## Monday, October 1, 2012

### Problem 809: Trisecting a Line Segment AB, Three Circles, Radius, Center

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 809.

#### 5 comments:

1. Let CD cut AB at I and BC cut DE at J
Note that ABD, ABC and CEF are equilateral triangles
We have DB//AC//CE => DBEC is a parallelogram
And J is the midpoint of BC
In triangle DBC, DJ and BI are medians => N is the centroid of DBC
So BN=2.IN
Due to symmetric we have AM=2.IN=MN=NB

2. Since FC//CB//AD, and FC = CB = AD,
so FADC is a parallelogram,
thus, FD and AC bisect each other.

Join AD and DC. Consider ΔADC,
DM bisects AC, and AB bisects DC.
So M is the centroid of ΔADC.

Let AB meets DC at H.
Then AM:MH = 2:1.

Similarly, HN:NB = 1:2.

Since MH = HN, so AM:MN:NB = 1:1:1.
As a result, AM = MN = NB.

3. Using vector to solve the problem.
Let vector(AC) = u, vector(BC) = v.
Then vector(DE) = 2u+v, vector(BA) = v-u.
Now let
vector(DN) = p*vector(DE) =p*(2u+v);
vector(BN) = q*vector(BA) = q*(v-u).

Since
vector(DN) = vector(DB) + vector (BN)
=> p*(2u+v) = u + q*(v-u)
=> p = q = 1/3.

Therefore BN = (1/3)BA
Symmetrical on the other side, q.e.d.

4. Since Tr.s AMD and BMF are similar BM = 2AM. Similarly AN = 2BN and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

5. Problem 809
Is AB=AC=AD=BD=CE=CF=BC then <ABE=90.AD intersects EB at K then <DBK=30=<DKB.
So DB=DK. But ΑD=DK and EB=BK so point N is centroid of triangle AEK.Then NB=AB/3.
Similar AM=AB/3.Therefore AM=MN=NB.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE