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Problem 1834

National Olympiad, first round

Geometry Difficulty 6.8 Prove it Vietnamese Mathematical Olympiad · Vietnam

Let $ABC$ be an acute, non-isosceles triangle with altitudes $AD$ , $BE$ and $CF$ . The circle ( $AD$ ) intersects $DE$ , $DF$ at $M$ , $N$ . Let $P$ , $Q$ be the points on $AB$ , $AC$ respectively such that $NP$ is perpendicular to $AB$ and $MQ$ is parallel to $AC$ . Let ( $I$ ) be the circumcircle of triangle $APQ$ .

a) Prove that ( $I$ ) is tangent to $EF$ .

b) Let $T$ be the tangency point of the circumcircle of triangle $APQ$ with $EF$ , $K$ be the intersection of $DT$ , $MN$ and $L$ be the reflection of $A$ through $MN$ . Prove that the circumcircle of triangle $DKL$ passes through the intersection of $MN$ , $EF$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

a) Let $T$ be the foot of $A$ on $EF$ . Note that $FC$ is the internal bisector of $\angle DFE$ so $FM$ , $FT$ are symmetric with respect to $AB$ . On the other hand, $\angle FNA = \angle FTA = 90^\circ$ then $N$ , $T$ are symmetric with respect to $AB$ . Therefore, $N$ , $P$ , $T$ are collinear and $TP \perp AB$ . Similarly, $TQ \perp AC$ and $T$ , $M$ , $Q$ are collinear. Hence, $AT$ is the diameter of $(I)$ and $AT \perp EF$ then $EF$ is tangent to $(I)$ .

b) Let $X$ , $Y$ , $Z$ and $J$ be the intersections of $MN$ with $EF$ , $AB$ , $AC$ and $AD$ . It is well-known that $M$ , $N$ are the reflections of $T$ through $AE$ , $AF$ respectively then $MN$ passes through the feet of altitudes from $E$ , $F$ in triangle $AEF$ . Thus, $EY \perp AB$ and $FZ \perp AC$ . On the other hand, we also have $AD$ is the perpendicular bisector of $MN$ then $J$ is the midpoint of $MN$ . Hence, $D(EF, TX) = -1$ . Projecting on $MN$ , we have $(MN, KX) = -1$ . Note that $J$ is the midpoint of $MN$ , we have
$\overline{JK} \cdot \overline{JX} = JM^2 = JN^2 = -\overline{JA} \cdot \overline{JD} = \overline{JL} \cdot \overline{JD}.$
Therefore, $DLKS$ is cyclic or $(DLK)$ passes through $S$ . $\square$

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.