circle geometry showing perpendicularity

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#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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#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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#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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#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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#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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