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## Fontene’s theorem and some corollaries

Fontene theorem and some corollaries

Problem by Le Duc Tam, A2k43 Highshool for gifted student, Hanoi University of Science:

Fontene’s application

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## An approach to prove the concurrency

This article was published in the magazine Mathematics and Youth 05/2010:

an approach to prove the concurrency

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## 8 incenters are concyclic

Problem: Given a quadrilateral $ABCD$ which is inscribed in $(O)$ such that $AC\perp BD$. The tangents of $(O)$ through $A, B, C, D$ intersect each other and make a circumscribed quadrilateral $XYZT$. $XZ$ intersects $YT$ at $P$. Prove that the incenters of $8$ triangles $XPY, YPZ, ZPT, TPX, XYZ, YZT, ZTX, TXY$ are concyclic.

Proof:

Denote $I_1, I_2, I_3, I_4, I_5, I_6, I_7, I_8$ the incenters of triangles $XPY, YPZ, ZPT, TPX, XYZ, YZT, ZTX, TXY$, respectively. It’s well-known that $AC, BD, XZ, YT$ concur at $P$.
Since $AC\perp BD$ we get $\angle TXY+\angle TZY=180^o$ or $XYZT$ is a cyclic quadrilateral.
Then $\angle XYP=\angle TZP$. But $\angle PAY=\angle PCZ$ so $\Delta PAY\sim \Delta PCZ$, which follows that $\frac{PY}{PZ}=\frac{AY}{CZ}=\frac{YB}{ZB}$. This means $PB$ is the bisector of $\angle YPZ$ or $I_2\in PB$.
Similarly, $I_3\in PC$.
We have $\frac{PI_3}{CI_3}=\frac{PZ}{CZ}=\frac{PZ}{BZ}=\frac{PI_2}{BI_2}$. Therefore $I_2I_3//BC$.
$\Rightarrow \angle I_1I_3I_2=\angle ACB=90^o-\frac{1}{2}\angle XYZ.$
Thus $\angle I_1I_3I_2+\angle I_1I_5I_2=90^o-\frac{1}{2}\angle XYZ+90^o+\frac{1}{2}\angle XYZ=180^o.$
It is equivalent to $I_1, I_2, I_3, I_5$ are concyclic. Similarly, $I_1, I_2, I_4, I_5$ are concyclic. So $I_5\in (I_1I_2I_3I_4)$. Similar for $I_6, I_7, I_8$. We are done.

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## A problem about Miquel point

Problem (by Lamoen (van) F. and Grinberg D.): A line intersects sides (or their extensions) $AB$,$BC$and $CA$ of triangle$ABC$ at points $C_{1}$,$B_{1}$ and $A_{1}$, respectively; let$O$,$O_{a}$,$O_{b}$and $O_{c}$ be the centers of the circumscribed circles of triangles $ABC$,$AB_{1}C_{1}$,$A_{1}BC_{1}$ and $A_{1}B_{1}C$ ,respectively; let $H$,$H_{a}$,$H_{b}$and $H_{c}$be the respective orthocenters of these triangles. Prove that:
a)$\Delta O_aO_bO_c \sim \Delta ABC$
b)The midperpendiculars to segments $OH$,$O_{a}H_{a}$,$O_{b}H_{b}$ and $O_{c}H_{c}$ meet at one point.

My proof:

a, Let $E$ be the Miquel point of the completed quadrilateral $ABCA_1B_1C_1$. We have $O_aO_b\perp EC_1, O_aO_c\perp EB_1$ then $\angle O_bO_aO_C=180^o- \angle C_1EB_1=\angle BAC$. Similarly we are done.

b,  First we will express two lemmas:
Lemma 1: Given two circles $(O_1)$ and $(O_2)$ intersect each other at $F$ and $G$. A line through $F$ and perpendicular to $FG$ intersects $(O_1)$ and $(O_2)$ at $B$ and $C$. Let $J,I$ be the arbitrary points on $(O_1)$ and $(O_2)$, respectively. The lines through $B$ and parallel to $FI$ cuts the lines through $C$ and parallel to $FJ$ at $A$. Let $H$ be the projection of $A$ on $BC$. Then $I, J, F, H$ are concyclic.
Proof:

Let $Q, R$ be the second intersections of $AB$ and $(O_1), AC$ and $(O_2)$. $FI$ cuts $GR$ at $X, FJ$ cuts $GQ$ at $Y$. Note that $FX//AB$ and $GQ\perp AB$ so $FX\perp GQ$.Similarly, $FY \perp GR$. This means $G$ is the orthocenter of triangle $XFY$.
Therefore $FG\perp XY$ or $XY//BC$.
Construct two parallelograms $BMXF$ and $FCNY$. Let $S$ be the projection of $F$ on $XY$. We have:
$YS.YM=YG.YQ=YJ.YF$. Thus $MSJF$ is a cyclic quadrilateral or $\angle MJF=90^o$. Similarly, $\angle NIF=90^o$.
We get $AH, MJ, NI$ are the altitudes of triangle $AMN$. Therefore they concur at $T$.
Hence $H, I, J$ lie on the circle with diameter $FT$. We are done.

Lemma 2: Given a completed quadrilateral $ABCA_1B_1C_1$. Let $H, H'$ be the orthocenters, $O$ and $O'$ be the circumcenters of triangles $ABC$ and $AB_1C_1$. $E, E'$ are the midpoints of $OH$ and $O'H'$. Then the lines through $E$ and perpendicular to $B_1C_1$, through $E'$ and perpendicular to $BC$ and $HH'$ concur at $M$. Let $N$ be the projection of $A$ on $HH'$ then $E, E', M, N$ are concyclic.

Proof:

We only solve the lemma in 1 case as in the figure.
For the concurrency see the topic IMO Shortlist 2009 – Problem G6.
Let $Q, R,Y,U$ be the midpoints of $AH', AH, BC, B_1C_1$. We have $R, Q$ lie on the 9-point circle of triangles $ABC, AB_1C_1$, respectively. So $\frac{ER}{RN}=\frac{R_{ABC}}{\frac{1}{2}AH}=\frac{R_{ABC}}{OY}=\frac{1}{\cos A}=\frac{E'Q}{QN}$.
On the other side, $\angle ERN=\angle ERH-\angle NRH, \angle E'QN=\angle HQN-\angle UQH'.$
But$\angle NRH+\angle NQH'=2\angle HAH'=\angle H'AO'+\angle HAO=\angle XQU+\angle YRV.$ Therefore $\angle ERN=\angle E'QN$. We obtain $\Delta NQE'\sim \Delta NRE$. Hence $\Delta ENE'\sim \Delta RNQ\sim \Delta HAH'$. This follows that$\angle NEE'=\angle AHH'=\angle E'MN$.
Thus $E',E,M, N$ are concyclic.

Back to our problem:

The Steiner’s theorem is stated that $H, H_a, H_b, H_c$ are collinear. Let $S$ be the intersection of the line through $E$ and perpendicular to $B_1C_1$ and the line through $E'$ and perpendicular to $BC$. Applying lemma 2 we have $S\in H_aH.$
Let $R$ be the intersection of the perpendicular bisector of $H_aO_a$ and the line through $S$ and perpendicular to $H_bH_c$. $(STE')$ intersects the line through $S$ and parallel to $AH_a$ at $E''$. From lemma 1 we can easy show that $E''$ lies on the circle with diameter $RH$. Applying lemma 2 we have $E,S,T,E'$ are concyclic so $E''\equiv E$.
Thus the perpendicular bisectors of $OH, O_aH_a$ and the line through $S$ and perpendicular to $HH_a$ concur at $R$. Similarly we are done.

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## My problems about concurrent circles

Problem 1: Given a triangle $ABC$. Let $P$ be an arbitrary point in the plane, $A_1B_1C_1$ be the pedal triangle of $P$ wrt $\Delta ABC$. Let $O$ be the center of $(A_1B_1C_1)$,$latex L$ be an arbitrary point on $PO$. $A_1L, B_1L, C_1L$ intersects $(A_1B_1C_1)$ at $A_2,B_2,C_2$, respectively. Prove that $AA_2, BB_2, CC_2$ are concurrent.
Note that this is the generalization of the Steinbart‘s point.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=366219

Problem 2: Given a triangle $ABC$. Let $P$ be a point in the plane such that the pedal triangle $A_1B_1C_1$ of $P$ wrt $\Delta ABC$ is the cevian triangle of $Q$ wrt $\Delta ABC$. Let $Q'$ be the isogonal conjugate of $Q$ wrt $\Delta ABC$. $PQ'$ intersects $BC, AC, AB$ at $A_2, B_2, C_2$ ,respectively. Prove that $(AA_1A_2), (BB_1B_2), (CC_1C_2), (ABC)$ are concurrent.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=366218

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## The 9-point center lies on the altitude

Problem (by Lym): Prove that the 9-point centers of the line (k) perpendicular to Euler line of triangle ABC forming 3 triangles with $\Delta ABC$ are on the altitudes of $\Delta ABC$.

My proof:
We only need to prove in case $k$ passes through $A$. Other cases we can use the homothetic to finish proof.
In the figure below, let $k$ intersect $OH$ at $P, BC$ at $Q$. The line through $E$ and perpendicular to $OH$ intersects $BC$ at $J$. The line through $J$ and perpendicular to $BH$ intersects $BH$ at $E_b$. We will show that $E_b$ is the 9-point center of triangle $ABQ$.
Since $E_b, E, H_a$ lie on the circle $(HJ)$ then $\angle E_bEH_a=\angle BHH_a=\angle M_cRH=\frac{1}{2}\angle M_cEH_a.$
Therefore $EE_b$ is the bisector of $\angle M_cEH_a$ or $E_bM_c=E_bH_a.$
On the other side, let $S$ be the intersection of $AQ$ and $M_bM_c$.
We have $\angle M_cE_bH_a=360^o-2\angle H_aE_bE=360^o-2\angle E_bJM_a=360^o-(360^o-2\angle HOM_a)$
Therefore $\angle M_cE_bH_a=2\angle HOM_a.$
But $\angle H_aSM_c=\angle ASM_c=\angle AQB=\angle HOM_a$ then $\angle M_cE_bH_a=2\angle H_aSM_c.$
This means $E_b$ is the center of 9-point circle of triangle $ABQ$. We are done.

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