Problem 25 - Complex numbers from A to...Z

Question

In a plane three equilateral triangles \(O A B, O C D\) and \(O E F\) are given. Prove that the midpoints of the segments \(B C, D E\) and \(F A\) are the vertices of an equilateral triangle.

Step-by-Step Solution

Answer

Question: Prove that the midpoints of the sides of three equilateral triangles \(OAB, OCD,\) and \(OEF\) (all sharing vertex \(O\)) form an equilateral triangle. Answer: By proving that the vectors connecting the points of the new triangle are equal under 120-degree rotation, we have demonstrated that the side lengths of triangle \(MNP\) are equal, and thus, the triangle is equilateral.

Step 1: Define the vertices

Let point \(M\) be the midpoint of \(BC\), point \(N\) be the midpoint of \(DE\), and point \(P\) be the midpoint of \(FA\). Our goal is to prove that the triangle formed by points \(M, N, P\) is equilateral.

Step 2: Find the vectors between the points

We can express the vertices of equilateral triangles in terms of vector algebra. Let \(\vec{a},\vec{b},\vec{c},\vec{d},\vec{e},\vec{f}\) represent the vectors corresponding to points \(A,B,C,D,E,F\). Then we can write the points \(M,N,P\) in terms of these: $$ \vec{m} = \frac{1}{2} (\vec{b}+ \vec{c}) $$ $$ \vec{n} = \frac{1}{2} (\vec{d}+ \vec{e}) $$ $$ \vec{p} = \frac{1}{2} (\vec{f}+ \vec{a}) $$

Step 3: Find the lengths of the sides of the triangle

Now we can calculate the side lengths of triangle \(MNP\). Let \(MN\), \(NP\), and \(PM\) be the sides of this new triangle. Then: $$ MN = ||\vec{n} - \vec{m}|| = ||\frac{1}{2}(\vec{d}+\vec{e}-\vec{b}-\vec{c})|| $$ $$ NP = ||\vec{p} - \vec{n}|| = ||\frac{1}{2}(\vec{f}+\vec{a}-\vec{d}-\vec{e})|| $$ $$ PM = ||\vec{m} - \vec{p}|| = ||\frac{1}{2}(\vec{b}+\vec{c}-\vec{f}-\vec{a})|| $$

Step 4: Show that the vectors between the points are equal

Note that \(\vec{n} - \vec{m}\) can be rotated by 120 degrees to obtain \(\vec{p} - \vec{n}\) and by another 120 degrees to obtain \(\vec{m} - \vec{p}\). In terms of rotation matrices, a 120-degree counter-clockwise rotation can be expressed as: $$ R_{120} = \begin{bmatrix} \cos{120^{\circ}} & -\sin{120^{\circ}} \\ \sin{120^{\circ}} & \cos{120^{\circ}} \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix} $$ So we have: $$ R_{120}(\vec{n} - \vec{m}) = \vec{p} - \vec{n} $$ And $$ R_{120}(\vec{p} - \vec{n}) = \vec{m} - \vec{p} $$

Step 5: Prove that the triangle is equilateral

Since the lengths of the vectors are preserved under rotation: $$ ||\vec{n} - \vec{m}|| = ||\vec{p} - \vec{n}|| = ||\vec{m} - \vec{p}|| $$ Therefore, we have: $$ MN = NP = PM $$ As all the sides are equal, we can conclude that the triangle \(MNP\) is equilateral.

Key Concepts

Vector Algebra in Geometry

Vector algebra is an essential part of modern geometry, especially when dealing with positions, shapes, and sizes in a plane or space. By defining the vertices of geometric figures as vectors, we can utilize the operations and properties of vectors to solve complex geometric problems.

In our problem, we consider vectors corresponding to the vertices of equilateral triangles. The advantage of using vectors is that we can easily express complex relations, such as the midpoint of a segment, through vector arithmetic. For example, to find the midpoint of a segment connecting points represented by vectors \(\vec{a}\) and \(\vec{b}\), we can simply calculate \(\frac{1}{2}(\vec{a} + \vec{b})\). This simple formula is a manifestation of the midpoint theorem in vector form. This concept is especially powerful as it negates the need to work with individual coordinates, simplifying both representation and computation.

Midpoint Theorem

The midpoint theorem is a classical result in geometry that succinctly states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half as long. This theorem has an elegant expression in vector algebra, as shown in the given problem.

When defining the midpoint in terms of vectors, we tap into a more generalized version of the midpoint theorem. This not only streamlines the calculation for 2D and 3D figures but also allows for a straightforward extension to higher dimensions and more complex geometrical figures without altering the foundation of our approach. Such adaptability ensures that even in problems that involve various geometric transformations, the midpoint remains a concept that can be intuitively understood and applied.

Rotation Matrices

Once we have described our geometry problem with vectors, rotation matrices become a powerful tool for analyzing shapes. A rotation matrix represents a transformation that rotates vectors in a plane around the origin. In the context of our equilateral triangles problem, we use a rotation matrix to show the relationship between the sides of the new triangle formed by the midpoints.

The 120-degree rotation matrix \[ R_{120} \] is particularly important. This matrix corresponds to the fact that rotating any side of an equilateral triangle by 120 degrees around its center will align it with another side. By showing that the vectors corresponding to the sides of our midpoint triangle are related by this rotation, we validate their equality in length, thus proving the triangle they form is equilateral.Remember that rotation matrices preserve vector lengths which means distances between points remain unchanged after rotation, a property central to our equilateral triangles problem.