# 2018 IMO Problems/Problem 6

A convex quadrilateral satisfies Point lies inside so that and Prove that

## Solution

We want to find the point Let and be the intersection points of and and and respectively. The poinx is inside so points follow in this order.

is cyclic lie on circle

Similarly, lie on circle

Point is the point of intersection of circles and

**Special case**

Let and

The points and are symmetric with respect to the circle **(Claim 1).**

The circle is orthogonal to the circle **(Claim 2).**

of of

**(Claim 3)**

Similarly,

**Common case **

Denote by the intersection point of and the perpendicular bisector of Let be a circle (red) with center and radius

We will prove using point symmetric to with respect to

The points and are symmetric with respect to **(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 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 circle orthogonal to

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

Hence we get

If then This is a special case.

In all other cases, the equality of the sines follows

* 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

* Lemma* The opposite sides of the quadrilateral intersect at points and ( lies on ). The circle centered at the point contains the ends of the diagonal The points and are symmetric with respect to the circle (in other words, the inversion with respect to maps into Then the circles and are symmetric with respect to

* Proof* We will prove that the point symmetric to the point with respect to belongs to the circle becouse

A circle containing points and symmetric with respect to is orthogonal to * (Claim 2)* and maps into itself under inversion with respect to the circle Hence, the point under this inversion passes to some point of the same circle

A straight line containing the point of the circle under inversion with respect to maps into the circle Hence, the inscribed angles of this circle are equal maps into and maps into Consequently, the angles These angles subtend the of the circle, that is, the point symmetric to the point with respect to belongs to the circle

**vladimir.shelomovskii@gmail.com, vvsss**