Difference between revisions of "2018 IMO Problems/Problem 6"
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<cmath>\frac {BX}{DY} = \frac {R^2}{OD \cdot OY} ,\frac {DX}{BY} = \frac{R^2}{OB \cdot OY}.</cmath> | <cmath>\frac {BX}{DY} = \frac {R^2}{OD \cdot OY} ,\frac {DX}{BY} = \frac{R^2}{OB \cdot OY}.</cmath> | ||
<cmath>\frac{BX}{DX} =\frac{DY}{BY} \cdot \frac {OB}{OD} = \frac{AD}{BC} \cdot \frac{BC}{CD} \cdot \frac{AB}{AD} = \frac{AB}{CD}.</cmath> | <cmath>\frac{BX}{DX} =\frac{DY}{BY} \cdot \frac {OB}{OD} = \frac{AD}{BC} \cdot \frac{BC}{CD} \cdot \frac{AB}{AD} = \frac{AB}{CD}.</cmath> | ||
− | [[File:2018 IMO 6 angles.png| | + | [[File:2018 IMO 6 angles.png|370px|right]] |
+ | [[File:2018 IMO 6 Claim 3.png|370px|right]] | ||
+ | [[File:2018 IMO 6a.png|430px|right]] | ||
+ | [[File:2018 IMO 60.png|430px|right]] | ||
Denote <math>\angle XAB = \angle XCD = \alpha, \angle BXA = \varphi, \angle DXC = \psi.</math> | Denote <math>\angle XAB = \angle XCD = \alpha, \angle BXA = \varphi, \angle DXC = \psi.</math> | ||
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<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 2</b></i> Let points <math>B</math> and <math>D</math> be symmetric with respect to the circle <math>\omega.</math> Then any circle <math>\Omega</math> passing through these points is orthogonal to <math>\omega.</math> | <i><b>Claim 2</b></i> Let points <math>B</math> and <math>D</math> be symmetric with respect to the circle <math>\omega.</math> Then any circle <math>\Omega</math> passing through these points is orthogonal to <math>\omega.</math> | ||
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Consequently, the angles <math>\angle AFC = \angle ADB – \angle FBD = \angle AGE - \angle CGE = \angle AGC.</math> | Consequently, the angles <math>\angle AFC = \angle ADB – \angle FBD = \angle AGE - \angle CGE = \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> | 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> | ||
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+ | <i><b>Attention</b></i> The uniqueness of the point <math>X</math> has not been proven. In the figure, the blue and green curves show the locus points with equal pairs of angles for one case. In the general case, one can divide the plane into parts and prove that the point <math>X</math> cannot be outside <math>ABCD.</math> It's not done here. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 09:58, 22 August 2022
A convex quadrilateral satisfies
Point
lies inside
so that
and
Prove that
Solution
Suppose point X is unique. We will construct point and prove that
coincides with the point
Special case
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
and the perpendicular bisector of
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
This means that point coincides with the point
.
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
Hence we get
If then
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
These angles subtend the
of the
circle, that is, the point
symmetric to the point
with respect to
belongs to the circle
Attention The uniqueness of the point has not been proven. In the figure, the blue and green curves show the locus points with equal pairs of angles for one case. In the general case, one can divide the plane into parts and prove that the point
cannot be outside
It's not done here.
vladimir.shelomovskii@gmail.com, vvsss