Project Euler Lab - Problem 363

#363 - Bézier Curves

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A cubic Bézier curve is defined by four points: \(P_0, P_1, P_2,\) and \(P_3\).

0363_bezier.png

The curve is constructed as follows:

On the segments \(P_0 P_1\), \(P_1 P_2\), and \(P_2 P_3\) the points \(Q_0, Q_1,\) and \(Q_2\) are drawn such that \(\dfrac{P_0 Q_0}{P_0 P_1} = \dfrac{P_1 Q_1}{P_1 P_2} = \dfrac{P_2 Q_2}{P_2 P_3} = t\), with \(t\) in \([0, 1]\).

On the segments \(Q_0 Q_1\) and \(Q_1 Q_2\) the points \(R_0\) and \(R_1\) are drawn such that
\(\dfrac{Q_0 R_0}{Q_0 Q_1} = \dfrac{Q_1 R_1}{Q_1 Q_2} = t\) for the same value of \(t\).

On the segment \(R_0 R_1\) the point \(B\) is drawn such that \(\dfrac{R_0 B}{R_0 R_1} = t\) for the same value of \(t\).

The Bézier curve defined by the points \(P_0, P_1, P_2, P_3\) is the locus of \(B\) as \(Q_0\) takes all possible positions on the segment \(P_0 P_1\).
(Please note that for all points the value of \(t\) is the same.)

From the construction it is clear that the Bézier curve will be tangent to the segments \(P_0 P_1\) in \(P_0\) and \(P_2 P_3\) in \(P_3\).

A cubic Bézier curve with \(P_0 = (1, 0), P_1 = (1, v), P_2 = (v, 1),\) and \(P_3 = (0, 1)\) is used to approximate a quarter circle.
The value \(v \gt 0\) is chosen such that the area enclosed by the lines \(O P_0, OP_3\) and the curve is equal to \(\dfrac{\pi}{4}\) (the area of the quarter circle).

By how many percent does the length of the curve differ from the length of the quarter circle?
That is, if \(L\) is the length of the curve, calculate \(100 \times \dfrac{L - \frac{\pi}{2}}{\frac{\pi}{2}}\)
Give your answer rounded to 10 digits behind the decimal point.

This problem is taken from Project Euler, Problem 363.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=363. Published Sunday, 18th December 2011, 10:00 am. Solved by 1,336 members at time of mirroring.

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