## Tuesday, January 28, 2014

### Geometry Problem 968: Equilateral Triangle, Rectangle, Common Vertex, Midpoint

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

Click the figure below to see the complete problem 968.

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

Let z(A)=0, z(F)=a, z(C)=a-bi
Then
z(B)=(1/2 a + √3/2 b)+(√3/2 a - 1/2 b)i
z(E)=a+(√3/2 a - 1/2 b)i
z(M)=(1/4 a + √3/4 b)+(√3/4 a - 1/4 b)i

Easy to check that
z(F) + (-1/2 + √3/2 i) z(E) + (-1/2 - √3/2 i) z(M) = 0

Hence, EMF is equilateral triangle.

2. Connect CM
Note that CM ⊥BM => MBEC is cocyclic
Since M is midpoint of AB => triangle EMF is isosceles
In circle MBEC , ∠MEC=∠MBC=60 => EMF is equilateral

3. CM is perpendicular to AB hence MBEC is con cyclic, so < MEC = < MBC = 60

Also ACFM is con cyclic so < EFM = < MAC = 60

So 2 angles of Tr. FEM have been proved to be 60 hence this triangle must be equilateral

Sumith Peiris
Moratuwa
Sri Lanka

4. AMFC and BMCE are concyclic. Thus follows the results