Given the distances from an interior point to three vertices of a rectangle, find the maximum and minimum possible areas of the rectangle.

A.Casares M.
January 6th, 2017

The original problem and the actual data:

plot of chunk fig1 For instance:
Let BX = 9, CX = 7, AX = 3

Given AX, BX and CX, is the 4th distance DX determined?

  • Answer: Yes, with Pythagoras' help.
  DX <- sqrt(BX^2+CX^2-AX^2)
  cat(sprintf('DX = %3.1f',DX))

DX = 11.0

The general question:

The point X moves along a circle with r=AX and center A.

When X moves, the rectangle adjusts its dimensions.

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To see it and examine the problem's behaviour:

  • You can choose the three data distances in a select box.
  • And vary the polar angle BAX through a slider control.
  • The figures react to these changes. You can see how the area keeps changing.

    Between what limits?

Curve of areas as function of polar angle.



plot of chunk fig2

Empirical procedure:

  • Areas computed in each point of an uniform sample.

  • The extreme areas may be easily computed from it.

  • The figure shows the answers for a set of distances

  • The minimum value of the area is at angle BAX = 0 degrees.

Analytical - Numerical Procedure

Finding the analytical function and its derivative is hard to do.

plot of chunk fig3

  • Looks for angles making the derivative null, without using the cumbersome function.

  • Numerical derivatives, shown in blue, exhibit good linear pattern.

  • A linear model fits with high R-square.

  • Answers are practically identical, and close to empirical ones.