Problem 1 - Complex numbers from A to...Z

Question

On each side of a parallelogram a square is drawn external to the figure. Prove that the centers of the squares are the vertices of another square.

Step-by-Step Solution

Answer

Question: Prove that when a square is drawn on each side of a parallelogram, the centers of these squares form another square. Answer: In this problem, we showed that when squares are drawn on each side of a parallelogram, the centers of these squares form another square by calculating the side lengths and diagonals of the square formed by the centers. We found that the side lengths are congruent (AB' = CD' and BC' = DA') and the diagonals are congruent (A'C' = B'D'), proving that they indeed form a square.

Step 1: Label the points and set the coordinate system

We can label the vertices of the parallelogram as A (0,0), B (a,b), C (a+c, b+d), and D (c,d). When we draw squares on each side of the parallelogram, they will have vertices A', B', C', and D' respectively, which are the midpoints of the sides of the squares.

Step 2: Determine the coordinates of the centers of the squares

In order to find the coordinates of the centers of the squares (A', B', C', and D'), we must first find the lengths of the diagonals of the squares. Since the squares are formed on each side of the parallelogram, their diagonals would have the same length, and to find this length we can utilize the Pythagorean theorem: Length of diagonal = \(\sqrt{(a-c)^2 + (b-d)^2}\) Now, we can determine the coordinates of the centers of the squares: A': midpoint of the diagonal of the square with vertices A and A', Coordinates of A': \(\left(\frac{-b}{2}, \frac{a}{2}\right)\) B': midpoint of the diagonal of the square with vertices B and B', Coordinates of B': \(\left(\frac{a-b}{2}, \frac{a+b}{2}\right)\) C': midpoint of the diagonal of the square with vertices C and C', Coordinates of C': \(\left(\frac{a+c+b}{2}, \frac{b+d-a}{2}\right)\) D': midpoint of the diagonal of the square with vertices D and D', Coordinates of D': \(\left(\frac{c+b}{2}, \frac{d-a}{2}\right)\)

Step 3: Show that the centers of the squares form a square

First, calculate the side lengths of the square formed by A', B', C', and D' using the distance formula: AB' = \(\sqrt{((\frac{a-b}{2}) - (\frac{-b}{2}))^2 + ((\frac{a+b}{2}) - (\frac{a}{2}))^2} = \frac{\sqrt{a^2+b^2}}{2}\) BC' = \(\sqrt{((\frac{a+c+b}{2}) - (\frac{a-b}{2}))^2 + ((\frac{b+d-a}{2}) - (\frac{a+b}{2}))^2} = \frac{\sqrt{c^2+d^2}}{2}\) CD' = \(\sqrt{((\frac{c+b}{2}) - (\frac{a+c+b}{2}))^2 + ((\frac{d-a}{2}) - (\frac{b+d-a}{2}))^2} = \frac{\sqrt{a^2+b^2}}{2}\) DA' = \(\sqrt{((\frac{-b}{2}) - (\frac{c+b}{2}))^2 + ((\frac{a}{2}) - (\frac{d-a}{2}))^2} = \frac{\sqrt{c^2+d^2}}{2}\) Now, show that the diagonals of the square formed by A', B', C', and D' are congruent: A'C' = \(\sqrt{((\frac{a+c+b}{2}) - (\frac{-b}{2}))^2 + ((\frac{b+d-a}{2}) - (\frac{a}{2}))^2} = \sqrt{(a-c)^2 + (b-d)^2}\) B'D' = \(\sqrt{((\frac{c+b}{2}) - (\frac{a-b}{2}))^2 + ((\frac{d-a}{2}) - (\frac{a+b}{2}))^2} = \sqrt{(a-c)^2 + (b-d)^2}\) Since the side lengths are congruent (AB' = CD' and BC' = DA') and the diagonals are congruent (A'C' = B'D'), the centers of the squares A', B', C', and D' form a square.

Key Concepts

Square Construction

When constructing a square externally on each side of a parallelogram, we create a geometric shape with new properties. Each side of the parallelogram acts as the base of a square.
From any given vertex, you draw lines perpendicular to the segment, ensuring all sides are equal. This symmetrical extension results in four squares.

• To create a square: draw a perpendicular line at the endpoint of a segment.
• Ensure the new line's length equals the original segment.
• Connect the endpoints to form a square.

The interesting result is that the centers of these squares form another geometric shape, also a square, highlighting the elegance of symmetry in geometry.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to use algebra to solve and prove geometric concepts by placing figures in a coordinate plane. Each point on a shape is assigned ordered pairs \((x,y)\).
In this problem, labeling the vertices of the parallelogram and the squares using coordinates simplifies the problem-solving process.

• Vertex A is at (0,0), providing a simple condition to start from.
• Other vertices, like B at (a,b), are expressed with variable coordinates.

Through these coordinates, calculations such as distances and midpoints become straightforward, allowing us to derive other complex geometric properties.

Pythagorean Theorem

The Pythagorean Theorem is essential when dealing with distances and diagonals in coordinate geometry. It states that in a right triangle, the square of the hypotenuse (longest side) equals the sum of the squares of the other two sides.
The formula is written as \(c^2 = a^2 + b^2\).
Applying this to find diagonal lengths of squares built on the parallelogram:

• Given coordinates \((x_1, y_1)\) and \((x_2, y_2)\), we find the diagonal's length: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
• This calculation helps determine congruent side lengths and diagonals of the newly formed squares.

These relationships cement the mathematical proof that the centers indeed form another square.

Midpoint Formula

The midpoint formula helps identify the center of any segment in coordinate geometry. By finding midpoints, we can easily locate the centers of the squares attached to the parallelogram.
The formula is given by \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\).
In this exercise, use the formula to find the center of each square:

• Point A': Calculate using vertices of square on side AB.
• Similar calculations for points B', C', and D'.

Through these midpoints, we logically demonstrate the centers of these squares form a perfect square. This step is crucial for verifying geometric symmetry in relation to the original parallelogram.