Involved conditional geo

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Source: Balkan MO 2024 Shortlist G4

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#1 h Apr 27, 2025, 10:27 PM • 1 Y

Y by RainbowJessa

Let $ABC$ be an acute-angled triangle with $AB < AC$ , orthocenter $H$ , circumcircle $\Gamma$ and circumcentre $O$ . Let $M$ be the midpoint of $BC$ and let $D$ be a point such that $ADOH$ is a parallellogram. Suppose that there exists a point $X$ on $\Gamma$ and on the opposite side of $DH$ to $A$ such that $\angle DXH + \angle DHA = 90^{\circ}$ . Let $Y$ be the midpoint of $OX$ . Prove that if $MY = OA$ , then $OA = 2OH$ .

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#2 h Apr 27, 2025, 11:02 PM • 2 Y

Y by RainbowJessa, pacelaaa

Complex bash with $\Gamma$ as the unit circle, so that
$|a|=|b|=|c|=1$ $h = a+b+c$ $o=0$ $m = \frac{b+c}2$ $d = a + o - h = -b-c$ Now we consider the coordinate of $X$ . We have $|x|=1$ and
$\left(\frac{d-x}{h-x}\right)\left(\frac{d-h}{a-h}\right) \in i\mathbb{R}$ $\frac{(b+c+x)(a+2b+2c)}{(b+c)(a+b+c-x)} = -\frac{(bc+bx+cx)(2ab+2ac+bc)}{(b+c)(abx+acx+bcx-abc)}$ $(a+2b+2c)(x+b+c)(abx+acx+bcx-abc) = (2ab+2ac+bc)(bx+cx+bc)(x-a-b-c)$ Expanding both sides out,
$x^2(a+2b+2c)(ab+ac+bc) + x(a+2b+2c)(ab^2+abc+ac^2+b^2c+bc^2) - abc(b+c)(a+2b+2c) = x^2(b+c)(2ab+2ac+bc) - x(2ab+2ac+bc)(ab+ac+b^2+bc+c^2) - bc(a+b+c)(2ab+2ac+bc)$ Combining coefficients gives
$x^2(a^2b+a^2c+abc+b^2c+bc^2) + x(3a^2b^2+5a^2bc+3a^2c^2+4ab^3+10ab^2c+10abc^2+4ac^3+3b^3c+5b^2c^2+3bc^3) + bc(a^2b+a^2c+abc+b^2c+bc^2) = 0$ We also know $MY = OA$ . Now
$y = \frac{o+x}2 = \frac{x}2$ and $|o-a| = 1$ . So
$|m-y| = \left|\frac{b+c-x}2\right| = 1$ $(b+c-x)\left(\frac1b+\frac1c-\frac1x\right) = 4$ $(x-b-c)(bx+cx-bc) + 4bcx = 0$ $x^2(b+c) - x(b^2-bc+c^2) + bc(b+c) = 0$ From these two equations combined, we have
$x + \frac{bc}x = -\frac{3a^2b^2+5a^2bc+3a^2c^2+4ab^3+10ab^2c+10abc^2+4ac^3+3b^3c+5b^2c^2+3bc^3}{a^2b+a^2c+abc+b^2c+bc^2} = \frac{b^2-bc+c^2}{b+c}$ Clearing denominators gives
$(b+c)(3a^2b^2+5a^2bc+3a^2c^2+4ab^3+10ab^2c+10abc^2+4ac^3+3b^3c+5b^2c^2+3bc^3) \\ + (b^2-bc+c^2)(a^2b+a^2c+abc+b^2c+bc^2) = 0$ This expands out to
$4a^2b^3+8a^2b^2c+8a^2bc^2+4a^2c^3+4ab^4+15ab^3c+19ab^2c^2+15abc^3+4ac^4+4b^4c+8b^3c^2+8b^2c^3+4bc^4 = 0$ We factor this as
$(b^2+bc+c^2)(4a^2b+4a^2c+4ab^2+11abc+4ac^2+4b^2c+4bc^2) = 0$ Now if $b^2+bc+c^2 = 0$ , then we have $(b+c)^2 = bc$ . So let $N$ be the midpoint of arc $BC$ not containing $A$ , so that $b+c = n$ and $bc = n^2$ . Then $d = -n$ , and we see that the equations for $x$ both become $(x+n)^2 = 0$ , so $x = -n = d$ as well. This is impossible since $D,X$ cannot be the same point. So instead we have
$4a^2b+4a^2c+4ab^2+11abc+4ac^2+4b^2c+4bc^2 = 0$ and so
$4(a+b+c)(ab+ac+bc) - abc = 0 \implies (a+b+c)\left(\frac1a+\frac1b+\frac1c\right) = \frac14 \implies |a+b+c| = |o-h| = \frac12$ as desired. $\blacksquare$

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#3 h Apr 28, 2025, 1:06 AM

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Let $M'$ be the reflection of $O$ over $M$ and let $\ell$ be the perpendicular bisector of $\overline{BC}$ . Further let $H'$ and $X'$ be the reflection of $H$ and $X$ over $\ell$ , respectively. Now if $\Gamma$ is the unit circle, then $d = a + o - h = -b - c$ and $m' = 2m - o = b + c$ , and therefore we can conclude that $D$ and $M'$ are reflections over $O$ . Moreover, observe that $XM' = 2YM = 2AO = 2XO$ by the midpoint theorem on $\triangle OXM'$ .

Claim: $DXH'HX'$ is cyclic.
Proof. It suffices to prove that $DXH'H$ is cyclic because $XH'HX'$ is an isosceles trapezium. But $\angle DXH = 90^\circ - \angle DHA = \angle DHH' = \angle DH'H$ as $AH \perp HH'$ and $\triangle DHH'$ is isosceles with apex $D$ , so we're done. $\square$

Let $\Gamma'$ be the reflection of $\Gamma$ over $BC$ , which passes through $H$ and $H'$ by orthocentre reflections, and let $\Omega$ be the circle with centre $O$ and radius $R / 2$ , where $R$ is the radius of $\Gamma$ . Clearly $M'$ is the centre of $\Gamma'$ . Recall the forgotten coaxiality lemma:

The forgotten coaxiality lemma wrote:

Let $\omega_1$ and $\omega_2$ be two circles, and let $k \neq 1$ be a real number. Then the locus $\omega_3$ of points $X$ satisfying $\operatorname{Pow}(X, \omega_1) = k \operatorname{Pow}(X, \omega_2)$ is a circle coaxial with $\omega_1$ and $\omega_2$ .

Set $\omega_1 = \Gamma'$ , $\omega_2 = \Omega$ and $k = 4$ in the above theorem.

Firstly, $\operatorname{Pow}(D, \Gamma') = DM'^2 - R^2 = 4(DO^2 - (R / 2)^2) = 4 \operatorname{Pow}(D, \Omega)$ .
Secondly, $\operatorname{Pow}(X, \Gamma') = XM'^2 - R^2 = 4(XO^2 - (R / 2)^2) = 4 \operatorname{Pow}(X, \Omega)$ .
Lastly, $\operatorname{Pow}(X', \Gamma') = 4 \operatorname{Pow}(X', \Omega)$ because of the above item and the fact that both $\Gamma'$ and $\Omega$ are symmetric over $\ell$ .

Hence, $\omega_3 = (DXH'HX')$ is coaxial with $\Gamma'$ and $\Omega$ . However, $(DXH'HX')$ and $\Gamma'$ intersect at $H$ and $H'$ exactly, so these points should lie on $\Omega$ too. Thus, $OH = R / 2$ , which solves the problem.

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#4 h Apr 28, 2025, 8:01 AM • 1 Y

Y by Lahmacuncu

Let $O'$ be the reflection of $O$ wrt $BC$ .Let $K=(DXH)\cap OD$ . Since $\angle DKH=\angle DXH=90-\angle AHD=90- \angle HDK$ $\angle DHK= 90$ . From Thales $O'X=2MY=2OA=2OX$ . So $\frac{XO}{XO'}=\frac{1}{2}$ . And since $AHON$ is a parallelogram $DO=AH=OO'$ .So $\frac{DO}{DO'}=\frac{1}{2}$ . Since $(DXKH)$ 's diameter is on $OO'$ and $\frac{XO}{XO'}=\frac{DO}{DO'}=\frac{1}{2}$ from the Apollionus setup $\frac{1}{2}=\frac{HO}{HO'}=\frac{HO}{AO}$ . So we are done.

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