Difference between revisions of "2018 IMO Problems/Problem 6"
(→Solution) |
(→Solution) |
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[[File:2018 IMO 6.png|490px|right]] | [[File:2018 IMO 6.png|490px|right]] | ||
[[File:2018 IMO 6 Claim 3.png|370px|right]] | [[File:2018 IMO 6 Claim 3.png|370px|right]] | ||
+ | [[File:2018 IMO 6a.png|490px|right]] | ||
<i><b>Special case</b></i> | <i><b>Special case</b></i> | ||
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<i><b>Claim 3</b></i> The sum of the arcs between the points of intersection of two perpendicular circles is <math>180^\circ.</math> | <i><b>Claim 3</b></i> The sum of the arcs between the points of intersection of two perpendicular circles is <math>180^\circ.</math> | ||
In the figure they are a blue and red arcs <math>\overset{\Large\frown} {CD}, \alpha + \beta = 180^\circ.</math> | In the figure they are a blue and red arcs <math>\overset{\Large\frown} {CD}, \alpha + \beta = 180^\circ.</math> | ||
+ | |||
+ | <i><b>Common case </b></i> | ||
+ | |||
+ | Denote by <math>O</math> the intersection point of the midpoint perpendicular of the segment <math>AC</math> and the line <math>BD.</math> Let <math>\omega</math> be a circle (red) with center <math>O</math> and radius <math>OA.</math> | ||
+ | |||
+ | The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega</math> <i><b>(Claim 1).</b></i> | ||
+ | |||
+ | The circles <math>BDF</math> and <math>BDE</math> are orthogonal to the circle <math>\omega</math> <i><b>(Claim 2).</b></i> | ||
+ | |||
+ | Circles <math>ACF</math> and <math>ACE</math> are symmetric with respect to the circle <math>\omega</math> <i><b>(Lemma).</b></i> |
Revision as of 14:51, 17 August 2022
A convex quadrilateral satisfies
Point
lies inside
so that
and
Prove that
Solution
Special case
We construct point and prove that
coincides with the point
Let and
Let and
be the intersection points of
and
and
and
respectively.
The points and
are symmetric with respect to the circle
(Claim 1).
The circle
is orthogonal to the circle
(Claim 2).
Let
be the point of intersection of the circles
and
(quadrilateral
is cyclic) and
(quadrangle
is cyclic). This means that
coincides with the point
indicated in the condition.
subtend the arc
of
subtend the arc
of
The sum of these arcs is
(Claim 3)..
Hence, the sum of the arcs is
the sum
Similarly,
Claim 1 Let and
be arbitrary points on a circle
be the middle perpendicular to the segment
Then the straight lines
and
intersect
at the points
and
symmetric with respect to
Claim 2 Let points and
be symmetric with respect to the circle
Then any circle
passing through these points is orthogonal to
Claim 3 The sum of the arcs between the points of intersection of two perpendicular circles is
In the figure they are a blue and red arcs
Common case
Denote by the intersection point of the midpoint perpendicular of the segment
and the line
Let
be a circle (red) with center
and radius
The points and
are symmetric with respect to the circle
(Claim 1).
The circles and
are orthogonal to the circle
(Claim 2).
Circles and
are symmetric with respect to the circle
(Lemma).