tag:blogger.com,1999:blog-42720068292357066952024-03-05T20:07:59.823-08:00Dao's blogAnonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.comBlogger103125tag:blogger.com,1999:blog-4272006829235706695.post-45254395158782429142015-04-18T04:50:00.002-07:002015-04-18T04:51:26.477-07:00102-Square in a square<div dir="ltr" style="text-align: left;" trbidi="on">
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQrsu6q2v4ZIdB2na2Ca6ZpsCkvaRZM7uAKniU_OADp8ByrcKsuXBTi2KXgOc_GfX4cuPo1Zg55WVgfTgc12jfjwPoT4emQEKlbukdxyQzJjIMmoXd0XU1OFXXVw1aNrnwxm_qrkYPQhY/s1600/1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQrsu6q2v4ZIdB2na2Ca6ZpsCkvaRZM7uAKniU_OADp8ByrcKsuXBTi2KXgOc_GfX4cuPo1Zg55WVgfTgc12jfjwPoT4emQEKlbukdxyQzJjIMmoXd0XU1OFXXVw1aNrnwxm_qrkYPQhY/s1600/1.png" height="420" width="640" /></a></div>
Let $ABCD$ be a parallelogram. $IG//AB//CD$, $FH//BC//AD$. Denote as the figure attachment, then show that $KJLM$ be a parallelogram. <br />
<b>Special case:</b><br />
if $ABCD$ be a rectangle then show that $KJLM$ be a rhombus<br />
If $ABCD$ be a square then show that $KJLM$ be a square, </div>
Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-62158532550434675072015-04-17T01:51:00.000-07:002015-04-17T01:51:33.448-07:00101-A property of six points lie on a conic<div dir="ltr" style="text-align: left;" trbidi="on">
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.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDGeS-ePZMfSe661IDDA1FUGQM8VPsfEPczTbwhZVWSpqUdgtGdKn4Bz9nflknELh57NWiKx6c2bmSiALy5XDaC4HYEirMb6M5axIT-H52_0-6wu4hEroxptSwbwQjps8MhEG4zaKjE7k/s1600/A+property+of+six+point+lie+on+a+conic.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDGeS-ePZMfSe661IDDA1FUGQM8VPsfEPczTbwhZVWSpqUdgtGdKn4Bz9nflknELh57NWiKx6c2bmSiALy5XDaC4HYEirMb6M5axIT-H52_0-6wu4hEroxptSwbwQjps8MhEG4zaKjE7k/s1600/A+property+of+six+point+lie+on+a+conic.png" height="640" width="590" /></a></div>
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-50852586761544224972015-03-31T03:21:00.002-07:002015-04-17T01:51:45.206-07:00100-Mở rộng định lý Napoleon kết hợp với một lục giác<div dir="ltr" style="text-align: left;" trbidi="on">
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<b>Mở rộng định lý Napoleon kết hợp với một lục giác:</b></div>
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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.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzMLdNckihJreDXG06ZP5TzoqULWlL8Ja2b6x2ouzemWgHiWg-cvy5MAaIjjfL4SqbWgNHOjx6EpcHgePTrXkG5lSduvtrVmt6shQV6acxS06fR8epUIbSEoQIczyxsG_cPn-mndH5-zg/s1600/New_Ge_Napoleon+theorem+associated+with+a+hexagon.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzMLdNckihJreDXG06ZP5TzoqULWlL8Ja2b6x2ouzemWgHiWg-cvy5MAaIjjfL4SqbWgNHOjx6EpcHgePTrXkG5lSduvtrVmt6shQV6acxS06fR8epUIbSEoQIczyxsG_cPn-mndH5-zg/s1600/New_Ge_Napoleon+theorem+associated+with+a+hexagon.png" height="368" width="400" /> </a></div>
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The theorem found and proved by Dr.Paul Yiu and me </div>
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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.</div>
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<i>Rõ ràng khi lục giác Suy biến thành một tam giác ta có định lý Napoleon.</i></div>
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-56407621645697148362015-02-26T03:54:00.000-08:002015-03-02T01:03:07.729-08:0099-Two conjectures in inequality<div dir="ltr" style="text-align: left;" trbidi="on">
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:<br />
<br />
\[\frac{\lambda_1f(x_1)+\lambda_2f(x_2)+....+\lambda_nf(x_n)}{f({\lambda_1x_1+\lambda_2x_2+....+\lambda_nx_n})} \geq \frac{\lambda_1f(y_1)+\lambda_2f(y_2)+....+\lambda_nf(y_n)}{f({\lambda_1y_1+\lambda_2y_2+....+\lambda_ny_n})}\]<br />
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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:<br />
<br />
$$\lambda_1f(x_1)+\lambda_2f(x_2)+....+\lambda_nf(x_n)-f({\lambda_1x_1+\lambda_2x_2+....+\lambda_nx_n})$$<br />
$$\geq \lambda_1f(y_1)+\lambda_2f(y_2)+....+\lambda_nf(y_n)-f({\lambda_1y_1+\lambda_2y_2+....+\lambda_ny_n})$$<br />
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-24040657475510888702015-02-11T07:28:00.003-08:002015-02-11T07:28:39.332-08:0098-Around Anticenter<div dir="ltr" style="text-align: left;" trbidi="on">
<b>Problem 1:</b> 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:<br />
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1-$FG$, and the line through $B$ perpendicular to $AG$, and the line through $C$ perpendicular to $DG$ are concurrent at $P_{bc}$.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5WiAZWM7L6AIfaCH3zWQcuaNCyTfnN2cmcUbsWZR8BDbF5LTJuU6xTXvKcWYH4QlG64nc0Ra6lB_-7x-HFYXF-kN1bZJ7hGK4QKA_VJd7p84o8OYnN6lX1e7HBEZ6-CCmaQfzX-WOPFk/s1600/2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5WiAZWM7L6AIfaCH3zWQcuaNCyTfnN2cmcUbsWZR8BDbF5LTJuU6xTXvKcWYH4QlG64nc0Ra6lB_-7x-HFYXF-kN1bZJ7hGK4QKA_VJd7p84o8OYnN6lX1e7HBEZ6-CCmaQfzX-WOPFk/s1600/2.png" height="490" width="640" /></a></div>
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2-If $ABCD$ be a cyclic quadrilateral, show that $P_{ab}P_{bc}P_{cd}P_{da}$ be a tangential quadrilateral<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVmncTHlVjdKJfFYUFvEIkQCvvuaJTMBoCCkSldhvaKsvDtJ6JZcNCrd-2CECZEL0VdD74B3z2_M3GJvFkrPgxS87f1-EslJdWwsG24uZhnO3v65d7qYSeAQMr4rspvGsUPIqi6CJH6iE/s1600/Tangential+quadrilateral.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVmncTHlVjdKJfFYUFvEIkQCvvuaJTMBoCCkSldhvaKsvDtJ6JZcNCrd-2CECZEL0VdD74B3z2_M3GJvFkrPgxS87f1-EslJdWwsG24uZhnO3v65d7qYSeAQMr4rspvGsUPIqi6CJH6iE/s1600/Tangential+quadrilateral.png" height="582" width="640" /></a></div>
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<b>Problem 2: </b>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.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8Nm_aJw6YnhEv3WylkZRtrKHNr3cssETslh8-PH-42sgHrw-Y6rG32oUh2eg0liLWqY38932y6bzISOmJXhufHWZw5sowfDpz7BjxTJCuHRjdOkt_4fkIKygthbA0pWMxu_1QwjE0yL0/s1600/3.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8Nm_aJw6YnhEv3WylkZRtrKHNr3cssETslh8-PH-42sgHrw-Y6rG32oUh2eg0liLWqY38932y6bzISOmJXhufHWZw5sowfDpz7BjxTJCuHRjdOkt_4fkIKygthbA0pWMxu_1QwjE0yL0/s1600/3.png" height="526" width="640" /></a></div>
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-14796327800190883992015-02-10T08:51:00.002-08:002015-02-11T07:36:23.229-08:0097-A generalization Zeeman-Gossard perpector<div dir="ltr" style="text-align: left;" trbidi="on">
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:<br />
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\[ \frac{\overline{D_aH_a}}{\overline{DaGa}}=\frac{\overline{D_bH_b}}{\overline{D_bG_b}}=\frac{\overline{D_cH_c}}{\overline{D_cG_c}} \space=\frac{\overline{DH}}{\overline{DG}}=t \]<br />
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Let $L_a,L_b,L_c$ = three lines through $D_a,D_b,D_c$ and parallel to $BC,CA,AB$ respectively.<br />
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<a href="http://tube.geogebra.org/student/m645553"><span style="color: red; font-size: x-large;">View in Geogebra</span></a><br />
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<b>Problem 1:</b> Then triangle bounded by three line $(L_a,L_b,L_c)$ are congruent<br />
and homothety to triangle ABC. The homothetic center lie on the line L.<br />
When $L$ is the Euler line or t = ∞ the problem is Zeeman-Gossard perpector<br />
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<b>Problem 2:</b> Newton lines of four quadrilateral $(AB,AC,L_a,L)$, $(BC,BA,L_b,L)$,<br />
$(CA,CB,L_c,L)$, $(AB,BC,CA,L)$ also through the homothetic center in problem 1<br />
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Where if $L_i$ $(i=1,2,...n)$ be a line, define $(L_1,L_2,....,L_n)$ = Polygon bound by $L_1,L_2,....,L_n$<br />
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-28299292943488017042015-02-10T08:48:00.000-08:002015-02-10T09:03:06.359-08:0096-Ten point circle associated X4240 in ETC<div dir="ltr" style="text-align: left;" trbidi="on">
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:<br />
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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.<br />
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Center of this circle is midpoint of $X(3)$ of $ABC$ and $O'$<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEil6wCSBLojKHkAvBdnPyUyU1tIA8YenO2Aishc6_go6GjdHtfSo-5EEegvMmEWxPIHRTA_lVeQtMqJfHAwSehs6dJB6APy5VwA2V9wuCfQG1tzBsrESFIwSQ3Kg064jUvmXRw0hKCmEEw/s1600/X4240.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEil6wCSBLojKHkAvBdnPyUyU1tIA8YenO2Aishc6_go6GjdHtfSo-5EEegvMmEWxPIHRTA_lVeQtMqJfHAwSehs6dJB6APy5VwA2V9wuCfQG1tzBsrESFIwSQ3Kg064jUvmXRw0hKCmEEw/s1600/X4240.png" height="640" width="614" /></a></div>
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-79575962338908640202015-01-31T08:25:00.000-08:002015-01-31T08:28:48.849-08:0095-Một bài toán về tính chất của đường thẳng đi qua tâm đường tròn ngoại tiếp<div dir="ltr" style="text-align: left;" trbidi="on">
<i><span lang="VI" style="font-size: 13.5pt; line-height: 20.7000007629395px;">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 $ABC$</span></i></div>
Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-13500527594167246952015-01-31T04:59:00.002-08:002015-01-31T05:00:25.532-08:0094-X(5463) and X(5464) in Kimberling center<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart4.html#X5463">X(5463) = REFLECTION OF X(13) IN X(2)</a><br />
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<span style="background-color: white;">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)</span><br />
<span style="background-color: white;"><br /></span>
<span style="background-color: white;"><a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart4.html#X5464">X(5464) = REFLECTION OF X(14) IN X(2)</a></span><br />
<span style="background-color: white;"><br /></span>
<span style="background-color: white;">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).</span><br />
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-25844810294073078392015-01-31T04:24:00.000-08:002015-01-31T05:00:12.657-08:0093-Four Jerabek Centers lie on a circle<div dir="ltr" style="text-align: left;" trbidi="on">
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Let $ABC$ be a triangle, let the Euler line of $ABC$ meets $BC,CA,AB$ at $A_1,B_1,C_1$ respectively.</div>
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Define $A_2=X_{125}$ of the triangle $AB_1C_1$. Define $B_2,C_2$ cyclically. </div>
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1-Then circumcenter of the triangle $A_2B_2C_2$ is Gossard perpector</div>
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2-Two triangle $A_2B_2C_2$ and ABC are similar and perpective, Which is the perpector?</div>
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3-$X(125)$ of ABC also lie on circumcenter of $A_2B_2C_2$ (Four point $X(125)$ lie on a circle)</div>
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4-Let $A_3B_3C_3$ be the paralogic triangle of ABC whose perpectrix is Euler line, then <span style="line-height: 1.25; word-spacing: normal;"> $</span><span id="yui_3_15_0_1_1422706721176_1037" style="line-height: 1.25; word-spacing: normal;">A_2B_2C_2$ perpective with</span><span id="yui_3_15_0_1_1422706721176_1603" style="line-height: 1.25; word-spacing: normal;"> $A_3B_3C_3$</span><span id="yui_3_15_0_1_1422706721176_1601" style="line-height: 1.25; word-spacing: normal;">, which is the perpector?</span></div>
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5-Circumcircle of $(A_3B_3C_3), (A_2B_2C_2)$ and $(ABC)$ concurrent at one point, which point?</div>
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-86181403576422901602015-01-29T04:49:00.003-08:002015-01-29T05:01:02.296-08:0092-Two new circle in a triangle<div dir="ltr" style="text-align: left;" trbidi="on">
<b>Lemma 1:</b> <i>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 $(POB)$.</i><br />
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<b>Lemma 2: </b>Symmendian point is Pole of Euler line respect to Kiepert hyperbola.<br />
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Application we can show that:<br /><br /><b>Theorem 3</b><i><b>:</b> In any triangle: The Circumcenter, the Nine point center, the Symmedian point and the Kiepert center lie on a circle</i><br />
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<b>Lemma 4:</b> <i>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 $(A'EM)$. </i><br />
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Application we can show that: <br /><br /><b>Theorem 5:</b> <i>In any triangle: The orthocenter, the Nine point center and Tarry point and midpoint of Brocard diameter lie on a circle. </i><br />
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<b>Lemma 6:</b> <i>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 $(FF'B)$. </i><br />
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Application we can show that: <br /><br />Lester circle theorem</div>
Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-69012156175070005112015-01-25T00:18:00.001-08:002015-01-30T00:01:23.324-08:0091-Dual problem of a generalization of Napoleon theorem<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="font-family: Verdana, Arial, Helvetica, sans-serif;"><span style="font-size: 11.5px; line-height: 22px;">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 $X(2)X(14)$)</span></span><br />
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-77414825696857275312015-01-24T04:54:00.002-08:002015-01-30T00:01:44.738-08:0090-A generalization Napoleon's theorem associated with the Kiepert hyperbola<div dir="ltr" style="text-align: left;" trbidi="on">
Very nice theorem: http://tube.geogebra.org/student/m542855<br />
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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.<br />
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<b>Lemma 1:</b>(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 $AO \perp PQ.$<br />
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<b>Telv Cohl's proof:</b><br />
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Let $ O' $ be the circumcenter of $ \triangle ACQ $ . Let $ M, N $ be the midpoint of $ CQ, CR $, respectively .<br />
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Easy to see $ R \in (O') $ .<br />
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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 $ ,<br />
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) $<br />
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From $ (1) $ and $ (2) $ we get $ \triangle AOO' \sim \triangle QPC $ , so from $ OO' \perp PC $ and $ AO' \perp QC \Longrightarrow AO \perp QP $ .<br />
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<b>Lemma 2:</b><br />
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Let $ D $ be a point out of $ \triangle ABC $ satisfy $ \angle DBC=\angle DCB=\theta $ .<br />
Let $ E $ be a point out of $ \triangle ABC $ satisfy $ \angle EAC=\angle ECA=90^{\circ}-\theta $ .<br />
Let $ F $ be a point out of $ \triangle ABC $ satisfy $ \angle FAB=\angle FBA=90^{\circ}-\theta $ . Then $ AD \perp EF $ .<br />
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<b>Proof of the lemma:</b><br />
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Let $ B' \in AF, C' \in AE $ satisfy $ AB=AB', AC=AC' $ and $ T=BC' \cap CB' $ .<br />
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Easy to see $ \triangle ABB' \cup F \sim \triangle ACC' \cup E \Longrightarrow EF \parallel B'C' $ .<br />
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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 $<br />
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From the lemma we get the following property about Kiepert triangle :<br />
The pedal triangle of the isogonal conjugate of $ K_{90-\phi} $ WRT $ \triangle ABC $ and the Kiepert triangle with angle $ \phi $ are homothetic .<br />
( Moreover, the homothety center of these two triangles is the Symmedian point of $ \triangle ABC $ ! )<b> (1)</b><br />
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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 $ )<br />
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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 $ .<br />
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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 <b>(1) </b>we get $ \triangle A_0B_0C_0 $ and the outer (or inner) Napoleon triangle are homothetic .<br />
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<b>Reference:</b><br />
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[1] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=622242<br />
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[2] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&t=621954<br />
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[3] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=148830</div>
Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-70695523357925186712015-01-21T19:54:00.002-08:002015-01-22T17:08:19.841-08:0089-A generalization of Parry circle<div dir="ltr" style="text-align: left;" trbidi="on">
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.<br />
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-63256526851800733482015-01-21T19:49:00.001-08:002015-01-22T17:08:43.651-08:0088-A generalization of Nine point circle<div dir="ltr" style="text-align: left;" trbidi="on">
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?<br />
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<b>Telv Cohl's proof:</b><br />
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Let $ A_B, A_C $ be the projection of $ A_0 $ on $ AB, AC $, respectively .<br />
Let $ B_C, B_A $ be the projection of $ B_0 $ on $ BC, BA $, respectively .<br />
Let $ C_A, C_B$ be the projection of $ C_0 $ on $ CA, CB $, respectively .<br />
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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.<br />
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<b>Luis González's proof:</b><br />
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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 $\varrho = \sqrt{\tfrac{1}{4}R^2+S \cdot \tan \theta}.$<br />
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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<br />
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${A_bA_c}^2=(XA_b+XA_c)^2={XA_b}^2+{XA_c}^2 +2 \cdot XA_b \cdot XA_c=$<br />
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$=4R^2-(AX^2+XU^2)+2 \cdot AX \cdot XU=AD^2-(AX-XU)^2=$<br />
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$=(AX+XU)^2+XM^2-(AX-XU)^2=XM^2+4 \cdot AX \cdot XU \Longrightarrow$<br />
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${NA_b}^2=NL^2+{LA_b}^2=NL^2+\tfrac{1}{4}XM^2+AX \cdot XU.$<br />
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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<br />
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${NA_b}^2=\tfrac{1}{2}HA \cdot HX+\tfrac{1}{4}OH^2+AX \cdot XU.$<br />
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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 $\varrho=\sqrt{\tfrac{1}{4}R^2+S \cdot \tan \theta}.$<br />
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-3781968354444459802015-01-21T10:21:00.001-08:002015-01-21T10:22:46.578-08:0087-Six center of Thebault circle lie on a conic<div dir="ltr" style="text-align: left;" trbidi="on">
<b>Problem:(Own) </b> 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.<br />
<a href="http://tube.geogebra.org/material/show/id/524977">Geogebra Check</a><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiuLBx-1WZ3mWWsTkaiz9JQ2YfSABsTNpg1wabKUSMXMjo2z49p8aIhizQkg1IwiYxFags3Xbn6Uylna1l06NXVDJ105tQ83r1S0yplRtM7DdVzUX_fcxOMWMeXQ60l72JvrcyAMyJyZ3g/s1600/1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiuLBx-1WZ3mWWsTkaiz9JQ2YfSABsTNpg1wabKUSMXMjo2z49p8aIhizQkg1IwiYxFags3Xbn6Uylna1l06NXVDJ105tQ83r1S0yplRtM7DdVzUX_fcxOMWMeXQ60l72JvrcyAMyJyZ3g/s1600/1.png" height="540" width="640" /></a></div>
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-15009461627427796022015-01-08T23:55:00.003-08:002015-01-08T23:55:46.970-08:0086-Rectangular hyperbola and Inscribed parabola of a triangle<div dir="ltr" style="text-align: left;" trbidi="on">
Please see:<br />
<br />
http://mathworld.wolfram.com/ChaslessPolarTriangleTheorem.html<br />
http://forumgeom.fau.edu/FG2004volume4/FG200427.pdf<br />
<br />
I proposed problem construction of a rectangular hyperbolar and a inscribed parabola as follows:<br />
<br />
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:<br />
<br />
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).<br />
<br />
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.)</div>
Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-85595119243353143112015-01-08T19:06:00.000-08:002015-01-08T19:11:09.712-08:0085-A pair equilateral triangle<div dir="ltr" style="text-align: left;" trbidi="on">
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:<br />
$\angle A0BP=\angle A0CP = 60^0$; define $B_0,C_0$ cyclically. Show that $A_0B_0C_0$ is a equilateral triangle (and A0B0C0 lie on cirumcircle).<br />
<br />
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<br />
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhwnAxnmd_fr7Pnsxn_R0_hDySg-6Joo89oQqY0tFfxxhAMYO4Brwfo_OruxaTAd8hdRoTUaarkSowmAE-aRCUzeUpZP4yIxhspg6IFYdnuqKor0q7RP5iy98Yg1X2bY2iT4LeA3cs8BJw/s1600/1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhwnAxnmd_fr7Pnsxn_R0_hDySg-6Joo89oQqY0tFfxxhAMYO4Brwfo_OruxaTAd8hdRoTUaarkSowmAE-aRCUzeUpZP4yIxhspg6IFYdnuqKor0q7RP5iy98Yg1X2bY2iT4LeA3cs8BJw/s1600/1.png" /></a></div>
<br />
<b>Telv Cohl's proof:</b><br />
<br />
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) $ .<br />
<br />
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 .</div>
Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-49936878061996758792014-12-31T19:05:00.003-08:002014-12-31T19:56:15.605-08:0084-New inequality in geometry<div dir="ltr" style="text-align: left;" trbidi="on">
<b>Theorem: </b>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: $BP_b+CP_c > AP_a$.<br />
<br />
I call result above is theorem because this theorem similarly: https://en.wikipedia.org/wiki/Pompeiu%27s_theorem<br />
<br />
<b>Leo Giugiuc's proof:</b><br />
<br />
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 $bc=1$.<br />
<br />
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 $z-\overline{z}=ki, k \in R$<br />
<br />
$2BP_b=|2b-c-1-z+c.\overline{z}|$ and $2CP_c=|2c-b-1-z+b.\overline{z}|$<br />
<br />
We have:<br />
<br />
$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)<br />
<br />
Since (1) and (2), the proof of theorem are complete.</div>
Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-73302160061021050712014-11-05T07:40:00.002-08:002014-11-06T19:16:38.165-08:0083-Six points lie on a circle associated Botema configuration<div dir="ltr" style="text-align: left;" trbidi="on">
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.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizbzezkIjgbOf-cueKmCj0cSPx7EGFOrUIZ1gdiI-vo9qGipoj7BcFfWrubBKMUJGyccfxGYsVtMpBPi-scS_yoieTbpi0-e4lh1Y9dq_1So0kRztFQA9_73PnZFS49YrcCCNZGny1GJA/s1600/Onthe+Vecten+configuration.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizbzezkIjgbOf-cueKmCj0cSPx7EGFOrUIZ1gdiI-vo9qGipoj7BcFfWrubBKMUJGyccfxGYsVtMpBPi-scS_yoieTbpi0-e4lh1Y9dq_1So0kRztFQA9_73PnZFS49YrcCCNZGny1GJA/s1600/Onthe+Vecten+configuration.png" height="408" width="640" /></a></div>
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-75099261660140202552014-11-05T06:57:00.003-08:002014-11-05T06:57:59.743-08:0082-A generalization Pythagoras' Theorem<div dir="ltr" style="text-align: left;" trbidi="on">
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDAS5VYPTLmhtXd5wpYCaeUNTwtYaOwfGSLvWu42eya6HSj2LEP2SI0WAroo76geNykv_XN4RxxAT1LUhFQxuNbsyZEpzgTi7uF9mN4tgLluU5HAQ-ZaAxbEimM41NDvXNU6_8uMVC_2M/s1600/Generalization+Pitago+theorem.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDAS5VYPTLmhtXd5wpYCaeUNTwtYaOwfGSLvWu42eya6HSj2LEP2SI0WAroo76geNykv_XN4RxxAT1LUhFQxuNbsyZEpzgTi7uF9mN4tgLluU5HAQ-ZaAxbEimM41NDvXNU6_8uMVC_2M/s1600/Generalization+Pitago+theorem.png" height="640" width="625" /></a></div>
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-65247988386690133632014-11-05T06:54:00.000-08:002014-11-05T06:54:31.361-08:0081-Two conic problem<div dir="ltr" style="text-align: left;" trbidi="on">
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<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfuhrCZ6Lm3H8ReZnbT7YOE1DizTsH4gU6_12J9osTRxJ13xKdlBQ1yLVAKHoOgViZ1EjquGD_sT9Veu_Mb9E6YQ84ZUy8vjMqNTi4DvcrAgEgggYlC4FqJZdq8z45WZYueJGpkjDKlXw/s1600/Two+Conics+Problems.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfuhrCZ6Lm3H8ReZnbT7YOE1DizTsH4gU6_12J9osTRxJ13xKdlBQ1yLVAKHoOgViZ1EjquGD_sT9Veu_Mb9E6YQ84ZUy8vjMqNTi4DvcrAgEgggYlC4FqJZdq8z45WZYueJGpkjDKlXw/s1600/Two+Conics+Problems.png" height="466" width="640" /></a></div>
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-83878795705343681212014-10-28T21:07:00.003-07:002015-09-01T23:16:05.277-07:00PUBLISH IN IN SOME JOURNALS<div dir="ltr" style="text-align: left;" trbidi="on">
<h2 style="background: none rgb(255, 255, 255); border-bottom-color: rgb(170, 170, 170); border-bottom-style: solid; border-bottom-width: 1px; font-family: 'Linux Libertine', Georgia, Times, serif; font-weight: normal; line-height: 1.3; margin: 1em 0px 0.25em; overflow: hidden; padding: 0px;">
<span style="background-color: white; color: red; font-family: sans-serif; font-size: large; font-weight: normal; line-height: 22.3999996185303px;"> I.In Forum Geometricorum Jounal</span></h2>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
<a href="http://forumgeom.fau.edu/" style="line-height: 22.3999996185303px;">Web site Forum Geometricorum</a></div>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
1- Đào Thanh Oai, Volum 14, Issue 10, <a href="http://forumgeom.fau.edu/FG2014volume14/FG201410.pdf">A simple proof of Gibert's generalization of the Lester circle theorem</a></div>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
2- Đào Thanh Oai, Volum 14, Issue 18, <a href="http://forumgeom.fau.edu/FG2014volume14/FG201418.pdf">Two pairs of Archimedean circles in the arbelos</a></div>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
3- Nikolaos Dergiades, Volum 14, Issue 24, <a href="http://forumgeom.fau.edu/FG2014volume14/FG201424.pdf">Dao's theorem on six circumcenters associated with a cyclic </a></div>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
4- Telv Cohl, Volum 14, Issue 29, <a href="http://forumgeom.fau.edu/FG2014volume14/FG201429.pdf">A Purely Synthetic Proof of Dao’s Theorem on Six Circumcenters Associated with a Cyclic Hexagon</a><br />
<br />
5-<a href="http://forumgeom.fau.edu/FG2015volume15/FG201509index.html">Dao Thanh Oai, Equilateral triangles and Kiepert perspectors in complex numbers, 105--114. </a><br />
<br />
6- Dao Thanh Oai and Paul Yiu, Some simple constructions of equilateral triangles associated with a triangle (<span class="short_text" id="result_box" lang="en"><span class="hps">accep</span></span>ted)<br />
<br />
7- Ngo Quang Duong, Two generalizations of the Simson line theorem (submited)<br />
<br />
8-Dao Thanh Oai, On the Jacobi Triangle (<span class="short_text" id="result_box" lang="en"><span class="hps">accep</span></span>ted)</div>
<h3 style="background: none rgb(255, 255, 255); border-bottom-style: none; font-family: sans-serif; line-height: 1.6; margin: 0.3em 0px 0px; overflow: hidden; padding-bottom: 0px; padding-top: 0.5em;">
<span class="mw-headline" id="T.E1.BA.A1p_ch.C3.AD_Crux-_T.E1.BA.A1p_ch.C3.AD_n.C3.A0y_c.E1.BB.A7a_h.E1.BB.99i_to.C3.A1n_h.E1.BB.8Dc_Canada" style="color: red; font-size: large; font-weight: normal;">II. In Crux Mathematicorum</span></h3>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
<a href="http://cms.math.ca/crux/">Web site Crux Mathematicorum</a></div>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
1- Problem 3845, Issue 5, Volum 39</div>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
2-<a href="https://cms.math.ca/crux/v39/n7/public_SolversProposers_39_7.pdf"> Problem 3869, Isue 7, Volum 39</a><br />
<span style="line-height: 22.3999996185303px;">3- </span><a href="https://cms.math.ca/crux/v39/n8/public_ProposersSolvers_39_8.pdf" style="line-height: 22.3999996185303px;">Problem 3878, Isue 8, Volum 39</a></div>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
<span style="line-height: 22.3999996185303px;">4- </span><a href="https://cms.math.ca/crux/v39/n9/public_SolversProposers_39_9.pdf" style="line-height: 22.3999996185303px;">Problem 3885, Isue 9, Volum 39</a><br />
5- <a href="https://cms.math.ca/crux/v39/n10/public_SolversProposers_39_10.pdf">Problem 3894(Dao Thanh Oai and Nguyen Minh Ha), Isue 10, Volum 39</a></div>
<h3 style="background: none rgb(255, 255, 255); border-bottom-style: none; font-family: sans-serif; line-height: 1.6; margin: 0.3em 0px 0px; overflow: hidden; padding-bottom: 0px; padding-top: 0.5em;">
<span style="background-color: white; font-family: sans-serif; font-weight: normal; line-height: 22.3999996185303px;"><span style="color: red; font-size: large;">III. In Global Journal of advanced research on classical and mordern geometries (Romania)</span></span></h3>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
<span style="line-height: 22.3999996185303px;"> </span><a href="http://gjarcmg.geometry-math-journal.ro/" style="line-height: 22.3999996185303px;">Web site GLOBAL JOURNAL OF ADVANCED RESEARCH ON CLASSICAL AND MODERN GEOMETRIES</a></div>
<div style="background-color: white; color: #252525; font-family: sans-serif; line-height: 22.3999996185303px; margin-bottom: 0.5em; margin-top: 0.5em;">
1- Trần Hoàng Sơn, Volum 3, Issue 2, <a href="http://gjarcmg.geometry-math-journal.ro/">A SYNTHETIC PROOF OF DAO’S GENERALIZATION OF GOORMAGHTIGH’S THEOREM</a><br />
<br />
2-<a href="http://diendantoanhoc.net/forum/index.php?app=core&module=attach&section=attach&attach_id=22603">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</a><br />
<br />
3-Nguyen Ngoc Giang, A proof of Dao's theorem (<span class="short_text" id="result_box" lang="en"><span class="hps">accep</span></span>ted)<br />
<span class="mw-headline" id="T.E1.BA.A1p_ch.C3.AD_Romania:_T.E1.BA.A1p_ch.C3.AD_th.E1.BB.A9_hai_b.C3.AAn_Romania" style="color: red; font-size: large; font-weight: normal;"> </span><br />
<span class="mw-headline" id="T.E1.BA.A1p_ch.C3.AD_Romania:_T.E1.BA.A1p_ch.C3.AD_th.E1.BB.A9_hai_b.C3.AAn_Romania" style="color: red; font-size: large; font-weight: normal;">IV. International jourbal of Geometry (Romania)</span></div>
<div>
Web site <a href="http://ijgeometry.com/">International jourbal of Geometry</a></div>
<div>
<span style="background-color: white; color: #252525; font-family: sans-serif; line-height: 19.7119998931885px; white-space: pre-wrap;"><br /></span></div>
<div>
<span style="background-color: white; color: #252525; font-family: sans-serif; line-height: 19.7119998931885px; white-space: pre-wrap;">Organized by the Department of Mathematics and Computer Science, Vasile Alecsandri National College of Bacau and Vasile Alecsandri University of Bacau,</span><br />
<span style="background-color: white; color: #252525; font-family: sans-serif; line-height: 19.7119998931885px; white-space: pre-wrap;"><br /></span><span style="background-color: white; color: #252525; font-family: sans-serif; line-height: 19.7119998931885px; white-space: pre-wrap;"><a href="http://ijgeometry.com/">Web site International Journal of Geometry</a></span></div>
<div>
<span class="mw-headline"><span style="background-color: white; color: #252525; font-family: sans-serif; line-height: 19.7119998931885px; white-space: pre-wrap;"><br /></span></span></div>
<div>
<span class="mw-headline"><span style="background-color: white; color: #252525; font-family: sans-serif; line-height: 19.7119998931885px; white-space: pre-wrap;">1- Telv Cohl, Volum 3, Isue 8, <a href="http://ijgeometry.com/wp-content/uploads/2014/10/8.pdf">DAO'S THEOREM ON CONCURRENCE OF THREE EULER LINES</a> </span></span><br />
<span class="mw-headline"><span style="background-color: white; color: #252525; font-family: sans-serif; line-height: 19.7119998931885px; white-space: pre-wrap;"><br /></span></span></div>
<span style="background-color: white; color: #252525; font-family: sans-serif; line-height: 19.7119998931885px; white-space: pre-wrap;">2- Đào Thanh Oai, Volum 3, </span><a href="http://ijgeometry.com/wp-content/uploads/2014/10/9.pdf" style="font-family: sans-serif; line-height: 19.7119998931885px; white-space: pre-wrap;">A SYNTHETIC PROOF OF A. MYAKISHEV'S GENERALIZATION OF VAN LAMOEN CIRCLE THEOREM AND AN APPLICATION </a><br />
<div>
<span class="mw-headline"><br /></span>
<br />
<h3 class="r" style="background-color: white; margin: 0px; overflow: hidden; padding: 0px; text-overflow: ellipsis;">
<span style="color: red; font-family: arial, sans-serif; font-size: large;"><span style="font-weight: normal; white-space: nowrap;">V. </span></span><span style="color: red; font-family: arial, sans-serif; font-size: large; font-weight: normal; white-space: nowrap;">ENCYCLOPEDIA OF TRIANGLE CENTERS</span></h3>
<br />
<span class="mw-headline" style="color: red; font-size: large;"><a href="http://faculty.evansville.edu/ck6/encyclopedia/ETC.html">Web site ENCYCLOPEDIA OF TRIANGLE CENTERS</a></span><br />
<br />
1- <a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html#X4240">X(4240) = DAO TWELVE EULER LINES POINT</a><br />
<br />
2- <a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart4.html#X5569">X(5569) = CENTER OF THE DAO 6-POINT CIRCLE</a><br />
<br />
3-<a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart4.html#X5607">X(5607) = CENTER OF 1st POHOATA-DAO-MOSES CIRCLE</a><br />
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4-<a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart4.html#X5608">X(5608) = CENTER OF 2nd POHOATA-DAO-MOSES CIRCLE</a><br />
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5-<span id="goog_1193319280"></span><span id="goog_1193319281"></span><a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart4.html#X6103">X(6103) = RADICAL CENTER OF THE DAO-MOSES-TELV CIRCLE, CIRCUMCIRCLE, AND NINE-POINT CIRCLE</a><a href="https://www.blogger.com/"></a><br />
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6-<a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart4.html#X6118">X(6118) = CENTER OF 1st DAO-VECTEN CIRCLE</a><br />
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7-<a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart4.html#X6119">X(6119) = CENTER OF 2nd DAO-VECTEN CIRCLE</a><br />
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8-<a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart4.html#X6188">X(6188) = DAO (a,b,c,R) PERSPECTOR</a><br />
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9-<a href="http://faculty.evansville.edu/ck6/encyclopedia/ETCPart5.html#X7668">DAO'S CONJUCTURE GENERALIZATION OF THE LESTER CIRCLE</a> <br />
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<h3 style="text-align: left;">
<span style="color: red; font-weight: normal;">VI. THE MATHEMATICAL GAZETTE</span></h3>
1-<a href="http://journals.cambridge.org/action/displayIssue?jid=MAG&volumeId=99&seriesId=0&issueId=544">Dao Thanh Oai, A family of Napoleon triangles associated with the Kiepert configuration, The Mathematical Gazette, Published online: 13 March 2015</a><br />
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2- Nguyen Le Phuoc, Nguyen Chuong Chi, A proof of Dao generalization of the Simson line theorem (<span class="short_text" id="result_box" lang="en"><span class="hps">accep</span></span>ted) <br />
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<h3 style="text-align: left;">
<span style="color: red; font-weight: normal;">VII-AMERICAN MATHEMATICAL MONTHLY</span></h3>
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<span style="color: red; font-weight: normal;"><br /></span></div>
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<span style="color: red;">1-<a href="http://www.jstor.org/discover/10.4169/amer.math.monthly.122.03.284?sid=21106386425863&uid=2&uid=3739320&uid=4">Problem 11830, The American Mathematical Monthly Vol. 122, No. 3 (March 2015), pp. 284-291 Published by: Mathematical Association of America</a></span><br />
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<span style="color: red;">VIII-SOMES ANOTHER NICE RESULT</span><br />
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<span style="color: red;">1-A generalization Gossard perspector theorem</span><br />
<span style="color: red;"><span style="color: black;"><br />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. </span></span><br />
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<b>Problem 1: </b>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 $P_1P_2$<br />
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<b>Problem 2: </b>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</div>
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<span style="color: red;"><span style="color: black;"><span style="color: blue;"><a href="http://www.artofproblemsolving.com/community/u192421h624341p5128577">Click to see in post AoPS</a></span> </span></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCXVn5_q4iH0nLhIup7LHmzIunTlh4dPuk2Q8u4uX1AOmehzydCyOqM1KIkMqURdzEcqnwFOTgVYtEFIcvBLvRHmBa8myLxeTsulGY8PAcXhtBkkr9-5Fx9cJVaJB29oqlGw5JVQoemhQ/s1600/Gerneralization%252BGossard%252Bperpector%252Bagain.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="416" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCXVn5_q4iH0nLhIup7LHmzIunTlh4dPuk2Q8u4uX1AOmehzydCyOqM1KIkMqURdzEcqnwFOTgVYtEFIcvBLvRHmBa8myLxeTsulGY8PAcXhtBkkr9-5Fx9cJVaJB29oqlGw5JVQoemhQ/s640/Gerneralization%252BGossard%252Bperpector%252Bagain.png" width="640" /></a></div>
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<span style="color: red;">2-A generalization of the Napoleon theorem associated with Kiepert hyperbola</span><br />
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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.<br />
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<a href="http://www.artofproblemsolving.com/community/u192421h622242p3721052">Please click to see a solution in AoPS</a><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4JBVaiyBza8Dsez3xpxRuwHOCnAR_zM1PDtMIcLTSg_gHXZE21rG_e6PaPhxFiCPzKRUvW7nnccv_atEzbnCC5oLTx1xiA2cpP6HaOupncYCwNkpcuR4qpjaXzwfgS5TMy1VKBwX4o2w/s1600/A+generalization+Napoleon+theorem.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="531" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4JBVaiyBza8Dsez3xpxRuwHOCnAR_zM1PDtMIcLTSg_gHXZE21rG_e6PaPhxFiCPzKRUvW7nnccv_atEzbnCC5oLTx1xiA2cpP6HaOupncYCwNkpcuR4qpjaXzwfgS5TMy1VKBwX4o2w/s640/A+generalization+Napoleon+theorem.png" width="640" /></a></div>
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<span style="color: red;">3-A generalization Simson line theorem 1</span><br />
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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.<br />
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<a href="http://www.artofproblemsolving.com/community/u192421h1075523p4696683">See two proof in AoPS</a> and <a href="https://xa.yimg.com/kq/groups/87514778/192082749/name/DAO.pdf">Click to get another proof in yahoo discustion Advanced Plane Geometry</a><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjgrk56VaI1_c7vNJxzII7zH82Drq7GxnUcr5W40fKAHhOi_wK0_W6vjkoLJ2W4bRsTN-YRGjpV5cQH6YlB8o5shm3rXLGu1CkZpM66ulrXbpbon0J7c46xzf2tYiEOZ5tuGuDdlsWVLPA/s1600/A_generalization_of_the_Simson_line.svg.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="483" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjgrk56VaI1_c7vNJxzII7zH82Drq7GxnUcr5W40fKAHhOi_wK0_W6vjkoLJ2W4bRsTN-YRGjpV5cQH6YlB8o5shm3rXLGu1CkZpM66ulrXbpbon0J7c46xzf2tYiEOZ5tuGuDdlsWVLPA/s640/A_generalization_of_the_Simson_line.svg.png" width="640" /></a></div>
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<span style="color: red;">4. 2nd Ageneralization Simson line theorem</span><br />
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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.<br />
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See two link in AoPS: <a href="http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47\&t=327661">Link 1 in AoPS</a> ; <a href="http://www.artofproblemsolving.com/community/u192421h560673p3264204">Link 2 in AoPS</a><br />
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<a href="http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=9834854&fulltextType=XX&fileId=S0025557215020549">See link in Geoff Smith's paper, publish in Mathematical Gazette</a><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9QEelUQgDHM3wu9PHTWRCHHACdoEVGSKm9Po57sN1a199pFHgaVqiOlpiITyBo40PoqK4xW9aVqekE-nZmQ92bzriY1muzrmBW7BSD773itOTrIvh2bDGQXM0a-F7EiVGMkdmXI3lm1Q/s1600/A_propjective_Simson_line.svg.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="392" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9QEelUQgDHM3wu9PHTWRCHHACdoEVGSKm9Po57sN1a199pFHgaVqiOlpiITyBo40PoqK4xW9aVqekE-nZmQ92bzriY1muzrmBW7BSD773itOTrIvh2bDGQXM0a-F7EiVGMkdmXI3lm1Q/s640/A_propjective_Simson_line.svg.png" width="640" /></a></div>
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<span style="color: red;">5-A generalization of Parry Circle</span><br />
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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 <b>Parry circle</b>. Furthemore K alway lie on the Jerabek hyperbola.<br />
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<a href="http://www.artofproblemsolving.com/community/c6h621979p3718488">Click to see post in AoPS</a><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgwvdJDzh5a4ctRDNA-GNFOesKnQ1G32rqThrIMPCy9GcFhlcMbakuUgIsCQV7igfoQlzHCquIgveIxSFh8diUFPTlC1VC1kPe4oErhuF8Qdvp_WqMKsw-dHLNZfFRpQI5eNQD4RhwT2u4/s1600/Parry+circle.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="410" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgwvdJDzh5a4ctRDNA-GNFOesKnQ1G32rqThrIMPCy9GcFhlcMbakuUgIsCQV7igfoQlzHCquIgveIxSFh8diUFPTlC1VC1kPe4oErhuF8Qdvp_WqMKsw-dHLNZfFRpQI5eNQD4RhwT2u4/s640/Parry+circle.png" width="640" /></a></div>
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<span style="color: red;"><br /></span><br />
<span style="color: red;">6. A generalization Steiner line and Miquel circle</span><br />
<span style="color: red;"><br /></span>
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.<br />
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<span style="color: red;">Generalization of the Steiner line: </span>Then show that: $A_1, B_1, C_1, P_1$ are collinear.<br />
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<span style="color: red;">A gneralization of Miquel circle:</span> 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.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgsjwYaGJlVmvrjKxP1jJ41eR5DUp4yW0viedsmdtlSJDcjS09qR-mLgLOKGiGYSVlVjnfmsOoE5tMJPIEgurjBchkBu2FTxyBlUIs1giX5WGhr9IVrriG1CNSXe0vXZBECBaDb8ry8-4c/s1600/Steiner+line+and+Miquel+circle.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="457" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgsjwYaGJlVmvrjKxP1jJ41eR5DUp4yW0viedsmdtlSJDcjS09qR-mLgLOKGiGYSVlVjnfmsOoE5tMJPIEgurjBchkBu2FTxyBlUIs1giX5WGhr9IVrriG1CNSXe0vXZBECBaDb8ry8-4c/s640/Steiner+line+and+Miquel+circle.png" width="640" /></a></div>
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http://tube.geogebra.org/material/show/id/1434055<br />
<span style="color: red;"><br /></span>
<span style="color: red;"><a href="http://www.artofproblemsolving.com/community/u192421h1119053p5137161">See post and two proof in AoPS</a></span><br />
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<span style="color: red;">7. A generalization specialcase of Brianchon theorem and Pascal theorem in one configuration</span><br />
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<span style="color: red;"><span style="color: black;">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.<br /><b> </b></span></span><br />
<span style="color: red;"><span style="color: black;"><b>1. (A generalization of Pascal theorem)</b> Then show that six points which they are intersection of<br />$(O_1)$, $(O_4)$; $(O_2)$, $(O_5)$; $(O_3)$, $(O_6)$ (if they are exist) lie on a circle.<br /><b> </b></span></span><br />
<span style="color: red;"><span style="color: black;"><b>2. (A generalization of Brianchon theorem) </b>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)</span><br /><br /><a href="http://tube.geogebra.org/m/1539785"><span style="color: black;">Applet in Geogebra website</span></a></span><br />
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<span style="color: red;"><span style="color: black;"> </span> </span><br />
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-91527458653607196242014-10-27T07:35:00.000-07:002014-10-29T02:25:54.693-07:0080-Another four conics theorem<div dir="ltr" style="text-align: left;" trbidi="on">
<b>Theorem:</b> 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.<br />
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<b>Special case:</b> Exist a conic touching with three given circles(The conic touching with one circle at two points)<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhQ-qQ0-9UCxzmIdkrkBO-mID9NffDhHKyO2e6KWzCiuBbIv3f6087rj5RN9xUz7cbExfX4CaK-n5eA0b0a7Om0sLJYXx-IUfGBq55h0FuWuN8AukuQXjje8SmaOBJaE9P_2dfHJ5M3Hs/s1600/Four+conic+theorem.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhhQ-qQ0-9UCxzmIdkrkBO-mID9NffDhHKyO2e6KWzCiuBbIv3f6087rj5RN9xUz7cbExfX4CaK-n5eA0b0a7Om0sLJYXx-IUfGBq55h0FuWuN8AukuQXjje8SmaOBJaE9P_2dfHJ5M3Hs/s1600/Four+conic+theorem.png" height="316" width="640" /></a></div>
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Anonymoushttp://www.blogger.com/profile/10779519291641232461noreply@blogger.com0tag:blogger.com,1999:blog-4272006829235706695.post-30834580577623061692014-09-29T09:09:00.000-07:002014-09-29T09:09:51.836-07:0079-On the Pascal line theorem<div class="separator" style="clear: both; text-align: center;">
<img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhno_sToLsdQs2omX8lxPLJEso5CIA3F0DHy6V-ycVWuBXSYU7vU2nWLOZ0CQp__qJTpjQHpAeHuRjaDP5C1iQpSA9HvhVvAiR83nCpX8rIuhkzTLIK2YnWWwfOQKPKSy8zCOemd3l68Cs/s1600/Untitled.png" /></div>
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<a href="http://www.cut-the-knot.org/Generalization/OverlookedPascal.shtml">Cite Course: Cut The Knot</a></div>
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