2008 JBMO Shortlist G9

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2008 JBMO Shortlist G9

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Source: 2008 JBMO Shortlist G9

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#1 h Oct 10, 2017, 4:27 PM • 1 Y

Y by Adventure10

Let $O$ be a point inside the parallelogram $ABCD$ such that $\angle AOB + \angle COD = \angle BOC + \angle AOD$ . Prove that there exists a circle $k$ tangent to the circumscribed circles of the triangles $\vartriangle AOB, \vartriangle BOC, \vartriangle COD$ and $\vartriangle DOA$ .

This post has been edited 1 time. Last edited by parmenides51, Oct 10, 2017, 8:00 PM
Reason: typo corrected in the given relation, thanks to aleksam for informing me

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#3 h Jun 2, 2022, 6:09 PM • 4 Y

Y by Ahmet_Kerim, AlperenINAN, erkosfobiladol, egxa

Let $A^* , B^* , C^* , D^*$ be the points $A , B , C , D$ under the inversion centered at O with lenght R. The image becomes much more simple. Then the claim of this question equals to this claim: "There exist a circle who is tangent to the sides of the quadrilateral $A^*B^*C^*D^*$ ". Let $O'$ be the point that provides $OA \parallel O'D$ and $OB \parallel O'C$ . Because $AB \parallel CD$ the triangles $AOB$ and $DO'C$ are similar. And since $AD \parallel BC$ , then $AD \parallel BC$ $\parallel OO'$ . And by using the information $\angle{AOB}+\angle{DOC}=180$ and $\angle{AOB}=\angle{DO'C}$ we get that the quadrilateral $DOCO'$ is cyclic. Then by angle chasing $\angle{ABO}=\angle{DCO'}=\angle{DOO'}=\angle{ADO}$ . Since by the properties of inversion the quadrilaterals $A^*ABB^*$ and $AA^*DD^*$ are cyclic thus $\angle{OA^*B^*}=\angle{ABO}=\angle{ADO}=\angle{OA^*D^*}$ . But we can do this for $B^*, C^*, D^*$ . Thus the angle bisectors of $A^*B^*C^*D^*$ meet at the point O which means there exist a circle that is centered at O and is tangent to the sides of the quadrilateral $A^*B^*C^*D^*$ thus the proof is finished.
Note: There must be a more basic one solution but this one is the only one I found. Sorry if there is a mistake I am new around here.

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