Saturday, June 11, 2011

Online Geometry Problem 620: Three Rectangles, Diagonals, Centers, Angles

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

Click the figure below to see the complete problem 620.

 Online Geometry Problem 620: Three Rectangles, Diagonals, Centers, Angles

4 comments:

  1. http://img830.imageshack.us/img830/474/problem620.png
    we have KL//FC and KJ//BI
    let FC cut BI at J
    We have angle (IEA)= angle (DEF)=35 both angles complement to angle AED
    AIBE and ECFD are cyclic quadrilaterals
    so angle (DCF)= angle (DEF)= 35 ( see picture)
    and angle (IBA)= angle (AEI)=35
    so x=180-55-55= 70
    Peter Tran

    ReplyDelete
  2. Comment(See Problem 620)
    More generally,
    Let ∆PQR be the orthocentric triangle of ∆EAD
    (formed by the feet P,Q,R of the perpendiculars from E,A,D upon the opposite sides AD,DE,AE respectively).
    Then ∆KJL is similar to ∆PQR with
    ∠JKL = 180°-2∠AED,
    ∠KJL = 180°-2∠EAD,
    ∠JLK = 180°-2∠ADE.

    ReplyDelete
  3. Pravin


    It is not clear to me about statement
    "Then ∆KJL is similar to ∆PQR with
    ∠JKL = 180°-2∠AED, "
    Please explain.

    Peter Tran

    ReplyDelete
  4. Let angle AED = y. Then your proof will be as follows:
    "we have KL//FC and KJ//BI
    let FC cut BI at M (You named it as J)
    We have angle (IEA)= angle (DEF)=90 - y both angles complement to angle AED
    AIBE and ECFD are cyclic quadrilaterals
    so angle (DCF)= angle (DEF)= 90 - y ( see picture)
    and angle (IBA)= angle (AEI)=90 - y
    So angle MBC=angle MCB=y
    so x=180-y-y= 180-2y"

    Now it is wellknown that in any triangle ABC,
    if D,E,F are the feet of the perpendiculars from A,B,C on the opposite sides, and triangle DEF is formed, then its angles are given by
    D=180-2A,etc

    In our problem we have triangle EAD and its orthic triangle is named by me as PQR
    So angle QPR = 180-2AED = angle JKL etc

    ReplyDelete