## Wednesday, April 10, 2013

### Problem 869: Triangle, Median, Three Equilateral Triangles, Collinear Points, Midpoint

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

This entry contributed by Ajit Athle.
Click the figure below to see the complete problem 869. 1. AB=BD
ABM=a
MBC=60-a
DBE=a
BM=BE
=> ABM~=DBE
=> BED=b=AMB
=> BMC=180-b
DE=AM=MC
BF=BC
FBE=60-a
=> BFE~=BCM
=> ABE=180-b
ABE+BED=180-b+b=180 => DEF collinear
EF=MC=AM
EF=DE => E midpoint of DF

2. Let z(P) be the complex number representing P.
Let ω=exp(iπ/3).

Then
z(M) = 1/2[z(A) + z(C)]
z(D) = z(B) + ω[z(A) − z(B)] = (1−ω)z(B) + ω z(A)
z(E) = z(B) + ω[z(M) − z(B)] = (1−ω)z(B) + ω z(M)
z(F) = z(B) + ω[z(C) − z(B)] = (1−ω)z(B) + ω z(C)

Obviously
z(E) = 1/2[z(D) + z(F)]

Thus E is the mid-point of D,F, and so D,E,F collinear.

3. Consider the rotation anti-clockwise with angle 60° about B.
B→B
A→D
M→E
C→F

Thus BAMC→BDEF, so they are congruent.

Since A,M,C collinear with mid-point M,
thus D,E,F collinear with mid-point E.

4. Let us change the statement:

Let tr ABC any triangle and let us rotate it 60° counterclockwise with center B obtaining tr A'BC'.
Prove that tr MBM' is equilateral, where M is mid point of AC and M' is midpoint of A'C'.

This becomes almost trivial, indeed <MBM'=60° (rotation angle is preserved by each line and its image) and MB=M'B.