Difference between revisions of "2019 IMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | [[File:2019 6 s1.png| | + | [[File:2019 6 s1.png|450px|right]] |
+ | [[File:2019 6 s2.png|450px|right]] | ||
<i><b>Step 1</b></i> | <i><b>Step 1</b></i> | ||
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<i><b>Step 2</b></i> | <i><b>Step 2</b></i> | ||
+ | |||
+ | We find a simplified way to define the point <math>Q.</math> | ||
+ | |||
+ | We define <math>\angle BAC = 2 \alpha \implies \angle AFE = \angle AEF = 90^\circ – \alpha \implies</math> | ||
+ | <math>\angle BFE = \angle CEF = 180^\circ – (90^\circ – \alpha) = 90^\circ + \alpha = \angle BIC</math> | ||
+ | <math>(AI, BI,</math> and <math>CI</math> are bisectrices). | ||
+ | We use the Tangent-Chord Theorem and get <math>\angle EPF = \angle AEF = 90^\circ – \alpha.</math> | ||
+ | |||
+ | <math>\angle BQC = \angle BQP + \angle PQC = \angle BFP + \angle CEP =</math> | ||
+ | <math>=\angle BFE – \angle EFP + \angle CEF – \angle FEP =</math> | ||
+ | <math>= 90^\circ + \alpha + 90^\circ + \alpha – (90^\circ + \alpha) = </math> | ||
+ | <math>90^\circ + \alpha = \angle BIC \implies</math> | ||
+ | |||
+ | points <math>Q, B, I,</math> and <math>C</math> are concyclic. |
Revision as of 13:03, 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.