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Archive for Tháng Mười Một, 2010

The generalization of Newton line

Newton

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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,BCand CA of triangleABC at points C_{1},B_{1} and A_{1}, respectively; letO,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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