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My new book is published

My new Vietnamese book “108 Euclidean geometry problems” is published by Vietnam National University press. You may contact me to get the print. Email: nguyenvanlinhkhtn@gmail.com

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108 Olympiad Geometry Problems

My new book is coming soon…

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Pascal theorem

Pascal theorem

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Problem is proposed by Dao Thanh Oai.

Brianchon generalization

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Problem (me): Given triangle ABC with its circumcircle (O). Let L be an arbitrary point on arc BC which does not contain A. Prove that the A-mixtilinear incircle of triangle ABC, P-mixtilinear incircles of triangles PAB and PAC have a common tangent.

Mixtilinear2Proof: See here

 

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Coaxal circles

Problem. Given a triangle ABC with its circumcenter O. Let d be an arbitrary line in the plane. d intersects BC, CA, AB at X, Y, Z, respectively. Let H be the projection of O on d. Prove that 3 circles (AXH), (BYH), (CZH) are coaxal.

Solution.

Lemma: Given a triangle ABC with its circumcircle (O). Let P be an arbitrary point in the plane. A line d through P and perpendicular to OP intersects (BPC), (CPA), (APB) again at X, Y, Z, respectively. Then AX, BY, CZ are concurrent.

Proof:


We prove this lemma in case d cuts (O) and I will leave another case for the readers.
Let J, K, L be the intersections of AZ and CX, BX and AY, BC and KJ, respectively.
According to Desargues’s theorem, AX, BY, CZ are concurrent iff J, K, L are collinear.
We have \angle ZAB=\angle ZPB=\angle BCJ thenJ \in (O). Similarly, K\in (O).
Let M, N be the intersections of d and (O). Note that P is the midpoint of MN.
Since LM.LN=LB.LC=LP.LX then (NMLX)=-1. We get XJ.XC=XK.XB=XM.XN=XL.XP or B,K,L,P and P,L,C,J are concyclic.
Then \angle PK=180^o-\angle PBX=180^o-\angle PCJ=180^o-\angle PLJ. This means K, L, J are collinear. We are done.
Back to our problem:


Let A', B', C' be the second intersections of (AHX), (BHY), (CHZ) and (O), respectively. The idea is to show that AA', BB', CC' concur at H'.
Let M, N be the conserve point of H and X, respectively.
The inversion with center O, power R^2 maps (AHX) to (AMN) then A'\in (AMN).
Let I be the intersection of MN and XH. We have \angle XNM=\angle XHM=90^o then IM.IN=IX.IH. This means I lies on the radical axis of (AHX) and (AMN), or lies on AA'.
Let K be the intersections of the tangents of (O) at B and C then M,N,K are collinear.
We get N\in (OBC). So XN.XO=XB.XC=XI.XH or I lies on (HBC). Applying the lemma above we obtain AA', BB', CC' concur at H'. Therefore 3 circles (AHX), (BHY), (CHZ) have the same radical axis HH'. We are done.

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Circumscribed quadrilateral

This is a nice problem I found a long time ago. It’s easy but nice. You can solve it by using inversion.

Problem:  Given a circumscribed quadrilateral ABCD. AC intersects (ABD) and (CBD) again at E, F, respectively. Prove that FBED is also a circumscribed quadrilateral.


Link to problem above: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=355881

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