Difference between revisions of "2018 IMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
[[File:2018 IMO 6.png|490px|right]] | [[File:2018 IMO 6.png|490px|right]] | ||
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<i><b>Special case</b></i> | <i><b>Special case</b></i> | ||
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Let <math>E</math> and <math>F</math> be the intersection points of <math>AB</math> and <math>CD,</math> and <math>BC</math> and <math>DA,</math> respectively. | Let <math>E</math> and <math>F</math> be the intersection points of <math>AB</math> and <math>CD,</math> and <math>BC</math> and <math>DA,</math> respectively. | ||
− | The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega = | + | The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega = ACEF</math> <i><b>(Claim 1).</b></i> |
The circle <math>\Omega = FBD</math> is orthogonal to the circle <math>\omega</math> <i><b>(Claim 2).</b></i> | The circle <math>\Omega = FBD</math> is orthogonal to the circle <math>\omega</math> <i><b>(Claim 2).</b></i> | ||
Let <math>X_0</math> be the point of intersection of the circles <math>\omega</math> and <math>\Omega.</math> | Let <math>X_0</math> be the point of intersection of the circles <math>\omega</math> and <math>\Omega.</math> | ||
− | Quadrilateral <math>AX_0CF</math> is cyclic <math>\implies</math> <cmath>\angle X_0AB = \frac { | + | Quadrilateral <math>AX_0CF</math> is cyclic <math>\implies</math> <cmath>\angle X_0AB = \frac {\overset{\Large\frown} {X_0CE}}{2} = \frac {360^\circ -\overset{\Large\frown} {X_0AFE}}{2} = 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>. | Similarly, quadrangle <math>DX_0BF</math> is cyclic <math>\implies \angle X_0BC = \angle X_0DA</math>. | ||
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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>. | ||
− | <math>\hspace{10mm} \angle FCX = \angle BCX = \frac { | + | <math>\hspace{10mm} \angle FCX = \angle BCX = \frac {\overset{\Large\frown} {XAF}}{2}</math> of <math>\omega.</math> |
− | <math>\hspace{10mm} \angle CBX = \angle XDA = \frac { | + | <math>\hspace{10mm} \angle CBX = \angle XDA = \frac {\overset{\Large\frown} {XBF}}{2}</math> of <math>\Omega.</math> |
The sum <math>\overset{\Large\frown} {XAF} + \overset{\Large\frown} {XBF} = 180^\circ</math> <i><b>(Claim 3)</b></i> <math>\implies</math> | The sum <math>\overset{\Large\frown} {XAF} + \overset{\Large\frown} {XBF} = 180^\circ</math> <i><b>(Claim 3)</b></i> <math>\implies</math> | ||
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<math>\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.</math> | <math>\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.</math> | ||
− | Similarly, <math>\angle AXD = 90^\circ \implies \angle BXA + \angle DXC = | + | Similarly, <math>\angle AXD = 90^\circ \implies \angle BXA + \angle DXC = 360^\circ -\angle AXD -\angle CXB = 180^\circ.</math> |
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[[File:2018 IMO 6a.png|430px|right]] | [[File:2018 IMO 6a.png|430px|right]] | ||
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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]] | ||
+ | [[File:2018 IMO 6 Claim 3.png|370px|right]] | ||
<i><b>Common case </b></i> | <i><b>Common case </b></i> | ||
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We make transformation and get <cmath>{\sin \psi} = \frac {CD}{DX} {\sin \alpha} = \frac{CD}{DX} \cdot {\frac{DX \cdot AB}{BX \cdot CD}} {\sin \alpha} = \frac {AB}{BX}\sin \alpha = \sin \psi.</cmath> | We make transformation and get <cmath>{\sin \psi} = \frac {CD}{DX} {\sin \alpha} = \frac{CD}{DX} \cdot {\frac{DX \cdot AB}{BX \cdot CD}} {\sin \alpha} = \frac {AB}{BX}\sin \alpha = \sin \psi.</cmath> | ||
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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> | ||
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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> | ||
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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> | ||
+ | In the figure they are a blue and red arcs <math>\overset{\Large\frown} {CD}, \alpha + \beta = 180^\circ.</math> |
Revision as of 10:13, 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 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 we obtain
By the law of sines for we obtain
We make transformation and get
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