Difference between revisions of "2018 IMO Problems/Problem 6"
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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> | ||
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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> | ||
[[File:2018 IMO 6 Claim 3.png|370px|right]] | [[File:2018 IMO 6 Claim 3.png|370px|right]] | ||
[[File:2018 IMO 6a.png|430px|right]] | [[File:2018 IMO 6a.png|430px|right]] | ||
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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> | ||
Revision as of 12:03, 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