Three circles are concurrent

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Source: RMM 2025 P5

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#1 h Feb 13, 2025, 12:37 PM • 3 Y

Y by cubres, puntre, soryn

Let triangle $ABC$ be an acute triangle with $AB<AC$ and let $H$ and $O$ be its orthocenter and circumcenter, respectively. Let $\Gamma$ be the circle $BOC$ . The line $AO$ and the circle of radius $AO$ centered at $A$ cross $\Gamma$ at $A’$ and $F$ , respectively. Prove that $\Gamma$ , the circle on diameter $AA’$ and circle $AFH$ are concurrent.
Proposed by Romania, Radu-Andrew Lecoiu

This post has been edited 4 times. Last edited by Twoisaprime, Feb 14, 2025, 6:14 AM

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#2 h Feb 13, 2025, 1:09 PM • 4 Y

Y by SomeonesPenguin, maria_gorgan, cubres, soryn

Nice problem! The $\sqrt{bc}$ reduction feels a bit underwhelming, though (and it might be how the problem was created?).

We $\sqrt{bc}$ invert. Then $H$ maps to $A'$ , while $O$ maps to the reflection of $A$ across $BC$ , which we label $R$ . Additionally, $F$ maps to the intersection between the circle centered at $A$ with radius $AR$ and $\left(BRC\right)$ , which we label $S$ .

Finally, the intersection of $\left(AFH\right)$ and $\left(BOC\right)$ maps to $K$ , the second intersection of $A'S$ and $\left(BRC\right)$ . The conclusion thus reduces to showing that $\angle AHK=90^{\circ}$ , or equivalently $KH\parallel BC$ . This further reduces to $\angle CSA'=\angle OAC$ . Now note that $A$ is the $A'$ -excenter of $\triangle A'BC$ .

Therefore (by phantom points), if rephrased with respect to $\triangle A'BC$ , the problem reduces to the following.

Rephrasing wrote:

Let $ABC$ be a triangle and let $I_A$ be its $A$ -excenter. Let $S$ be a point so that $\angle BSA=\frac{1}{2}\angle ACB$ and $\angle CSA=\frac{1}{2}\angle ABC$ , on the same side of $BC$ as $A$ . Show that $I_AR$ is twice the $A$ -exradius.

Let $U$ be the reflection of $I_A$ across $AC$ and let $V$ be the reflection of $I_A$ across $AB$ . Then we wish to show that $I_A$ is the circumcenter of $\triangle VSU$ . Since $I_A$ lies on the perpendicular bisector of $UV$ , it suffices to show that $\angle VI_AU=360^{\circ}-2\angle VSU$ .

Now it easily follows that $\triangle ACI_A\equiv\triangle ACU$ , so $\angle AUC=\frac{1}{2}\angle ABC$ . Therefore $ASUC$ is cyclic, so
$\angle CSU=\angle CAU=\angle I_AAC=\frac{1}{2}\angle BAC.$
Similarly $\angle VSB=\frac{1}{2}\angle BAC$ , hence
$\begin{align*} \angle VSU&=\angle VSB+\angle BSC+\angle CSU\\ &=90^{\circ}+\frac{1}{2}\angle BAC\\ &=180^{\circ}-\angle BI_AC\\ &=180^{\circ}-\frac{1}{2}\angle VI_AU, \end{align*}$ as desired.

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#3 h Feb 13, 2025, 1:11 PM • 1 Y

Y by cubres

(Diagram to follow...)

Let $D$ be the reflection of $O$ over $A$ . Suppose $\Gamma$ and the circle on diameter $AA'$ intersect at $P\neq A'$ .

Invert about $(ABC)$ . $\Gamma$ swaps with $BC$ , so the image of $A'$ lies on $BC$ , which implies $P^*$ is the foot of the altitude from $A$ to $BC$ . $D^*$ is the midpoint of $AO$ , and $F^*$ is the intersection of the perpendicular bisector of $AO$ with $BC$ . So $P^*$ and $D^*$ lie on the circle with diameter $AF^*$ , i.e. $DAFP$ is cyclic. So it suffices to show that $H$ also lies on this circle.

Let the bisector of $\angle{A}$ intersect $(ABC)$ at $M\neq A$ , and let $X$ , $Y$ be the centres of $(BHC)$ and $\Gamma$ respectively. These circles swap under root- $b$ - $c$ inversion, as $O$ is sent to the reflection of $A$ over $BC$ , which lies on $(BHC)$ . Thus, $AM$ bisects $\angle{XAY}$ .

Reflecting $AO$ over $AM$ and $AY$ gives lines $AH$ and $AF$ respectively, so we can compute $\angle{HAF}=\angle{XAY}$ . So it suffices to show that $\angle{HDF}=\angle{XAY}$ .

A homothety of factor $\frac 12$ centred at $A$ takes $(BHC)$ to the nine-point circle of $\triangle{ABC}$ , so the midpoint of $AX$ is $N_9$ . But this is also the midpoint of $OH$ , so $AOXH$ is a parallelogram. Hence, $ADHX$ is also a parallelogram, i.e. $DH\parallel AX$ . But also $DF\parallel AY$ as they are both perpendicular to $OF$ , so $\angle{HDF}=\angle{XAY}$ as desired.

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#4 h Feb 13, 2025, 1:56 PM • 2 Y

Y by Funcshun840, cubres

Rename the point $A'$ in the problem to $D$ . We will start with a $\sqrt{bc}$ inversion. This gives us the following problem:

Triangle $\triangle ABC$ with orthocenter $H$ and circumcenter $O$ . Let $AO$ intersect $(BOC)$ at $D\neq O$ and let $A'$ be the reflection of $A$ in $BC$ . $F\neq A'$ is on $(A'BHC)$ such that $AF=AA'$ . Prove that the line through $H$ parallel to $BC$ and the line $DF$ intersect at the circle $(A'BHC)$ .

Because of Thales this is equivalent to proving $\angle A'FD=90^\circ$ . Now let $AF$ intersect $(A'BHCF)$ again at $X$ . From inversion things we get that $AH\cdot AD=AB\cdot AC=AO\cdot AA'$ and that implies $A'D\parallel HO$ . Also obviously $A'F\parallel HX$ because of isosceles trapezoid. Now we see $\frac{AX}{AF}=\frac{AH}{AA'}=\frac{AO}{AD}$ . This implies that triangles $\triangle HOX$ and $\triangle A'DF$ are similar with parallel sides. Therefore we want to show that $\angle HXO=90^\circ$ . Now we have reduced the problem to the following lemma:

Triangle $\triangle ABC$ with circumcenter $O$ and orthocenter $H$ . $X\neq H$ is on $(BHC)$ such that $AX=AH$ . Prove that $\angle HXO=90^\circ$ .

Let $M$ be the reflection of $O$ in $BC$ , we know that $M$ is the center of $(BHC)$ . Let $X'$ be the point such that $AMX'O$ is an isosceles trapezoid. We have $AX'=OM=AH$ and $MX'=AO=BO=MB$ , so $X'=X$ . Now look at the radical axis of circle $(BHC)$ and the circle centered at $A$ through $H$ . Obviously $HX$ is that radax so $HX\perp AM$ . Also $AM\parallel XO$ , so $HX\perp XO$ and we are done.

This post has been edited 1 time. Last edited by ErTeeEs06, Feb 13, 2025, 1:57 PM
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#5 h Feb 13, 2025, 3:03 PM • 1 Y

Y by cubres

no complex bashes yet?

Invert around $O$ . Then $A'$ goes to $A'^\ast=AO\cap BC$ . Then $(AA')$ goes to $(AA'^\ast)$ and $\Gamma$ goes to $BC$ , so their other intersection is the foot of the perpendicular from $A$ to $BC$ , call it $X$ .
$F$ goes to the intersection of $BC$ and the perpendicular bisector of $AO$ , call it $Y$ .
Let $H$ go to $Z$ , we can directly compute it later.
Obviously $\angle AXY=90^{\circ}$ so it suffices to prove $\angle YZA=90^{\circ}$ .

Now let's bash. I did this on paper so I will only provide a summary here. For $y$ , our two equations simplify to
$bc\overline{y}+y=b+c$ $a^2\overline{y}+y=a.$ Obviously
$z=\frac{1}{\overline{a+b+c}}=\frac{abc}{ab+bc+ca}.$ Now after only five lines of computation, we get
$\frac{y-z}{z-a}=\frac{a(b^2+bc+c^2)}{(b+c)(bc-a^2)}$ which we can check to be equal to its negative conjugate.

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#6 h Feb 13, 2025, 3:28 PM • 6 Y

Y by OronSH, khina, cubres, NCbutAN, MS_asdfgzxcvb, ehuseyinyigit

$\sqrt{bc}$ inversion gives the following problem.

Quote:

Let $\triangle ABC$ be a triangle with orthocenter $H$ and circumcenter $O$ . Line $AO$ meet $\odot(BOC)$ at $A'$ . Let $A_1$ be the reflection of $A$ across $BC$ . Point $F$ lies on $\odot(BHC)$ such that $AA_1=AF$ . Prove that $A'F$ , $\odot(BHC)$ , and line through $H$ parallel to $BC$ are concurrent.

To prove this, let $O_1$ be the center of $\odot(BHC)$ . Note that $A'A_1\parallel OH$ , so since $AO_1$ bisects $OH$ , it follows that $AO_1$ bisects $A_1A'$ . Now, if $M$ is the midpoint of $A_1A'$ , then $MF=MA_1=MA'$ , which implies $\angle A_1FA'=90^\circ$ . Thus, if $A'F$ meet $\odot(BHC)$ again at $P$ , then $\angle PHA_1=\angle PFA_1=90^\circ$ , so $HP\parallel BC$ .

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#7 h Feb 13, 2025, 4:16 PM • 3 Y

Y by OronSH, cubres, ehuseyinyigit

When the easiest problem at RMM is the p5 geo..

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#8 h Feb 13, 2025, 4:23 PM • 3 Y

Y by cubres, NCbutAN, amirhsz

Let $AHF$ meet $\Gamma$ at $X$ . Let the circle with center $A$ and radius $AO$ meet $AO$ at $E \neq O$ . we need to prove $\angle AXA' = 90$ . Note that $\angle AXA' = \angle AXF+\angle FXA' = \angle AXF+\angle FOA$ . Since $\angle EFO=90$ we need to prove $\angle AEF=\angle AXF$ so we need to prove $E$ lies on $AHFX$ . Let $O'$ be reflection of $O$ about $BC$ and $N$ be the midpoint of $OH$ . Let $T$ be the reflection of $A$ about $OF$ . Let $AT$ meet $OO'$ at $K$ . First note that $K$ lies on perpendicular bisectors of $BC$ and $FO$ so $K$ is the center of $BOC$ . Note that since $EA=AO$ and $HN=NO$ then $\angle EHA=\angle HAN=\angle OO'A$ and $\angle EFA=\angle FEA=\frac{\angle FAO}{2}=\angle TAO=\angle ATO$ so we need to prove $AOO'T$ is cyclic. we have that $KT.KA=KO^2-R^2$ where $R$ is the radius of $ABC$ and we need to prove $KT.KA=KO'.KO$ so we need to prove $KO.(KO-KO')=R^2$ or $KO.OO'=R^2$ which is true since $OCO'$ and $OKC$ are similar.
we're done.

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#9 h Feb 13, 2025, 4:26 PM • 1 Y

Y by cubres

Perform $\sqrt{bc}$ inversion.
New Problem Statement: $ABC$ is a triangle with circumcenter $O$ and orthocenter $H$ . Let $AO\cap (BOC)=R,\ A'$ be the reflection of $A$ over $BC$ . Let $W$ be the circumcenter of $(BHC)$ and $F$ be the reflection of $A'$ over $AW$ . Let $K\in (BHC)$ and $HK\parallel BC$ . Prove that $R,K,F$ are collinear.
Work on the complex plane. Let $(ABC)$ be the unit circle. Assume $a=1$ . We have $w=b+c,\ a'=b+c-bc$ hence $k=bc+b+c$ .
$f=\frac{(b+c-a)(\frac{1}{b}+\frac{1}{c}-\frac{a}{bc})+\frac{a}{b}+\frac{a}{c}-\frac{b+c}{a}}{\frac{1}{b}+\frac{1}{c}-\frac{1}{a}}=\frac{b^2+c^2+2bc+1-b-c-b^2c-bc^2}{b+c-bc}$ $r=\frac{bc+1}{b+c}, \ k=bc+b+c$ $\frac{k-r}{k-f}=\frac{b+c+bc-\frac{1+bc}{b+c}}{b+c+bc-\frac{b^2+c^2+2bc+1-b-c-b^2c-bc^2}{b+c-bc}}=\frac{(b+1)(c+1)(b+c-1)(b+c-bc)}{(b+c)(-b^2c^2+b^2c+bc^2+b+c-1)}$ $\overline{\frac{(b+1)(c+1)(b+c-1)(b+c-bc)}{(b+c)(-b^2c^2+b^2c+bc^2+b+c-1)}}=\frac{(\frac{b+1}{b})(\frac{c+1}{c})(\frac{b+c-bc}{bc})(\frac{b+c-1}{bc})}{(\frac{b+c}{bc})(\frac{-1+b+c+bc^2+b^2c-b^2c^2}{b^2c^2})}=\frac{(b+1)(c+1)(b+c-1)(b+c-bc)}{(b+c)(-b^2c^2+b^2c+bc^2+b+c-1)}$ As desired. $\blacksquare$

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polishedhardwoodtable

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polishedhardwoodtable

#10 h Feb 13, 2025, 5:11 PM • 2 Y

Y by OronSH, cubres

Let $O'$ be the reflection of $O$ over $BC$ and let $(AOO')$ and $(ABC)$ intersect at point $D$ .

Observe that line $BC$ bisects circle $(AOO')$ , which is equivalent to $BC$ and $(AOO')$ forming right angles with one another. then, inverting at $O$ with radius $AO$ sends $(AOO')$ to line $AD$ and sends $BC$ to $(BOC)$ , inversion preserves angles so these new two curves must still be perpendicular to each other, thus line $AD$ bisects $(BOC)$ .

Now recall that $AF=AO$ , so the perpendicular bisector of $FO$ both passes through $A$ and bisects $(BOC)$ implying that this is the same line as $AD$ , thus $D$ lies on the perpendicular bisector so $FD=DO=AO=AF$ . Thus $AODF$ is a rhombus.

Reflect $H$ across $BC$ to point $H'$ lying on the circumcircle. Clearly $AH'||OO'$ , $AO=H'O=HO'$ , and $\angle OO'H=\angle H'OO'=\angle OH'A=\angle H'AO$ (using the fact that triangle $AOH'$ is isosceles) thus $AOO'H'$ is a parallelogram. Thus $\vec{OA}=\vec{DF}=\vec{O'H}$ .

Now define point $E$ to be the reflection of $O$ over $A$ . Clearly, $\vec{AE}=\vec{OA}$ , so if we shift cyclic quadrilateral $AODO'$ by vector $\vec{OA}$ , we get quadrilateral $EAFH$ , so $E$ lies on $(AFH)$ .

Let $(AFH)$ intersect with $(BFC)$ at $X$ , we now have that $\angle AXA'=\angle AXF+\angle FXA'=\angle AEF+\angle FOA'=\angle OEF+\angle FEO+\angle OFE=\angle OFE$ which is clearly a right angle, thus the circle with diameter $AA'$ indeed passes through $X$ as desired.

This post has been edited 1 time. Last edited by polishedhardwoodtable, Feb 13, 2025, 5:14 PM
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#11 h Feb 13, 2025, 5:54 PM • 1 Y

Y by cubres

same as jatloe

Let $\Omega = (ABC),$ and let $X,$ $Y,$ and $D$ be on $BC$ such that $AX = OX$ , $Y$ lies on $AO,$ and $D$ lies on $AH.$ Invert about $\Omega$ ; let the point $H$ get sent to $H'.$ Then $\Gamma$ is sent to $BC$ and the circle with diameter $AA'$ is sent to the circle with diameter $AY.$ Furthermore, the circle with center $A$ passing through $O$ is sent to the perpendicular bisector of $AO,$ so $F$ is sent to $X.$ We would like to show that the circle with diameter $AY$ intersects $(AH'X)$ on $BC,$ or equivalently $AH'XD$ is cyclic. Since $\angle ADX = 90^\circ,$ it suffices to show that $\angle AH'X = 90^\circ$ as well.

We now use complex numbers with $\Omega$ as the unit circle. Then $h = a+b+c,$ so $h' = \frac{1}{\overline{a+b+c}} = \frac{abc}{ab+bc+ca}.$ Now we find $x.$ Let $\omega = \frac{1}{2} + \frac{\sqrt{3}}{2} i$ , so that the perpendicular bisector of $AO$ passes through $a\omega$ and $a\overline{\omega}.$ Thus by the complex intersection formula, we get $x = \frac{a \omega \cdot a\overline{\omega} (b+c) - bc(a\omega + a\overline{\omega})}{a\omega \cdot a\overline{\omega} - bc} = \frac{a^2 b + a^2 c - abc}{a^2 - bc}.$ Then we would like to show that $z = \frac{h'-a}{h'-x} = \frac{\frac{abc}{ab+bc+ca} - a}{\frac{abc}{ab+bc+ca} - \frac{a^2 b + a^2 c - abc}{a^2 - bc}}$ is pure imaginary, which follows by noting that $\overline{z} = -z$ (we see this by clearing denominators and expanding).

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SomeonesPenguin

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SomeonesPenguin

#12 h Feb 13, 2025, 7:12 PM • 2 Y

Y by trigadd123, cubres

We $\sqrt{bc}$ invert and denote by $X'$ the inverse of $X$ . Clearly $O'$ is the reflection of $A$ over $BC$ , $H'$ is actually $A'$ (from the problem), $F'$ lies on $(BCO')$ such that $AO'=AF'$ . Circle $(AFH)$ maps to $F'A'$ , the circle with diameter $AA'$ maps to the line parallel to $BC$ through $H$ and the circle $(BOC)$ maps to the circle $(BCO')$ . Let $T$ be the point on $A'F'$ such that $TH\parallel BC$ . We wish to show that $HTF'O'$ is cyclic, or equivalently that $\angle TF'O'=90^\circ$ . This is further equivalent to $AH$ and $F'A'$ intersecting on the circle with center $A$ and radius $AO'$ . Looking at the diagram before the inversion, it would suffice to prove that the reflection of $O$ over $A$ lies on $(AHF)$ . Denote this point by $E$ .

Now we proceed with complex numbers. Let $(ABC)$ be the unit circle. Clearly $e=2a$ and $h=a+b+c$ . $\left\vert a-f\right\vert=1\iff \overline f=\frac{f}{af-a^2}$ $\frac{b-f}{c-f}\cdot\frac{c}{b}\in\mathbb R\iff\frac{bc-cf}{bc-bf}=\frac{1-b\overline{f}}{1-c\overline{f}}$ $\iff f\overline{f}(b-c)(b+c)-f(b-c)-bc\overline{f}(b-c)=0\iff \overline{f}=\frac{f}{f(b+c)-bc}$ Therefore, $f=\frac{a^2-bc}{a-b-c}$ . Now we just need to check that $\frac{a}{b+c}\cdot\frac{a+b+c-\frac{a^2-bc}{a-b-c}}{2a-\frac{a^2-bc}{a-b-c}}\in\mathbb R\iff \frac{ab^2+ac^2+abc}{(b+c)\left(a^2-2ab-2ac+bc\right)}=\frac{\frac{1}{a}}{\frac{1}{b}+\frac{1}{c}}\cdot\frac{\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{bc}}{\frac{1}{a^2}-\frac{2}{ab}-\frac{2}{ac}+\frac{1}{bc}}$ $\iff \frac{ab^2+ac^2+abc}{a^2-2ab-2ac+b}=\frac{\frac{b}{c}+\frac{c}{b}+1}{\frac{1}{a}-\frac{2}{b}-\frac{2}{c}+\frac{a}{bc}}$ Which is clear. $\blacksquare$

diagram

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OronSH

#13 h Feb 13, 2025, 8:32 PM • 2 Y

Y by cubres, ihatemath123

Force overlay invert. Let $T,Z$ be the reflections of $A,O$ over $BC$ and let $S$ be the reflection of $T$ over $AZ$ . The problem becomes showing that $(BHC)$ , line $A'S$ and the line through $H$ parallel to $BC$ concur.

We claim the point is the antipode $X$ of $T$ on $(BHC)$ . Clearly $X\in(BHC)$ and $XH\perp HT\perp BC$ . Thus we want $X,A',S$ collinear, and since $XS\perp ST\perp AZ$ this is the same as $AZ\parallel A'S$ , or $[AZA']=[AZS]=-[AZT]=-[AOT]$ .

Now we get that $T=b+c-\frac{bc}a$ and $A'=\frac{bc+a^2}{b+c}$ so computation gives $\begin{vmatrix}1&a&\frac1a\\1&0&0\\1&b+c-\frac{bc}a&\frac1b+\frac1c-\frac a{bc}\end{vmatrix}=\frac ba+\frac ca-\frac ab-\frac ac+\frac{a^2}{bc}-\frac{bc}{a^2}=-\begin{vmatrix}1&a&\frac1a\\1&b+c&\frac1b+\frac1c\\1&\frac{bc+a^2}{b+c}&\frac{bc+a^2}{a^2(b+c)}\end{vmatrix}$ as desired.

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#14 h Feb 14, 2025, 11:05 PM • 2 Y

Y by rstenetbg, cubres

We will $\sqrt{bc}$ invert to get rid of the weird circle $(AFH)$ ; also it seems that this approach after the inversion has not been posted above.

After the inversion, if $A_1$ is the reflection of $A$ in $BC$ , $P \in (BHC)$ is such that $HP \parallel BC$ and $Q\in (BHC)$ is such that $AA_1=AQ$ , then we have to show that $P, A', Q$ are collinear. We will show that $\angle A'PA_1=\angle QHA_1=\angle QPA_1$ , which is sufficient.

We first claim that $OPA'A_1$ is cyclic. Indeed, since $\angle PA_1H=90^{\circ}-\angle HPA_1=90^{\circ}-\angle HCA_1=\beta-\gamma=\angle OAH,$ we have that if $T=PA_1 \cap AA'$ , then $TA=TA_1$ , i.e. $T \in BC$ . Thus, by power of point, $TP \cdot TA_1=TB \cdot TC=TO \cdot TA'$ , so $OPA'A_1$ is indeed cyclic, so we have shown $\angle A'PA_1=\angle A'OA_1$ , and hence we need $\angle A'OA_1=\angle QHA_1$ .

If $\angle AA_1Q=\angle AQA_1=\varphi$ , then $\angle QHA_1=180^{\circ}-\angle HQA_1-\varphi=90^{\circ}+\beta-\gamma-\varphi$ and $\angle A'OA_1=\angle OAH+\angle OA_1A=\beta-\gamma+\angle OA_1A,$ so we need $\angle OA_1A=90^{\circ}-\varphi$ , which is equivalent to the circumcenter of $\triangle AA_1Q$ lying on $OA_1$ . Indeed, if $OA_1 \cap BC=X$ , we have to show that $X$ lies on the perpendicular bisector of $A_1Q$ . But this perpendicular bisector is the line through $A$ and the circumcenter of $BHC$ and since this line is the reflection of $OA_1$ in $BC$ , we obtain that $X$ lies on it, which finishes the problem.

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#15 h Feb 17, 2025, 6:46 PM • 3 Y

Y by cubres, OronSH, numbersandnumbers

Let $D$ be the foot from $A$ to $BC$ and let $X$ denote $(AA')\cap (BOC)$ . Inversion about $(ABC)$ swaps $D$ and $X$ , so $O,D$ , and $X$ are collinear.

Next, let $Y=(AHX)\cap ODX$ . We have
$DY\cdot DX=DH\cdot DA=DB\cdot DC=DO\cdot DX,$ so $Y$ is the reflection of $O$ over $D$ .

Now, we claim $AHFY$ is an isosceles trapezoid, which suffices as it will then be cyclic. First, if we let $H'$ be the reflection of $H$ over $D$ , we have
$HY=OH'=R=AF,$ so it suffices to prove
$YF=AH\Longleftrightarrow ND=OM,$ where $N,M$ are the midpoints of $OF$ and $BC$ respectively.

To finish, let $O_1$ denote the center of $(BOC)$ , so $N$ is the foot from $O$ to $AO_1$ , and define variables $OO_1=x$ and $OM=y$ . Let $P$ and $Q$ be the feet from $N$ to $AD$ and $OM$ respectively, and define $\theta=\angle AO_1O$ .

We have
$\begin{align*} DP &= y-OQ = y-x \sin^2\theta \\ OA&=OB=\sqrt{OM\cdot 2OO_1}=\sqrt{2xy}\\ AN&=\sqrt{OA^2-ON^2}=\sqrt{2xy-x^2\sin^2\theta}\\ PN&=AN\sin\theta=\sqrt{2xy\sin^2\theta-x^2\sin^4\theta}\\ DN&=\sqrt{DP^2+PN^2}=\sqrt{(y-x\sin^2\theta)^2+2xy\sin^2\theta-x^2\sin^4\theta}=y=OM, \end{align*}$ as desired.

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#16 h Feb 21, 2025, 12:48 AM • 2 Y

Y by Radmandookheh, X.Luser

Let the circle passing through points $B, O, C$ (with center $Q$ ) and the circle with diameter $AA'$ intersect at $S \neq A'$ . To prove that $AFHS$ is concyclic, it suffices to show $\angle ASF = \angle AHF$ .

Observe that:
$\angle ASF = 90^\circ - \angle FSA' = 90^\circ - \angle AOF = \angle QAF$
Let $HF$ intersect $AQ$ at $T$ . To prove $\angle AHT = \angle FAT$ , it suffices to show:
$\left( \frac{AF}{AH} \right)^2 = \frac{TF}{TH}$
We compute:
$\frac{TF}{TH} = \frac{AF \sin \angle FAT}{AH \sin \angle HAT} = \frac{AF \sin \angle QAO}{AH \sin \angle AQO} = \frac{AF}{AH} \cdot \frac{OQ}{AO}$
Thus, it suffices to show:
$\frac{OQ}{AO} = \frac{AF}{AH} \quad \text{or equivalently} \quad AO^2 = OQ \cdot AH$ which is trivial.

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#17 h Feb 25, 2025, 9:05 AM

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We perform a $\sqrt{bc}$ inversion.

Let $O \mapsto A_1$ , where $A_1$ is the reflection of $A$ across $BC$ , and $H \mapsto A'$ (proven via trivial angle chasing).
Similarly, let $F \mapsto F'$ , where $F'$ lies on $(BHC)$ such that $AA_1 = AF'$ . Define $P$ as a point on $(BHC)$ such that $PH \parallel BC$ .

The problem reduces to proving that $P, F', A'$ are collinear.

Observe that $A_1P$ is the diameter of $(BHC)$ , so it suffices to show that $\angle H_1F'A' = 90^\circ$ .

Let $O_1$ be the circumcenter of $(BHC)$ , and let $A_0$ be the reflection of $A$ over $O_1$ .
Define $R$ as the second intersection of $A_0$ with $(BHC)$ such that $AR > A_0R$ .
Through inversion, we obtain $H_0 \parallel A_1R$ .
Furthermore, since $AA_1 = AF'$ , it follows that $A_0A_1 \perp A_1F'$ , which implies $A_1R = RF'$ .
Hence, it remains to show that $R$ is the midpoint of $A_1A'$ .

By homothety, this is equivalent to proving that $A_0$ bisects $OH$ .
Let the perpendicular bisector of $O$ intersect $BC$ at $T$ .
By a well-known lemma, we have $AH = 2(OT)$ .
Moreover, it is easy to see that $O_1$ is the reflection of $O$ over $BC$ , implying $OO_1 = 2(OT) = AH$ , which confirms the desired result.

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#18 h Mar 7, 2025, 10:50 AM

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Nice problem. This problem can be generalized using isogonal conjugate as follows my post

This post has been edited 2 times. Last edited by kaede_Arcadia, Mar 10, 2025, 2:36 AM

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#19 h Mar 10, 2025, 2:27 AM

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Let $K$ the reflection of $A$ over $BC$ and let $L$ a point on $(BHC)$ such that $AK=AL$ , from $\sqrt{bc}$ invert, as we need is that $\angle KLA'=90$ , so now let $N$ midpoint of $KA'$ and $O'$ reflection of $O$ over $BC$ , it happens to be center of $(BHC)$ but also from homothety $AHO'O$ is a parallelogram so because of inversion we have $OH \parallel KA'$ and therefore $A,O',N$ are colinear but this means $NA'=NK=NL$ and therefore $N$ is center lying on $KA'$ of $(KA'L)$ which finishes thus we are done :cool:

:cool:

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This post has been edited 1 time. Last edited by MathLuis, Mar 10, 2025, 2:28 AM

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#20 h Mar 12, 2025, 5:06 PM • 3 Y

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Claim. Let the refleciton of $O$ over $A$ be $O'$ . Then $O'AFH$ is cyclic.
Proof. Let $N$ be the nine-point center, and let $K$ be the circumcenter of $\triangle BOC$ . Recall that $AK$ and $AN$ are isogonal; hence, one obtains $\measuredangle AFO' = \measuredangle FO'A = \measuredangle FO'O = \measuredangle KAO = \measuredangle HAN.$ Now reflect $H$ across $A$ to $H'$ , and observe $\measuredangle AHO' = \measuredangle AH'O = \measuredangle HH'O = \measuredangle HAN = \measuredangle AFO'$ , as desired. $\square$

From here, let $(AFH)$ meet $(BOC)$ again at $X$ . We have $\measuredangle FXA' = \measuredangle FOA' = \measuredangle FOA = 90^\circ - \measuredangle AO'F = 90^\circ - \measuredangle AXF \implies \measuredangle AXA' = 90^\circ,$ as desired.

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#21 h Mar 15, 2025, 1:35 AM

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Let $\Gamma$ and the circle with diameter $AA'$ meet at $D$ , we want to prove $AFHD$ are concyclic.

Realize $\angle{ADH}=90-\angle{FDA'}=90-\angle{AOF}$ , denote the center of $(BOC)$ as $O'$ , note $AO'$ is perpendicular to bisector of $OF$ by radax, so we have $\angle{ADH}=\angle{OAO'}=\angle{FAO'}$

Now we want to have $\angle{ADH}=\angle{AHF}=\angle{FAO'}$

Realize $\frac{AH}{AO}=\frac{AH}{AF}=\frac{\sin(\angle{AFH})}{\sin(\angle{AHF})}=\frac{\sin(\angle{AFH}+\angle{FHA})}{\sin(\angle{AHF})}=2\cos(A)$ by property.

Now we can also have $\frac{\sin(\angle{FAO'}+\angle{HAF})}{\sin(\angle{FAO'})}=\frac{\sin(\angle{OO'A})}{\sin(\angle{OAO'})}=\frac{OA}{OO'}=\frac{OC}{OO'}=2\cos(A)=\frac{AH}{AF}$ which implies $\angle{AHF}=\angle{FAO'}$ , and it yields the desired result.

This post has been edited 1 time. Last edited by Bluesoul, Mar 15, 2025, 1:36 AM

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L13832

#22 h Mar 18, 2025, 5:41 PM • 1 Y

Y by alexanderhamilton124

solution

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