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Problem 1862

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.8 Prove it VN IMO Booklet · Vietnam

Let $ABC$ be an acute, non-isosceles triangle with $H, O, O'$ as its orthocenter, circumcenter, nine-point center, and $D, E, F$ as the midpoints of the segments $BC, CA, AB$ , respectively. $P$ is an arbitrary point inside triangle $DEF$ . Let $DP, EP, FP$ intersect $(O')$ again at $D', E', F'$ , respectively. $A'$ is the reflection of $A$ through $D'$ . We define points $B', C'$ similarly.

a. Assume that $PO = PO'$ , prove that the circle $(A'B'C')$ passes through $O$ .

b. Let $X$ be the reflection of $A'$ with respect to the line $OD$ . We define $Y, Z$ similarly. Suppose that $XH, YH, ZH$ intersect $BC, CA, AB$ at $M, N, K$ respectively. Prove that $M, N, K$ are collinear.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

(a) Let $I$ be the reflection of $O$ with respect to $P$ . Since $O'$ is the midpoint of $OH$ , it follows that $O'P \parallel IH$ . Moreover, we have $PO = PO'$ , thus $IO = IH$ .
Let $S, G$ be midpoints of $AI$ and $AH$ , respectively. We have
$SP = \frac{1}{2}AO = \frac{1}{2}R = O'D$
and $SP \parallel AO \parallel O'D$ , thus $O'SPD$ is a parallelogram. It follows that $DP \parallel O'S$ .
Furthermore, we have $SP = \frac{1}{2}R = O'D'$ , therefore $SD'PO'$ is an isosceles trapezoid which leads to $O'P = SD'$ and
$IH = 2O'P = 2SD' = IA'.$
Thus $IA' = IH = IO$ which implies $A'$ lies on the circle $(I, IO)$ . Similarly, $B'$ and $C'$ also lie on $(I, IO)$ . This leads to the conclusion of (a).

(b) Let $R$ be the radius of the circle $(O)$ . It is obvious that $GD = R$ . Consider the homothetic transformation with center $A$ and ratio $\frac{1}{2}$ which sends $B, C, A', X, H, M$ and the perpendicular bisector of $BC$ to $F, E, D', U, G, M'$ and the perpendicular bisector $EF$ , respectively. Then $\frac{MB}{MC} = \frac{M'F}{M'E}$ and $U$ is the reflection of $D'$ with respect to $EF$ . Thus
$\frac{MB}{MC} = \frac{M'F}{M'E} = \frac{GF}{GE} \cdot \frac{UF}{UE} = \frac{\sqrt{R^2 - DF^2}}{\sqrt{R^2 - DE^2}} \cdot \frac{D'E}{D'F}.$
Similarly, we can calculate $\frac{NC}{NA}$ and $\frac{KA}{KB}$ .

Since $DD'$ , $EE'$ , $FF'$ are concurrent, it follows
$\frac{D'F}{D'E} \cdot \frac{F'E}{F'D} \cdot \frac{E'D}{E'F} = 1.$
Thus $\frac{MB}{MC} \cdot \frac{NC}{NA} \cdot \frac{KA}{KB} = 1$ , which implies $M, N, K$ are collinear. This is the desired result.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.