Four midpoints are cyclic

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Four midpoints are cyclic

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Source: Czech-Polish-Slovak Match 2019 P6

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#1 h Jul 15, 2019, 12:27 PM • 1 Y

Y by Adventure10

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.

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#2 h Jul 19, 2019, 10:07 PM • 2 Y

Y by Adventure10, Mango247

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. $\square$

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#3 h Jul 23, 2019, 9:45 AM • 2 Y

Y by AlastorMoody, Adventure10

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 $\angle BAC = 120^\circ$ .

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#4 h Aug 6, 2019, 8:19 PM • 3 Y

Y by Iamsohappy, Smkh, Adventure10

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.
$\square$

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#5 h Aug 11, 2019, 2:24 AM • 3 Y

Y by top1csp2020, Adventure10, Mango247

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

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#6 h Aug 11, 2019, 10:04 AM • 1 Y

Y by Adventure10

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.

This post has been edited 1 time. Last edited by huricane, Aug 11, 2019, 10:19 AM
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