A Beautiful Concurrency in Quadrilaterals

Consider a quadrilateral \(ABCD\) (maybe convex or concave). Construct the circle passing through points \(B, C, D\), denoted \(\Gamma_A\). Next, construct the circle through \(C, D, A\), denoted \(\Gamma_B\). Similarly define \(\Gamma_C\) and \(\Gamma_D\) as shown below.

For each \(i \in \{A, B, C, D\}\), shrink the circle \(\Gamma_i\) by half about point \(i\), resulting in circle \(\gamma_i\). This is equivalent to a homothety with center \(i\) and scale factor \(\frac{1}{2}\). Construct the circles \(\gamma_A, \gamma_B, \gamma_C, \gamma_D\) and remove the parent circles \(\Gamma_A, \Gamma_B, \Gamma_C, \Gamma_D\) to get the figure below.

What do you observe in the diagram? Is there something peculiar?

All circles \(\gamma_A, \gamma_B, \gamma_C, \gamma_D\) appear to concur at a point. How do we prove this?

Let \(M_{AB}, M_{BC}, M_{CD}, M_{DA}, M_{AC}, M_{BD}\) be the midpoints of \(AB, BC, CD, DA, AC, BD\) respectively. The circle \(\gamma_A\) passes through \(M_{AB}, M_{AC}, M_{AD}\) because scaling \(\Gamma_A\) by \(\frac{1}{2}\) about \(A\) maps \(B, C, D\) to these midpoints. Similar reasoning applies to \(\gamma_B, \gamma_C, \gamma_D\). Define \(K\) as the intersection of \(\gamma_A\) and \(\gamma_C\) (other than \(M_{AC}\)), and prove \(K\) lies on \(\gamma_B, \gamma_D\).

Let \(\angle DAB = A\), \(\angle ABC = B\), \(\angle BCD = C\), \(\angle CDA = D\). Consider point \(A'\) on \(\Gamma_A\) such that \(K\) is its transformation on \(\gamma_A\) (doubling the distance from \(A\)). Similarly, \(C'\) on \(\Gamma_C\) transforms to \(K\) on \(\gamma_C\). Then, \(\angle M_{AB}KM_{AD} = \angle BA'D\), and since scaling preserves angles, \(\angle BA'D = \angle BCD = C\). Thus, \(\angle M_{AB}KM_{AD} = C\). Similarly, \(\angle M_{BC}KM_{CD} = A\). For \(K\) to lie on \(\gamma_B\), we need \(\angle M_{AB}KM_{BC} = \angle ADC = D\).

Since \(AA'\) and \(CC'\) bisect each other at \(K\), \(ACA'C'\) is a parallelogram. Thus, \(\angle C'AC + \angle A'CA = 180^\circ\), so \(\angle C'AD + \angle A'CD = 180^\circ - (\angle DAC + \angle DCA) = D\). As \(A' \in \Gamma_A\) and \(C' \in \Gamma_C\), \(\angle C'AD = \angle C'BD\) and \(\angle A'CD = \angle A'BD\). Hence, \(\angle C'BA' = D\). Since \(KM_{BC} \parallel C'B\) and \(KM_{AB} \parallel A'B\) by the Midpoint Theorem in \(\triangle CC'B\) and \(\triangle AA'B\), \(\angle M_{AB}KM_{BC} = D\), implying \(\angle M_{CD}KM_{AD} = B\). Therefore, \(\boxed{K \in \{\gamma_B, \gamma_D\}}\).

There’s also a proof using complex numbers, and perhaps an elegant synthetic solution exists. Think about it!