Monday, October 15, 2012

Problem 814: Two Equilateral Triangles with a common Vertex, Angle

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

Click the figure below to see the complete problem 814.

Online Geometry Problem 814: Two Equilateral Triangles with a common Vertex, Angle

Posted by Antonio Gutierrez at 8:48 AM
Labels: , , , ,

8 comments:

  1. W FungOctober 15, 2012 at 9:27 AM

    Observe that triangle BCF is congruent to triangle BAD by rotation.
    Let angle BCF = x, then angle BAD = x,
    Angle AGF = angle ACF - angle CAG = (60+x)-(x-60)=120

    ReplyDelete
  2. Bogdan RusuOctober 15, 2012 at 4:24 PM

    @W Fung: The proof is forced. We do not know anything about angles FBC and DBA or sides (CF) and (AD) so that we can say that triangle (BAD) is obtained from the triangle (BCF) using rotation.

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    Replies
    1. W FungOctober 15, 2012 at 5:59 PM

      To Rusu:

      I skipped the short proof of congruence as i believe it is too obvious.

      For your information,

      BA = BC ; BD = BF ; angle ABD = 60 - angle DBC = angle FBC
      So by S-A-S, triangle ABD and CBF are congruent. Hence the rotation at point B exists.

      Delete
  3. Jacob HA (EMK2000)October 15, 2012 at 6:05 PM

    Rotation clockwise 60° about B.

    B→B
    C→A
    F→D
    ΔBCF→ΔBAD

    ∴ΔBCF≡ΔBAD

    ---
    ΔBCF≡ΔBAD
    ⇒ ∠BCF = ∠BAD
    ⇒ ∠BCF = ∠BAG
    ⇒ B,A,G,C concyclic
    ⇒ ∠AGC = 180° - ∠ABC = 120°
    ⇒ ∠AGF = 120°

    ReplyDelete
  4. AnonymousDecember 29, 2012 at 5:04 PM

    solution to the problem 814( without rotation), by Michael Tsourakakis from Greece

    angle x+angle ω=60 and angle y+angle ω=60, so,angle x=angle y
    The triangles ABD, BCF ,are equal ,why have, AB=BC ,BD=BF and angle x=angle y Therefore,angle BDA=angle BFC .So, the quadrilateral BCDF ,is , cyclic quadrilateral. Therefore,angleFGD=angleDBF=60. So, angle AGF=120
    Image: http://img4.imageshack.us/img4/4129/56463515.png

    ReplyDelete
  5. Μιχάλης ΝάννοςOctober 10, 2013 at 6:51 AM

    http://www.mathematica.gr/forum/viewtopic.php?f=20&t=31981&p=148019

    ReplyDelete
  6. Sumith PeirisJanuary 29, 2016 at 8:55 AM

    < ABD = < CBF so Tr.s ABD and CBF are congruent SAS

    Hence < ADB = < BFG .= say @

    So < AGF = < ADF + DFG = @+60 + 60-@ = 120

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Sumith PeirisJanuary 29, 2016 at 9:01 AM

    Further it follows that BGDF and ABCG are cyclic

    ReplyDelete

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