Thursday, September 26, 2013

Geometry Problem 928: Triangle, Midpoint, Median, Cevian, Concurrency, Area

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 928.

Online Geometry Problem 928: Triangle, Midpoint, Median, Cevian, Concurrency, Area

Posted by Antonio Gutierrez at 9:12 AM
Labels: , , , , ,

4 comments:

  1. Jacob HA (EMK2000)September 26, 2013 at 6:20 PM

    Denote the area of ΔABC by S(ABC).

    Obviously, S(AME)=S(CME). i.e. S5=S6.

    By Ceva's theorem, BF/AF=BD/CD. i.e. FD//AC.
    Thus, S(FAC)=S(DAC), which implies S1=S4.

    Lastly, S(BAM)=S(BCM), which implies S1=S4.

    Hence, S1+S3+S5 = S2+S4+S6 = S/2.

    ReplyDelete
  2. nakaSeptember 26, 2013 at 10:34 PM

    By affine transformation to any isosceles triangle, all the properties in the problem is preserved.
    After the transformation,
    S1=S4 ; S2=S3 ; S5=S6
    As affine preserved area ratio, the proof is completed.

    ReplyDelete
  3. Μιχάλης ΝάννοςOctober 9, 2013 at 8:37 AM

    http://www.mathematica.gr/forum/viewtopic.php?f=22&t=40193&p=187258#p187258

    ReplyDelete
  4. c.t.e.o.January 19, 2017 at 11:52 AM

    S6=S5 => h1=h2 (h1=AA', h2=CC' alt)
    => h3=h4 (h3=FF', h4=DD') because Right tr AA'B~ tr FF'B and BCC'~BFF'
    => S2=S3 => S1=S4

    ReplyDelete

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