A nice problem associated with refelection

by daothanhoai, Oct 31, 2021, 4:35 AM

Problem 1: Let $ABC$ be a triangle, let a line $(\ell)$ meets three sidelines $BC$ , $CA$ , $AB$ at $A_0$ , $B_0$ , $C_0$ respectively. Let circle $(P)$ which center $P$ lies on the line $(\ell)$ . Denote $A_1$ , $B_1$ , $C_1$ are reflection of $A_0$ , $B_0$ , $C_0$ in $(P)$ then $AA_1$ , $BB_1$ , $CC_1$ are concurrent.

Problem 2: Let $ABC$ be a triangle, let $P$ be a point in the plane, let a line $(\ell)$ through $P$ and meets three sidelines $BC$ , $CA$ , $AB$ at $A_0$ , $B_0$ , $C_0$ respectively. Denote $A_1$ , $B_1$ , $C_1$ are reflection of $A_0$ , $B_0$ , $C_0$ in $P$ .

1) Then $AA_1$ , $BB_1$ , $CC_1$ are concurrent, let the point of concurrence is $Q$ . When line $(l)$ around the point $P$ then locus of $Q$ is a circumconic with center $P$ . If $P$ is the $X(3)$ the circumconic is the circumcircle.

2) Let $M_a$ , $M_b$ , $M_c$ be the midpoint of $BC$ , $CA$ , $AB$ respectively. Then three line through $A_1$ , $B_1$ , $C_1$ and parallel to $PM_a$ , $PM_b$ , $PM_c$ meets three sidelines $BC$ , $CA$ , $AB$ at three points $A'$ , $B'$ , $C'$ are collinear. Let the line is $(d)$ , then four lines $BC$ , $CA$ , $AB$ , $(d)$ formed a quadrilateral circumscribed about an ellipse. If $P$ is $X(8)$ of ABC then the ellipse is incircle of $ABC$ .

See also:

[1]- https://artofproblemsolving.com/community/c2961h1053033_three_lines_concurrents

[2]-https://forumgeom.fau.edu/FG2016volume16/FG201608.pdf

This post has been edited 1 time. Last edited by daothanhoai, Oct 31, 2021, 4:37 AM

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A generalization of the Simson line theorem

by daothanhoai, Apr 8, 2015, 1:35 AM

Problem 1:
Let $ABC$ be a triangle, let a line $L$ through circumecenter, let a point $P$ lie on circumcircle. Let $AP,BP,CP$ meets $L$ at $A_P, B_P, C_P$ . Denote A_0,B_0,C_0 are projection (mean perpendicular foot) of $A_P, B_P, C_P$ to $BC,CA,AB$ respectively. Then $A_0,B_0,C_0$ are collinear.
Problem 2:
The new line $\overline {A_0B_0C_0}$ bisect the orthocenter and $P$

- When $L$ pass through $P$ , this line is Simson line.

Reference:

[1] https://www.geogebra.org/student/m527653

[2] T. O. Dao, Advanced Plane Geometry, message 1781, September 26, 2014.

A generalization Brianchon theorem

by daothanhoai, Mar 3, 2015, 8:58 AM

A generalization Brianchon theorem

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Ten point circle associated X(4240) in ETC

by daothanhoai, Feb 10, 2015, 4:34 PM

Let $ABC$ be a triangle, Let $(W)$ be the circle through $X(3),X(110)$ and $X(4240)$ in $ETC$ . I found 7 points also lie on $(W)$ as follows:

Let the Euler line of ABC meets the sidelines $BC,CA,AB$ at $A_0,B_0,C_0$ . Six points $X(3), X(110)$ of three triangles $AB_0C_0, BA_0C_0, CB_0A_0$ also lie on $(W)$ . The circumcenter $X(3)$ of Paralogic triangle whose perpectrix is the Euler line of $ABC$ also lie on $(W)$ . So $10$ points lie on this circle.

Center of this circle is midpoint of $X(3)$ of $ABC$ and $O'$

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This post has been edited 3 times. Last edited by daothanhoai, Feb 10, 2015, 5:03 PM

Rectangular circumhyperbola and circle

by daothanhoai, Feb 1, 2015, 9:23 AM

1. Let a rectangular hyperbola with center $O$ , let $P$ be a point on the plain such that the polar of $P$ to the hyperbola meets the hyperbola at $A, B$ . Then $AB$ is common tangent of two circle $(POA)$ and $(POB)$ .

Application we can show that: In any triangle: The Circumcenter, the Nine point center, the Symmedian point and the Kiepert center lie on a circle

2. Let a rectangular hyperbola with center $O$ , let $P$ be a point on the plain such that the polar of $P$ to the hyperbola meets the hyperbola at $A, B$ . Let $A'$ be the reflection of $A$ in $O$ . Let $M$ be the midpoint of $AB$ and $D$ be the reflection of $M$ in $B$ . Let $E$ be the midpoint of $PB$ . Then $MD$ is common tangent of two circle $(A'ED)$ and $(A'EM)$ .

Application we can show that: In any triangle: The orthocenter, the Nine point center and Tarry point and midpoint of Brocard diameter lie on a circle.

3. Let a rectangular hyperbola, $F$ lie on the rectangular hyperbola, $F'$ be reflection of $F$ in center of the hyperbola. AB be two point lie on one brach of the rectangular hyperbola such that $FF'$ through midpoint of $AB$ . Then $AB$ is common tangent of two circle $(FF'A)$ and $(FF'B)$ .

Application we can show that: Lester circle

4.Let ABC be a rectangular circumhyperbola, let P be a point on the plain. The polar of P relative to the hyperbola meets the hyperbola at HG. if E be a point on the rectangular hyperbola(or E lie on isogonal Conjugate of the hyperbola ), And E' be the inverse of E in circle whose diameter is HG. Then four points: The center of the hyperbola, and E,E' and P lie on a circle. The circle which diameter HG is a generalization of orthocentroidal circle. The circle X(1)X(4) has all properties similarly with the orthocentroidal circle.

Attachments:

Two circles through four triangle centers.pdf (179kb)

This post has been edited 2 times. Last edited by daothanhoai, Feb 3, 2015, 5:58 PM

A property of a line through circumcenter

by daothanhoai, Jan 29, 2015, 12:19 PM

1-Problem: Let $ABC$ be a triangle, let a line L through circumcenter and the line meets $BC,CA,AB$ at $A_0,B_0,C_0$ . Let $A_b,C_b$ be projection of $A_0$ on $AC$ and $AB$ respectively. Let $A_a$ be the midpoint of $A_bA_c$ define $B_b,C_c$ cyclically. show that $A_b,C_b,C_c$ and Nine point center of $ABC$ are collinear.

Note that: If the line L don't through circumcenter then $A_a,B_b,C_c$ also are collinear

2-TelvCohl's proof:

Lemma

Given $\triangle ABC$ and a line $\ell$ .
Let $P_1, P_2, P_3$ be three points on $\ell$ .
Let $\triangle D_1E_1F_1$ be the pedal triangle of $P_1$ WRT $\triangle ABC$ .
Let $G_1$ be the centroid of $\triangle D_1E_1F_1$ and $R_1$ be a point on ray $P_1G_1$ such that $P_1G_1:G_1R_1=2:1$ .
Define $\triangle D_2E_2F_2, G_2, R_2$ for $P_2$ and $\triangle D_3E_3F_3, G_3, R_3$ for $P_3$ similarly .

Then $R_1, R_2, R_3$ are collinear .
____________________________________________________________
Proof:

Let $M_1, M_2, M_3$ be the midpoint of $E_1F_1, E_2F_2, E_3F_3$ , respectively .

Since $E_1E_2:E_2E_3=P_1P_2:P_2P_3=F_1F_2:F_2F_3$ ,
so from ERIQ lemma $\Longrightarrow M_1, M_2, M_3$ are collinear and $M_1M_2:M_2M_3=P_1P_2:P_2P_3$ .

Since $D_1G_1:G_1M_1=D_2G_2:G_2M_2=D_3G_3:G_3M_3=2:1$ ,
so from ERIQ lemma $\Longrightarrow G_1, G_2, G_3$ are collinear and $G_1G_2:G_2G_3=P_1P_2:P_2P_3$ .

Since $P_1G_1:G_1R_1=P_2G_2:G_2R_2=P_3G_3:G_3R_3=2:1$ ,
so from ERIQ lemma $\Longrightarrow R_1, R_2, R_3$ are collinear and $R_1R_2:R_2R_3=P_1P_2:P_2P_3$ .
____________________________________________________________
Back to the main problem:

Let $P$ be a moving point on $L$ .
Let $P_a, P_b, P_c$ be the projection of $P$ on $BC, CA, AB$ , respectively .
Let $G$ be the centroid of $\triangle P_aP_bP_c$ and $R$ on the ray $PG$ such that $PG:GR=2:1$ .

When $P$ coincide with $A_0 \Longrightarrow R \equiv A_a$ . ... (1)
When $P$ coincide with $B_0 \Longrightarrow R \equiv B_b$ . ... (2)
When $P$ coincide with $C_0 \Longrightarrow R \equiv C_c$ . ... (3)

From (1), (2), (3) and the lemma we get $A_a, B_b, C_c$ are collinear .

If $L$ pass through the circumcenter $O$ of $\triangle ABC$ :
Consider the case $P \equiv O \Longrightarrow R$ coincide with the 9-point center $N$ of $\triangle ABC$ ,
so from the lemma we get $A_a, B_b, C_c, N$ are collinear

3-Luis González's proof:

In general, the applications $B_0 \mapsto B_b$ and $C_0 \mapsto C_c$ are affine homographies, thus when $\ell$ varies, $B_b$ and $C_c$ describe two line $\ell_b$ and $\ell_c.$ Since $B_0 \mapsto C_0$ is a perspectivity, then $B_b \mapsto C_c$ is also a perspectivity between $\ell_b,\ell_c,$ because when $B_0 \equiv C_0 \equiv A,$ then $B_b \equiv C_c$ is the midpoint of the A-altitude $\Longrightarrow$ $B_bC_c$ goes through a fixed point.

Assume the case when $C_0 \equiv B_c$ is the midpoint of $AB.$ From cyclic $BCB_0O$ and $BB_aB_0C_0,$ we get $\angle COB_0=\angle B_aBB_0=\angle B_aC_0B_0$ $\Longrightarrow$ $C_0B_bB_a \parallel CO$ $\Longrightarrow$ $C_0B_b$ goes through the 9-point center $N$ of $\triangle ABC.$ Further, the pencil formed by $C_0C_a,C_0C_b,C_0B_a$ and the parallel from $C_0$ to $C_aC_b$ and the pencil formed by $CB,CA,$ the tangent of $(O)$ at $C$ and the C-symmedian are projective, as they have corresponding perpendicular rays, thus $C_0B_a$ passes trough the midpoint $C_c$ of $C_aC_b$ $\Longrightarrow$ $N \in B_bC_c$ and similarly we'll have $N \in B_bC_c$ when $B_0$ is the midpoint of $AC.$ Analogously $C_cA_a \in N$ $\Longrightarrow$ $A_a,B_b,C_c,N$ are collinear for all $\ell.$

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This post has been edited 2 times. Last edited by daothanhoai, Jan 30, 2015, 4:24 AM

A generalization Napoleon theorem associated with the Kiepert

by daothanhoai, Jan 24, 2015, 10:14 AM

Very nice theorem: http://tube.geogebra.org/student/m542855

Let ABC be a triangle, $F$ be the first (or secon) Fermat point, let $K$ be the point on the Kiepert hyperbola. Let $P$ be the point on line FK. The line through P and perpendicular to $BC$ meet $AK$ at $A_0$ . Define $A_0,B_0,C_0$ cyclically. Show that $A_0B_0C_0$ is an equilateral triangle. This triangle homothety to the outer(or inner) Napoleon triangle.

http://www.artofproblemsolving.com/Forum/blog.php?page=download&mode=download&id=4288

Lemma 1: (USA TST 2006, Problem 6) Let $ABC$ be a triangle. Triangles $PAB$ and $QAC$ are constructed outside of triangle $ABC$ such that $AP = AB$ and $AQ = AC$ and $\angle{BAP}= \angle{CAQ}$ . Segments $BQ$ and $CP$ meet at $R$ . Let $O$ be the circumcenter of triangle $BCR$ . Prove that $AO \perp PQ.$

Telv Cohl's proof:

Let $O'$ be the circumcenter of $\triangle ACQ$ . Let $M, N$ be the midpoint of $CQ, CR$ , respectively .

Easy to see $R \in (O')$ .

Since $O', M, N, C$ are concyclic , so we get $\angle AO'O=\angle QCP$ . ... $(1)$ Since $\angle RO'O=\angle BQC, \angle O'OR=\angle CBQ$ ,
so we get $\triangle ORO' \sim \triangle BCQ$ , hence $\frac{O'A}{CQ}=\frac{O'R}{CQ}=\frac{O'O}{QB}=\frac{O'O}{CP}$ . ... $(2)$

From $(1)$ and $(2)$ we get $\triangle AOO' \sim \triangle QPC$ , so from $OO' \perp PC$ and $AO' \perp QC \Longrightarrow AO \perp QP$ .

Lemma 2:

Let $D$ be a point out of $\triangle ABC$ satisfy $\angle DBC=\angle DCB=\theta$ .
Let $E$ be a point out of $\triangle ABC$ satisfy $\angle EAC=\angle ECA=90^{\circ}-\theta$ .
Let $F$ be a point out of $\triangle ABC$ satisfy $\angle FAB=\angle FBA=90^{\circ}-\theta$ . Then $AD \perp EF$ .

Proof of the lemma 2:

Let $B' \in AF, C' \in AE$ satisfy $AB=AB', AC=AC'$ and $T=BC' \cap CB'$ .

Easy to see $\triangle ABB' \cup F \sim \triangle ACC' \cup E \Longrightarrow EF \parallel B'C'$ .

From $\triangle AB'C \sim \triangle ABC' \Longrightarrow \angle BTC=180^{\circ}-(90^{\circ}-\theta )=90^{\circ}+\theta$ , so combine with $\angle DBC=\angle DCB=\theta$ we get $D$ is the circumcenter of $\triangle BTC$ , hence from lemma 1, we get $AD \perp B'C'$ . i.e. $AD \perp EF$

From the lemma 2 we get the following property about Kiepert triangle :
The pedal triangle of the isogonal conjugate of $K_{90-\phi}$ WRT $\triangle ABC$ and the Kiepert triangle with angle $\phi$ are homothetic .
( Moreover, the homothety center of these two triangles is the Symmedian point of $\triangle ABC$ ! ) (1)

Let $H_b, H_c$ be the orthocenter of $\triangle FCA, \triangle FAB$ , respectively . ( $H_b, H_c$ also lie on the Kiepert hyperbola of $\triangle ABC$ )

Easy to see all $\triangle A_0B_0C_0$ are homothetic with center $K$ , so it is suffices to prove the case when $P$ coincide with $F$ .

From Pascal theorem (for $CKBH_cFH_b$ ) we get $AF \perp B_0C_0$ .
Similarly, we can prove $BF \perp C_0A_0$ and $CF \perp A_0B_0$ ,
so $\triangle A_0B_0C_0$ and the pedal triangle of the isogonal conjugate of $F$ WRT $\triangle ABC$ are homothetic , hence from (1) we get $\triangle A_0B_0C_0$ and the outer (or inner) Napoleon triangle are homothetic .

Dual problem:

Let ABC be a triangle, let P be a point on the line $X(13)X(15)$ (or $X(14)X(16)$ ). Three line through $P$ and perpendicular to $BC$ meets the line $AX(13)$ (or $AX(16$ ) at $A0$ , define $B0,C0$ cyclically. Show that A0B0C0 are an equilateral triangle homothety to Napoleon triangle and the homothetic center on the line $X(2)X(13)$ (or $X(2)X(14)$ )

http://www.artofproblemsolving.com/Forum/download/file.php?id=52768&mode=view

Reference:

[1] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=622242

[2] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=621954

[3] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=148830

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This post has been edited 13 times. Last edited by daothanhoai, Jan 25, 2015, 4:49 AM

A generalization of Parry circle

by daothanhoai, Jan 22, 2015, 4:02 AM

Let a rectangular circumhyperbola of ABC, let L is the isogonal conjugate line of the rectangular hyperbola. The tangent line of the hyperbola at X(4) meets L at point K. The line through K and center of the hyperbola. meet the hyperbola at $F_+,F_-$ . Let $I_+,I_-,G$ are isogonal conjugate of $F_+,F_-$ and $K$ respectively. Show that: $I_+,I_-,G,X(110)$ alway lie on a circle, this circle is a generalization of Parry circle.

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This post has been edited 2 times. Last edited by daothanhoai, Jan 22, 2015, 8:03 AM

A generalization of Nine point circle

by daothanhoai, Jan 22, 2015, 4:01 AM

Let ABC be a triangle, let $A_0B_0C_0$ be Kiepert triangle of ABC. The circle with diameter $AA_0$ meets $BC$ at $A_bA_c$ . Define $B_a,B_c,C_a,C_b$ cyclically. Show that six point $A_bA_c$ , $B_a,B_c$ , $C_a,C_b$ lie on a circle. When the Kiepert triangle is median triangle of ABC the circle is Nine point circle. The result is well-known?

Telv Cohl's proof:

Let $A_B, A_C$ be the projection of $A_0$ on $AB, AC$ , respectively .
Let $B_C, B_A$ be the projection of $B_0$ on $BC, BA$ , respectively .
Let $C_A, C_B$ be the projection of $C_0$ on $CA, CB$ , respectively .

Easy to see $B_A \in \odot (BB_0)$ and $C_A \in \odot (CC_0)$ .
From $Rt \triangle AB_0B_A \sim Rt \triangle AC_0C_A \Longrightarrow AB_A:AC_A=AB_0:AC_0=AC:AB$ ,
so we get $AB_c \cdot AB_a=AB \cdot AB_A=AC \cdot AC_A=AC_a \cdot AC_b \Longrightarrow B_c, B_a, C_a, C_b$ are concyclic .
Similarly, we can prove $C_a, C_b, A_b, A_c$ are concyclic and $A_b, A_c, B_c, B_a$ are concyclic ,
so from Davis theorem we get $A_b, A_c, B_c, B_a, C_a, C_b$ are concyclic.

Luis González's proof:

If one goes for the center and radius of the circle the result is quite nice. If $\theta$ denotes the Kiepert angle and $R$ and $S$ denote the circumradius and area of $\triangle ABC,$ respectively, we prove that these 6 points lie on a circle with center the 9-point center $N$ and radius $\varrho = \sqrt{\tfrac{1}{4}R^2+S \cdot \tan \theta}.$

Let $O,H$ denote the circumcenter and orthocenter of $\triangle ABC.$ $X$ is the projection of $A$ on $BC,$ $M$ is the midpoint of $BC$ and $L$ is the midpoint of $XM.$ If $AX$ cuts the circle with diameter $AD$ again at $U,$ then by symmetry $A_bUDA_c$ is an isosceles trapezoid. In the cyclic $AA_bUA_c$ with perpendicular diagonals, we have

${A_bA_c}^2=(XA_b+XA_c)^2={XA_b}^2+{XA_c}^2 +2 \cdot XA_b \cdot XA_c=$

$=4R^2-(AX^2+XU^2)+2 \cdot AX \cdot XU=AD^2-(AX-XU)^2=$

$=(AX+XU)^2+XM^2-(AX-XU)^2=XM^2+4 \cdot AX \cdot XU \Longrightarrow$

${NA_b}^2=NL^2+{LA_b}^2=NL^2+\tfrac{1}{4}XM^2+AX \cdot XU.$

Substituting $XM^2=OH^2-(HX-OM)^2,$ $NL=\tfrac{1}{2}(OM+HX)$ and $OM=\tfrac{1}{2}HA$ into the latter expression and factoring yields

${NA_b}^2=\tfrac{1}{2}HA \cdot HX+\tfrac{1}{4}OH^2+AX \cdot XU.$

Now, substituting $HA \cdot HX=\tfrac{1}{2}(R^2-OH^2)$ and $XU=MD=BM \cdot \tan \theta$ into the latter expression, we obtain ${NA_b}^2=\tfrac{1}{4}R^2+S \cdot \tan \theta,$ which is obviously a symmetric expression. Thus we conclude the 6 described points lie on a circle with center $N$ and radius $\varrho=\sqrt{\tfrac{1}{4}R^2+S \cdot \tan \theta}.$

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This post has been edited 1 time. Last edited by daothanhoai, Jan 23, 2015, 1:00 AM

Center of six Thebault's circle lie on a conic

by daothanhoai, Jan 21, 2015, 6:19 PM

Problem:(Own) Let ABC be a triangle, P be a point on the plaine, please show that three pair Thebault circle respecto AP,BP,CP alway lie on a conic.

http://tube.geogebra.org/material/show/id/524977

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Can I ask you a question on the Simson live/Essay triangle?
by Notacademic, Jul 13, 2019, 7:41 AM
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by AlastorMoody, Feb 23, 2019, 4:41 PM
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by daothanhoai, Jun 27, 2014, 11:18 AM
OK man !!
This is some stuff !!

Great Blog
by utkarshgupta, Jun 22, 2014, 9:02 AM
by daothanhoai, Apr 13, 2014, 3:38 PM
I think no one here has a greater interest in Geometry than you have.
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Thank to ThirdTimeLucky.
by daothanhoai, Feb 11, 2014, 7:06 AM
Very nice problems, will surely try them after my exams!
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