Question

Sum of isosceles distances a. An isosceles triangle has a base of length 4 and two sides of length \(2 \sqrt{2} .\) Let \(P\) be a point on the perpendicular bisector of the base. Find the location \(P\) that minimizes the sum of the distances between \(P\) and the three vertices. b. Assume in part (a) that the height of the isosceles triangle is \(h>0\) and its base has length \(4 .\) Show that the location of \(P\) that gives a minimum solution is independent of \(h\) for \(h \geq \frac{2}{\sqrt{3}}\)

Short answer

Answer: The point P that minimizes the sum of the distances between P and the three vertices of the given isosceles triangle lies at its centroid. Moreover, this location is independent of the height h for \(h\geq\frac{2}{\sqrt{3}}\).

Step-by-step solution

Draw and label the triangle

Draw the isosceles triangle with base length 4 and the two sides with length \(2\sqrt{2}\). Label the vertices A, B, and C, where A and B are the base vertices and C is the vertex with the height from C to the midpoint of the base. Label the midpoint of the base M and draw the altitude h from C to M.

Find the location of point P

Let P be a point on the perpendicular bisector of the base (line segment CM). Let x be the distance from M to P, where x can be positive or negative. Let D, E, and F be the feet of the perpendiculars from P to sides AB, AC, and BC, respectively.

Calculate distances PD, PE, and PF

From similar triangles, we have that \(\frac{PD}{2}=x\) and \(\frac{PE}{h}=x\), and \(\frac{PF}{h}=-x\). Therefore, \(PD=2x\), \(PE=hx\), and \(PF=-hx\).

Calculate the sum of distances

The sum of the distances between P and the three vertices is PA + PB + PC. Using triangle inequality, we get PA + PB + PC >= PD + DE + PF + EF + PC >= PD + PF + PC. Therefore, the sum PA + PB + PC >= PD + PF + PC, which is greater or equal to \(2x-hx-hx=-2hx\).

Minimize the sum of distances

To minimize the sum of distances PA+PB+PC subject to \(2x-hx-hx \geq -2hx\), we need to find the minimum value of \(-2hx\). Since h > 0, the minimum value of \(-2hx\) occurs when x = 0. When x = 0, P coincides with the centroid of triangle ABC.

Show the minimum location is independent of h

To show that the location of P giving a minimum solution is independent of h for \(h\geq \frac{2}{\sqrt{3}}\), observe that the minimum location of P is the centroid of triangle ABC, and that this location does not change even when h changes.

Conclusion

The location P that minimizes the sum of the distances between P and the three vertices of the isosceles triangle is the centroid when P lies on the perpendicular bisector of the base. This location is independent of h for \(h\geq\frac{2}{\sqrt{3}}\).

Key concepts

Perpendicular Bisector

A perpendicular bisector is a line that divides a line segment into two equal parts at a 90-degree angle. In the context of our isosceles triangle, the base of the triangle is bisected by the perpendicular bisector. Let's break down this concept further:

• The base of the isosceles triangle is 4 units long.
• The perpendicular bisector divides the base into two equal segments of 2 units each.
• This bisector is also the line of symmetry for the isosceles triangle, meaning that it passes through the vertex opposite the base (vertex C) and the midpoint M of the base.

Understanding the role of the perpendicular bisector helps in finding point P because P must lie on this line. As point P moves along the perpendicular bisector, the distances to the vertices of the triangle change. Our aim is to find the position on this line that minimizes the sum of these distances.

Triangle Centroid

The centroid of a triangle is the point where all three medians intersect. It is often referred to as the "center" or "balance point" of the triangle. For our isosceles triangle, the centroid is significant when solving the optimization problem of minimizing distances. Here’s why:

• The centroid can be found using the formula: \[G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)\]where \(x_1, y_1\), \(x_2, y_2\), and \(x_3, y_3\) are the coordinates of the triangle's vertices.
• In an isosceles triangle, the centroid lies on the perpendicular bisector as it is symmetric, aligning it perfectly with our criteria for point P.
• When point P is precisely at the centroid, the sum of the distances from P to the vertices is minimized. This is a key property leveraged in the given optimization task.

Thus, locating the centroid is essential because it offers a reliable point that balances all sides of the triangle perfectly.

Triangle Inequality

The triangle inequality is a fundamental principle in geometry that states that for any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. This principle is crucial for ensuring the possibility of triangle formation.

• For our specific scenario involving the isosceles triangle, the triangle inequality helps us set bounds for distances PA, PB, and PC.
• This inequality support the arrangement of vertices such that the sum of the segment lengths forms a valid closed triangle.
• When calculating the sum of distances from point P to each vertex, the inequality also ensures that indirect paths are considered less optimal than direct paths, leading us closer to the correct positioning of point P.

By applying the triangle inequality, we ensure that when minimizing the distances, the sums are valid within the geometric constraints, ensuring an accurate and applicable solution.