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
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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