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.

## Wednesday, April 10, 2013

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

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AB=BD

ReplyDeleteABM=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

Let z(P) be the complex number representing P.

ReplyDeleteLet 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.

Consider the rotation anti-clockwise with angle 60° about B.

ReplyDeleteB→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.

Let us change the statement:

ReplyDeleteLet 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.