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.ạ 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.

    5. No need of Trigonometry to prove CD = 2a.

      Since AC = AF, triangles AKC and ABD are congruent ASA.
      Hence KC = AB = a.
      So CD = 2a

  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.

  4. Extend AB to meet DC at P and form the equlateral triangle APD
    Since given AD = 3a and AB = a
    we have BP = 2a, PC = a, CD = 2a and since m(BPC) = 60
    => BPC is a 30-60-90 triangle
    Hence BC = √3a

    Since m(BCP) = 90 => m(BCD) = 90 (Hence BC _|_ CD)
    Applying pythogrous to triangle BCD
    => BD^2 = BC^2+CD^2
    => BD^2 = 3a^2+4a^2
    => BD = √7a = AC

    Observe that the triangles PAC and ADB are congruent per SSA (since PA = AD, AC=BD and m(APC) = m(BAD) = 60)
    Let m(ADB)=m(PAC)= α
    => m(PCA) = 120-α => m(BCA) = 30-α
    and m(BDC) = 60-α => m(DBC) = 30+α (Corrected this from m(DBA) to m(DBC))
    ∴ in the triangle BEC, m(BEA) = 120
    => m(BEA) = 60
    So triangle EAB is similar to ADB
    ∴ AB/BE = BD/AB
    => BE = a^2/√7a
    Hence BE = a/√7

    Similarly AE/BE = AD/AB = a/3a = 1/3
    Hence AE = 3BE

    Finally, Area of Equilateral triangle APD = √3(9a^2)/4
    Area of right triangle PBC = √3(a^2)/2
    ∴ Area ABCD = Area APD - Area PBC = √3(a^2)[9/4-1/2] = 7√3(a^2)/4