2018 IMO Problems/Problem 6
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 AX0CF is cyclic) and ∠X0BC = ∠X0DA (quadrangle DX0BF is cyclic) are equal to the property of opposite angles of cyclic quadrangles. This means that X0 coincides with the point X indicated in the condition. The angle ∠FCX = ∠BCX subtend the arc XF of ω, ∠CBX = ∠XDA subtend the arc XF of Ω. The sum of these arcs is 180° (Claim 3 for orthogonal circles). 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°.