Difference between revisions of "2019 IMO Problems/Problem 6"
(→Solution) |
(→Solution) |
||
Line 40: | Line 40: | ||
Points <math>Q, B, I,</math> and <math>C</math> are concyclic. | Points <math>Q, B, I,</math> and <math>C</math> are concyclic. | ||
+ | |||
+ | <i><b>Step 3</b></i> | ||
+ | |||
+ | We perform inversion around <math>\omega.</math> The straight line <math>PST</math> maps onto circle <math>PITS.</math> We denote this circle <math>\Omega.</math> We prove that the midpoint of <math>AD</math> lies on the circle <math>\Omega.</math> | ||
+ | In the diagram, the configuration under study is transformed using inversion with respect to <math>\omega.</math> The images of the points are labeled in the same way as the points themselves. Points D,E,F,P,S,G have saved their position. Vertices A, B and C have moved to the midpoints of the segments EF, FD, and DE, respectively. | ||
+ | Let <math>M</math> be the midpoint <math>AD. ABDC</math> is parallelogram <math>\implies M</math> is midpoint <math>BC.</math> | ||
+ | We define <math>\angle MID = \beta, \angle MDI = \gamma \implies \angle IMA = \angle MID + \angle MDI = \beta + \gamma = \varphi.</math> |
Revision as of 13:35, 29 August 2022
Problem
Let be the incenter of acute triangle
with
. The incircle
of
is tangent to sides
,
, and
at
,
, and
, respectively. The line through
perpendicular to
meets
again at
. Line
meets ω again at
. The circumcircles of triangles
and
meet again at
.
Prove that lines
and
meet on the line through
perpendicular to
.
Solution
Step 1
We find an auxiliary point
Let be the antipode of
on
where
is radius
We define
is cyclic
An inversion with respect
swap
and
is the midpoint
Let meets
again at
We define
Opposite sides of any quadrilateral inscribed in the circle meet on the polar line of the intersection of the diagonals with respect to
and
meet on the line through
perpendicular to
The problem is reduced to proving that
Step 2
We find a simplified way to define the point
We define
and
are bisectrices).
We use the Tangent-Chord Theorem and get
Points and
are concyclic.
Step 3
We perform inversion around The straight line
maps onto circle
We denote this circle
We prove that the midpoint of
lies on the circle
In the diagram, the configuration under study is transformed using inversion with respect to
The images of the points are labeled in the same way as the points themselves. Points D,E,F,P,S,G have saved their position. Vertices A, B and C have moved to the midpoints of the segments EF, FD, and DE, respectively.
Let
be the midpoint
is parallelogram
is midpoint
We define