tag:blogger.com,1999:blog-6933544261975483399.post2870573171928813708..comments2022-12-02T06:23:48.816-08:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1233: Triangle, Euler Line, Orthic Axis, Perpendicular, 90 Degrees, Orthic TriangleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-21716240283059755572016-07-08T23:44:45.535-07:002016-07-08T23:44:45.535-07:00https://goo.gl/photos/AJ3pAb2CxNFo9j26A
New improv...https://goo.gl/photos/AJ3pAb2CxNFo9j26A<br />New improved solution to problem 1233<br />Draw nine- points circle of triangle ABC. The center N of this circle will be on Euler line NGH.<br />This nine- points circle will pass through A1 , B1, C1, and midpoints of sides of triangle ABC.<br />Quadrilateral ACA1C1 is cyclic => B2C. B2A=B2A1. B2C1 (power of B2 to circle ACA1C1= power of B2 to circle O= power of B2 to circle N)<br />So B2 have the same power to both circles O or N => B2 will be on the radical axis of circles O and N.<br />Similarly ABA1B1 and BCB1C1 are cyclic and C2B.C2A=C2A1. C2B1 ,A2B.A2C= A2C1. A2B1 => C2 and A2 are on radical axis of circles N and O<br />Radical axis A2C2B2 will perpendicular to line connected centers of circles O and N<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83861307302043342472016-07-08T16:22:07.794-07:002016-07-08T16:22:07.794-07:00https://goo.gl/photos/hm7ruYYeKsGei8867
Draw 9 po...https://goo.gl/photos/hm7ruYYeKsGei8867<br /><br />Draw 9 points circle of triangle ABC. The center N of this circle will be on Euler line NGH.<br />This circle will pass through A1 , B1, C1, and midpoints of sides of triangle ABC.<br />Draw new circle diameter ẠC <br />From B2 draw a tangent line to circle O1 at D.<br />Since( A,C, B1, B2)= -1 => DB1 ⊥ AC <br />Power of B2 to circle O= B2C . B2A=BD^2<br />But BD^2= B2B1. B2O1 (relationship in right triangle)= power of B2 to circle N<br />So B2 is on the radial axis of circles N and Ọ<br />Similarly C2 and A2 are on radial axis of circles N and O<br />Radial axis A2C2B2 will perpendicular to line connected centers of circles O and N<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com