## 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.

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

Let AB meet DC at K
AKD is an equilateral triangle
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
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)
=> 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