Geometry Problem

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

Click the figure below to see the complete problem 1145.

## Tuesday, August 4, 2015

### Geometry Problem 1145: Three Squares, Midpoints

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

ReplyDeletez(C₁)—z(B₁) = i [z(A₁)—z(B₁)]

z(C₂)—z(B₂) = i [z(A₂)—z(B₂)]

summing up, we have

z(C)—z(B) = i [z(A)—z(B)]

Thus, AB=BC and ∠ABC=90°.

Similar argument for others.

Hence, ABCD is a square.

Let O is the center of spiral similarity to transform square A1B1C1D1 to A2B2C2D2

ReplyDeleteWe will have ∆A1OA2 ~ ∆C1OC2~ ∆B1OB2~ ∆D1OD2… (properties of spiral similarity transformation)

With medians OA, OB, OC, OD

Due to similarity of these triangles we also have

1. ∠ (A1OA)= ∠ (C1OC)= ∠ (B1OB)= ∠ (D1OD)=θ … (angles forms between corresponding sides to corresponding medians of similar triangles)

2. OA/OA1=OB/OB1=OC/OC1= OD/OD1=k …( ratio of corresponding medians to corresponding sides of similar triangles)

So square ABCD will be the image of square A1B1C1D1 in the spiral similarity transformation ( center O, angle of rotation= θ and factor of dilation = k)

Since A1B1C1D1 is a square so the image ABCD will be a square