Connect Midpoints of Sides of Squares to Get Another Square

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Source: ILL 1970 - Problem 11.

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#1 h May 24, 2011, 3:10 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ and $A'B'C'D'$ be two arbitrary squares in the plane that are oriented in the same direction. Prove that the quadrilateral formed by the midpoints of $AA',BB',CC',DD'$ is a square.

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#2 h May 24, 2011, 3:24 AM • 2 Y

Y by Adventure10, Mango247

Rijul saini wrote:

Take two squares $ABCD$ and $PQRS$ in the plane. Prove that the quadrangle formed by the midpoints of $AP,BQ,CR,DS$ is itself a square. (Solutions with complex numbers are discouraged).
Extra: Prove or disprove it for two squares in space. (Frankly, I don't know, but it's interesting.)

KrazyFK wrote:

Can I just bash it with vectors?

Let $A,B,C,D,P,Q,R,S$ have vector positions denoted by the same letters. Let $W,X,Y,Z$ be the midpoints in question.

Then we have:

$X-W = \frac{B+Q-A-P}{2} = \frac{(B-A)+(Q-P)}{2}$

$Z-Y = \frac{D+S-C-R}{2} = \frac{(D-C) + (S-R)}{2}$

But since $ABCD$ and $PQRS$ are squares, we have $B-A = D-C$ and $Q-P = S-R$ , so from the above we get $X-W = Z-Y$ .

Similarly, $Y-X = W-Z$ , so $WXYZ$ is a rhombus. It remains to show that it has a right angle, for which we need to show that:

$(X-W) \cdot (Y-X) = 0$

$\Longleftrightarrow ((B-A)+(Q-P)) \cdot ((C-B)+(R-Q)) = 0$

$\Longleftrightarrow (B-A) \cdot (C-B) + (B-A) \cdot (R-Q) + (Q-P) \cdot (C-B) + (Q-P) \cdot (R-Q) = 0$

$\Longleftrightarrow (B-A) \cdot (R-Q) + (Q-P) \cdot (C-B) = 0$

$\Longleftrightarrow |B-A||R-Q| \cos \theta + |Q-P||C-B| \cos \phi = 0$ (with \theta and \phi being the obvious angles)

But we have that $\cos \theta = - \cos \phi$ and so this is obviously true.

Rijul saini wrote:

Nice Job KrazyFK! I think Your Solution works for the case of the squares being in space.
Here's my solution for the plane case.
Let $W,X,Y,Z$ be the midpoints of $AP,BQ,CR,DS$ respectively.
Consider the composition of rotations
$\text{Rot}(Q,-90^{\circ}) \ \circ \ \text{Rot}(Y,180^{\circ}) \ \circ \ \text{Rot}(B,90^{\circ}) \ \circ \ \text{Rot}(W,180^{\circ})$
Since the angles add up to $360^{\circ}$ , therefore, the composition must be a translation. But this translation keeps $A$ fixed ( $A \rightarrow P \rightarrow R \rightarrow C \rightarrow A$ ), so it is the identity transformation. Therefore, we have,
$\text{Rot}(B,90^{\circ}) \ \circ \ \text{Rot}(W,180^{\circ}) = \text{Rot}(Y,180^{\circ}) \ \circ \ \text{Rot}(Q,90^{\circ})$
Let both of these be equal to $\text{Rot}(K,270^{\circ})$ . This implies that $\triangle BWK$ and $\triangle QYK$ are both isosceles right angled triangles, right angled at $W,Y$ respectively. Now, consider the composition of rotations
$\text{Rot}(Y,90^{\circ}) \ \circ \ \text{Rot}(W,90^{\circ}) \ \circ \ \text{Rot}(X,180^{\circ})$
Again, this is easily seen to be a translation, as the sum of the angles is $360^{\circ}$ , which keeps $Q$ fixed ( $Q \rightarrow K \rightarrow B \rightarrow Q$ ), and hence is the identity transformation. Therefore,
$\text{Rot}(Y,90^{\circ}) \ \circ \ \text{Rot}(W,90^{\circ}) = \text{Rot}(X,180^{\circ})$
And incidentally, this implies that $\triangle YXW$ to be an isosceles right angled triangle, right angled at $X$ . Similarly, getting $\triangle YZW$ also to be isosceles right triangle, right angled at $Z$ , completes the proof.

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ftong

#3 h May 24, 2011, 11:38 PM • 2 Y

Y by Adventure10, Mango247

This is also a consequence of the "Mean Geometry Theorem," which states that if polygons $A = A_1 A_2 \cdots A_n$ and $B = B_1 B_2 \cdots B_n$ are directly similar,
then the polygon whose vertices are the midpoints of $A_i B_i$ is directly similar to both $A$ and $B$ .

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#4 h May 25, 2011, 12:52 AM • 1 Y

Y by Adventure10

Quote:

Proposed Problem. Let $ABCD$ and $A'B'C'D'$ be two arbitrary parallelograms in the plane that are oriented in the
same direction. Prove that the quadrilateral formed by the midpoints of $[AA']$ , $[BB']$ , $[CC']$ , $[DD']$ is a parallelogram.

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#5 h May 25, 2011, 6:05 PM • 1 Y

Y by Adventure10

ftong wrote:

This is also a consequence of the "Mean Geometry Theorem," which states that if polygons $A = A_1 A_2 \cdots A_n$ and $B = B_1 B_2 \cdots B_n$ are directly similar,
then the polygon whose vertices are the midpoints of $A_i B_i$ is directly similar to both $A$ and $B$ .

But the two squares may not be directly similar. "Oriented in the same direction" just means the the sets of vertices run in the same order - both clockwise or anticlockwise.

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#6 h May 25, 2011, 6:17 PM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula wrote:

Quote:

Proposed Problem. Let $ABCD$ and $A'B'C'D'$ be two arbitrary parallelograms in the plane that are oriented in the
same direction. Prove that the quadrilateral formed by the midpoints of $[AA']$ , $[BB']$ , $[CC']$ , $[DD']$ is a parallelogram.

Let $A,B,C,D,A',B',C',D'$ denote the vector positions of the obvious points.

Let $P,Q,R,S$ be the midpoints of $AA', BB', CC', DD'$ respectively.

Then $P-Q = \frac{A+A'-B-B'}{2} = \frac{(A-B) + (A'- B')}{2}$

And also $R-S = \frac{C+C'-D-D'}{2} = \frac{(C-D) + (C'-D')}{2}$

But since $ABCD$ and $A'B'C'D'$ are parallelograms, we have $A-B = D-C$ and $A'-B'= D'-C'$ , so from the above we get:

$P-Q = S-R$

And so $PQRS$ is a parallelogram.

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