orthocenter, intersection of 2 circles, 2 midpoints of arcs, are concyclic

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orthocenter, intersection of 2 circles, 2 midpoints of arcs, are concyclic

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Source: Turkey JBMO TST 2018 p3

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#1 h Sep 16, 2018, 2:07 PM • 4 Y

Y by mathematicsy, giveandtake, Adventure10, Mango247

Let $H$ be the orthocenter of an acute angled triangle $ABC$ . Circumcircle of the triangle $ABC$ and the circle of diameter $[AH]$ intersect at point $E$ , different from $A$ . Let $M$ be the midpoint of the small arc $BC$ of the circumcircle of the triangle $ABC$ and let $N$ the midpoint of the large arc $BC$ of the circumcircle of the triangle $BHC$ Prove that points $E, H, M, N$ are concyclic.

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#2 h Sep 16, 2018, 2:16 PM • 1 Y

Y by Adventure10

Let $P$ be the midpoint of $\overline{BC}$ . It is well-known that $P$ lies on $\overline{EH}$ and obvious that $P$ lies on $\overline{MN}$ . As the circumcircles of $BHC$ and $ABC$ are reflections over $BC$ , it follows that $PM \cdot PN = \text{pow}(P, (BHC))$ . But if $A'$ is the antipode of $A$ on $(ABC)$ , then $\text{pow}(P, (ABC)) = PA' \cdot PE = PH \cdot PE$ by orthocenter reflections, so as $P$ lies on the radical axis of $(BHC)$ and $(ABC)$ we're done.

This post has been edited 2 times. Last edited by wu2481632, Sep 16, 2018, 2:17 PM

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#3 h May 17, 2020, 7:18 PM • 1 Y

Y by giveandtake

After realizing $PM\cdot PN= pow(P,(BHC))= PB\cdot PC= PB^2$ , then we must prove $PB$ is tangent to $(EHB)$ at $B$ which is equivalent to $\angle HEB=\angle HBC$ . But this follows from the fact that $EH$ lies on the diameter of $(ABC)$

This post has been edited 1 time. Last edited by electrovector, May 17, 2020, 7:21 PM

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#4 h Jul 17, 2020, 11:20 AM

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Solution by complex numbers is possible...

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#5 h Apr 30, 2021, 9:29 AM

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electrovector wrote:

After realizing $PM\cdot PN= pow(P,(BHC))= PB\cdot PC= PB^2$ , then we must prove $PB$ is tangent to $(EHB)$ at $B$ which is equivalent to $\angle HEB=\angle HBC$ . But this follows from the fact that $EH$ lies on the diameter of $(ABC)$

Nice solution!

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llplp

#6 h May 1, 2021, 12:12 AM

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WolfusA wrote:

Solution by complex numbers is possible...

Indeed! Toss on the complex plane with $a = x^2$ , $m = -yz$ etc. and $(ABC)$ the unit circle, then $h = a+b+c = x^2+y^2+z^2$ . Since $(ABC)$ and $(BHC)$ are symmetric wrt to $BC$ , the point $N$ is symmetric with the centre of the arc $BC$ containing $A$ on $(ABC)$ . This point is represented by $yz$ , so $n = b+c-bc\overline{(yz)} = y^2+z^2-yz$ . Furthermore, it is pretty well-known that $E,H$ , the midpoint of $BC$ and the antipode of $A$ on $(ABC)$ are collinear, so $e = \frac{h-(-a)}{-a \overline{h} - 1} = -\frac{y^2 z^2 (2x^2+ y^2+z^2)}{x^2 y^2+x^2 z^2+2 y^2 z^2}.$ Now it remains to compute $\frac{e-m}{e-h} : \frac{n-m}{n-h}$ . Firstly $n-m = y^2+z^2$ and $n-h = -(x^2+yz)$ . The rest is computation:
$e - m = -\frac{y^2 z^2 (2x^2+ y^2+z^2)}{x^2 y^2+x^2 z^2+2 y^2 z^2} - yz = yz \frac{yz(2x^2+y^2+z^2) + (x^2 y^2+x^2 z^2+2 y^2 z^2)}{x^2 y^2+x^2 z^2+2 y^2 z^2} =- yz \frac{(y+z)^2 (x^2 + yz)}{x^2 y^2+x^2 z^2+2 y^2 z^2},$ $e - h = -\frac{y^2 z^2 (2x^2 + y^2+z^2) -(x^2+y^2+z^2)(x^2 y^2+x^2 z^2+2 y^2 z^2)}{x^2 y^2+x^2 z^2+2 y^2 z^2} = - \frac{(x^2+y^2)(x^2+z^2)(y^2+z^2)}{x^2 y^2+x^2 z^2+2 y^2 z^2},$ so:
$\frac{e-m}{e-h} : \frac{n-m}{n-h} = \frac{yz(y-z)^2 (x^2+ yz)}{(x^2+y^2)(x^2+z^2)(y^2+z^2)} : \frac{(y^2+z^2)}{(x^2+yz)}$ $=\frac{yz (y-z)^2 (x^2+yz)^2}{(x^2+y^2)(x^2+z^2) (y^2+z^2)^2},$ which is clearly self-conjugate as required. 'twas not even that painful

This post has been edited 1 time. Last edited by llplp, May 1, 2021, 12:51 AM

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#7 h Jan 2, 2025, 12:14 PM

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Easy problem my solution was the same as @above

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