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]] | ||
+ | [[File:2018 IMO 6 Claim 3.png|370px|right]] | ||
<i><b>Special case</b></i> | <i><b>Special case</b></i> | ||
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<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>. | <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 | + | Hence, the sum of the arcs <math>\overset{\Large\frown} {XF}</math> is <math>180^\circ \implies</math> |
− | < | + | the sum <math>\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.</math> |
− | 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. | + | Similarly, <math>\angle AXD = 90^\circ \implies \angle BXA + \angle DXC = 180^\circ.</math> |
+ | |||
+ | <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. | ||
+ | |||
+ | <i><b>Claim 2</b></i> Let points B and D be symmetric with respect to the circle ω. Then any circle Ω passing through these points is orthogonal to ω. | ||
+ | |||
+ | <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 CD, α + β = 180°. |
Revision as of 14:20, 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 is
the sum
Similarly,
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.
Claim 2 Let points B and D be symmetric with respect to the circle ω. Then any circle Ω passing through these points is orthogonal to ω.
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 CD, α + β = 180°.