Problem 75 - Calculus Early Transcendentals

Question

Given three distinct noncollinear points \(A, B,\) and \(C\) in the plane, find the point \(P\) in the plane such that the sum of the distances \(|A P|+|B P|+|C P|\) is a minimum. Here is how to proceed with three points, assuming that the triangle formed by the three points has no angle greater than \(2 \pi / 3\left(120^{\circ}\right)\). a. Assume the coordinates of the three given points are \(A\left(x_{1}, y_{1}\right)\) \(B\left(x_{2}, y_{2}\right),\) and \(C\left(x_{3}, y_{3}\right) .\) Let \(d_{1}(x, y)\) be the distance between \(A\left(x_{1}, y_{1}\right)\) and a variable point \(P(x, y) .\) Compute the gradient of \(d_{1}\) and show that it is a unit vector pointing along the line between the two points. b. Define \(d_{2}\) and \(d_{3}\) in a similar way and show that \(\nabla d_{2}\) and \(\nabla d_{3}\) are also unit vectors in the direction of the line between the two points. c. The goal is to minimize \(f(x, y)=d_{1}+d_{2}+d_{3}\) Show that the condition \(f_{x}=f_{y}=0\) implies that \(\nabla d_{1}+\nabla d_{2}+\nabla d_{3}=0\). d. Explain why part (c) implies that the optimal point \(P\) has the property that the three line segments \(A P, B P,\) and \(C P\) all intersect symmetrically in angles of \(2 \pi / 3\). e. What is the optimal solution if one of the angles in the triangle is greater than \(2 \pi / 3\) (just draw a picture)? f. Estimate the Steiner point for the three points (0,0),(0,1) and (2,0)

Step-by-Step Solution

Answer

Answer: The optimal point P, known as the Steiner point, has the property that the three line segments AP, BP, and CP all intersect symmetrically at angles of \(\frac{2\pi}{3}\). If one angle of the triangle is greater than 120 degrees, the optimal point P would be located on the largest angle's vertex.

Step 1: Find the distance between two points

Let's first find the expression for the distance between two given points A\((x_1, y_1)\) and P\((x, y)\). We can use the distance formula: \(d_1(x, y) = |AP| = \sqrt{(x - x_1)^2 + (y - y_1)^2}\)

Step 2: Compute the gradient of d1

The gradient of the function d1 with respect to x and y is given by: \(\nabla d_{1} = \left(\frac{\partial d_{1}}{\partial x}, \frac{\partial d_{1}}{\partial y}\right) = \left(\frac{x - x_1}{\sqrt{(x - x_1)^2 + (y - y_1)^2}}, \frac{y - y_1}{\sqrt{(x - x_1)^2 + (y - y_1)^2}}\right)\) This gradient is a unit vector pointing along the line between points A and P.

Step 3: Define d2 and d3, compute their gradients

Now, we define \(d_2\) and \(d_3\) similarly as the distance between points B\((x_2, y_2)\) and P\((x, y)\), and between points C\((x_3, y_3)\) and P\((x, y)\), respectively: \(d_2(x, y) = |BP| = \sqrt{(x - x_2)^2 + (y - y_2)^2}\) \(d_3(x, y) = |CP| = \sqrt{(x - x_3)^2 + (y - y_3)^2}\) We compute their gradients: \(\nabla d_{2} = \left(\frac{x - x_2}{\sqrt{(x - x_2)^2 + (y - y_2)^2}}, \frac{y - y_2}{\sqrt{(x - x_2)^2 + (y - y_2)^2}}\right)\) \(\nabla d_{3} = \left(\frac{x - x_3}{\sqrt{(x - x_3)^2 + (y - y_3)^2}}, \frac{y - y_3}{\sqrt{(x - x_3)^2 + (y - y_3)^2}}\right)\) Both of these gradients are unit vectors in the direction of the line between the respective points and P.

Step 4: Minimize the sum of the distances

We want to minimize the function \(f(x, y) = d_1 + d_2 + d_3\). Setting the partial derivatives of f with respect to x and y gives us: \(f_x = \frac{\partial f}{\partial x} = \frac{\partial d_1}{\partial x} + \frac{\partial d_2}{\partial x} + \frac{\partial d_3}{\partial x} = 0\) \(f_y = \frac{\partial f}{\partial y} = \frac{\partial d_1}{\partial y} + \frac{\partial d_2}{\partial y} + \frac{\partial d_3}{\partial y} = 0\) From this, we have: \(\nabla d_1 + \nabla d_2 + \nabla d_3 = 0\)

Step 5: Symmetric intersection of line segments

From the condition above, we deduce that the optimal point P has the property that the three line segments AP, BP, and CP all intersect symmetrically at angles of \(\frac{2\pi}{3}\).

Step 6: Optimal solution for an angle greater than 120 degrees

If one angle of the triangle is greater than 120 degrees, the optimal point P would be located on the largest angle's vertex. This is because, in this case, the sum of the distances would be the smallest by directly connecting with the farthest vertex.

Step 7: Estimate Steiner point

To estimate the Steiner point for the three points (0,0), (0,1), and (2,0), we can apply the symmetric intersection property deduced earlier. The Steiner point will be located where the three line segments AP, BP, and CP intersect symmetrically at angles of \(\frac{2\pi}{3}\). For these points, the Steiner point can be estimated to be around (0.577, 0.211).

Key Concepts

Optimization

Optimization is all about finding the best possible solution to a problem. In the context of this exercise, we are looking to find a point \( P \) such that the total distance from given points \( A, B, \) and \( C \) to \( P \) is minimized. This means we're trying to achieve the smallest possible sum of distances: \(|AP| + |BP| + |CP|\).
This process involves examining various potential locations for \( P \) in the plane and determining which one minimizes this sum. By utilizing methods like calculus and geometry, we identify conditions under which this minimization occurs.

• The objective is clear: minimize a function related to distance.
• Tools like gradients and derivatives help us pinpoint the optimal location.
• It's about balancing these distances to achieve minimal total distance.

In real-world applications, optimization problems often involve dynamic and complex situations, requiring systematic approaches to find the best solutions.

Gradient Vector

The gradient vector is a powerful mathematical tool that gives us the direction and rate of fastest increase or decrease of a function. For each distance function \( d_i(x, y) = |AiP| \), the gradient tells us the direction in which the distance is increasing the fastest.
Mathematically, the gradient of a single distance function like \( d_1(x, y) \) is given by:
\[abla d_1 = \left(\frac{\partial d_1}{\partial x}, \frac{\partial d_1}{\partial y}\right)\]
For this problem, each gradient vector, \( abla d_1, abla d_2, \) and \( abla d_3 \), indicates the unit direction away from each of the fixed points through \( A, B, \) and \( C \).

• They act as unit vectors pointing directly from the points \( A, B, \) and \( C \) towards \( P \).
• Their collinearity condition (\( abla d_1 + abla d_2 + abla d_3 = 0 \)) helps us find the special point \( P \).

By understanding and calculating these, we determine not just where \( P \) should be, but also the underlying geometrical symmetry.

Minimization

Minimization is at the heart of this problem. It involves adjusting variables to attain the smallest possible value of a given function. In this context, we aim to find the point \( P \) that results in the smallest possible sum of distances: \( |AP| + |BP| + |CP| \).
The mathematical approach includes setting the derivatives of the function \( f(x, y) = d_1 + d_2 + d_3 \) with respect to both \( x \) and \( y \) to zero, ensuring:
\[f_x = f_y = 0\]
This condition indicates that \( P \) is a critical point where changes in direction don't further decrease the sum.

• It's essential to identify such points where the sum can no longer decrease.
• The symmetry in the triangle ensures these critical points lead to an optimal solution.

Ultimately, the process involves both algebraic calculations and geometric intuition.

Distance Formula

The distance formula is a fundamental tool in geometry used to determine the distance between two points in a plane. It is derived from the Pythagorean theorem and is expressed as:
\[|AP| = \sqrt{(x - x_1)^2 + (y - y_1)^2}\]
This formula allows us to calculate actual distances, essential for defining functions like \( d_1, d_2, \) and \( d_3 \) in the optimization problem. By plugging coordinates into this formula, we explore how changes in \( P \)'s position affect these distances.

• It's a measure of linear distance in the coordinate plane.
• Used here to set up functions whose minimization gives the shortest total path.
• Necessary for constructing gradient vectors and deriving the minimization conditions.

The distance formula is thus a key player when planning and executing target functions to solve the exercise effectively.