Thursday, November 3, 2016

Geometry Problem 1282 Squares, Common Vertex, Midpoint, Perpendicular, 90 Degrees

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 1282.

Geometry Problem 1282 Three Squares, Common Vertex, Midpoint, Perpendicular, 90 Degrees.

3 comments:

  1. https://goo.gl/photos/QQWPcN1k1LiZB8Jt8

    Let N, Q, E, F, P and R are the projection of B, D, M, B1, B2 and D2 over CA1 ( see sketch)
    See sketch for positions of K, G , L and H.
    Let u= D2R=A1P
    v=B2P=A1R
    v1=BN=CQ
    u1=CN=DQ
    w=FB1=FA1=FC=FD1
    1. Since M is the midpoint of BB2=> E is the midpoint of NP
    By observation we have EF=MG=NF-1/2.NP= ½(u1-u)
    ME=1/2(v-v1)=GF
    GB1= w+1/2(v-v1)
    D2K=QR= v+2w-v1= 2.GB1
    DK=DQ-QK=u1-u=2.MG
    So MG/DK=GB1/D2K= ½
    2. Triangle MGB1 similar to DKD2 ( case SAS)
    So ∠ (MB1G)= ∠ (DD2K)=> D2,L, H, B1 are cocyclic
    And ∠ (B1HD2)= ∠ (B1LD2)= 90 degrees
    So B1M ⊥DD2

    ReplyDelete
    Replies
    1. Thanks Peter
      Another conclusion in your step 2:
      DD2 = 2(B1M)

      Delete
    2. BY CONGRUENCE D1D2=B2B1, BB1=DD1 AND D1D2 ┴ B2B1 , BB1 ┴ DD1
      Let X be the intersection of D1D2 AND B2B1,Y be the intersection of BB1 AND DD1
      THEN ( C, Y, B1, A1, X, D1 ):CONCYCLIC
      ∠(CYD1)=45=∠(XA1D1) => ∠(XD1A1)=∠(YB1C). SO,∠(BB1B2)=90 ISW ∠(DD1D2)=90

      Triangle BB1B2 ≡ DD1D2 ( SAS )
      M : triangle BB1B2 with the circumcenter
      So BB2=DD2=2*MB1


      Let W be the intersection of MB1 AND DD2
      ∠YDD2= ∠B1BB2= ∠MB1B THEN (D.Y.W.B1) CONCYCLIC => ∠DYB1=∠DWB1=90

      Delete