Difference between revisions of "2018 IMO Problems/Problem 6"
(→Solution) |
(→Solution) |
||
Line 17: | Line 17: | ||
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> | ||
− | |||
− | <math>\angle | + | 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} {AXE}) = 180^\circ - \angle X_0CE = \angle X_0CE.</cmath> | ||
+ | |||
+ | 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> indicated in the condition. | ||
+ | |||
+ | <math>\angle FCX = \angle BCX = \frac {1}{2} \overset{\Large\frown} {XAF}</math> of <math>\omega.</math> | ||
+ | <math>\angle CBX = \angle XDA = \frac {1}{2} \overset{\Large\frown} {XF}</math> of <math>\Omega.</math> | ||
+ | The sum <math>\overset{\Large\frown} {XAF} + \overset{\Large\frown} {XAF} = 180^\circ</math> <i><b>(Claim 3).</b></i>. | ||
Hence, the sum of the arcs <math>\overset{\Large\frown} {XF}</math> is <math>180^\circ \implies</math> | Hence, the sum of the arcs <math>\overset{\Large\frown} {XF}</math> is <math>180^\circ \implies</math> |
Revision as of 05:01, 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 indicated in the condition.
of of The sum (Claim 3)..
Hence, the sum of the arcs is
the sum
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).