Geometry Problem

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

Click the figure below to see the complete problem 809.

## Monday, October 1, 2012

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

Labels:
circle,
segment,
trisection

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Let CD cut AB at I and BC cut DE at J

ReplyDeleteNote 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

Since FC//CB//AD, and FC = CB = AD,

ReplyDeleteso 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.

Using vector to solve the problem.

ReplyDeleteLet 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.

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

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Problem 809

ReplyDeleteIs 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