Reflections meet on circumcircle of ABC

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Enthurelxyz
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Reflections meet on circumcircle of ABC

Unread post by Enthurelxyz » Sun Jan 24, 2021 7:56 pm

Let $l$ is the Euler line of $\triangle ABC$. $l_a,l_b,l_c$ are the reflections of $l$ from $BC,AC,AB$ respectively. Prove that $l_a,l_b,l_c$ meet on the circumcircle of $ABC$.
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Anindya Biswas
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Re: Reflections meet on circumcircle of ABC

Unread post by Anindya Biswas » Tue Jan 26, 2021 2:56 pm

Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$ respectively.
Let $X, Y, Z$ be the intersection point of pairs of lines $(l, l_a), (l, l_b), (l, l_c)$ respectively.
Since $l, l_a$ are reflections with respect to line $BC$, so the line $BC$ must go through $X$. Similarly, the lines $CA, AB$ goes through the points $Y,Z$ respectively.
It is a well known fact that the reflection of the orthocenter with respect to a side of the triangle lies on the circumcircle of that triangle.
Let $D,E,F$ be the reflections of $H$ with respect to $BC, CA, AB$ repectively. So, $D,E,F$ must lie on the circumcircle of $\triangle ABC$. Also, $D,E,F$ lies on $l_a, l_b, l_c$ respectively since $H$ lies on $l$.
Let $P$ be the intersection point of pairs of lines $(l_b, l_c)$. Let $\measuredangle (l_b,l_c)$ be the directed angle mod $\pi$ that rotates the line $l_b$ to $l_c$.
Now, $\measuredangle (l_b,l_c)=\measuredangle EPF=\measuredangle YPZ=-\measuredangle PZY-\measuredangle ZYP$
$=\measuredangle YZF+\measuredangle EYZ=2(\measuredangle YZA+\measuredangle AYZ)=2\measuredangle YAZ=2\measuredangle CAB=\measuredangle EAF$.
So, $EPFA$ is a cyclic quadrilateral. So, $P$ lies on the circumcircle.
Let $Q$ be the intersection point of pairs of lines $(l_c, l_a)$. By the similar argument above, we can show that $Q$ also lies on the circumcircle.
So, $l_c$ intersects the circumcircle at $F, P, Q$. Since a line can only intersect a circle on at most $2$ distinct points, either $Q=P$ or $Q=F$.
If $Q=F$, then $l_a$ would be the line joining $F,D$. In this case, $l_a, l_b$ both has $2$ intersection points with the circle and non of them are intersection of $l_a$ and $l_b$. But this contradicts the fact that $(l_a, l_b)$ must have a intersection point on the circle.
So, it must be the case that $Q=P$. That means, $l_a, l_b, l_c$ goes through $P$ which lies on the circumcircle of $\triangle ABC$.
Q.E.D.
Reflection of Euler line with respect to the sides.png
Figure : $l_a, l_b, l_c$ are reflections of $l$ with respect to $BC, CA, AB$ respectively
Reflection of Euler line with respect to the sides.png (180.85KiB)Viewed 8642 times
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Anindya Biswas
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Re: Reflections meet on circumcircle of ABC

Unread post by Anindya Biswas » Tue Jan 26, 2021 10:54 pm

Another remark : I only used the fact that $l$ goes through $H$. Any line going through $H$ has this nice property, not just the very very special Euler Line!
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Mehrab4226
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Re: Reflections meet on circumcircle of ABC

Unread post by Mehrab4226 » Wed Jan 27, 2021 1:26 am

I didn't read your proof yet, Aninda Biswas. So it might be the same. If so then sorry :(
In the figure, the pink line is $l_C$ and the orange line is $l_B$
Let, $H'_1$ and $H'_2$ be the reflection of H over AB and AC respectively[Both of them lie on the circle which can be proved by a bit of angle chasing]. Let us join $H'_1, B$ and $l_C$ and $l_B$ intersect the circle at points say, F and F' respectively. And the rest is corresponding to the figure. (Please don't make me type out all the other points I constructed since there are a lot of them). Okay, now we can start,

$\angle HH'_2D = \angle HH'_2F' = \angle H'_2HD$[Because of reflection]
$\angle H'_2HD = \angle EHB$[Vertically Opposite]
Now $H'_1B$ is the reflection of $HB$ over AB and the pink line is the reflection of HE over AB.
$\therefore \angle EHB = \angle BH'_1F$[Reflection]
$\therefore \angle HH'_2F' = \angle BH'_1F$
That means F and F' are the same points.
So two reflected lines of the Euler line intersect at the circle. So by pairing $l_C$ with $l_A$ they must intersect each other at either $F$ or $H'_1$. Similarly, with $l_B$ and $l_A$ they must intersect at F or $H'_2$. If we assume that $l_A$ do not intersect at F it must pass through $H'_1$ and $H'_2$. But $l_A$ also passes through the reflection of H over BC which also lies on the circle, which is not possible for a straight line. CONTRADICTION.
So the lines concur at the circumcircle. $\square$
Attachments
geogebra-export.png
Some of the points mentioned not explicitly mentioned in the proof are shown here.
geogebra-export.png (292.28KiB)Viewed 8630 times
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