Let $ABC$ be an acute triangle with $AB<AC$ and $\angle BAC=60^{\circ}$. Denote its altitudes by $AD,BE,CF$ and its orthocenter by $H$. Let $K,L,M$ be the midpoints of sides $BC,CA,AB$, respectively. Prove that the midpoints of segments $AH, DK, EL, FM$ lie on a single circle.
Problem
Source: Czech-Polish-Slovak Match 2019 P6
Tags: geometry
20.07.2019 01:07
Let $O$ and $N$ be the circumcenter and nine-point center of $ABC$, respectively. It is well known that $H$ and $O$ are symmetric wrt the bisector of $BAC$. Since $N$ is the midpoint of $OH$, it lies on that bisector. Let $S$ be the orthocenter of $AHO$. Claim 1. $S$ and $N$ are isogonal conjugates wrt $ABC$. Proof of Claim 1. We have $\measuredangle BHC = 120^{\circ}=\measuredangle BOC$, so the points $B$, $H$, $O$, $C$ are concyclic. It is clear that $SO$ is tangent to the circle passing through those four points. By symmetry wrt the bisector of $BAC$ we see that $SH$ is also tangent to that circle. Hence the lines $BS$ and $BN$ are isogonal wrt $HBO$. From here we can see that they are also isogonal wrt $ABC$. Analogously, the lines $CS$ and $CN$ are isogonal wrt $ACB$, proving the claim. $\square$ Let $OH$ intersect $AB$ and $AC$ at $P$, $Q$ respectively. It is easy to see that $APQ$ is an equilateral triangle. Let $\omega_{1}$ be the common pedal circle of $N$ and $S$ wrt $ABC$ and $\omega_{2}$ be the nine point circle of $ABC$. Claim 2. The midline of $APQ$ parallel to $PQ$ is the radical axis of $\omega_{1}$, $\omega_2$. Proof of Claim 2. Let the lines $HF$ and $OM$ intersect $AN$ at $X$, $Y$, respectively and let $G$ be the centroid of $APQ$. Let $PN=QN=a$, $HN=ON=x$ and $\measuredangle HAN=\varphi$. We have $$\cot \varphi=\frac{AN}{HN}=\frac{a\sqrt{3}}{x},$$so $$\frac{AG+\frac{GY\cdot GX}{GN}}{2x}=\frac{AG+\frac{GN^2-NX^2}{GN}}{2x}=\frac{AN-\frac{(\frac{x\sqrt{3}}{3})^2}{\frac{a\sqrt{3}}{3}}}{2x}=\frac{a\sqrt{3}-\frac{x^2}{a\sqrt{3}}}{2x}=\frac{3a^2-x^2}{2ax\sqrt{3}}=\frac{(\frac{a\sqrt{3}}{x})^2-1}{2\cdot \frac{a\sqrt{3}}{x}}=\cot 2\varphi=\frac{AS}{2x}=\frac{AG+GS}{2x}.$$From here we get that $$GX\cdot GY=GS\cdot GN.$$Let $G'$, $N'$, $S'$ denote the orthogonal projections of the points $G$, $N$, $S$, respectively onto $AB$. Since $G'$, $N'$, $S'$, $M$, $F$ are the orthogonal projections of the points $G$, $N$, $S$, $Y$, $X$, respectively onto $AB$, we have $$G'M\cdot G'F=G'S'\cdot G'N'.$$This means that $G'$ has equal powers wrt $\omega_1$, $\omega_2$. However, $G'$ is the midpoint of $AP$. By analogy, the midpoint of $AQ$ has equal powers wrt $\omega_1$, $\omega_2$. From here the claim clearly follows. $\square$ From Claim 2. we get that the midpoint of $AH$ lies on the radical axis of $\omega_1$, $\omega_2$. The afromentioned midpoint lies on $\omega_2$, so it must also lie on $\omega_1$. It is obvious that the midpoints of $DK$, $EL$, $FM$ also lie on $\omega_1$, concluding the proof. $\square$
23.07.2019 12:45
This problem was proposed by Burii. Let $O$ and $N$ be the circumcenter and nine-point center of $ABC$, respectively. Then the midpoints of $DK$, $EL$ and $FM$ are vertices of pedal triangle of $N$. Recall that the circumcircle of pedal triangle of $X$ passes through the center of rectangular hyperbola $ABCHX$. Hence it is enough to prove that the midpoint of $AH$ is the center of hyperbola $ABCHN$. Recall that the midpoint of hyperbola $ABCHX$ is the midpoint of segment $HY$ where $Y$ is the fourth common point of hyperbola $ABCHX$ and the circle $(ABC)$. Hence it is enough to prove that hyperbola $ABCHN$ is tangent to the circle $(ABC)$. But this is equivalent to the statement that the line isogonal to $ABCHN$ is parallel to $BC$. Let $S$ be the isogonal conjugate of $N$ with respect to $ABC$. We need to prove that $SO \parallel BC$. We shall prove that $S$ is the orthocenter of $AHO$. This will imply that $SO \perp AH \perp BC$, i.e. $SO \parallel BC$. By assumption $\angle BAC = 60^\circ$, we get $AH=AO$. Since $N$ is the midpoint of $OH$, $AN \perp OH$. Since $AO$ and $AH$ are isogonal lines in the angle $BAC$, we see that $AN$ is the bisector of angle $BAC$. Consequently, $S$ lies on $AN$. Hence $AS \perp OH$. Since $N$ lies on $OH$, $S$ lies on hyperbola $ABCHO$. Hence $S$ lies on a rectangular hyperbola passing through the vertices of triangle $AHO$ and $AS \perp OH$. It follows that $S$ is the orthocenter of $AHO$ and we are done. Remark. The thesis of the problem remains true if $\angle BAC = 120^\circ$.
06.08.2019 23:19
Let $O, N$ denote the circumcenter and nine-point center of $\triangle ABC$ respectively. Let $B', C'$ be the points on rays $AB, AC$ such that $AB' = AC' = \frac{b+c}{2}$. Let $A'$ be the point such that $ADA'K$ is a parallelogram. By dilation at $A$ with factor $2$, it suffices to prove that $B'HC'A'$ is cyclic. As $B'HOC'$ is an isosceles trapezoid, it suffices only to show that $B'OC'A'$ is cyclic. Observe that point $K$ lies on $B'C'$ (by Menelaus, say) and is clearly on $OA'.$ Therefore, by Power of the Point it suffices only to show that $$KB' \cdot KC' = KO \cdot KA'.$$Observe that $$KO \cdot KA' = KO \cdot AD = \frac{1}{\sqrt{3}} \cdot BK \cdot AD = \frac{1}{\sqrt{3}} \cdot \Delta, \qquad (1)$$where $\Delta$ denotes the area of $\triangle ABC.$ Since $KM || AC'$ with $K \in B'C', M \in B'A$, we've that $\triangle B'KM \sim \triangle B'C'A,$ which means that $\triangle B'KM$ is equilateral. This means that $B'K = MB' = \frac{b+c}{2} - AM = \frac{b}{2}$, and analogously $C'K = \frac{c}{2}.$ This means that $$KB' \cdot KC' = \frac{bc}{4} = \frac{1}{\sqrt3} \cdot \Delta, \qquad (2)$$where we used that $\Delta = \frac12 \cdot b \cdot c \cdot \sin \angle A = \frac12 \cdot b \cdot c \cdot \frac{\sqrt{3}}{2}.$ Comparing $(1), (2),$ we are done. $\square$
11.08.2019 05:24
P-Euler point, N,J,T-projection of P wrt ABC, K,L,M midpoint, D,E,F attitude, G midpoint AH. Notice JN is midline of ALEM, I be midpoint of AO then IGJN isoceles trapezoid. Let Q midpoint LM then A,K,Q and Q,P,T collinear. But 8QI.QT=AH.AD=AB.AF=2ME.FL=8QJ.QN so JNIT cyclic hence q.e.d
11.08.2019 13:04
Let $O,N_9$ be the circumcenter and the nine point center of $\triangle{ABC}.$ Denote by $P,N_A,N_B,N_C$ the midpoints of $\overline{AH},\overline{AN_9},\overline{BN_9}$, and $\overline{CN_9}$ respectively. Observe that $AH=AO$, so $N_9$ lies on the bisector of $\angle{BAC}$, and that $BHOC$ is cyclic, hence $\angle{BHN_9}=150^0$ and $\angle{CHN_9}=30^0.$ Now, $\angle{MPN_A}+\angle{MN_BN_A}=\angle{BHN_9}+\angle{BAN_9}=150^0+30^0=180^0$, so $P$ lies on the NPC of $\triangle{ABN_9}$. In the same way we prove that $P$ also lies on the NPC of $\triangle{ACN_9}.$ Now taking into consideration that $P$ also lies on the NPC of $\triangle{ABC}$, it follows that $P$ is the Poncelet point of $ABCN_9.$ Therefore $P$ lies on the circumcircle of the pedal triangle of $N_9$ wrt $\triangle{ABC}$, which is what we wanted.