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

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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.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

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.

DeleteMake 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.

Delete