Subject: Calculus

# Regiomontanus Angle Maximization Problem

Johannes Müller, called Regiomantus after his birthplace of Königsberg, posed the question, “At what horizontal distance from an elevated rod would a person have to stand such that the appearance of the rod should be a maximum?” in 1471 to professor Christian Roder of Erfurt. This problem, which in itself is not difficult, nevertheless is of note as being the first extremal problem in mathematics since antiquity.

Presenting this problem in terms of data analysis can easily be done by measuring the inscribed angles formed between the “ground” and the rod and constructing a table of values with x the horizontal distance from the bottom of the suspended rod, and y the measure of the angle. This can be represented by the following figure

Working equation: \theta = \tan^{-1}\frac{AO}{X} - \tan^{-1}\frac{BO}{X}, where \tan{\beta}=\frac{AO}{X} and \tan{\alpha }=\frac{BO}{X}. AO and BO are given quantities. Define AO = A, and BO = B. However, a dispute over the existence of the solution by Regiomontanus after Eli Moar was not able to verify it. Moar does however offer the following argument based on elementary methods that would have been available to Regiomontanus.

Replacing \cot{\beta} with \frac{X}{A}, and \cot{\alpha} = \frac{X}{B}. We can therefore express \cot{\theta} as

The advantage of expressing \cot{\theta} as the sum of two terms is to take advantage of a theorem from algebra that states that the arithmetic mean of two positive numbers is never smaller than their geometric mean. That is, (c + d)/2 is greater than or equal to \sqrt{cd}. This can be seen if we note (\sqrt{c}- \sqrt{d})^2 is greater than or equal to zero, in other words the square of a real number can never be negative. If we let \frac{X}{A- B} = c and \frac{AB}{(AX - BX} = d, then \cot{\theta}=c+d. But recall, the cotangent function decreases over the interval (0_o, 90_o).

Therefore, a maximum angle--which is what we’re looking for--will occur for a minimum value of \cot{\theta}. \frac{c + d}{2} has its minimum when it equals \sqrt{cd} which can only occur when c = d, in which case, \frac{X}{A - B} = \frac{AB}{AX - BX}, giving X =\sqrt{AB}.. In other words, the optimum distance for viewing is the geometric mean of A and B.

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