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.
Tuesday, July 5, 2016
Geometry Problem 1230: Quadrilateral, 60 Degrees, Congruence, Metric Relations
Labels:
60 degrees,
congruence,
metric relations,
quadrilateral
Subscribe to:
Post Comments (Atom)
https://goo.gl/photos/TSKinFA2FzwTcMNU6
ReplyDeleteLet 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
https://goo.gl/photos/baBUwbeKwDLUEN1V9
Delete(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
There is no explanation why cd=2a
DeleteNote that
Delete- triangle AKD is equilateral
- FD= a
- triangle FAC is isosceles ( FA=BD=AC)
from above information we can calculate CD= 2a
Oops, you used trigonometry. Ah well.
DeleteNo need of Trigonometry to prove CD = 2a.
DeleteSince AC = AF, triangles AKC and ABD are congruent ASA.
Hence KC = AB = a.
So CD = 2a
Make F such that AFD is equilateral (true as angle BAD and ADC are 60 degrees).
ReplyDeleteMake 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)
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.
ReplyDeleteSo, 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.
whoops, at the end, I meant to say "AE has length of 3*a/sqrt(7)" which is really thrice of BE.
DeleteExtend AB to meet DC at P and form the equlateral triangle APD
ReplyDeleteSince 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