Dao's blog
Thứ Bảy, 18 tháng 4, 2015
Thứ Sáu, 17 tháng 4, 2015
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:
$$\lambda_1f(x_1)+\lambda_2f(x_2)+....+\lambda_nf(x_n)-f({\lambda_1x_1+\lambda_2x_2+....+\lambda_nx_n})$$
$$\geq \lambda_1f(y_1)+\lambda_2f(y_2)+....+\lambda_nf(y_n)-f({\lambda_1y_1+\lambda_2y_2+....+\lambda_ny_n})$$
\[\frac{\lambda_1f(x_1)+\lambda_2f(x_2)+....+\lambda_nf(x_n)}{f({\lambda_1x_1+\lambda_2x_2+....+\lambda_nx_n})} \geq \frac{\lambda_1f(y_1)+\lambda_2f(y_2)+....+\lambda_nf(y_n)}{f({\lambda_1y_1+\lambda_2y_2+....+\lambda_ny_n})}\]
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.
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 $L_1,L_2,....,L_n$
\[ \frac{\overline{D_aH_a}}{\overline{DaGa}}=\frac{\overline{D_bH_b}}{\overline{D_bG_b}}=\frac{\overline{D_cH_c}}{\overline{D_cG_c}} \space=\frac{\overline{DH}}{\overline{DG}}=t \]
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 $O'$
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).
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
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
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 $\varrho=\sqrt{\tfrac{1}{4}R^2+S \cdot \tan \theta}.$
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
Geogebra Check
Thứ Năm, 8 tháng 1, 2015
86-Rectangular hyperbola and Inscribed parabola of a triangle
Please see:
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.)
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 .
$\angle A0BP=\angle A0CP = 60^0$; 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.
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
Thứ Ba, 28 tháng 10, 2014
PUBLISH IN IN SOME JOURNALS
I.In Forum Geometricorum Jounal
1- Đào Thanh Oai, Volum 14, Issue 10, A simple proof of Gibert's generalization of the Lester circle theorem
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)
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
4- Problem 3885, Isue 9, Volum 39
5- Problem 3894(Dao Thanh Oai and Nguyen Minh Ha), Isue 10, Volum 39
5- Problem 3894(Dao Thanh Oai and Nguyen Minh Ha), Isue 10, Volum 39
III. In Global Journal of advanced research on classical and mordern geometries (Romania)
1- Trần Hoàng Sơn, Volum 3, Issue 2, A SYNTHETIC PROOF OF DAO’S GENERALIZATION OF GOORMAGHTIGH’S THEOREM
2-Dao Thanh Oai-Nguyen Minh Ha, AN INTERESTING APPLICATION OF THE BRITISH FLAG THEOREM, Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, Vol.4, (2015), Issue 1, pp.31-34
3-Nguyen Ngoc Giang, A proof of Dao's theorem (accepted)
IV. International jourbal of Geometry (Romania)
2-Dao Thanh Oai-Nguyen Minh Ha, AN INTERESTING APPLICATION OF THE BRITISH FLAG THEOREM, Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, Vol.4, (2015), Issue 1, pp.31-34
3-Nguyen Ngoc Giang, A proof of Dao's theorem (accepted)
IV. International jourbal of Geometry (Romania)
Web site International jourbal of Geometry
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
Web site International Journal of Geometry
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 20152- Nguyen Le Phuoc, Nguyen Chuong Chi, A proof of Dao generalization of the Simson line theorem (accepted)
VII-AMERICAN MATHEMATICAL MONTHLY
1-Problem 11830, The American Mathematical Monthly Vol. 122, No. 3 (March 2015), pp. 284-291 Published by: Mathematical Association of America
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
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 two link in AoPS: Link 1 in AoPS ; Link 2 in AoPS
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
Thứ Hai, 29 tháng 9, 2014
Thứ Ba, 23 tháng 9, 2014
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
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
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)]
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
{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
Thứ Bảy, 13 tháng 9, 2014
71-A line through H
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 $A',B',C'$
1-$A_0B_0C_0$ are similar and perpective with $ABC$. Which is the locus of perpector?
2- The circumcenter and the orthocenter $A_0B_0C_0$ , also lie on the circle with diameter $X(3)X(4)$ center $X(5)$
3-I don't know when the $A_0B_0C_0$ become to a point, but I known $A_0B_0C_0$ become a point three time when we moved $A',B',C'$ on circumcircle(such that $AA'//BB'//CC'$), these points also lie on the circle with diameter $X(3)X(4)$ center $X(5)$
69-New or old? Three lines concurrents
Let $ABC$ be a triangle, let three points $A_0,B_0,C_0$ lie on $BC,CA,AB$ such that $A_0,B_0,C_0$ are collinear. Let circle $(P)$ which center $P$ lie on the line $\overline{A_0B_0C_0}$. Denote $A_1,B_1,C_1$ are reflection of $A_0,B_0,C_0$ in $(P)$ show that $AA_1,BB_1,CC_1$ are concurrent.
67-A generalization of Christipher Zeeman's theorem
Let $ABC$ be a triangle, let a line $(d)$ cut the sidelines $BC,CA,AB$ at $A_1,B_1,C_1,$ denote $H_A,H_B,H_C$ are the orthocenter of $AB_1C_1, BC_1A_1$, $CA_1B_1$ ; Denote $A',B',C'$ are reflection of $A$ in midpoint of $B_1C_1$; $B$ in midpoint of $C_1A_1$; C in midpoint of $A_1B_1$ respectively.
Theorem 1: Three line through $H_A,H_B,H_C$ and parallel to $BC,CA,AB$ form a triangle congruent and inversely homothety to $ABC$
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
66-A Generalization Gossard perspector and X(110)
Let $ABC$ be a triangle, let a line $(d)$ cut the sidelines $BC,CA,AB$ at $A_1,B_1,C_1$, denote $H_A,H_B,H_C$ are the orthocenter of $AB_1C_1$, $BC_1A_1$, $CA_1B_1$ ; Denote $A',B',C'$ are reflection of $A$ in midpoint of $B_1C_1$; $B$ in midpoint of $C_1A_1$; $C$ in midpoint of $A_1B_1$ respectively.
Theorem 1: Three line through $H_A,H_B,H_C$ and parallel to $BC,CA,AB$ form a triangle congruent and inversely homothety to $ABC$
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)=$x^n+a_{n−1}x^{n−1}+⋯+a_2x^2+a_1x+a_0$
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:
$f(x)+f(y)=f(z)$
63-Một tính chất quan trọng của hyperbol chữ nhật
Giới thiệu: Như ta đã biết luôn có một conic đi qua năm điểm phân biệt trên mặt phẳng. Nhưng với dấu hiệu nào ta có thể nhận ra đường conic là hyperbol, ellipse, hoặc parabol? Trong bài viết này chỉ ra một dấu hiệu về tồn tại một Hyperbol chữ nhật qua năm điểm cho trước. Bài viết này là tổng quát của các vấn đề sau:
- Tổng quát của đường thẳng Droz-Farny,
- Tổng quát vấn đề 3878 trên tạp chí Crux mathematicorum,
issue 8, volum 39
- Tổng quát định lý về tương ứng vuông góc
- Tổng quát của đường thẳng Droz-Farny,
- Tổng quát vấn đề 3878 trên tạp chí Crux mathematicorum,
issue 8, volum 39
- Tổng quát định lý về tương ứng vuông góc
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:
\begin{equation}y=\frac{p}{x}
\end{equation}
(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:
\begin{equation}x-\frac{p}{a.e}y_{DA_1}=d-\frac{p^2}{e.a.d}\end{equation}
Tương tự phương
trình đường thẳng $DB_1,DC_1$ là:
\begin{equation}x-\frac{p}{b.e}y_{DB_1}=d-\frac{p^2}{e.b.d}\end{equation}
\begin{equation}x-\frac{p}{c.e}y_{DA_1}=d-\frac{p^2}{e.c.d}\end{equation}
Phương trình đường
thẳng $BC,CA,AB$ lần lượt là:
\begin{equation}y_{BC}=-\frac{p}{c.b}x+\frac{p}{b}+\frac{p}{c}\end{equation}
\begin{equation}y_{CA}=-\frac{p}{c.a}x+\frac{p}{c}+\frac{p}{a}\end{equation}
\begin{equation}y_{BC}=-\frac{p}{a.b}x+\frac{p}{a}+\frac{b}{c}\end{equation}
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:
$\overrightarrow{A_1C_1}=\frac{(c-a)(b-d)p}{d.p^2+e.a.c.b.d}(p,
ed)$
$\Rightarrow$
\begin{equation}\overrightarrow{A_1B_1}= \frac{(b-a)(c-d)}{(c-a)(b-d)}
\overrightarrow{A_1C_1} \end{equation}
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.
Email address:
daothanhoai@hotmail.com
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