Wednesday, February 4, 2009

Elearn Geometry Problem 244: Squares, Diagonals, Centers, Distance

Problem: Square, Diagonals, Centers, Distance

See complete Problem 244 at:
gogeometry.com/problem/p244_square_centers_diagonal_vertex.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 1:29 PM
Labels: , , , , ,

7 comments:

  1. AnonymousFebruary 8, 2009 at 9:46 AM

    Imagine initially that the smaller square had GC on BC, then, that it made a counter-clockwise rotation of θ , around C. Every point on the smaller square, including M and E, will describe an arc of circle of θ, with center in C.

    Now, look at triangles OMC and DEC. They are similar, because angle BCG = angle OCM = angle DCE = θ, and corresponds to the amount of the rotation done by the smaller square relatively to the bigger one. By the same token, angle COM = angle CDE = φ, and corresponds to the angle described by OM and DE, relatively to their initial positions OC and DC, respectively.

    Therefore, CD/CO = CE/CM = DE/OM = √2

    Then, DE = (√2)*OM, or b = (√2)*a q.e.d.

    ReplyDelete
  2. c .t . e. oJanuary 11, 2010 at 11:21 AM

    draw DH // a & DH = a
    join H to M and to E

    have to prove HE = a & ang DHE = 90

    tr COM and tr HME have
    1) HM = OC ( HM = OD )
    2) ang OCM = ang HME ( 90 - MCK & 90-FMH, CK perpe CO)
    3) CM = ME ( CM = ME, 1/2 diagonals)

    => an COM = MHE & HE = a
    => an DHE = DHM + MHE or DOP + COM = 90

    ReplyDelete
  3. AnonymousJuly 6, 2010 at 5:31 AM

    Hy, I don't understand where is point H. Help please.

    ReplyDelete
  4. c .t . e. oAugust 16, 2010 at 9:13 AM

    H is a point in a line // to OM and in a distance a
    from D
    We had to achieve in conclusion
    DEH is a right isoceles triangle
    in it from pythagore theorem a² + a² = b² or b = a√2

    ReplyDelete
  5. Sumith PeirisSeptember 28, 2019 at 5:59 AM

    <OCM = < DCE and CD/CO = CE/CM = sqrt2

    Hence Triangles COM and DCE are similar and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. rv.littlemanOctober 27, 2020 at 7:46 AM

    See the drawing

    Define H such as ΔHOB is the rotation of ΔMOC of -Π
    => OM ⊥ OH, OM=OH=a, MC ⊥ HB
    MC ⊥ ME, MC ⊥ HB => ME // HB
    Hence MEBH is a parallelogram
    => HM=BE=b
    => b is the hypothenuse of a rectangle triangle with side=a
    Therefore 2a^2=b^2

    ReplyDelete
  7. MarcoMarch 18, 2023 at 8:03 AM

    Let length of ABCD & CEGF be x & y respectively, also let <DCF=m
    By Pyth. Theorem,
    OC=(sqrt2)*x/2, CM=(sqrt2)*y/2

    Consider triangle OCM,
    a^2=OC^2+CM^2-2(OC)(CM)cos<OCM
    =x^2/2+y^2/2-xycos<(45+m)
    2a^2=x^2+y^2-2xycos(45+m)-------(1)

    Consider triangle CDE
    b^2=x^2+y^2-2xycos<DCE
    =x^2+y^2-2xycos(45+m)
    =2a^2 (by (1))
    So, b=a(sqrt2)

    ReplyDelete

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