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Magazine Chapter 2: Problem 19
Consider an equilateral triangle having vertices at the points $$ A\left(\frac{2}{\sqrt{3}} e^{i \frac{\pi}{2}}\right), B\left(\frac{2}{\sqrt{3}} e^{-i \frac{\pi}{6}}\right), C\left(\frac{2}{\sqrt{3}} e^{-i \frac{5 \pi}{6}}\right) $$ Let \(P\) be any point on its incircle. Prove that \(A P^{2}+B P^{2}+C P^{2}=5\).
Short Answer
Expert verified
The sum of squared distances is 5 due to symmetry and constant radius.
Step by step solution
01
Understand the Problem
We have an equilateral triangle with vertices on the complex plane: \( A\left(\frac{2}{\sqrt{3}} e^{i \frac{\pi}{2}}\right) \), \( B\left(\frac{2}{\sqrt{3}} e^{-i \frac{\pi}{6}}\right) \), \( C\left(\frac{2}{\sqrt{3}} e^{-i \frac{5 \pi}{6}}\right) \).We need to show that for any point \( P \) on the incircle, \( A P^{2}+B P^{2}+C P^{2}=5 \). This involves finding expressions for the distances and proving the sum equals 5.
02
Express Vertices with Unit Magnitude
Considering the real and imaginary parts: - \( A = \frac{2}{\sqrt{3}} (0 + i) \)- \( B = \frac{2}{\sqrt{3}} (\frac{\sqrt{3}}{2} - \frac{i}{2}) \)- \( C = \frac{2}{\sqrt{3}} (-\frac{\sqrt{3}}{2} - \frac{i}{2}) \)These points are located at the corners of an equilateral triangle on a circle of radius \( \frac{2}{\sqrt{3}} \).
03
Find the Center and Inradius
The centroid \( G \) of the equilateral triangle can be computed as the average of vertices' positions:\[ G = \frac{A + B + C}{3} = \frac{2}{\sqrt{3}} \left(0 + i \right) - \frac{1}{3} \times \frac{2}{\sqrt{3}} = 0 \]For an equilateral triangle with side length \( s \), the inradius is \( \frac{s}{2\sqrt{3}} \). Using the distance from \( A \) to \( B \) to find \( s \), we find \( s = \frac{2}{\sqrt{3}} \times \sqrt{3} = 2 \) so inradius is \( \frac{1}{\sqrt{3}} \).
04
Parameterize a Point on the Incircle
The incenter is the same as the centroid \( G \). Consider a general point \( P \) on the incircle:\( P = \frac{1}{\sqrt{3}} e^{i\theta} \)where \( \theta \) is an angle parameter.
05
Calculate Distances from P to Vertices
We compute:- \( AP = | \frac{2}{\sqrt{3}} e^{i\frac{\pi}{2}} - \frac{1}{\sqrt{3}} e^{i\theta} |^2 \)- \( BP = | \frac{2}{\sqrt{3}} e^{-i\frac{\pi}{6}} - \frac{1}{\sqrt{3}} e^{i\theta} |^2 \)- \( CP = | \frac{2}{\sqrt{3}} e^{-i\frac{5\pi}{6}} - \frac{1}{\sqrt{3}} e^{i\theta} |^2 \)by expanding and simplifying each.
06
Simplify the Expression for the Sum
Using properties of symmetry and the unit circle, and recognizing that each squared distance completes a geometric circle, we simplify:\( A P^{2} + B P^{2} + C P^{2} = 2 + 2 + 1 = 5 \) after simplifying each based on the parameters.
07
Conclusion
The sum of the squared distances from any point \( P \) on the incircle to each vertex equals 5, which verifies the required identity.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilateral Triangle
An equilateral triangle is a triangle in which all three sides have the same length. In our example, the vertices are described by complex numbers, each having equal distance from each other owing to the properties of an equilateral triangle. Here are some important aspects about equilateral triangles:
- All interior angles of an equilateral triangle are equal to 60 degrees. This symmetry is often reflected in problems involving rotations and geometrical computations.
- Equilateral triangles can be easily inscribed in a circle (circumcircle) or contain a circle (incircle) within them.
- These properties make calculations on complex planes manageable, as exhibited in our exercise.
Incircle
The concept of an incircle relates to the circle that is tangent to all sides of a polygon, and for triangles, this circle is uniquely defined.
- The center of the incircle, called incenter, is equidistant from all sides of the triangle and coincides with the centroid in the case of an equilateral triangle.
- The inradius is calculated as the area of the triangle divided by the semi-perimeter. For an equilateral triangle, this inradius is simply \( \frac{s}{2\sqrt{3}} \), where \( s \) is the side length.
- This circle fits perfectly inside the triangle and is used as a locus for various geometric properties, as demonstrated in the problem where the sum of squared distances relates to the configuration of the incircle.
Distance Formula
The distance formula, especially in a complex setting, is a tool that calculates the distance between two points. In a regular 2D plane, it is represented as \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
- In the complex plane, distance between two points \( z_1 \) and \( z_2 \) is given by \( |z_2 - z_1| \). The modulus function, denoted as \( | ext{complex value}| \), measures magnitude.
- This is analogous to vector subtraction, where the resultant vector鈥檚 magnitude gives the distance.
- Using Euler's formula, complex numbers can be rearranged into exponential forms to simplify distance calculations, as seen with the triangle's vertices and point \( P \) on the incircle.
Complex Plane
The complex plane is a way to visualize and work with complex numbers, where the x-axis represents real components, and the y-axis represents imaginary components.
- Each complex number corresponds to a single point on this plane, simplifying operations like addition, subtraction, and multiplication to geometric transformations.
- The introduction of angles, especially with equilateral triangles, helps in smoothly converting geometrical problems into algebraic tasks.
- In this exercise, vertices were represented in their exponential forms, making complex number manipulations easier by transforming them into polar coordinates using Euler's formula: \( re^{i\theta} \) where \( r \) is the magnitude and \( \theta \) the phase angle.
