A geometry problem based on circles.

Question

Question:- Consider three equal circles $S_1$, $S_2$, and $S_3$ each of which passes through a given point $H$. Other than that, $S_1$ and $S_2$ intersect at $A$, $S_2$ and $S_3$ intersect at $B$, $S_3$ and $S_1$ intersect at $C$. Show that $H$ is the orthocenter of triangle $ABC$. (Note:- I have assumed that $O_1$ is the centre of $S_1$, $O_2$ is the centre of $S_2$ and $O_3$ is the centre of $S_3$. Also, I have made an observation that quadrilaterals $O_1CO_3H$ and $O_3BO_2H$ are rhombuses. Is this fact going to help in the proof? )

Answer 1

We have that $H$ is the radical centre of the circles $S_1, S_2, S_3$, so $AH\perp O_2 O_3$ and the midpoint of $AH$ is also the midpoint $M_A$ of $O_2 O_3$. In particular, $H$ is the circumcenter of $O_1 O_2 O_3$, hence it is the orthocenter of its medial triangle $M_A M_B M_C$. Since $ABC$ and $M_A M_B M_C$ are homothetic with respect to a dilation centered at $H$, $H$ is also the orthocenter of $ABC$.

We also have that the circumradius of $M_A M_B M_C$ is at the same time half the circumradius of $ABC$ (by homothety) and half the circumradius of $O_1 O_2 O_3$, since $M_A M_B M_C$ is the medial triangle of $O_1 O_2 O_3$. It follows that the circumradius of $ABC$ has the same length as the radius of $S_1,S_2$ or $S_3$: this is known as Johnson's theorem.

Answer 2

This is a known property of Johnson circles. See Property 6 here and the proof that follows it.

Answer 3

As the circles have equal radii, $\angle HAC$ and $\angle CBH$, which are both over the same secant $HC$, are equal. Writing down all symmetry variants of the observation, we have $\delta_1:=\angle HAC=\angle CBH$, $\delta_2:=\angle HBA=\angle ACH$, $\delta_3:=\angle HCB=\angle BAH$. From the sum of interior angles, we conclude $\delta_1+\delta_2+\delta_3=90^\circ$. Moreover, the angle between $AB$ and $HC$ equals $\angle BAH+\angle HAC+\angle ACH$, which also is $\delta_1+\delta_2+\delta_3$.