In a triangle $ABC$ with $AB<AC<BC$, the perpendicular bisectors of $AC$ and $BC$ intersect $BC$ and $AC$ at $K$ and $L$, respectively. Let $O$, $O_1$, and $O_2$ be the circumcentres of triangles $ABC$, $CKL$, and $OAB$, respectively. Prove that $OCO_1O_2$ is a parallelogram.
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Tags: geometry, parallelogram, modular arithmetic, circumcircle, power of a point, radical axis, geometry proposed
Luis González
02.11.2010 18:59
From the isosceles $\triangle ACK,\triangle BCL$ with apices $K,L,$ we have $\angle AKB =2\angle ACB$ and $\angle ALB=2\angle ACB$ $\pmod\pi$ $\Longrightarrow$ $A,B,O,K,L$ are concyclic. Now, since lines $AB$ and $KL$ are antiparallel with respect to $CA,CB,$ it follows that $CO_1 \perp AB$ and $CO \perp KL,$ i.e. $CO \parallel O_1O_2$ and $CO_1 \parallel OO_2$ $\Longrightarrow$ $OCO_1O_2$ is a parallelogram.
FantasyLover
02.11.2010 23:28
We prove that $OCO_1O_2$ is a parallelogram by proving that $O_1O_2\parallel OC$ and $\overline{O_1O_2}=\overline{OC}$.
First, we prove $O_1O_2\parallel OC$ by showing that both of $O_1O_2$ and $OC$ are perpendicular to $KL$.
We claim that $A, B, O, K, L$ are concyclic. $\angle AKO=\angle OKC=\angle OLC$. Hence, $\angle AKO+\angle ALO=\angle AKO+180^{\circ}-\angle OLC=180^{\circ}$, and therefore $AKOL$ is cyclic. Analogously we can prove that $BKOL$ is also cyclic, and we indeed have that $A, B, O, K, L$ are concyclic.
Now, since circles $O_1$ and $O_2$ share two points $K$ and $L$, $KL$ is the radical axis of $O_1$ and $O_2$, and this gives that $O_1O_2\perp KL$. Also, we know that $O$ is the orthocenter of $\triangle KLC$ and therefore $OC\perp KL$. Hence, $O_1O_2\parallel OC$.
It remains to prove that $\overline{O_1O_2}=\overline{OC}$.
We prove that $r_1=r_2$, where $r_1$ is the circumradius of $ABOKL$ and $r_2$ is the circumradius of $CKL$. Using the fact the the area of triangle $ABC$ is given by $\frac{abc}{4R}$, we have $\frac{AL}{CL}=\frac{[AKL]}{[CKL]}=\frac{\frac{AK\cdot KL\cdot LA}{4r_1}}{\frac{CK\cdot KL\cdot LC}{4r_2}}=\frac{AL\cdot r_2}{CL\cdot r_1}$. Hence, $r_1=r_2$, as claimed.
Now, if we prove that $\angle BO_1O=\angle O_1LO_2$, $\triangle BO_1K\cong \triangle O_1LO_2$, and we will be done.
If we let $\angle O_1LO_2$ be $180-2\alpha$, we have the following:
$\angle O_1LO_2=180-2\alpha\iff \angle LO_1O_2=\alpha\iff\angle LO_1K=2\alpha$ $\iff \angle LBK=\alpha\iff \angle BLO=90-\alpha\iff \angle BO_1O=180-2\alpha$.
Thus, $\triangle BO_1K\cong \triangle O_1LO_2$, and $\overline{O_1O_2}=\overline{OC}$.
Since $O_1O_2\parallel OC$ and $\overline{O_1O_2}=\overline{OC}$, $OCO_1O_2$ is a parallelogram, as desired. $\blacksquare$
xeroxia
19.01.2013 16:39
Since $\angle ALB = 2\cdot \angle ACB = \angle AKB = \angle AOB$, $A,L,O,K,B$ are cyclic, and their circumcenter is $O_2$. And we have $\angle CLK = \angle ABC$. $O_1O_2$ is perpendicular to $LK$ at its midpoint. Similarly, $OO_2$ is perpendicular to $AB$ at its midpoint. $\angle CO_1O_2 = \angle CO_1K + \angle KO_1O_2 = 2\angle CLK + \angle KCL = 2\angle ABC + \angle ACB$ Similarly, $\angle COO_2 = \angle COA + \angle AOO_2 = 2\angle CBA + \angle ACB$ So we have $\angle COO_2 = \angle CO_1O_2 = 2\angle CBA + (180^\circ - \angle CBA - \angle CAB) = 180^\circ - (\angle CAB - \angle CBA)$ Also, $\angle O_1CO = \angle O_1CK - \angle OCK = 90^\circ - \angle CBA - (90^\circ - \angle CAB) = \angle CAB - \angle CBA$ Since $\angle CO_1O_2 = \angle COO_2 = 180^\circ - \angle O_1CO$, the quadrilateral $O_1O_2OC$ is a parallelogram.