Difference between revisions of "2018 IMO Problems/Problem 6"
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Quadrilateral <math>AX_0CF</math> is cyclic <math>\implies</math> <cmath>\angle X_0AB = \frac {1}{2}\overset{\Large\frown} {X_0CE} = \frac {1}{2} (360^\circ -\overset{\Large\frown} {X_0AE}) = 180^\circ - \angle X_0CE = \angle X_0CD.</cmath> | Quadrilateral <math>AX_0CF</math> is cyclic <math>\implies</math> <cmath>\angle X_0AB = \frac {1}{2}\overset{\Large\frown} {X_0CE} = \frac {1}{2} (360^\circ -\overset{\Large\frown} {X_0AE}) = 180^\circ - \angle X_0CE = \angle X_0CD.</cmath> | ||
− | + | Similarly, quadrangle <math>DX_0BF</math> is cyclic <math>\implies \angle X_0BC = \angle X_0DA</math>. | |
This means that point <math>X_0</math> coincides with the point <math>X</math>. | This means that point <math>X_0</math> coincides with the point <math>X</math>. | ||
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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> | ||
− | [[File:2018 IMO 6a.png| | + | [[File:2018 IMO 6a.png|430px|right]] |
+ | [[File:2018 IMO 6bb.png|430px|right]] | ||
<i><b>Common case </b></i> | <i><b>Common case </b></i> | ||
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Denote by <math>X_0</math> the point of intersection of the circles <math>BDF</math> and <math>ACE.</math> | Denote by <math>X_0</math> the point of intersection of the circles <math>BDF</math> and <math>ACE.</math> | ||
− | Quadrangle <math>BX_0DF</math> is cyclic | + | Quadrangle <math>BX_0DF</math> is cyclic <math>\implies \angle X_0BC = \angle X_0DA</math> (see Special case). |
− | + | Similarly, quadrangle <math>AX_0CE</math> is cyclic <math>\implies \angle X_0AB = \angle X_0CD = \alpha.</math> | |
+ | |||
The required point <math>X = X_0</math> is constructed. | The required point <math>X = X_0</math> is constructed. | ||
+ | |||
+ | Denote by <math>Y</math> the point of intersection of circles <math>BDF</math> and <math>ACF.</math> | ||
+ | |||
+ | Quadrangle <math>BYDF</math> is cyclic <math>\implies \angle CBY = \angle ADY.</math> | ||
+ | |||
+ | Quadrangle <math>AYCD</math> is cyclic <math>\implies \angle YAD = \angle BCY.</math> | ||
+ | |||
+ | The triangles <math>\triangle YAD \sim \triangle YCD</math> by two angles, so <cmath>frac {BC}{AD} = \frac {CY}{AY} = \frac {BY} {DY}.</cmath> |
Revision as of 07:04, 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,
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 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