Geometry Problem

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

Click the figure below to see the complete problem 968.

## Tuesday, January 28, 2014

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

Labels:
common vertex,
equilateral,
midpoint,
rectangle,
triangle

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

ReplyDeleteLet 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.

Connect CM

ReplyDeleteNote 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

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

ReplyDeleteAlso 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