## Monday, October 1, 2012

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

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

Click the figure below to see the complete problem 810. 1. Let CD cut AB at I
Note 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

2. Since DE is diameter with center A,

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.

1. Typo: Therefore, M is the centroid of ΔEDB.

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

Hence 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.

4. This is a proof using the result in problem 809.
*(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.

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

Let'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)

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

Similarly AN = 2BN and the result follows

Sumith Peiris
Moratuwa
Sri Lanka