Problem 68 - Calculus with Concepts in Calculus

Question

Show that the midpoints of the sides of any rectangle are the vertices of a rhombus (a quadrilateral with all sides of equal length). (Hint: Let the vertices of the rectangle be \((0,0),(a, 0),(0, b)\), and \((a, b) .)\)

Step-by-Step Solution

Answer

The midpoints form a rhombus as all sides are equal in length.

Step 1: Identify Midpoints

Consider a rectangle with vertices at points \((0, 0)\), \((a, 0)\), \((a, b)\), and \((0, b)\). To find the midpoints of the sides, calculate: \(M_1\) is midpoint of \((0, 0)\) and \((a, 0)\), which is \(\left(\frac{a}{2}, 0\right)\); \(M_2\) is midpoint of \((a, 0)\) and \((a, b)\), which is \(\left(a, \frac{b}{2}\right)\); \(M_3\) is midpoint of \((a, b)\) and \((0, b)\), which is \(\left(\frac{a}{2}, b\right)\); \(M_4\) is midpoint of \((0, b)\) and \((0, 0)\), which is \(\left(0, \frac{b}{2}\right)\).

Step 2: Calculate Side Lengths

To show that the midpoints form a rhombus, verify that all sides are equal. Compute the distances between consecutive midpoints: \(M_1\) and \(M_2\), \(M_2\) and \(M_3\), \(M_3\) and \(M_4\), \(M_4\) and \(M_1\). Each distance is computed as \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

Step 3: Compute Distance for M_1 to M_2

Calculate the distance between \(M_1\left(\frac{a}{2}, 0\right)\) and \(M_2\left(a, \frac{b}{2}\right)\):\[d_{1,2} = \sqrt{\left(a - \frac{a}{2}\right)^2 + \left(\frac{b}{2} - 0\right)^2} = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2}\]Simplify this:\[d_{1,2} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \sqrt{\frac{a^2 + b^2}{4}} = \frac{\sqrt{a^2 + b^2}}{2}\]

Step 4: Compute Distance for M_2 to M_3

Calculate the distance between \(M_2\left(a, \frac{b}{2}\right)\) and \(M_3\left(\frac{a}{2}, b\right)\):\[d_{2,3} = \sqrt{\left(\frac{a}{2} - a\right)^2 + (b - \frac{b}{2})^2} = \sqrt{\left(-\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2}\]Simplify this:\[d_{2,3} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \frac{\sqrt{a^2 + b^2}}{2}\]

Step 5: Compute Distance for M_3 to M_4

Calculate the distance between \(M_3\left(\frac{a}{2}, b\right)\) and \(M_4\left(0, \frac{b}{2}\right)\):\[d_{3,4} = \sqrt{(0 - \frac{a}{2})^2 + (\frac{b}{2} - b)^2} = \sqrt{\left(-\frac{a}{2}\right)^2 + \left(-\frac{b}{2}\right)^2}\]Simplify this:\[d_{3,4} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \frac{\sqrt{a^2 + b^2}}{2}\]

Step 6: Compute Distance for M_4 to M_1

Calculate the distance between \(M_4\left(0, \frac{b}{2}\right)\) and \(M_1\left(\frac{a}{2}, 0\right)\):\[d_{4,1} = \sqrt{(\frac{a}{2} - 0)^2 + (0 - \frac{b}{2})^2} = \sqrt{\left(\frac{a}{2}\right)^2 + \left(-\frac{b}{2}\right)^2}\]Simplify this:\[d_{4,1} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \frac{\sqrt{a^2 + b^2}}{2}\]

Step 7: Conclude all Sides Equal

Since each side length between consecutive midpoints \((d_{1,2}, d_{2,3}, d_{3,4}, d_{4,1})\) is \(\frac{\sqrt{a^2 + b^2}}{2}\), all sides are equal. The midpoints form a quadrilateral with all sides equal, confirming that they are vertices of a rhombus.

Key Concepts

midpoints of a rectangle

When tackling geometry problems, midpoints of a rectangle are incredibly useful for various proofs. For a rectangle, you can easily find midpoints if you know the coordinates of the vertices. Consider a rectangle with vertices at \(0, 0\), \(a, 0\), \(a, b\), and \(0, b\). These points are positioned on a Cartesian plane, making initial calculations straightforward.

To find the midpoints of a rectangle's sides, use the midpoint formula: \(M = rac{{(x_1 + x_2)}}{2}, rac{{(y_1 + y_2)}}{2}\). Applying this to our rectangle's sides gives us four midpoints:

• \(M_1 = rac{a}{2}, 0\) - midpoint between \(0, 0\) and \(a, 0\)
• \(M_2 = a, rac{b}{2}\) - midpoint between \(a, 0\) and \(a, b\)
• \(M_3 = rac{a}{2}, b\) - midpoint between \(a, b\) and \(0, b\)
• \(M_4 = 0, rac{b}{2}\) - midpoint between \(0, b\) and \(0, 0\)

By understanding and calculating these midpoints, we can advance into more complex shapes and prove other geometrical properties.

properties of rhombus

A rhombus is a special type of polygon distinguished by its equal-length sides. More specifically, all four sides of a rhombus are of the same length, which makes it quite unique. Due to this property, the diagonals intersect at right angles and bisect each other, which helps in various geometrical proofs and constructions.

In the context of the rectangle's midpoint problem, you arrive at a rhombus because when you calculate distances from one midpoint to the next (using the distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)), all the distances are equal. This equal-length property confirms that the shape formed is indeed a rhombus.

Each set of consecutive points is equidistant, demonstrating:

• The congruency of all four sides \(d_{1,2}, d_{2,3}, d_{3,4}, d_{4,1} = \frac{\sqrt{a^2 + b^2}}{2}\)
• That any quadrilateral with equal side lengths is a rhombus

Understanding these properties further implicates a wide range of applications in both pure mathematics and practical fields like architecture and engineering.

coordinate geometry

Coordinate geometry merges algebra and geometry using a coordinate system, facilitating the analysis of geometric shapes. This branch is pivotal in providing a powerful toolset for solving geometrical problems, like our rectangle and rhombus discovery.

Using coordinates allows us to define points and prove shapes' properties through formulae and equations rather than relying solely on visual inspection. The establishment of midpoints and proving of quadrilaterals like a rhombus is streamlined by coordinate geometry's methods.

• **Midpoints**: Easily calculated using midpoint formulas based on coordinates.
• **Distances**: Precise and derived using the distance formula to ascertain equality between sides.
• **Utility**: Allows exact calculations of inclinations (slopes), midpoints, lengths, and area.

This method elucidates the necessity for precise calculations, paving the way for complex problem solving. It provides a robust framework for drawing conclusions as seen in our rectangle-to-rhombus example.