Problem of the Month for November, 2005

Problem:

Four houses, labeled A, B, C, and D, are located at the corners of a convex quadrilateral. The occupants of these houses wish to share a gazebo. Is there a unique place to build the gazebo so that the sum of the distances from the gazebo to their houses is as small as possible? If so, find such a place and explain why it minimizes the sum of the distances. If not, find more than one place where the sum of the distances is minimized and explain why each of those places minimizes the sum of the distances.

Note:

A purely geometric approach (perhaps using, for example, line segments, distances, triangles, intersections, bisections, etc) is probably a better way to go than attempting to use Calculus on this problem.

Possible generalizations you might consider:

If one of the four houses is inside the triangle formed by the other three, where should the gazebo be built to minimize the sum of the distances from the gazebo to the houses? What if there are only three houses? Where should the gazebo be in that case?

Answer:

There is a unique place to build the gazebo so that the sum of the distances from the gazebo to their houses is as small as possible. Draw a diagram with A, B, C, and D as shown below. Now draw the diagonals AC and BD as shown. The intersection P of these diagonals is the place to build the gazebo.


Let Q be any point in the plane (as an example to visualize, take Q to be a point inside the quadrilateral ABCD but not the same as P). Since the shortest distance between any two points is along a straight line, we can say that, for instance, d(AQ)+d(QC)≥d(AC), where this is saying that the distance from A to Q plus the distance from Q to C is greater than or equal to the distance from A to C (we are considering the triangle AQC when we think about this). In fact, we would have d(AQ)+d(QC)=d(AC) if and only if Q lies along the diagonal AC.

In a similar way, d(BQ)+d(QD)≥d(BD) with equality if and only if Q lies along the diagonal BD. Adding these two inequalities gives d(AQ)+d(QC)+d(BQ)+d(QD)≥d(AC)+d(BD)=d(AP)+d(PC)+d(BP)+d(PD) with equality if and only if Q = P (Q lies on both diagonals). This shows that P minimizes the sum of the distances from the gazebo at P to their houses at A, B, C, and D.

If one house is inside the triangle formed by the other three, put the gazebo at the same spot as the inner house. If there are three houses, put the gazebo at the so-called Fermat-Torricelli point. I.ll let you look this up on the Internet.