A Solution to Crux Mathematicorum Problem 4941

Problem 4941

Proposed by Michel Bataille

Let a non-equilateral triangle $ABC$ be inscribed in a circle $\Gamma$ with center $O$ . Its incircle $\gamma$ , with center $I$ , touches $BC$ , $CA$ , and $AB$ at $D$ , $E$ , and $F$ , respectively. The circle $\Gamma_A$ passes through $A$ and is tangent to $ID$ at $I$ ; the circles $\Gamma_B$ and $\Gamma_C$ are defined cyclically. Prove that $\Gamma_A$ , $\Gamma_B$ , $\Gamma_C$ , and the line $OI$ have a common point other than $I$ .

Figure 1. The three circles and their second common point on the line OI.Select the diagram to enlarge it.

Main idea

Invert with respect to the incircle. The three circles become three lines, while $A$ , $B$ , and $C$ become the midpoints of the sides of the contact triangle. A negative homothety then reveals the point at which those lines meet.

Solution

Proof

We first treat the scalene case, in which all three circles $\Gamma_A$ , $\Gamma_B$ , and $\Gamma_C$ are nondegenerate.

Let

P=IA\cap EF, \qquad Q=IB\cap FD, \qquad R=IC\cap DE,

and let $J$ be the circumcenter of $\triangle PQR$ .

Since $AE=AF$ and $IE=IF$ , both $A$ and $I$ lie on the perpendicular bisector of $EF$ . Thus $P$ is the midpoint of $EF$ . Moreover, $FP$ is the altitude to the hypotenuse $AI$ of the right triangle $AIF$ , and hence

IA\cdot IP=IF^2.

Cyclically,

IB\cdot IQ=ID^2, \qquad IC\cdot IR=IE^2.

Therefore, inversion in $\gamma$ maps $P$ , $Q$ , and $R$ to $A$ , $B$ , and $C$ , respectively.

The contact triangle DEF and its medial triangle PQR. The circumcenter J of PQR lies on OI; the lines JP, JQ, and JR are respectively parallel to ID, IE, and IF. — Figure 2. The contact triangle, its medial triangle, and the concurrent lines used in the inversion.Select the diagram to enlarge it.

The points $P$ , $Q$ , and $R$ are the midpoints of $EF$ , $FD$ , and $DE$ . The homothety centered at the common centroid of $\triangle DEF$ and $\triangle PQR$ , with ratio $-\tfrac12$ , maps

(D,E,F)\longmapsto(P,Q,R).

It also maps the circumcenter $I$ of $\triangle DEF$ to the circumcenter $J$ of $\triangle PQR$ . Consequently,

JP\parallel ID, \qquad JQ\parallel IE, \qquad JR\parallel IF.

The circumcircle of $\triangle PQR$ is mapped by the inversion to $\Gamma$ . The centers of two inverse circles are collinear with the center of inversion; hence $I$ , $J$ , and $O$ are collinear.

The points $I$ and $O$ are distinct. Indeed, if the incenter and circumcenter coincided, the three sides of $\triangle ABC$ would be chords at the same distance from the center of $\Gamma$ , so they would have equal lengths. The triangle would then be equilateral. We also have $J\ne I$ : otherwise the inverse of the circumcircle of $\triangle PQR$ would be centered at $I$ , forcing $O=I$ .

None of the lines $JP$ , $JQ$ , and $JR$ passes through $I$ . For example, if $I\in JP$ , then $JP=IP=IA$ . Since $JP\parallel ID$ and both $IA$ and $ID$ pass through $I$ , the lines $IA$ and $ID$ would coincide. The angle bisector from $A$ would then also be perpendicular to $BC$ , forcing $AB=AC$ , contrary to the scalene assumption. The other two cases are cyclic.

Let $J^\ast$ be the inverse of $J$ with respect to $\gamma$ . A line not passing through the center of an inversion is mapped to a circle through that center, and the tangent to the image circle there is parallel to the original line. Thus $JP$ , $JQ$ , and $JR$ are mapped to $\Gamma_A$ , $\Gamma_B$ , and $\Gamma_C$ , respectively. The line $OI$ is mapped to itself.

All four lines $JP$ , $JQ$ , $JR$ , and $OI$ pass through $J$ . Their images therefore pass through $J^\ast$ . Since $J\ne I$ , the point $J^\ast$ is distinct from $I$ , which proves the result.

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