Difference between revisions of "2018 IMO Problems/Problem 6"
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[[File:2018 IMO 6bb.png|430px|right]] | [[File:2018 IMO 6bb.png|430px|right]] | ||
[[File:2018 IMO 6c.png|430px|right]] | [[File:2018 IMO 6c.png|430px|right]] | ||
+ | [[File:2018 IMO 6d.png|430px|right]] | ||
<i><b>Common case </b></i> | <i><b>Common case </b></i> | ||
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Quadrangle <math>AYCD</math> is cyclic <math>\implies \angle YAD = \angle BCY.</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> | + | The triangles <math>\triangle YAD \sim \triangle YCD</math> by two angles, so <cmath>\frac {BC}{AD} = \frac {CY}{AY} = \frac {BY} {DY} \hspace{10mm} (1).</cmath> |
The points <math>X</math> and <math>Y</math> are symmetric with respect to the circle <math>\omega</math>, since they lie on the intersection of the circles <math>ACF</math> and <math>ACE</math> symmetric with respect to <math>\omega</math> and the orthogonal <math>\omega</math> circle <math>BDE.</math> | The points <math>X</math> and <math>Y</math> are symmetric with respect to the circle <math>\omega</math>, since they lie on the intersection of the circles <math>ACF</math> and <math>ACE</math> symmetric with respect to <math>\omega</math> and the orthogonal <math>\omega</math> circle <math>BDE.</math> | ||
The point <math>C</math> is symmetric to itself, the point <math>X</math> is symmetric to <math>Y</math> with respect to <math>\omega \implies \frac{CX}{CY} = \frac {R^2}{OC \cdot OY} , \frac {AX}{AY} = \frac {R^2}{OA \cdot OY}.</math> | The point <math>C</math> is symmetric to itself, the point <math>X</math> is symmetric to <math>Y</math> with respect to <math>\omega \implies \frac{CX}{CY} = \frac {R^2}{OC \cdot OY} , \frac {AX}{AY} = \frac {R^2}{OA \cdot OY}.</math> | ||
+ | Usung <math>(1)</math> and the equality <math>OA = OC,</math> we get <cmath>\frac{CY}{AY} = \frac {CX}{AX} = \frac{BC}{AD}.</cmath> | ||
+ | The point <math>C</math> is symmetric to itself, the point <math>B</math> is symmetric to <math>D</math> with respect to <math>\omega \implies</math> | ||
+ | <cmath>\triangle OBC \sim \triangle OCD \implies \frac {OB}{OC} = \frac {BC}{CD} = \frac {OC}{OD},</cmath> | ||
+ | <cmath>\frac {OB}{OD} = \frac {OB}{OC} \cdot \frac {OC}{OD} = \frac{BC^2}{CD^2} = \frac{BC}{CD} \cdot \frac {AB}{AD}.</cmath> | ||
+ | The point <math>B</math> is symmetric to <math>D</math> and the point <math>X</math> is symmetric to <math>Y</math> with respect to <math>\omega,</math> hence | ||
+ | <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> | ||
+ | Denote <math>\angle XAB = \angle XCD = \alpha, \angle BXA = \varphi, \angle DXC = \psi.</math> | ||
+ | |||
+ | By the law of sines for the <math>\triangle ABX,</math> we obtain <math>\frac {AB}{\sin \varphi} = \frac{BX}{\sin \alpha}.</math> |
Revision as of 08:56, 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
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 orthogonal
circle
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 the we obtain