Tuesday, July 5, 2016

Geometry Problem 1230: Quadrilateral, 60 Degrees, Congruence, Metric Relations

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1230.

Geometry Problem 1230: Quadrilateral, 60 Degrees, Congruence, Metric Relations


  1. https://goo.gl/photos/TSKinFA2FzwTcMNU6

    Let AB meet DC at K
    AKD is an equilateral triangle
    Draw BF // AD and let KE meet AD at X
    We have KB=2.a and CD=2.a
    We have BD^2= a^2+9a^2- 2.3a.แบก cos(60)= 7.a^2 => BD= a√7
    Apply Ceva’s theorem in triangle AKD we have
    BA/BK x CK/CD x XD/XA = 1 => XD/XA= 4
    We also have ED/EB= XD/XA + CD/CK= 4+ 2= 6
    So ED/6=EB/1=BD/7 => EB= BD/7= a/√7

    1. https://goo.gl/photos/baBUwbeKwDLUEN1V9

      (1) KBF is equilateral and AC=AF=BF= a√7
      Apply Van Aubel theorem for point E in triangle KAD
      AE/EC=BA/BK+XA/XD= ¾
      So AE/3=EC/4=AC/7=a/√7
      And AE= 3.a/√7 => BE/AE=1/3
      (2) .. In equilateral triangle KBF, the median BC perpendicular to KF
      (3)…BC= KB. √3/2= a√3
      (4).. S( ABCD)=S(KAD)- S(KBC)
      S(KAD)= AD^2. √3/4
      S(KBC)=SB^2. √3/8
      Replace AD= 3.a and SB=2.a we get S(ABCD)=7/4. a^2. √3

    2. There is no explanation why cd=2a

    3. Note that
      - triangle AKD is equilateral
      - FD= a
      - triangle FAC is isosceles ( FA=BD=AC)
      from above information we can calculate CD= 2a

    4. Oops, you used trigonometry. Ah well.

  2. Make F such that AFD is equilateral (true as angle BAD and ADC are 60 degrees).
    Make H on FD such that AH = BD, BH is parallel to AD. (Feasible as triangle AFD is equilateral which are symmetrical about their altitudes).
    Make G on FD such that AG is perpendicular to FD. As AFD and ACH show properties of isosceles triangles, G is bisector of CH and FD.
    Hence, CD = 2a.

    Mass points: Let E be the center of mass if fixed masses are placed on A, F, D (lines are frames which are strong and massless).
    Let D have mass 1. As E is center of mass, B (which is concurrent with E and D) must be center of mass of AF. Similarly, C is center of mass of FD.
    As CD = 2a, FC must be a, so the mass on F must be 2.
    As AB = a, BF must be 2a and therefore the mass on A is 4.
    So the total mass of AF is 6. This means that mass ratio of B:D is 6:1, so the length ratio of BE:ED = 6:1
    This indicates that BE has 1/7 of length of BD.

    BD = AC, AC^2 = AG^2 + CG^2 = AD^2 - GD^2 + CG^2 = 7*a^2
    Hence BD = sqrt(7) a
    That means that BE = a/sqrt(7)

  3. I said in an earlier post that F has mass 3, D is mass 1 and A is mass 3. This indicates that FD has total mass of 3.
    So, mass ratio A:C = 4:3, which means that AE:EC = 3:4
    Hence, AE is 3/7 of AC.

    I also found in that other post that AC is sqrt(7) a. This means at AE has length of 3*sqrt(7)*a, which is thrice of BE.

    1. whoops, at the end, I meant to say "AE has length of 3*a/sqrt(7)" which is really thrice of BE.