Problem 57 - University Calculus: Early Transcendentals

Question

Suppose that \(A, B,\) and \(C\) are vertices of a triangle and that \(a, b\) and \(c\) are, respectively, the midpoints of the opposite sides. Show that \(\overrightarrow{A a}+\overrightarrow{B b}+\overrightarrow{C c}=0\).

Step-by-Step Solution

Answer

\(\overrightarrow{A a} + \overrightarrow{B b} + \overrightarrow{C c} = 0\).

Step 1: Understanding the Midpoints

Let's consider the positions of the points and their vectors. In a triangle, point \(a\) is the midpoint of side \(BC\), point \(b\) is the midpoint of side \(AC\), and point \(c\) is the midpoint of side \(AB\). Therefore, the position vectors of these midpoints can be given as follows: \(\overrightarrow{a} = \frac{\overrightarrow{B} + \overrightarrow{C}}{2}\), \(\overrightarrow{b} = \frac{\overrightarrow{A} + \overrightarrow{C}}{2}\), and \(\overrightarrow{c} = \frac{\overrightarrow{A} + \overrightarrow{B}}{2}\).

Step 2: Calculating Vector \(\overrightarrow{Aa}\)

Calculate the vector \(\overrightarrow{Aa}\) by subtracting the position vector of \(A\) from \(a\). Thus, we have: \[\overrightarrow{Aa} = \overrightarrow{a} - \overrightarrow{A} = \frac{\overrightarrow{B} + \overrightarrow{C}}{2} - \overrightarrow{A} = \frac{-2\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{2}\].

Step 3: Calculating Vector \(\overrightarrow{Bb}\)

In a similar manner, calculate the vector \(\overrightarrow{Bb}\) using the position vectors: \[\overrightarrow{Bb} = \overrightarrow{b} - \overrightarrow{B} = \frac{\overrightarrow{A} + \overrightarrow{C}}{2} - \overrightarrow{B} = \frac{\overrightarrow{A} - 2\overrightarrow{B} + \overrightarrow{C}}{2}\].

Step 4: Calculating Vector \(\overrightarrow{Cc}\)

Finally, calculate the vector \(\overrightarrow{Cc}\) like the others: \[\overrightarrow{Cc} = \overrightarrow{c} - \overrightarrow{C} = \frac{\overrightarrow{A} + \overrightarrow{B}}{2} - \overrightarrow{C} = \frac{\overrightarrow{A} + \overrightarrow{B} - 2\overrightarrow{C}}{2}\].

Step 5: Add the Three Vectors

Now, we add the vectors \(\overrightarrow{Aa} + \overrightarrow{Bb} + \overrightarrow{Cc}\):\[\frac{-2\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{2} + \frac{\overrightarrow{A} - 2\overrightarrow{B} + \overrightarrow{C}}{2} + \frac{\overrightarrow{A} + \overrightarrow{B} - 2\overrightarrow{C}}{2}\]Simplify the expression:\[= \frac{-2\overrightarrow{A} + \overrightarrow{A} + \overrightarrow{A}}{2} + \frac{\overrightarrow{B} - 2\overrightarrow{B} + \overrightarrow{B}}{2} + \frac{\overrightarrow{C} + \overrightarrow{C} - 2\overrightarrow{C}}{2}\]\[= \frac{0\overrightarrow{A}}{2} + \frac{0\overrightarrow{B}}{2} + \frac{0\overrightarrow{C}}{2}\]\[= \overrightarrow{0}\].

Step 6: Conclusion: Verify the Result

From the calculations, it is evident that \(\overrightarrow{A a} + \overrightarrow{B b} + \overrightarrow{C c} = \overrightarrow{0}\). This is consistent with the fact that the centroid of the triangle divides each median in a 2:1 ratio, affirming that the sum of vectors to these midpoints is zero.

Key Concepts

Triangle Midpoints

In vector geometry, understanding midpoints is crucial for solving many problems related to triangles. A midpoint in a triangle is the point that divides a side into two equal segments. Consider a triangle with vertices at points \(A\), \(B\), and \(C\).
For each side of the triangle, there is a midpoint:

• The midpoint of side \(BC\) is often denoted as \(a\).
• The midpoint of side \(AC\) is \(b\).
• The midpoint of side \(AB\) is \(c\).

Mathematically, the position vectors for these midpoints can be stated as follows:

• \(\overrightarrow{a} = \frac{\overrightarrow{B} + \overrightarrow{C}}{2}\)
• \(\overrightarrow{b} = \frac{\overrightarrow{A} + \overrightarrow{C}}{2}\)
• \(\overrightarrow{c} = \frac{\overrightarrow{A} + \overrightarrow{B}}{2}\)

These formulas represent how each midpoint averages the coordinates of the endpoints of a side, effectively locating the "middle" in terms of vector addition.

Vector Addition

Vector addition is a fundamental operation in vector geometry where two vectors are combined to form a resultant vector. The result is obtained by summing each component of the vectors.
To solve our problem involving triangle midpoints, we use vector addition to find simple expressions for several vectors:

• \(\overrightarrow{Aa} = \overrightarrow{a} - \overrightarrow{A} = \frac{-2\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{2}\)
• \(\overrightarrow{Bb} = \overrightarrow{b} - \overrightarrow{B} = \frac{\overrightarrow{A} - 2\overrightarrow{B} + \overrightarrow{C}}{2}\)
• \(\overrightarrow{Cc} = \overrightarrow{c} - \overrightarrow{C} = \frac{\overrightarrow{A} + \overrightarrow{B} - 2\overrightarrow{C}}{2}\)

Vector addition allows us to combine these vectors to verify that their total sum is zero. By calculating \(\overrightarrow{Aa} + \overrightarrow{Bb} + \overrightarrow{Cc}\), the components effectively cancel out, illustrating the neat property of symmetry within the triangle.

Centroid of Triangle

The centroid of a triangle is a key concept in geometry. It is the point where all the medians of the triangle intersect, and it acts as the center of balance for the triangle.
In any triangle, the centroid has an important characteristic: it divides each median in a 2:1 ratio, where the larger segment is between the vertex and the centroid.
From the vector perspective, when you add the vectors pointing to the midpoints, like \(\overrightarrow{A a}\), \(\overrightarrow{B b}\), and \(\overrightarrow{C c}\), the sum is \(\overrightarrow{0}\). This outcome implies that the centroid serves as a balancing point where these vectors cancel out.
This balance confirms that the triangle has an intrinsic equilibrium, reflected in how its medians and centroid interact. Understanding this concept provides deep insight into both the geometry and algebra related to triangles.