Geometry Problem

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

Click the figure below to see the complete problem 810.

## Monday, October 1, 2012

### Problem 810: Trisecting a Line Segment AB, Two Circles, Radius, Center. Diameter, Chord

Labels:
circle,
radius,
segment,
trisection

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Let CD cut AB at I

ReplyDeleteNote that ABD, ABC and CBF are equilateral triangles

And G is the midpoint of BC and AF

In triangle DBC, DG 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 DE is diameter with center A,

ReplyDeletewe have EA = AD.

Since CADB is a rhombus, AC//DB,

hence, EH = HB.

Now in ΔEDB, A and H are mid-points,

which means BA and DH are medians.

Therefore, M is the centroid of ΔEAB.

Thus, AM:MB = 1:2.

Similarly, N is the centroid of ΔFAD.

We also have AN:NB = 2:1.

As a result, AM = MN = NB.

Typo: Therefore, M is the centroid of ΔEDB.

DeleteObserve that triangle CAG, triangle AGB, triangle FGB are congruent and they are 30-60-90 triangles. (proof is skipped)

ReplyDeleteHence AF is the perpendicular bisector of BC.

Now insert coordinate system with the midpoint of AB be O(0,0), and auxillarily let B(1,0).

Then D = (0,-sqrt3) ; C = (0,sqrt3) ; G is the midpoint hence = (1/2, (sqrt3)/2)

Line DG : y + sqrt3 = 3*sqrt3*x

When DG cuts AB, y = 0, x = 1/3. => N(1/3,0)

Hence ON is one-thrid of OB.

Symmetrically on the other side,

q.e.d.

This is a proof using the result in problem 809.

ReplyDelete*(using the points in problem 809)

Let the point where BC cuts DE named P.

Then since BC = CE, and CE//BD,

trianglePCE is congruent to trianglePBD.

Hence PB=PC.

So point P is acutally the point G in this problem.

First ABC is an equilateral triangle and G is the midpoint of CB, H is the midpoint of CA.

ReplyDeleteLet's call P the point where HG(extended) cuts BF.

. HG=GP (because CBF is an equilateral triangle)

. therefore using TALETE (HG || AB) follows that MN=NB and so MN=NB=AM (by simmetry)

AB = 2AH = BD and since Tr.s AHM and BDM are similar BM = 2AM

ReplyDeleteSimilarly AN = 2BN and the result follows

Sumith Peiris

Moratuwa

Sri Lanka