Saturday, July 31, 2010

Problem 488: Triangle, Cevian, Concurrency, Circle, Circumcircle

Geometry Problem
Click the figure below to see the complete problem 488 about Triangle, Cevian, Concurrency, Circle, Circumcircle.

Problem 488: Triangle, Cevian, Concurrency, Circle, Circumcircle
See also:
Complete Problem 488

Level: High School, SAT Prep, College geometry

2 comments:

  1. Let n=BA’ .BA” = BC’.BC” ( circumcircle A’B’C’ and secants BC’C” , BA”A’ )
    and m=AB’.AB”=AC’.AC” ; p=CA’.CA”=CB”.CB’ ( same reasons as above)
    1. Apply Ceva’s theorem for triangle ABC and point of concurrent D
    (A’B/A’C) * (B’C/B’A) *(C’A/C’B)=1

    2. Replace A’B=m/A”B ; A’C=p/A”C etc…. in above equation we have :
    (m/A”B)/(p/A”C) * (p/B”C)/(m/B”A) * ( m/C”A)/(n/C”B) =1
    3. Simplify above equation we get (A”C/A”B )*(B”A/B”C)*(C”B/C”A) =1
    Per Ceva’s theorem AA”, BB”, CC” are concurrent.

    Peter Tran
    vstran@yahoo.com

    ReplyDelete
  2. The concurrency point is called "the cyclocevian conjugate of P".

    ReplyDelete

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