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

AMFC and BMCE are concyclic. Thus follows the results

ReplyDelete