A Solution to Crux Mathematicorum Problem 5031

The original problem statement is intentionally not reproduced here. This article follows its notation and presents only the solution and diagram. The complete statement is available in the official issue of Crux Mathematicorum.

Solution

Proof

Invert with respect to the incircle of $\triangle ABC$ , and denote the image of any point $Y$ by $Y'$ . Basic inversion properties give the following:

$A'$ is the midpoint of $EF$ .
$C'$ is the midpoint of $DE$ .

Let $X$ be the intersection of $IP$ and $A'C$ .

Triangle ABC with incircle contact points D, E, and F. A prime and C prime are the midpoints of EF and DE; P lies on DE and on the line through A parallel to EF; X is the intersection of A prime C and IP. — Figure 1. The incircle and auxiliary lines used in the inversion.Select the diagram to enlarge it.

Under the inversion, the line $A'C$ maps to the circumcircle of $\triangle AC'I$ , whereas the line $IP$ maps to itself.

Since $AP\parallel EF\perp AI$ , we have $\angle IAP=90^\circ$ . Also, $P,C'\in DE$ and $IC'\perp DE$ , so $\angle IC'P=90^\circ$ . Thus the points $A$ , $C'$ , $I$ , and $P$ are concyclic. Therefore, we have $X'=P$ .

Consequently, the points $A$ , $A'$ , $P$ , and $P'(=X)$ are concyclic; indeed, if $r$ is the inradius, then $IA\cdot IA'=IP\cdot IP'=r^2$ .

Since $AA'\perp AP$ , we have $\angle A'AP=90^\circ$ . Hence $\angle A'XP=90^\circ$ . Since $A'$ , $X$ , $C$ are collinear and $I$ , $X$ , $P$ are collinear, it follows that $CA'\perp IP$ , which completes the proof.

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