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.

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.

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, suppose that $I\in JP$ . Since $P\in IA$ and $P\ne I$ , the lines $JP$ and $IA$ pass through the distinct points $I$ and $P$ , so they coincide. Since $JP\parallel ID$ , the line $IA$ has the same direction as $ID$ ; as both pass through $I$ , they 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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Main idea

Solution

Diagrams and complete source