Difference between revisions of "2018 IMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
+ | [[File:2018 IMO 6.png|490px|right]] | ||
<i><b>Special case</b></i> | <i><b>Special case</b></i> | ||
We construct point <math>X_0</math> and prove that <math>X_0</math> coincides with the point <math>X.</math> | We construct point <math>X_0</math> and prove that <math>X_0</math> coincides with the point <math>X.</math> | ||
− | Let <math>AD = CD</math> and <math>AB = BC \implies AB \cdot CD = BC \cdot DA.</math> 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>AD = CD</math> and <math>AB = BC \implies AB \cdot CD = BC \cdot DA.</math> | ||
+ | |||
+ | 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 = EACF</math> <i><b>(Claim 1).</b></i> | The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega = EACF</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> <math>\angle X_0AB = \angle X_0CD</math> (quadrilateral | + | Let <math>X_0</math> be the point of intersection of the circles <math>\omega</math> and <math>\Omega.</math> <math>\angle X_0AB = \angle X_0CD</math> (quadrilateral <math>AX_0CF</math> is cyclic) and <math>\angle X_0BC = \angle X_0DA</math> (quadrangle <math>DX_0BF</math> is cyclic). This means that <math>X_0</math> coincides with the point <math>X</math> indicated in the condition. |
− | + | ||
+ | <math>\angle FCX = \angle BCX</math> subtend the arc <math>\overset{\Large\frown} {XF}</math> of <math>\omega, \angle CBX = \angle XDA</math> subtend the arc <math>\overset{\Large\frown} {XF}</math> of <math>\Omega.</math> The sum of these arcs is <math>180^\circ</math> <i><b>(Claim 3).</b></i>. | ||
+ | |||
+ | Hence, the sum of the arcs XF is 180°, the sum of the angles ∠XСВ + ∠XВС = 90°, ∠СХВ = 90°. Similarly, ∠AXD = 90°, that is, ∠BXA + ∠DXC = 180°. | ||
+ | |||
+ | <i><b>Claim 1</b></i> | ||
+ | |||
+ | Let A, C, and E be arbitrary points on a circle ω, l be the middle perpendicular to the segment AC. Then the straight lines AE and CE intersect l at the points B and D, symmetric with respect to ω. |
Revision as of 14:04, 17 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) and
(quadrangle
is cyclic). This means that
coincides with the point
indicated in the condition.
subtend the arc
of
subtend the arc
of
The sum of these arcs is
(Claim 3)..
Hence, the sum of the arcs XF is 180°, the sum of the angles ∠XСВ + ∠XВС = 90°, ∠СХВ = 90°. Similarly, ∠AXD = 90°, that is, ∠BXA + ∠DXC = 180°.
Claim 1
Let A, C, and E be arbitrary points on a circle ω, l be the middle perpendicular to the segment AC. Then the straight lines AE and CE intersect l at the points B and D, symmetric with respect to ω.