A Solution to Crux Mathematicorum Problem 5031

Problem 5031

Proposed by Tran Quang Hung

Given a triangle $ABC$ with incircle $(I)$ touching the sides $BC$ , $CA$ , and $AB$ at $D$ , $E$ , and $F$ , respectively. The line through $A$ parallel to $EF$ intersects $DE$ at $P$ . Prove that the line through $C$ and perpendicular to $IP$ bisects the segment $EF$ .

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.

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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Solution

Diagrams and complete source