bisector of the angle 
. Thus, the line 
passes through the (fixed) point 
which is the reflection of the point
w.r.t. the line , i.e. the quadrilateral 
is a square.
Proposed problem 2 (from Mihai Miculita). Let
, 
, 
and 
be four circles with common point so that the pairs
of circles and , and are each exterior tangent. Denote , 
, 
and the second intersection point of
circles and , and , and , and respectively. Prove that 
.
Proof. Apply the inversion with pole pole and the power 
, i.e. 
and the circles , , ,
become four lines 
, 
, 
, 
so that 
and 
. Thus 
is a parallelogram. Using the relations
a.s.o. the relation 
becomes 
, what is truly because 
and 
.
Proposed problem 3. Let
be a fixed line and let be a fixed point so that 
. Let 
be two points
so that
(constant). Prove that the circumscircle of the triangle 
is allway tangent to a fixed circle.
Proof. Denote the projection of the point on the line and 
. The circle with diameter ![$ [FD]$](//latex.artofproblemsolving.com/e/1/7/e1754659141217748e234606561a9761ca31d969.png)
meets again the lines 
, 
in the points 
, 
respectively. Observe that ![$ [XY] = k\cdot\sin\phi$](//latex.artofproblemsolving.com/6/d/8/6d8c597bac0ea77670f1c2069c914d98fc517ec1.png)
(constant). Thus, the geometrical locus of the midpoint of the segment ![$ [XY]$](//latex.artofproblemsolving.com/2/7/1/271f927ba3da0d6a7fd173ceb8e4817046c5c32b.png)
is a fixed circle 
which is concentrically with the circle and the mobile line 
is allway tangent to circle . The inversion 
with the center and the power 
takes the line into the circumcircle 
of 
because 
, i.e 
. Thus, the inversion of the circle is a fixed circle which is tangent to the any circle because any line is tangent to the circle .
A similar problem Given are a circle
and a fixed point 
. For two mobile points 
so that 
(constant) define
the second intersections , of the circle with 
, respectively. Prove that the circumcircle 
is allway tangent to a fixed circle.
Proposed problem 4.The incircle of a non-isosceles
has center 
and touches 
, 
and 
in 
, 
and
respectively. Denote 
, 
and 
. Prove that 
.
Proof. Denote
. From the well-known relations 
and 
results 
and
, i.e. 
is the image of the circumcircle the diameter ![$[IP]$](//latex.artofproblemsolving.com/e/a/9/ea973f41f63c92568ec369e5487bc59ac5800094.png)
through the inversion with the pole and the power 
. Therefore, .
be a 
-right triangle with 
. Denote the midpoint 
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and for a mobile point 
denote the othogonal projections
, 
of the point 
on 
, 
respectively. Is given a line 
so that 
and 
. Denote 
and the projection 
of the point on 
.1) Prove that
are collinearly.2) Find the place of the point
for which the area of 
is maximum.3) Prove that for any point
the line 
pass through a fixed point.Proof.
Observe that 
is cyclically (the circle 
with diameter 
). Hence 
and 
.Since
(by symmetry) obtain 
. Since 
is cyclically obtain
. From three upper relations results 
. But 
.Remark. Denote
. The point is the orthocenter of the triangle 
and 
. The line 
is imageby the inversion of the circle
with the pole 
and the power 
because 
.
Since 
belongs to the circle with the diameter ![$ [AB]$](http://latex.artofproblemsolving.com/d/7/a/d7a8027c238eec9cf67de0f7ec6cb1df4df49a61.png)
, then the area ![$ [ABH]$](http://latex.artofproblemsolving.com/1/b/f/1bf71adb3e4982c95be2c9acb77aa2aed9721d5a.png)
is maximum iffthe distance of the point
to the diameter is maximum, i.e. 
what means 
.
The point belongs to the circumcircle (fixed) of the triangle 
. Since 
obtain that the ray
is the bisector of the angle 
. Thus, the line 
passes through the (fixed) point 
which is the reflection of the point
w.r.t. the line , i.e. the quadrilateral 
is a square.Proposed problem 2 (from Mihai Miculita). Let
, 
, 
and 
be four circles with common point so that the pairsof circles
and , and are each exterior tangent. Denote , 
, 
and the second intersection point ofcircles
and , and , and , and respectively. Prove that 
.Proof. Apply the inversion with pole pole
and the power 
, i.e. 
and the circles , , , become four lines 
, 
, 
, 
so that 
and 
. Thus 
is a parallelogram. Using the relations
a.s.o. the relation 
becomes 
, what is truly because 
and 
.Proposed problem 3. Let
be a fixed line and let be a fixed point so that 
. Let 
be two pointsso that
(constant). Prove that the circumscircle of the triangle 
is allway tangent to a fixed circle. Proof. Denote the projection
of the point on the line and 
. The circle with diameter ![$ [FD]$](http://latex.artofproblemsolving.com/e/1/7/e1754659141217748e234606561a9761ca31d969.png)
meets again the lines 
, 
in the points 
, 
respectively. Observe that ![$ [XY] = k\cdot\sin\phi$](http://latex.artofproblemsolving.com/6/d/8/6d8c597bac0ea77670f1c2069c914d98fc517ec1.png)
(constant). Thus, the geometrical locus of the midpoint of the segment ![$ [XY]$](http://latex.artofproblemsolving.com/2/7/1/271f927ba3da0d6a7fd173ceb8e4817046c5c32b.png)
is a fixed circle 
which is concentrically with the circle and the mobile line 
is allway tangent to circle . The inversion 
with the center and the power 
takes the line into the circumcircle 
of 
because 
, i.e 
. Thus, the inversion of the circle is a fixed circle which is tangent to the any circle because any line is tangent to the circle .A similar problem Given are a circle
and a fixed point 
. For two mobile points 
so that 
(constant) definethe second intersections
, of the circle with 
, respectively. Prove that the circumcircle 
is allway tangent to a fixed circle. Proposed problem 4.The incircle of a non-isosceles
has center 
and touches 
, 
and 
in 
, 
and
respectively. Denote 
, 
and 
. Prove that 
.Proof. Denote
. From the well-known relations 
and 
results 
and
, i.e. 
is the image of the circumcircle the diameter ![$[IP]$](http://latex.artofproblemsolving.com/e/a/9/ea973f41f63c92568ec369e5487bc59ac5800094.png)
through the inversion with the pole and the power 
. Therefore, .This post has been edited 21 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:25 PM
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