## Wednesday, November 6, 2019

### Dynamic Geometry Problem 1444: The Asymmetric Propeller Theorem, Equilateral Triangles

Geometry Problem, GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. #### 1 comment:

1. Lemma 1:
Assume that It exist a position of 1 red and 3 green equilateral triangles such that triangle M1M2M3 is equilateral .We will prove that If one green triangle BB1B2 is modified by rotational and dilation to become BB1’B2’ and M1, M3 become M1’, M3’ , then new triangle M2M3’M1’ is also equilateral.
See link below for sketch of problem 1444 #1
https://photos.app.goo.gl/EfQAEFrPTZ32iHZe9
Triangle BB2B2’ congruent to BB1B1’ ( case SAS)
So B1B1’=B2B2’ and angle(B2B2’, B1B1’)= 60 degrees
And M1M1’=M3M3’=1/2 B1B1’ and angle(M1’M1, M3’M3)= 60
Quad M1DM2M3 is cyclic => angle(M1D, M1M2)=angle(M3M2, M3D)
So triangle M2M1M1’ congruent to M2M3M3’ => triangle M2M1’M3’ is equilateral
Complete the prove of Lemma 1
Back to the original problem 1444,
We start with 3 green congruent equilateral per sketch 2 below . Due to symmetry, triangle M1M2M3 is equilateral
https://photos.app.goo.gl/fKKvVH6ALd76WbbD6
Now we modify one green triangle at a time. In each case ,triangle M1M2M3 is always equilateral
So M1M2M3 is always equilateral for any positions of equilateral Triangles AA1A2, BB1B2, CC1C2