## Thứ Bảy, 18 tháng 4, 2015

### 102-Square in a square

Let $ABCD$ be a parallelogram. $IG//AB//CD$, $FH//BC//AD$.  Denote as the figure attachment, then show that $KJLM$ be a parallelogram.
Special case:
if $ABCD$ be a rectangle then show that $KJLM$ be a rhombus
If $ABCD$ be a square then show that $KJLM$ be a square,

## Thứ Sáu, 17 tháng 4, 2015

### 101-A property of six points lie on a conic

Let $A,B,C,D,E,F$ lie on a conic. Let $H=AC \cap BF$, $I = EC \cap BD$, $K=FD \cap AE$ $O=KI \cap CF$, $N=HI \cap AD$, $P=HK \cap BE$ , Then show that $O,N,P$ are collinear.

## Thứ Ba, 31 tháng 3, 2015

### 100-Mở rộng định lý Napoleon kết hợp với một lục giác

Mở rộng định lý Napoleon kết hợp với một lục giác:

Cho $ABCDEF$ là một lục giác bất kỳ, dựng ba tam giác đều $AGB$, $CHD$, $EIF$ cùng ra ngoài hoặc cùng vào trong(hình vẽ đính kèm là dựng ra ngoài). Ta gọi $A_1,B_1,C_1$ lần lượt là trọng tâm của các tam giác $FGC, BHE, DIA$ và $A_2,B_2,C_2$ lần lượt là trọng tâm của các tam giác $EGD, AHF, CIB$. Khi đó hai tam giác $A_1B_1C_1$ và $A_2B_2C_2$ là các tam giác đều và chúng thấu xạ.

The theorem found and proved by Dr.Paul Yiu and me

Let ABCDEF be a  hexagon, constructed three equilaterals $AGB, CHD, EIF$  all externally or internally (as in the figure).  Let $A_1,B_1,C_1$ be then the centroid of  $FGC, BHE, DIA$ respectively. Let $A_2,B_2,C_2$ be the centroid of  $EGD, AHF, CIB$ respectively. Then show that $A_1B_1C_1$, and $A_2B_2C_2$ form an equilateral triangle and them perpective.

Rõ ràng khi lục giác Suy biến thành một tam giác ta có định lý Napoleon.

## Thứ Năm, 26 tháng 2, 2015

### 99-Two conjectures in inequality

1, Let $\lambda_1+\lambda_2+...+\lambda_n=1$ , $\lambda_i>0$   for  $i=1,2,..,n$; f,f' are real positive continuous function that is concave up in [a,b], If $x_1, . . . , x_n$ and $y_1, . . . , y_n$ are numbers in [a,b] such that $(x_1, . . . , x_n)$ majorizes $(y_1, . . . , y_n)$, then:

$\frac{\lambda_1f(x_1)+\lambda_2f(x_2)+....+\lambda_nf(x_n)}{f({\lambda_1x_1+\lambda_2x_2+....+\lambda_nx_n})} \geq \frac{\lambda_1f(y_1)+\lambda_2f(y_2)+....+\lambda_nf(y_n)}{f({\lambda_1y_1+\lambda_2y_2+....+\lambda_ny_n})}$

2, Let $\lambda_1+\lambda_2+...+\lambda_n=1$ , $\lambda_i>0$   for  $i=1,2,..,n$; f,f' are  real continuous function that is concave up in [a,b], If $x_1, . . . , x_n$ and $y_1, . . . , y_n$ are numbers in [a,b] such that $(x_1, . . . , x_n)$ majorizes $(y_1, . . . , y_n)$, then:

$$\lambda_1f(x_1)+\lambda_2f(x_2)+....+\lambda_nf(x_n)-f({\lambda_1x_1+\lambda_2x_2+....+\lambda_nx_n})$$
$$\geq \lambda_1f(y_1)+\lambda_2f(y_2)+....+\lambda_nf(y_n)-f({\lambda_1y_1+\lambda_2y_2+....+\lambda_ny_n})$$

## Thứ Tư, 11 tháng 2, 2015

### 98-Around Anticenter

Problem 1: Let $ABCD$ be a quadrilateral, $E$ be the midpoint of $BC$, $F$ be the midpoint of $AD$. The line though $E$ and perpendicular to $BC$ meets the line through $F$ and perpendicular $AD$ at $G$. Show that:

1-$FG$, and the line through $B$ perpendicular to $AG$, and the line through $C$ perpendicular to $DG$ are concurrent at $P_{bc}$.

2-If $ABCD$ be a cyclic quadrilateral, show that $P_{ab}P_{bc}P_{cd}P_{da}$ be a tangential quadrilateral

Problem 2: Let $ABCD$ be a tangential quadrilateral, let $A_1,B_1,C_1,D_1$ be midpoint of $AB,BC,CD,DA$ respectively. Let $A_2,B_2,C_2,D_2$ lie on $AB,BC,CD,DA$ respectively. Such that $A_1C_2, B_1D_2,C_1A_2,D_1B_2$ perpendicular to $CD,DA,AB,BC$ respectively. Then the quadrilateral bounded by $A_1C_2, B_1D_2,C_1A_2,D_1B_2$ is a tangential quadrilateral.

## Thứ Ba, 10 tháng 2, 2015

### 97-A generalization Zeeman-Gossard perpector

Let $ABC$ be a triangle, let a line $L$ meets the Euler line at $D$ and the line $L$ meets three sidelines $BC,CA,AB$ at $A_0,B_0,C_0$ respectively. Let $H_a, H_b,H_c$ be the orthocenter of three triangles $AB_0C_0, BC_0A_0, CA_0B_0$ respectively.  Let $G_a,G_b,G_c$ be the centroid of three triangles $AB_0C_0, BC_0A_0, CA_0B_0$ respectively. Let three points  $D_a,D_b,D_c$ lie on the Euler line of $AB_0C_0, BC_0A_0, CA_0B_0$ respectively such that:

$\frac{\overline{D_aH_a}}{\overline{DaGa}}=\frac{\overline{D_bH_b}}{\overline{D_bG_b}}=\frac{\overline{D_cH_c}}{\overline{D_cG_c}} \space=\frac{\overline{DH}}{\overline{DG}}=t$

Let $L_a,L_b,L_c$ = three lines through $D_a,D_b,D_c$ and parallel to $BC,CA,AB$ respectively.

View in Geogebra

Problem 1: Then triangle bounded by three line $(L_a,L_b,L_c)$ are congruent
and homothety to triangle ABC. The homothetic center lie on the line L.
When $L$ is the Euler line or t = ∞ the problem is Zeeman-Gossard perpector

Problem 2: Newton lines of four quadrilateral $(AB,AC,L_a,L)$, $(BC,BA,L_b,L)$,
$(CA,CB,L_c,L)$, $(AB,BC,CA,L)$ also through the homothetic center in problem 1

Where if $L_i$ $(i=1,2,...n)$ be a line, define $(L_1,L_2,....,L_n)$ = Polygon bound by $L_1,L_2,....,L_n$

### 96-Ten point circle associated X4240 in ETC

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 $O'$

## Thứ Bảy, 31 tháng 1, 2015

### 95-Một bài toán về tính chất của đường thẳng đi qua tâm đường tròn ngoại tiếp

Cho tam giác $ABC$ nội tiếp $(O)$. Một đường thẳng $d$ qua $O$ cắt $BC,CA,AB$ tại $A_0,B_0,C_0$. Gọi $A_b,A_c$ là hình chiếu của $A_0$ trên $AB,AC, B_a,B_c$ là hình chiếu của $B0$ lên $BA,BC. Ca,Cb$ là hình chiếu của $C0$ lên $CA,CB$. Gọi $A_a,B_b,C_c$ là trung điểm của $A_bA_c,B_aB_c, C_a,C_b$. Chứng minh $A_a,B_b,C_c$ nằm trên cùng một đường thẳng và đường thẳng đó đi qua tâm đường tròn Euler $N$ của $ABC$

### 94-X(5463) and X(5464) in Kimberling center

X(5463) = REFLECTION OF X(13) IN X(2)

Each of the following sets of 4 points are concyclic: {X(13), X(15), X(5463), X(5464)}, {X(14), X(16), X(5463), X(5464)}, {X(2), X(110), X(5463), X(5464)}, {X(3), X(15), X(110), X(5464)}, {X(3), X(16), X(110), X(5463)}. Moreover, the line X(13)X(5463) is tangent to both of the circles of {X(13), X(14), X(15)} and {X(14), X(15), X(5463)}; likewise, the line X(14)X(5464) is tangent to both of the circles of {X(13), X(14), X(16)} and {X(13), X(16), X(5464). (Dao Thanh Oai, ADGEOM #1237, April 7, 2014). X(5463) is the center of the equilateral antipedal triangle of X(13)

X(5464) = REFLECTION OF X(14) IN X(2)

Each of the following sets of 4 points are concyclic: {X(13), X(15), X(5463), X(5464)}, {X(14), X(16), X(5463), X(5464)}, {X(2), X(110), X(5463), X(5464)}, {X(3), X(15), X(110), X(5464)}, {X(3), X(16), X(110), X(5463)}. Moreover, the line X(13)X(5463) is tangent to both of the circles of {X(13), X(14), X(15)} and {X(14), X(15), X(5463):; likewise, the line X(14)X(5464) is tangent to both of the circles of {X(13), X(14), X(16)} and {X(13), X(16), X(5464). (Dao Thanh Oai, ADGEOM #1237, April 7, 2014).

### 93-Four Jerabek Centers lie on a circle

Let $ABC$ be a triangle, let the Euler line of $ABC$ meets $BC,CA,AB$ at $A_1,B_1,C_1$ respectively.
Define $A_2=X_{125}$ of the triangle $AB_1C_1$. Define $B_2,C_2$ cyclically.

1-Then circumcenter of the triangle $A_2B_2C_2$ is Gossard perpector

2-Two triangle $A_2B_2C_2$ and ABC are similar and perpective, Which is the perpector?

3-$X(125)$ of ABC also lie on circumcenter of $A_2B_2C_2$ (Four point $X(125)$ lie on a circle)

4-Let $A_3B_3C_3$ be the paralogic triangle of ABC whose perpectrix is Euler line, then  $A_2B_2C_2$ perpective with $A_3B_3C_3$, which is the perpector?

5-Circumcircle of $(A_3B_3C_3), (A_2B_2C_2)$ and $(ABC)$ concurrent at one point, which point?

## Thứ Năm, 29 tháng 1, 2015

### 92-Two new circle in a triangle

Lemma 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 $(POB)$.

Lemma 2: Symmendian point is Pole of Euler line respect to Kiepert hyperbola.

Application we can show that:

Theorem 3: In any triangle: The Circumcenter, the Nine point center, the Symmedian point and the Kiepert center lie on a circle

Lemma 4: 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:

Theorem 5: In any triangle: The orthocenter, the Nine point center and Tarry point and midpoint of Brocard diameter lie on a circle.

Lemma 6: 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 theorem

## Chủ Nhật, 25 tháng 1, 2015

### 91-Dual problem of a generalization of Napoleon theorem

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 draw the line $X(2)X(13)$ (or $X(2)X(14)$)

## Thứ Bảy, 24 tháng 1, 2015

### 90-A generalization Napoleon's theorem associated with the Kiepert hyperbola

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.

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:

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 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 .

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

## Thứ Tư, 21 tháng 1, 2015

### 89-A generalization of Parry circle

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 meets 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.

### 88-A generalization of Nine point circle

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 $\varrho=\sqrt{\tfrac{1}{4}R^2+S \cdot \tan \theta}.$

### 87-Six center of Thebault circle lie on a conic

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.
Geogebra Check

## Thứ Năm, 8 tháng 1, 2015

### 86-Rectangular hyperbola and Inscribed parabola of a triangle

http://mathworld.wolfram.com/ChaslessPolarTriangleTheorem.html
http://forumgeom.fau.edu/FG2004volume4/FG200427.pdf

I proposed problem construction of a rectangular hyperbolar and  a inscribed parabola as follows:

Let ABC be a triangle, let a circle with radii R, A0B0C0 is the triangle bounded by the polar of A,B,C to the circle. A1 is the intersection of the polar of A and BC; define B1,C1 are cyclically. By Chasless theorem we have A1,B1,C1 are collinear.  We call A1B1C1 is the Chasless line. Now by Desargues' theorem we have:  ABC and A0B0C0 are perpective. Denote the perpector is P. Show that:

1-P lie on a rectangular hyperbola through center of the circle when R changed. (I mean six point A,B,C, the orthocenter, P and center of the circle lie on a hyperbola with any R).

2-The Chasless line are tangent with a inscribed parabola when R changed (or center of the circle be moved on the line through the orthocenter of the triangle ABC and original location of center of the circle.)

### 85-A pair equilateral triangle

Let $ABC$ be a triangle, let $P$ be the point $X(15)$ or $X(16)$. Let $A_0$ be a point on the plain such that:
$\angle A0BP=\angle A0CP = 60^0$; define $B_0,C_0$ cyclically. Show that $A_0B_0C_0$ is a equilateral triangle (and A0B0C0 lie on cirumcircle).

Let $A_1B_1C_1$ be a triangle such that $A_0B_0C_0$ are median triangle of $A_1B_1C_1$. Show that ABC and $A_1B_1C_1$ are perpective. Please see the figure attachment

Telv Cohl's proof:

Since $\angle BCA_0=\angle BAP, \angle A_0BC=\angle PAC$ , so $A_0$ is the intersection of $AP$ and $\odot (ABC)$ . Similarly, $B_0=BP \cap \odot (ABC), C_0=CP \cap \odot (ABC)$ .

Since $\angle A_0B_0C_0=\angle PAB+\angle BCP=60^{\circ},\angle B_0C_0A_0=\angle PBC+\angle CAP=60^{\circ}$ , so we get $\triangle A_0B_0C_0$ is an equilateral triangle and $\odot(A_0B_0C_0)$ is the incircle of $\triangle A_1B_1C_1$ , hence from Steinbart theorem we get $AA_1, BB_1, CC_1$ are concurrent .

## Thứ Tư, 31 tháng 12, 2014

### 84-New inequality in geometry

Theorem: Let $ABC$ be an equilateral triangle, let $P$ be a point on the plain, let $P_a,P_b,P_c$ be projection foot of $P$ on three sidelines $BC,CA,AB$ respectively. Then: $BP_b+CP_c > AP_a$.

I call result above is theorem because this theorem similarly: https://en.wikipedia.org/wiki/Pompeiu%27s_theorem

Leo Giugiuc's proof:

Let $b=e^\frac{2\pi i}{3}$ and $c=e^\frac{4\pi i}{3}$ then $b^3=c^3=1$ and $b^2=c$ and $c^2=b$ and $b+c=-1$ and $bc=1$.

We choose $A=1, B=b, C=c$  and $P=z$ where $b,c$ define above. Let $R_a$ be the reflection of $P$ in $BC$, then $\frac{R_a-b}{c-b}=\frac{\overline{z-b}}{c-b}$ $\Rightarrow$  $\frac{R_a-b}{c-b}=\frac{\frac{1}{b}-\overline{z}}{\frac{1}{b}-\frac{1}{c}}$ $\Rightarrow$  $R_a=b+c-bc.\overline{z}=-1-\overline{z}$. Since $P_a$ is the midpoint of $PR_a$ show that: $P_a=\frac{-1+z-\overline{z}}{2}$, similarly we have: $P_b=\frac{1+c+z-c\overline{z}}{2}$  and $P_c=\frac{1+b+z-b\overline{z}}{2}$  $\Rightarrow$  $2AP_a=|3-z+\overline{z}|=|3-ki|=\sqrt{9+k^2}$ (1), where $z-\overline{z}=ki, k \in R$

$2BP_b=|2b-c-1-z+c.\overline{z}|$ and $2CP_c=|2c-b-1-z+b.\overline{z}|$

We have:

$2BP_b+2CP_c=|2b-c-1-z+c.\overline{z}|+|2c-b-1-z+b.\overline{z}|\\ =|c|.|2b-c-1-z+c.\overline{z}|+|b|.|2c-b-1-z+b.\overline{z}|\\= |2bc-c^2-c-zc+c^2.\overline{z}|+|2bc-b^2-b-bz+b^2.\overline{z}|\\=|2-b-c-zc+b.\overline{z}|+|2-c-b-bz+c.\overline{z}| \\= |3-zc+b.\overline{z}|+|3-bz+c.\overline{z}| \ge |6+z-\overline{z}|=\sqrt{36+k^2}$,  (2)

Since (1) and (2), the proof of theorem are complete.

## Thứ Tư, 5 tháng 11, 2014

### 83-Six points lie on a circle associated Botema configuration

Let $ABC$ be a triangle $M_a$ is midpoint of $BC$, $H$ is the orthocenter of the triangle $ABC$. Let $V_A$, $V_B$, $V'_A$, $V'_B$ are center of four squares. Show that six points $V_A,$ $V'_a,$ $M_a,$ $H_a,$ $V_B,$ $V'_B$  lie on a circle.

### 81-Two conic problem

Let two conics through four common points $A,B,C,D$. Tangent line of the first conic  at $B,D$ meets the second conic at $G,H$ respectively. Show that $AC,BD,GH$ are concurrent

## I.In Forum Geometricorum Jounal

2- Đào Thanh Oai, Volum 14, Issue 18,   Two pairs of Archimedean circles in the arbelos
3- Nikolaos Dergiades, Volum 14, Issue 24, Dao's theorem on six circumcenters associated with a cyclic
4- Telv Cohl, Volum 14, Issue 29, A Purely Synthetic Proof of Dao’s Theorem on Six Circumcenters Associated with a Cyclic Hexagon

5-Dao Thanh Oai, Equilateral triangles and Kiepert perspectors in complex numbers, 105--114.

6- Dao Thanh Oai and Paul Yiu, Some simple constructions of equilateral triangles associated with a triangle (accepted)

7- Ngo Quang Duong, Two generalizations of the Simson line theorem (submited)

8-Dao Thanh Oai, On the Jacobi Triangle (accepted)

### II. In Crux Mathematicorum

1- Problem 3845, Issue 5, Volum 39

### III. In Global Journal of advanced research on classical and mordern geometries (Romania)

Organized by the Department of Mathematics and Computer Science, Vasile Alecsandri National College of Bacau and Vasile Alecsandri University of Bacau,

Web site International Journal of Geometry

1- Telv Cohl, Volum 3, Isue 8, DAO'S THEOREM ON CONCURRENCE OF THREE EULER LINES

2- Đào Thanh Oai, Volum 3, A SYNTHETIC PROOF OF A. MYAKISHEV'S GENERALIZATION OF VAN LAMOEN CIRCLE THEOREM AND AN APPLICATION

### V. ENCYCLOPEDIA OF TRIANGLE CENTERS

Web site ENCYCLOPEDIA OF TRIANGLE CENTERS

1- X(4240) = DAO TWELVE EULER LINES POINT

2- X(5569) = CENTER OF THE DAO 6-POINT CIRCLE

3-X(5607) = CENTER OF 1st POHOATA-DAO-MOSES CIRCLE

4-X(5608) = CENTER OF 2nd POHOATA-DAO-MOSES CIRCLE

5-X(6103) = RADICAL CENTER OF THE DAO-MOSES-TELV CIRCLE, CIRCUMCIRCLE, AND NINE-POINT CIRCLE

6-X(6118) = CENTER OF 1st DAO-VECTEN CIRCLE

7-X(6119) = CENTER OF 2nd DAO-VECTEN CIRCLE

8-X(6188) = DAO (a,b,c,R) PERSPECTOR

9-DAO'S CONJUCTURE GENERALIZATION OF THE LESTER CIRCLE

### VI. THE MATHEMATICAL GAZETTE

1-Dao Thanh Oai, A family of Napoleon triangles associated with the Kiepert configuration, The Mathematical Gazette, Published online: 13 March 2015

2- Nguyen Le Phuoc, Nguyen Chuong Chi, A proof of Dao generalization of the Simson line theorem (accepted)

### VII-AMERICAN MATHEMATICAL MONTHLY

VIII-SOMES ANOTHER NICE RESULT

1-A generalization Gossard perspector theorem

Let $ABC$ be a triangle, Let $P_1,P_2$ be two points on the plane, the line $P_1P_2$ meets $BC, CA, AB$ at $A_0,B_0,C_0$ respectively. Let $A_1$ be a point on the plane such that $B_0A_1$ parallel to $CP_1$, $C_0A_1$ parallel to  $BP_1$. Define $B_1, C_1$ cyclically. Let $A_2$ be a point on the plane such that $B_0A_2$ parallel to $CP_2$, $C_0A_2$ parallel to  $BP_2$. Define $B_2, C_2$ cyclically.

Problem 1: The triangle bounded by three lines  $A_1A_2,B_1B_2,C_1C_2$ homothety and congruent to $ABC$, the homothetic center $Q$ lie on $P_1P_2$

Problem 2: Newton lines of four quadrilateral $(AB,AC,A_1A_2,L)$, $(BC,BA,B_1B_2,L)$, $(CA,CB,C_1C_2,L)$, $(AB,BC,CA,L)$ also through the homothetic center in problem 1. Where if Li (i=1,2,...n) be a line, define (L1,L2,....,Ln) = Polygon bound by L1,L2,L3...,Ln

2-A generalization of the Napoleon theorem associated with Kiepert hyperbola

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.

Please click to see a solution in AoPS

3-A generalization Simson line theorem 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. The new line $\overline {A_0B_0C_0}$ bisects the orthocenter and $P$. When $L$ pass through $P$, this line is Simson line.

See two proof in AoPS and  Click to get another proof in yahoo discustion Advanced Plane Geometry

4. 2nd Ageneralization Simson line theorem

Let a circumconic of the triangle ABC, let Q, P be two points on the plane. Let PA,PB,PC intersect the conic at A1,B1,C1 respectively. QA1 intersects BC at A2, QB1 intersects AC at B2, QC1 intersects AB at C2. Then four points A2,B2,C2,D are colinear if only if Q lie on the conic.

See link in Geoff Smith's paper, publish in Mathematical Gazette

5-A generalization of Parry Circle

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 meets the hyperbola at $F_+,F_-$.  Let  $I_+,I_-,G$ be the isogonal conjugate of $F_+,F_-$ and $K$ respectively. Let F be the inverse point of  G with respect to the circumcircle of ABC.  Show that: $I_+,I_-,G, X(110), F$ alway lie on a circle, this circle is a generalization of Parry circle. Furthemore K alway lie on the Jerabek hyperbola.

Click to see post in AoPS

6. A generalization Steiner line and Miquel circle

Let $ABC$ be a triangle, Let $P_1$ be any point on the plane. Let a line $L$ meets $BC, CA, AB$ at $A_0,B_0,C_0$ respectively. Let $A_1$ be a point on the plane such that $B_0A_1$ parallel to $CP_1$, $C_0A_1$ parallel to $BP$. Define $B_1, C_1$ cyclically.

Generalization of the Steiner line: Then show that: $A_1, B_1, C_1, P_1$ are collinear.

A gneralization of Miquel circle: Define $A_2,B_2,C_2,P_2$ be the isogonal conjugates $A_1,B_1,C_1,P_1$ respect to $AB_0C_0, BC_0A_0, CA_0B_0$ and $ABC$ (respectively). Then show that $A_2,B_2,C_2,P_2$ lie on a circle.

http://tube.geogebra.org/material/show/id/1434055

See post and two proof in AoPS

7. A generalization specialcase of Brianchon theorem and Pascal theorem in one configuration

Let six circles $(O_1), (O_2), (O_3), (O_4), (O_5), (O_6)$. Let  $(O_i), (O_{i+1})$ cut at $A_i, A'_i$ for $i=1, 2, 3, 4, 5,$6 (Here we take modulo 6).  Let $A_1, A_2, A_3, A_4, A_5, A_6$ lie on a circle and  $A'_1, A'_2, A'_3, A'_4, A'_5, A'_6$  lie on another circle.

1. (A generalization of Pascal theorem) Then show that six points which they are intersection of
$(O_1)$, $(O_4)$; $(O_2)$, $(O_5)$; $(O_3)$, $(O_6)$ (if they are exist) lie on a circle.

2. (A generalization of Brianchon theorem) Three lines $O_1O_4$, $O_2O_6$, $O_3O_5$ are concurrent (Problem 3845, Proposed by Dao Thanh Oai, Kien Xuong, Thai Binh, Viet Nam, Crux Mathematicorum, Volum 39; Solution by Luis Gonzalez)

Applet in Geogebra website

## Thứ Hai, 27 tháng 10, 2014

### 80-Another four conics theorem

Theorem: Let three conics and one of them touch with fourth conic at two points. Then remaining pair of common tangents intersect at three collinear points.

Special case: Exist a conic touching with three given circles(The conic touching with one circle at two points)

## Thứ Ba, 23 tháng 9, 2014

### 78-Cevian triangle associated with Jacobi Configuration

Let ABC be a triangle, let $\triangle J_aJ_bJ_c$, $\triangle J'_aJ'bJ'_c$ be two Jacobi triangle, $A_0=J_aJ'_a \cap BC$, define $B_0,C_0$ cyclically. Then $AA_0,BB_0,CC_0$ are concurrent

## Thứ Hai, 22 tháng 9, 2014

### 77-Transfomation line to line on the Pascal line

Let $A,B,C,C_1,B_1A_1$ lies on a conic. Let $A_2=BC_1 \cap B_1C, B_2=AC_1 \cap A_1C; C_2=AB_1 \cap A_1B$. Let $A_3,B_3,C_3$ lie on sidelines $BC,CA,AB$ such that $A_3,B_3,C_3$ are collinear. Let $A_4=A_3A_2 \cap B_1C_1, B_4=B_3B_2 \cap A_1C_1; C_4=A_3B_3 \cap A_1B_1$ . Then $A_4,B_4,C_4$ are collinear.

## Chủ Nhật, 21 tháng 9, 2014

### 76-A generalization of Simson line theorem

Problem 1: Let $ABC$ be a triangle, let a line $(L)$ through circumecenter and 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.

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

Problem 2: The new line $\overline {A_0B_0C_0}$ bisect the orthocenter and P

## Thứ Bảy, 20 tháng 9, 2014

### 75-Two similar rectangles theorem

If ABCD and $A_1B_1C_1D_1$ are two similar rectangles (mean $AB:AD=A_1B_1:A_1D_1)$, then

Red=Yellow <=> $area(AA_1B_1B)+area(CC_1D_1D)=area(AA_1D_1D)+area(CC_1B_1B)$

I found this result when I research the British flag theorem.

## Thứ Tư, 17 tháng 9, 2014

### 74-A circle tangent to circumcircle(Similar with the third Fontené)

Colling theorem: Let a line $(L)$ through the orthocener of the triangle ABC, reflection of $(L)$ in three sidelines concurrent be a point on the circumcircle(Colling point of $(L)$

Nice problem: Let $ABC$ be a triangle, let a line $(L)$ through the orthocenter of the triangle $ABC$ and $(L)$ cuts $AB, AC$ at $P, Q$. Denote $K$ be the intersection of two lines perpendicular to $AB,AC$ at $P,Q$. These line cut $BC$ at $M,N$. The circle through $KMN$ tangent the circumcircle at colling point of (L).

Note: $X_3$ of $ABC$ is $X_4$ of median triangle

### 73-X(4240) =12 EULER LINES POINT

X(4240) in Kimberling Center

Click to downloand detail defined and the figure

Barycentrics   f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (b4 + c4 - 2a4 + a2b2 + a2c2 - 2b2c2)/[(b2 - c2)(b2 + c2 - a2)]
X(4240) is the point of intersection of the Euler lines of nine triangles, constructed as in the next three paragraphs.
Let E be the Euler line of a triangle ABC. Let A1 = E∩BC, and define B1 and C1 cyclically. Let AB be the reflection of A in B1, and define BC and CA cyclically. Let AC be the reflection of C in B1, and define BA and CB cyclically. The Euler lines of the four triangles ABC, AABAC, BBCBA, CCACB concur in X(4240). (Dao Thanh Oai, Problem 1 in attachment to ADGEOM #1709, September 15, 2014). See also Telv Cohl, 'Dao's Theorem on the Concurrency of Three Euler Lines,' International Journal of Geometry 3 (2014) 70-73.
Continuing, let A*B*C* be the paralogic triangle of ABC whose perspectrix is E. Then X(4240) lies on the Euler line of A*B*C*. (Dao Thanh Oai, noted just after Figure 1 in attachment to ADGEOM #1709, September 15, 2014).
Continuing, redefine AB as the point on line AC and AC as the point on line AB such that B1, A1, AB, AC line on a circle and A1, AB, AC are collinear. Define BC and BA cyclically, and define CA and CB cyclically. Let A2 = BABC∩CACB and define B2 and C2 cyclically. The Euler lines of the five triangles ABC, A2B2C2, AABAC, BBCBA, CCACB concur in X(4240). (Dao Thanh Oai, Problem 2 in attachment to ADGEOM #1709, September 15, 2014).
X(4240) lies on these lines:
{2, 3}, {107, 110}, {112, 1302}, {476, 1304}, {925, 1301}, {2407, 3233

## Thứ Hai, 15 tháng 9, 2014

### 72-Concurrency on the Tucker circle

Let a Tucker circle cut sidelines of given triangle $ABC$ at $A_b,A_c, B_c,B_a,C_a,C_b$. If $A_2A_bA_c, B_2B_aB_c, C_2C_aC_b$ are three similar isosceles triangles then $AA_2BB_2CC_2$ are concurrent

# Let $ABC$ be a triangle, $B_a,C_a$ lie on $BC; C_b,B_c$ lie on $CA,BA$ respectively. Such that two circles $(BB_aB_c),(CC_aC_b)$ tangent at $P$ on the circumcircle and $B_cB_a \perp AC, C_aC_b \perp AB$. Denote $A_b=B_aB_c \cap AC, A_c=C_aC_b \cap AB$. Show that: $A_bA_c$ through the orthoenter of the triangle $ABC$. $M=B_cC_b \cap AO$, $M$ lie on circumcircle of $ABC$

### 70-Some problem on the Simson's line

Let $ABC$ be a triangle, let three points $A',B',C'$ on the circumcircle, such that $AA'//BB'//CC'$. Then the triangle $A_0B_0C_0$ form(bounded) by three Simson line of $A',B',C'$

# Theorem 2: Three line through $A',B',C'$ and parallel to $BC,CA,AB$ form a triangle congruent and homothety to $ABC$, the homothety at infinity

### 65-A conjecture on polynomial function

Let a polynomial function:
f(x)=$x^n+a_{n−1}x^{n−1}+⋯+a_2x^2+a_1x+a_0$
Where n is positive integer, $n \geq 3$ and $a_0, a_1, a_2, ..., a_n$ are constant coefficients   $\in Z$ and $f(x)=0$ has only solution $p \in Z$. Then has only $x=z,y =p or y=z,x =p or x=y=z=p$ with $x,y,z \in Z$ can satisfy the equation:

$f(x)+f(y)=f(z)$

# A Generalization Lester theorem follow:

Theorem: Let two point $A,B$ lie on one branch of a rectangular hyperbola. The line pass through midpoint of $AB$, and pass through center of the hyperbola cut hyperbola at two point $F_1,F_2$. Let four points $A,M,B,N$ are harmonic range then four point $F_1,F_2,M,N$ lie on a circle.

# Định lý 1: Cho $A,B,C,D,E$ nằm trên một đường hyperbol chữ nhật, ba đường thẳng qua $D$ và vuông góc với $EA,EB,EC$ giao với $BC,AC,AB$ lần lượt tại  $A_1,B_1,C_1$. Khi đó $A_1,B_1,C_1$ thẳng hàng và đường thẳng này vuông góc với $DE$.

Chứng minh:
Không giảm tổng quát, trong tọa độ Đề Các ta có thể giả sử đường hyperbol chữ nhật có phương trình là:
$$y=\frac{p}{x}$$
(Lưu ý: Bất cứ đường hyperbol chữ nhật nào đều có thể đưa về dạng trên)
Và $A(a,\frac{p}{a})$, $B(b,\frac{p}{b})$, $C(c,\frac{p}{c})$, $D(d,\frac{p}{d})$, $E(e,\frac{p}{e})$. Phương trình đường thẳng qua  $D$ và vuông góc với $DA$ là:
$$x-\frac{p}{a.e}y_{DA_1}=d-\frac{p^2}{e.a.d}$$
Tương tự phương trình đường thẳng $DB_1,DC_1$ là:
$$x-\frac{p}{b.e}y_{DB_1}=d-\frac{p^2}{e.b.d}$$
$$x-\frac{p}{c.e}y_{DA_1}=d-\frac{p^2}{e.c.d}$$
Phương trình đường thẳng  $BC,CA,AB$ lần lượt là:
$$y_{BC}=-\frac{p}{c.b}x+\frac{p}{b}+\frac{p}{c}$$
$$y_{CA}=-\frac{p}{c.a}x+\frac{p}{c}+\frac{p}{a}$$
$$y_{BC}=-\frac{p}{a.b}x+\frac{p}{a}+\frac{b}{c}$$
Trong tọa độ điểm $A_1$ là nghiệm của phương trình (2) và (5).

$x-\frac{p}{a.e}y_{DA_1}=d-\frac{p^2}{e.a.d} \\ y_{BC}=-\frac{p}{c.b}x+\frac{p}{b}+\frac{p}{a}$
Giải hệ phương trình tuyến tính hai ẩn ở trên ta có:

$x=\frac{d.c.p^2+d.b.p^2-c.b.p^2+e.a.c.b.d^2}{d.p^2+e.a.c.b.d} \\ y= \frac{e.a.c.p.d+e.a.b.p.d+p^3+e.a.p.d^2}{d.p^2+e.a.c.b.d}$

Do đó tọa độ điểm $A_1$ là:
$A_1(\frac{d.c.p^2+d.b.p^2-c.b.p^2+e.a.c.b.d^2}{d.p^2+e.a.c.b.d}, \frac{e.a.c.p.d+e.a.b.p.d+p^3-e.a.p.d^2}{d.p^2+e.a.c.b.d})$

Tương tự tọa độ các điểm $B_1,C_1$

$B_1(\frac{d.c.p^2+d.a.p^2-c.a.p^2+e.a.c.b.d^2}{d.p^2+e.a.c.b.d}, \frac{e.b.c.p.d+e.a.b.p.d+p^3-e.b.p.d^2}{d.p^2+e.a.c.b.d})$
$C_1(\frac{d.a.p^2+d.b.p^2-a.b.p^2+e.a.c.b.d^2}{d.p^2+e.a.c.b.d}, \frac{e.a.c.p.d+e.c.b.p.d+p^3-e.c.p.d^2}{d.p^2+e.a.c.b.d})$

$A_1,B_1,C_1$ thẳng hàng nếu và chỉ nếu: $\overrightarrow{A_1B_1}=k\overrightarrow{B_1C_1}$

$\overrightarrow{A_1B_1}=\frac{1}{d.p^2+e.a.c.b.d}(d.c.p^2+d.a.p^2-c.a.p^2+e.a.c.b.d^2-d.c.p^2-d.b.p^2 \\+ c.b.p^2-e.a.c.b.d^2 , e.b.c.p.d+e.a.b.p.d+p^3-e.b.p.d^2-e.a.c.p.d-e.a.b.p.d-p^3+e.a.p.d^2) \\ = \frac{1}{d.p^2+e.a.c.b.d}(d.a.p^2-c.a.p^2-d.b.p^2+c.b.p^2, e.b.c.p.d+e.b.p.d^2-e.a.c.p.d-c.a.p.d^2) \\ =\frac{(b-a)(c-d)p}{d.p^2+e.a.c.b.d}(p,ed)$

Tương tự như vậy chúng ta có:

$\overrightarrow{A_1C_1}=\frac{(c-a)(b-d)p}{d.p^2+e.a.c.b.d}(p, ed)$
$\Rightarrow$  $$\overrightarrow{A_1B_1}= \frac{(b-a)(c-d)}{(c-a)(b-d)} \overrightarrow{A_1C_1}$$

Do đo $A_1,B_1,C_1$ thẳng hàng. Hệ số góc của $DE$ là: $\frac{\frac{p}{e}-\frac{p}{d}}{e-d}=-\frac{p}{ed}$, hệ số góc của đường thẳng $\overline{A_1B_1C_1}$ là: $\frac{ed}{p}$. Do đó đường thẳng $\overline{A_1B_1C_1}$ vuông góc với đường thẳng $DE$. Định lý 1 chứng  hoàn tất.

Hệ quả 2 [1]:  Cho tam giác $ABC$, H là trực tâm, D là điểm bất kỳ trong mặt phẳng, ba đường thẳng qua $H$ và vuông góc với $DA,DB,DC$ giao với $BC,AC,AB$ lần lượt tại  $A_1,B_1,C_1$. Khi đó $A_1,B_1,C_1$ thẳng hàng và đường thẳng này vuông góc với $HD$.

Chứng minh:

Với điểm D bất kỳ thì A,B,C,H,D luôn nằm trên một hyperbol chữ nhật nên ta có điều phải chứng minh.

Hệ quả 3 [2]:  Cho tam giác $ABC$ và một điểm $P$ trên mặt phẳng, ba đường thẳng vuông góc với $PA,PB,PC$ tại $P$ lần lượt cắt ba cạnh $BC,CA,AB$ của tam giác tại $A_0,B_0,C_0$ thẳng hàng.

Chứng minh:
Với điểm P bất kỳ thì ta có 5 điểm A,B,C,P,P luôn nằm trên một hyperbol chữ nhật nên ta có điều phải chứng minh.

Định lý 4: Cho tam giác ABC, P,Q là hai điểm trên mặt phẳng sao 5 điểm A,B,C,P,Q nằm trền một hyperbol chữ nhật. Kẻ đường thẳng vuông góc AP qua Q, đường thẳng này cắt AB,AC tại $A_c,A_b$. Định nghĩa các điểm $B_a,B_c$,$C_b,C_a$ tương tự. Khi đó sáu điểm $A_c,A_b, B_a,B_c$,$C_b,C_a$ nằm trên một đường Conic.

Chứng minh:  Định lý 4 được chứng minh trực tiếp từ định lý 1 và định lý Pascal.

Tham khảo:

[1]-Đào Thanh Oai, Vấn đề 3878 tạp chí Crux mathematicorum, issue 8, volum 39

[2]- Gibert, B. "Orthocorrespondence and Orthopivotal Cubics." Forum Geom. 3, 1-27, 2003. http://forumgeom.fau.edu/FG2003volume3/FG200301index.html.

Dao Thanh Oai: Kien Xuong, Thai Binh, Viet Nam.