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Let $ABCD$ be a bicentric quadrilateral, let $O$ be its circumcenter and $I$, its incenter. Let $M$ and $N$ be the midpoints of $AC$ and $BD$ repectively. Let $X$ be the intersection of $AC$ and $BD$. Let $P$ be the second intersection of circles (IAC) and (IBD). Prove that P is the Miquel point of the quadrilateral $OMXN$.

My work:

I have managed to prove that $P$,$X$,$I$ and $O$ are colinear using properties of the radical axis $IP$. We also know that $M$,$N$ and $I$ are colinear because of the Newton line of a circumscriptible quardrilateral. So $I$ is actually the intersection point of the diagonals of our quadrialteral $OMXN$. We also know that the quadrilateral $OMXN$ is cyclic and the diameter of its circumcircle is $OX$ so we know that its Miquel point is located on $FG$ where {$F$}=$OM\cap XN$ and {$G$}=$ON\cap XM$. Let $M_q$ be the Miquel point of $OMXN$. Because $FM\perp MX$ and $GN\perp NX$ we know that $X$ is the orthocenter of $\triangle OFG$ so $OX\perp FG$ but $XM_q\perp FG$ (because $XNGM_q$ is cyclic by definition and $XN\perp NG$) so $O$,$X$ and $M_q$ are collinear. Now we can define $M_q$ as the intersection of $OX$ and $FG$ so if we prove that $P$ also lies on $FG$ we are done

Edit: From Brocard's theorem we know that $FG$ is the polar line of $I$ with respect to the circumcircle of $OMXN$ so we only need to prove that $P$ lies on this polar line which translates to proving that $(O,X;I,P)$ is a harmonic division which seems way easier to prove than the initial statement but I think that it is still pretty tough.

Source:

This is actually something I have dicovered while playing with bicentric quadrilaterals on Geogebra. It seems like it is true so I hope it is lol. However I am quite sure that it is true and I am curious why is that so

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  • $\begingroup$ I have proved it, it is a really hard problem... $\endgroup$
    – JetfiRex
    Jan 29, 2022 at 23:57

1 Answer 1

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First, we will proof some lemma:

Lemma 1: Suppose a line $l$ pass $A,B,C,D$ four points, let $B',C',D'$ be the image when doing inversion of $A$. Then $(A,C;B,D)$ is a harmonic division if and only if $C'$ is the midpoint of $B'D'$.

Proof: Let $A$ be zero, and $B,C,D$ has the coordinate $b,c,d$. So we have $(A,C;B,D)$ is a harmonic division if and only if $(c-b)d=b(d-c)$, if and only if $1/b+1/d=2/c$, if and only if $C'$ is the midpoint of $B'D'$.

Then we use inversion to finish your proof:

enter image description here

We use inversion w.r.t. the incircle of $ABCD$. Let four tangent points of $AD,DC,CB,BA$ to be $Q,R,S,T$. Let $E,F,G,H$ be the midpoint of $QT,TS,SR,RQ$. So $E,F,G,H$ are images of $A,B,C,D$ respectively. Furthermore, let $K$ be the intersection of $EG$ and $HF$, so $K$ is the image of $P$. Let $L$ to be the image of $O$. Let the circumcircle of $HIF$ and $EIG$ be $J$. So we only need to proof, by lemma 1, $JK=KL$. enter image description here Now we seek for another description of $L$. Notice that $O$ is the intersection of the perpendicular bisector of $BD$ and the perpendicular bisector of $AC$. Let $HIFU$ be the harmonic quadrilateral, so $L$ is on the circle passing $I$, $U$, and orthogonal to the circumcircle of $HIF$. So we know that $\angle ULI=90^\circ-\angle UJI$. Similarly, let $V$ be the point such that $EIGV$ is the harmonic quadrilateral, so $\angle VLI+\angle VGI=90^\circ$.

Notice that $I,J,K,L$ are collinear* (this is because $O,X,I,P$ collinear), and $K$ is the intersection of $IK$ and circumcircle of $FIH$. So $JU\parallel HF$. Similarly, $VJ\parallel EG$. Since $K$ is the midpoint of $EG$ and $VJ\parallel EG$, we have $KJ=KV$. Similarly, $KJ=KU$. So $K$ is the circumcenter of $UJV$. Since $I,J,L$ are collinear, and also $\angle VLI+\angle VGI=90^\circ$ and $\angle ULI=90^\circ-\angle UJI$, we have $\angle JUL=\angle JVL=90^\circ$. So $J,V,L,U$ are on the same circle, and since $\angle JVL=90^\circ$, $JL$ is the diameter. Since $K$ is the circumcenter of $UJV$, we have $KJ=KL$. So the claim is proved.

  • If we don't know this is collinear, we can first using radical axis of circumcircles of $HIF$, $EIG$ and $EFGH$ to proof $I,J,K$ collinear, then we can proof that $U,I,V$ collinear (this can be derived from $\angle EIH+\angle HIG=180^\circ$). Then proof $K$ to be the circumcircle of $JUV$. Then, we let $L'$ to be the intersection of the perpendicular line of $JU$ passing $U$ and the perpendicular line of $JV$ passing $V$. We have $J,U,L',V$ on the same circle and $I,J,K,L'$ are collinear. Then we have $\angle UL'K=90^\circ-\angle UJK=\angle ULK$ and $\angle VL'K=90^\circ-\angle VJK=\angle VLK$, so $L=L'$. So $I,J,K,L$ are collinear,
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  • $\begingroup$ Wow, nice proof. Honestly, I wasn't expecting it to be this hard but I think that it is a pretty nice property with a cool proof $\endgroup$
    – alien2003
    Jan 30, 2022 at 13:01

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