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Coaxal Circles
fattypiggy123   30
N 26 minutes ago by Ilikeminecraft
Source: China TSTST Test 2 Day 1 Q3
Let $ABCD$ be a quadrilateral and let $l$ be a line. Let $l$ intersect the lines $AB,CD,BC,DA,AC,BD$ at points $X,X',Y,Y',Z,Z'$ respectively. Given that these six points on $l$ are in the order $X,Y,Z,X',Y',Z'$, show that the circles with diameter $XX',YY',ZZ'$ are coaxal.
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fattypiggy123
Mar 13, 2017
Ilikeminecraft
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circle geometry showing perpendicularity
Kyj9981   4
N Apr 24, 2025 by cj13609517288
Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$. A line through $B$ intersects $\omega_1$ and $\omega_2$ at points $C$ and $D$, respectively. Line $AD$ intersects $\omega_1$ at point $E \neq A$, and line $AC$ intersects $\omega_2$ at point $F \neq A$. If $O$ is the circumcenter of $\triangle AEF$, prove that $OB \perp CD$.
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Kyj9981
Mar 18, 2025
cj13609517288
Apr 24, 2025
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circle geometry showing perpendicularity
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Kyj9981
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Kyj9981
#1 h  Mar 18, 2025, 11:53 AM • 1 Y
Y by Rounak_iitr
Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$. A line through $B$ intersects $\omega_1$ and $\omega_2$ at points $C$ and $D$, respectively. Line $AD$ intersects $\omega_1$ at point $E \neq A$, and line $AC$ intersects $\omega_2$ at point $F \neq A$. If $O$ is the circumcenter of $\triangle AEF$, prove that $OB \perp CD$.
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Retemoeg
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Retemoeg
#2 h  Mar 18, 2025, 4:54 PM • 1 Y
Y by Kyj9981
Could be resolved nicely through Carnot's theorem.
Denote $M, N$ midpoints of segments $AE, AF$.
We'd have that $ON \perp AF$, $OM \perp AE$.
\[ MA^2 - MD^2 + BD^2 - BC^2 + NC^2 - NA^2 = 0 \]Notice that, by power of a point and Pythagorean's: $MA^2 - MD^2 = (MA - MD)(ME + MD) = -DA\cdot DE = -DB\cdot DC$.
Similarly, $NC^2 - NA^2 = CB\cdot CD$. Thus, the above sum translates to
\[ -DB\cdot DC + CB\cdot CD + BD^2 - BC^2 = (DB - DC)\cdot CD - (DB - DC)\cdot CD = 0 \]And we are done.
This post has been edited 2 times. Last edited by Retemoeg, Mar 18, 2025, 4:55 PM
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Double07
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Double07
#3 h  Mar 18, 2025, 7:26 PM • 2 Y
Y by Calamarul, Kyj9981
We will solve this with Moving Points Method:

Fix the $\omega_1$ and $\omega_2$ circles, $O_1$ and $O_2$ their centers and $A$ and $B$ their intersections and we will move point $C$ along $\omega_1$.

First of all, we will prove that $B, O_1, O_2, O$ are concyclic.

Notice that $OO_1$ is the bisector of segment $AE$ and $OO_2$ is the bisector of segment $AF$.

$\widehat{O_1OO_2}=180^\circ-\widehat{EAF}=180^\circ-\widehat{CAD}$.

We want to prove that $\widehat{O_1OO_2}=180^\circ-\widehat{O_1BO_2}\iff \widehat{CAD}=\widehat{O_1BO_2}=\widehat{O_1AO_2}$.

But $\widehat{O_1AO_2}=180^\circ-\widehat{AO_1O_2}-\widehat{AO_2O_1}=180^\circ-\frac{1}{2}\widehat{AO_1B}-\frac{1}{2}\widehat{AO_2B}=180^\circ-\widehat{ACB}-\widehat{ADB}=\widehat{CAD}\quad\blacksquare$

Let now $X$ be the intersection of the perpendicular in $B$ on the $CD$ line with the $(BO_1O_2)$ circle (other than $B$).

We will show there exist projective maps $C\to O$ and $C\to X$ and then we will just have to prove that $O=X$ for $3$ points $C\in\omega_1$.

Since $C\to AC$ is projective ($A$ is fixed), $AC\to O_2O$ is projective ($O_2O$ is the perpendicular from $O_2$ to $AC$) and $O_2O\to O$ is projective (the $(BO_1O_2)$ circle is fixed and $O$ is the intersection of $O_2O$ with it), we have $C\to O$ - projective.

Since $C\to BC$ is projective ($B$ is fixed), $BC\to BX$ is projective ($BX$ is the perpendicular in $B$ on line $BC$) and $BX\to X$ is projective (the $(BO_1O_2)$ circle is fixed and $X$ is the intersection of $BX$ with it), we have $C\to X$ - projective.

Now we are just left to show that $O=X$ for $3$ points $C$.

1. $C\to B\implies BC$ becomes the tangent to $\omega_1$ in $B\implies BX\to BO_1\implies X\to O_1$.
$C\to B\implies AC\to AB\implies O_2O\to O_2O_1\implies O\to O_1\implies X=O$.

2. $C$ is the antipode of $A$ in $\omega_1\implies BC\parallel O_1O_2\implies BX\to BA\implies X=BA\cap (BO_1O_2)$.
But $\widehat{O_1XO_2}=180^\circ-\widehat{O_1BO_2}=180^\circ-\widehat{O_1AO_2}$.
Since $\widehat{O_1XO_2}=180^\circ-\widehat{O_1AO_2}$ and $XA\perp O_1O_2$, we can conclude that $A$ is the orthocenter of $\Delta XO_1O_2\implies O_1A\perp XO_2$.
We also have $AC=AO_1\implies O_2O\perp O_1A\implies X=O$.

3. $C\to A\implies AC$ becomes the tangent in $A$ at $\omega_1\implies OO_2\parallel O_1A$ and we similarly get $OO_1\parallel O_2A\implies O_1AO_2O$ is a parallelogram. Since $A$ is the reflection of $B$ across $O_1O_2$, is well-known that $O$ is the reflection of $B$ across the bisector of segment $O_1O_2$.
$C\to A\implies BC\to AB\implies BX\parallel O_1O_2\implies O_1O_2XB$ is an isoscelles trapezoid, so $X$ is also the reflection of $B$ across the bisector of segment $O_1O_2$, so $X=O$.

So projective functions $C\to O$ and $C\to X$ are equal in $3$ different points $C$, so $O=X$ for all points $C\in\omega_1$.
This means that $OB\perp CD$.
This post has been edited 1 time. Last edited by Double07, Mar 18, 2025, 7:30 PM
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JollyEggsBanana
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JollyEggsBanana
#4 h  Apr 24, 2025, 1:25 PM • 1 Y
Y by Kyj9981
Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively.

The main claim is that $O_1O_2BO$ is cyclic. Note $O_1O \perp AE$ and $O_2O \perp AF$ by radax. This implies $\measuredangle O_1OO_2 = \measuredangle DAC = \measuredangle O_1BO_2$.

From here we can angle chase
\[\measuredangle OBC = \measuredangle OBO_1 + \measuredangle O_1BC = \measuredangle FAB + 90 - \measuredangle CAB = 90\]
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cj13609517288
1919 posts
cj13609517288
#5 h  Apr 24, 2025, 2:41 PM • 1 Y
Y by Kyj9981
https://www.geogebra.org/calculator/vz7v5yep

Let $G=CE\cap DF$ which lies on $(AEF)$ by triangle Miquel. Then $\angle CBE=\angle DBF=\angle G$, so $\angle EOF=2\angle G=180^\circ-\angle EBF$, so $EBFO$ cyclic. Now we can access the argument of $OB$, so the problem dies:
\[\angle OBC=\angle OBE+\angle EBC=\angle OFE+\angle EAC=90^\circ.\]
This post has been edited 1 time. Last edited by cj13609517288, Apr 24, 2025, 2:41 PM
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