Difference between revisions of "2018 IMO Problems/Problem 6"
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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 {1}{2}\overset{\Large\frown} {X_0CE} = \frac {1}{2} (360^\circ -\overset{\Large\frown} { | + | 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> |
Analogically, quadrangle <math>DX_0BF</math> is cyclic <math>\implies \angle X_0BC = \angle X_0DA</math>. | Analogically, 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>. |
− | <math>\angle FCX = \angle BCX = \frac {1}{2} \overset{\Large\frown} {XAF}</math> of <math>\omega.</math> | + | <math>\hspace{10mm} \angle FCX = \angle BCX = \frac {1}{2} \overset{\Large\frown} {XAF}</math> of <math>\omega.</math> |
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− | |||
− | + | <math>\hspace{10mm} \angle CBX = \angle XDA = \frac {1}{2} \overset{\Large\frown} {XBF}</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> | |
+ | |||
+ | <math>\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.</math> | ||
Similarly, <math>\angle AXD = 90^\circ \implies \angle BXA + \angle DXC = 180^\circ.</math> | Similarly, <math>\angle AXD = 90^\circ \implies \angle BXA + \angle DXC = 180^\circ.</math> |
Revision as of 06:43, 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
Analogically, 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 midpoint perpendicular of the segment
and the line
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).