Friday, July 8, 2016

Geometry Problem 1233: Triangle, Euler Line, Orthic Axis, Perpendicular, 90 Degrees, Orthic Triangle

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1233.


Geometry Problem 1233: Triangle, Euler Line, Orthic Axis, Perpendicular, 90 Degrees, Orthic Triangle

2 comments:

  1. https://goo.gl/photos/hm7ruYYeKsGei8867

    Draw 9 points circle of triangle ABC. The center N of this circle will be on Euler line NGH.
    This circle will pass through A1 , B1, C1, and midpoints of sides of triangle ABC.
    Draw new circle diameter ẠC
    From B2 draw a tangent line to circle O1 at D.
    Since( A,C, B1, B2)= -1 => DB1 ⊥ AC
    Power of B2 to circle O= B2C . B2A=BD^2
    But BD^2= B2B1. B2O1 (relationship in right triangle)= power of B2 to circle N
    So B2 is on the radial axis of circles N and Ọ
    Similarly C2 and A2 are on radial axis of circles N and O
    Radial axis A2C2B2 will perpendicular to line connected centers of circles O and N

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  2. https://goo.gl/photos/AJ3pAb2CxNFo9j26A
    New improved solution to problem 1233
    Draw nine- points circle of triangle ABC. The center N of this circle will be on Euler line NGH.
    This nine- points circle will pass through A1 , B1, C1, and midpoints of sides of triangle ABC.
    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)
    So B2 have the same power to both circles O or N => B2 will be on the radical axis of circles O and N.
    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
    Radical axis A2C2B2 will perpendicular to line connected centers of circles O and N

    ReplyDelete