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 $ABC$
Thứ Bảy, 31 tháng 1, 2015
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 .
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