in a quadrilateral $ABCD$ digonals are perpendicular to each other. let $S$ be the intersection of digonals. $K$,$L$,$M$ and $N$ are reflections of $S$ to $AB$,$BC$,$CD$ and $DA$. $BN$ cuts the circumcircle of $SKN$ in $E$ and $BM$ cuts the circumcircle of $SLM$ in $F$. prove that $EFLK$ is concyclic.(20 points)
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Tags: geometry, geometric transformation, reflection, circumcircle, homothety, ratio, power of a point
vladimir92
07.08.2010 16:31
goodar2006 wrote: in a quadrilateral $ABCD$ digonals are perpendicular to each other. let $S$ be the intersection of digonals. $K$,$L$,$M$ and $N$ are reflections of $S$ to $AB$,$BC$,$CD$ and $DA$. $BN$ cuts the circumcircle of $SKN$ in $E$ and $BM$ cuts the circumcircle of $SLM$ in $F$. prove that $EFLK$ is concyclic.(20 points) I liked this problem, thank's a lot.
It's obvious that $AS=AK=AN$ meaning that $A$ is circumcenter of $\odot(NSK)$, similary $C$ is circumcenter of $\odot(MSL)$, thus, $\odot(NSK)$ and $\odot(MSL)$ are tangent to each other at $S$ and tangent to $BD$ at $S$. Then $BD$ is the radical axis of $\odot(NSK)$ and $\odot(MSL)$, it follow that $BE.BN=BF.BM$ then $NEFM$ is cyclic. Using that, we get $\angle{EFL}=180-\angle{MSL}+\angle{BNM}$ and we have $\angle{EKL}=\angle{NKL}-\angle{NKS}-\angle{SNB}$ summing those two and using the fact that $\angle{MSL}=\angle{SKL}+\angle{SNM}$ we find that $\angle{EKL}+\angle{EFL}=180$ from which follow that $KEFL$ is cyclic.
On the same configuration, we can also prove an easier result: $KNML$ is cyclic.
ProofProject $S$ on $AB$ , $BC$ , $CA$ and $DA$ at $S_1$ , $S_2$ , $S_3$ and $S_4$, of course quadrilaterals $SS_1BS_2$ , $SS_2CS_3$ ,... are all concyclic, then:
$\angle{S_4S_1S_2}=\angle{S_4S_1S}+\angle{SS_1S_2}=\angle{DAC}+\angle{DBC}$. and $\angle{S_4S_3S_2}=\angle{ACB}+\angle{ADB}$ then: $\angle{S_4S_3S_2}+\angle{S_4S_1S_2}=180$ therefore, the quadrilateral $S_1S_2S_3S_4$ is cyclic. An homotheti with center $S$ and radius $2$ maps $S_1$ , $S_2$ , $S_3$ and $S_4$ to $K$ , $L$ , $M$ and $N$, then $KLMN$ is cyclic.
mathVNpro
07.08.2010 21:27
Quote: in a quadrilateral $ABCD$ digonals are perpendicular to each other. let $S$ be the intersection of digonals. $K$,$L$,$M$ and $N$ are reflections of $S$ to $AB$,$BC$,$CD$ and $DA$. $BN$ cuts the circumcircle of $SKN$ in $E$ and $BM$ cuts the circumcircle of $SLM$ in $F$. prove that $EFLK$ is concyclic.(20 points) Let $K',$ $L',$ $M'$ and $N'$ respectively be the projections of $S$ onto $AB,$ $BC,$ $CD$ and $DA$. We have $\angle K'N'M'=$ $\angle K'N'S+$ $\angle SN'M'$ $=$ $\angle BAS$ $+\angle CDS$. With the same argument, we also gain $\angle K'L'M'=$ $\angle ABS$ $+\angle SCD$. Therefore, $\angle K'N'M'+$ $\angle K'L'M'$ $=180^{\circ}$. Thus, we deduce $K'L'M'N'$ is concylic. The homothety through center $S$, ratio $k=2$; $\mathcal {H}(S,k):$ $K'L'M'N'$ $\mapsto$ $KLMN$. Hence, $KLMN$ is also concyclic. Note that $(SLM)$ and $(SKN)$ share $BD$ as their comment tangent through the point $S$. As a result, the inversion through pole $B$, power $k'=BS^2$ will map $M$ into $F$, $N$ into $E$, $K$ and $L$ into themselves. Therefore, we conlude that $EFKL$ is concycli. Our proof is completed then. $\square$
mousavi
08.08.2010 17:29
it is clear that BK and BL are tangents to circle SKN and SLM with centers A and C respectively and B is on the radical axis of these two circles. we can use inversion with center B .translate E and F to N and M ,and K,L don't move.so here we should prove that KLMN is concyclic and it is just playing with angles!
mathVNpro
08.08.2010 18:46
mousavi wrote: it is clear that BK and BL are tangents to circle SKN and SLM with centers A and C respectively and B is on the radical axis of these two circles. we can use inversion with center B .translate E and F to N and M ,and K,L don't move.so here we should prove that KLMN is concyclic and it is just playing with angles! It seems like you have just the same idea with me !?!?! Please look at my previous post