Difference between revisions of "2018 IMO Problems/Problem 6"
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[[File:2018 IMO 6c.png|430px|right]] | [[File:2018 IMO 6c.png|430px|right]] | ||
[[File:2018 IMO 6d.png|430px|right]] | [[File:2018 IMO 6d.png|430px|right]] | ||
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<i><b>Common case </b></i> | <i><b>Common case </b></i> | ||
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In all other cases, the equality of the sines follows <math>\psi = 180° – \varphi \implies \varphi + \psi = 180°.</math> | In all other cases, the equality of the sines follows <math>\psi = 180° – \varphi \implies \varphi + \psi = 180°.</math> | ||
+ | [[File:2018 IMO 6 Claim 3.png|370px|right]] | ||
+ | [[File:2018 IMO 6a.png|430px|right]] | ||
<i><b>Claim 1</b></i> Let <math>A, C,</math> and <math>E</math> be arbitrary points on a circle <math>\omega, l</math> be the perpendicular bisector to the segment <math>AC.</math> Then the straight lines <math>AE</math> and <math>CE</math> intersect <math>l</math> at the points <math>B</math> and <math>D,</math> symmetric with respect to <math>\omega.</math> | <i><b>Claim 1</b></i> Let <math>A, C,</math> and <math>E</math> be arbitrary points on a circle <math>\omega, l</math> be the perpendicular bisector to the segment <math>AC.</math> Then the straight lines <math>AE</math> and <math>CE</math> intersect <math>l</math> at the points <math>B</math> and <math>D,</math> symmetric with respect to <math>\omega.</math> | ||
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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> | ||
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+ | <i><b>Lemma</b></i> The opposite sides of the quadrilateral <math>ABCD</math> intersect at points <math>E</math> and <math>F</math> (<math>E</math> lies on <math>AB</math>). The circle <math>\omega</math> centered at the point <math>O</math> contains the ends of the diagonal <math>AC.</math> The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega</math> (in other words, the inversion with respect to <math>\omega</math> maps <math>B</math> into <math>D).</math> Then the circles <math>ACE</math> and <math>ACF</math> are symmetric with respect to <math>\omega.</math> | ||
+ | <i><b>Proof</b></i> | ||
+ | We will prove that the point <math>G,</math> symmetric to the point <math>E</math> with respect to <math>\omega,</math> belongs to the circle <math>ACF.</math> For this, we will prove the equality <math>\angle AGC = \angle AFC.</math> | ||
+ | |||
+ | A circle <math>BDE</math> containing points <math>B</math> and <math>D</math> symmetric with respect to <math>\omega,</math> is orthogonal to <math>\omega</math> (Claim 2) and maps into itself under inversion with respect to the circle <math>\omega.</math> Hence, the point <math>E</math> under this inversion passes to some point <math>G,</math> of the same circle <math>BDE.</math> | ||
+ | |||
+ | A straight line <math>ABE</math> containing the point <math>A</math> of the circle <math>\omega,</math> under inversion with respect to <math>\omega,</math> maps into the circle <math>OADG.</math> Hence, the inscribed angles of this circle are equal <math>\angle ADB = \angle AGE.</math> | ||
+ | <math>\angle OCE = \angle CGE (CE</math> maps into <math>CG)</math> and <math>\angle OCE = \angle BCD (BC</math> maps into <math>DC).</math> | ||
+ | Consequently, the angles <math>\angle AFC = \angle ADB – \angle FBD = \angle AGE - </math>\angle CGE = <math>\angle AGC.</math> | ||
+ | These angles subtend the <math>\overset{\Large\frown} {AC}</math> of the <math>ACF</math> circle, that is, the point <math>G,</math> symmetric to the point <math>E</math> with respect to <math>\omega,</math> belongs to the circle <math>ACF.</math> |
Revision as of 13:01, 19 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
Similarly, quadrangle is cyclic
.
This means that point coincides with the point
.
of
of
The sum (Claim 3)
Similarly,
Common case
Denote by the intersection point of the perpendicular bisector of
and
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).
Denote by the point of intersection of the circles
and
Quadrangle
is cyclic
(see Special case).
Similarly, quadrangle
is cyclic
The required point is constructed.
Denote by the point of intersection of circles
and
Quadrangle is cyclic
Quadrangle is cyclic
The triangles by two angles, so
The points and
are symmetric with respect to the circle
, since they lie on the intersection of the circles
and
symmetric with respect to
and the circle
orthogonal to
The point is symmetric to itself, the point
is symmetric to
with respect to
Usung
and the equality
we get
The point
is symmetric to itself, the point
is symmetric to
with respect to
The point
is symmetric to
and the point
is symmetric to
with respect to
hence
Denote
By the law of sines for we obtain
By the law of sines for we obtain
We make transformation and get
If
then
Given that
This is a special case.
In all other cases, the equality of the sines follows
Claim 1 Let and
be arbitrary points on a circle
be the perpendicular bisector 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
Lemma The opposite sides of the quadrilateral intersect at points
and
(
lies on
). The circle
centered at the point
contains the ends of the diagonal
The points
and
are symmetric with respect to the circle
(in other words, the inversion with respect to
maps
into
Then the circles
and
are symmetric with respect to
Proof
We will prove that the point
symmetric to the point
with respect to
belongs to the circle
For this, we will prove the equality
A circle containing points
and
symmetric with respect to
is orthogonal to
(Claim 2) and maps into itself under inversion with respect to the circle
Hence, the point
under this inversion passes to some point
of the same circle
A straight line containing the point
of the circle
under inversion with respect to
maps into the circle
Hence, the inscribed angles of this circle are equal
maps into
and
maps into
Consequently, the angles
\angle CGE =
These angles subtend the
of the
circle, that is, the point
symmetric to the point
with respect to
belongs to the circle