1
$\begingroup$

Today while using the Geogebra program I discovered a new engineering feature, I don't know if it was already known or not

enter image description here

Data: $R$ is a circle with center $O$, $A,B,C,D∈R$, $AD∩BC=E$, $AD∩BC=F$, $R_1$ is the passing circle passing through the points $A,B,E$, its center is $M$, and $R_2$ is the passing circle passing through the points $A,C,D$, its center is $N$, $R_3$ is the passing circle passing through the points $B,C,F$, its center is $P$, and $R_4$ is the passing circle passing through the points $A,D,F$, its center is $Q$, and $R_1∩R_2∩R_3∩R_4=S$. Required: Prove that the points $M,N,O,P,Q,S$ lie on one circle.

But I don't know how to prove it.

Whoever can help, please kindly

If the feature is already known, please mention a source that talks about it, and thank you

$\endgroup$
7
  • $\begingroup$ Interesting. First hint I can suggest is to get rid of R3 and R4 as if you prove it for R1 and R2 it will be straightforward then to finish with the circles 3 and 4 $\endgroup$
    – Martigan
    Commented May 7 at 18:58
  • $\begingroup$ It seems that five of them are known to lie on one circle, the rest of the proof that the center of the circle R belongs to the same circle artofproblemsolving.com/wiki/index.php/Miquel%27s_point $\endgroup$ Commented May 7 at 19:08
  • $\begingroup$ Are you sure there is no additional condition in your case? In the link you gave it seems the center of the initial triangle (called O in your case) would not be cocyclic with the other centers. Maybe BCD isocele? $\endgroup$
    – Martigan
    Commented May 7 at 19:39
  • 1
    $\begingroup$ Fun Fact: $S$ is the foot of the perpendicular from $O$ to line $EF$. Moreover, this perpendicular passes through $AC \cap BD$. $\endgroup$
    – Blue
    Commented May 7 at 19:50
  • $\begingroup$ The link i provided talks about 4 points in the general case, while here it talks about 4 points that lie on a circle @Martigan $\endgroup$ Commented May 8 at 0:02

1 Answer 1

0
$\begingroup$

COMMENT.-The problem can be stated in another equivalent way.

From a point outside the given circle $R$ we draw two lines that determine the points $A,B,C,D$ of the circle which in turn determines the point $E$.

The center $Q$ of $R_4$ is the intersection of the bisectors of the sides $AF$ and $FD$ of the triangle $\triangle{AFD}$ and the center $P$ of $R_3$ is the intersection of the bisectors of the sides $BF$ and $FC$ of the triangle $\triangle{BFC}$.

The points $O,P,Q$ determine an unique circle (which will be the red circle of the statement).

A procedure exactly the same as the one we have just given determines the centers $M$ of $R_1$ and $N$ of $R_2$ and then we can optionally verify that $M$ and $N$ belong to the red circle or see that the equation of the circumcircle of the triangle $\triangle{OMN}$ coincides with the preceding one of the triangle $\triangle{OPQ}$.

enter image description here

$\endgroup$
1
  • 1
    $\begingroup$ WARNING.- The red circle of the statement IS NOT the red circle of the attached figure. $\endgroup$
    – Piquito
    Commented May 15 at 15:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .