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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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## Proof of the generalization of Brianchon theorem

Problem is proposed by Dao Thanh Oai.

Brianchon generalization

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## 3 mixtilinear incircles have a common tangent

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.

Proof: 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 $\angle ZAB=\angle ZPB=\angle BCJ$ then$J \in (O)$. 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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