Approximating Pi From the Law of Sines
One correct approach to approximate pi is to inscribe triangles and apply the “law of sines”.
Start with the unit circle, and draw the diameter as a horizontal line. Find the centre of the circle; each half of the diameter is the radius: “r=d/2=1/2”.
Draw a vertical diameter. Draw a line from the endpoints of these diameters, in the upper right quadrant of the circle. That is a right-angle isosceles triangle. The sides are in a “1-1-root(2)” proporation. Each short side is a radius with the length 1/2, so the hypoteneuse is the square root of one-half, or “(1/2)(1/2)“.
(The Pythagorean Theorem for an isosceles right triangle is “r2 + r2 = h2“. Since “r=1/2”, we have “r2=(1/2)2=1/4″. Then “r2 + r2 = 1/4 + 1/4 = 1/2 = h2“. Therefore “h=1/21/2=0.707…”).
We can only fit four of these triangles into the circle. The perimeter of the square formed by their hypoteneuses is “4*(1/2)1/2=4*0.707=2.828…”.
Then we can draw more isosceles triangles from the centre to the circle, with the angle at the centre smaller than 90 degrees. By repeatedly bisecting the angle, we would have triangles that need to be copied 8, 16, 32…2n times to “fill” the circle.
We could measure the length of the side of the triangle that forms a closer perimeter, and determine how closely we have approximated pi.
Let’s calculate the length by using the law of sines.
The Law of Sines
Let triangle ABC have angles <A, <B and <C at those vertices; and sides of length ‘a’, ‘b’ and ‘c’ opposite the vertices ‘A’, ‘B’ and ‘C’ respectively.
The Law of Sines states that a/sine(A) = b/sine(B) = c/sine(C) for any triangle.
Use the Law of Sines on Isosceles Triangles in a Circle
Let ‘C’ be the centre of the circle, and <C is the acute angle at the centre for triangle ABC where ‘A’ and ‘B’ are points on the circle.
The length of each side ‘a’ and ‘b’ is a radius of the circle, already known as ‘r’, so “c/sine(C)=r/sine(A)=r/sine(B)”.
The interior angles of a triangle sum to 180 degrees, so “A+B+C=180”, and therefore “A+B=180-C”.
Since angles ‘A’ and ‘B’ are equal, “2*A=180-C”. Thus “A=(180-C)/2”.
Then “c/sine(C)=r/sine(A)=r/sine((180-C)/2)”. We solve for ‘c’ by multiplying both sides by sine(C).
“c=r*sine(C)/sine((180-C)/2)”. Since the radius is one unit, this simplifies to “c=sine(C)/sine((180-C)/2)”.
This spreadsheet shows the values of ‘C’ (“Angle”), sine(C) (“Sine(Angle)”, ‘c’ (“Length”), the number of “Sides”, and the product of the number of sides times the length (“Perimeter”). Note how the length of the perimeter of the multi-triangle figure approaches the value of the circumference, pi.
The final column shows that the rounded error of the “perimeter – π” does shrink as the angle decreases.
Conclusion of the Circled Square Math Problem
The troll’s math problem argued that one process of altering a unit square could make it resemble a unit circle with an arbitrarily small error. Since the new figure’s perimeter remained constant length, 4, the troll concluded that the circumference of the circle must be 4.
That argument is flawed. The simplest disproof is to measure the circumference of the circle. This math problem also suffers from the flaw that successive iterations give the same value, which is not the value derived from the actual geometric figure. The sequence implies that the limit value is four; yet we know that the actual limit of the smooth geometric shape is π.
Finally, we can use a similar iterative process to approximate the value of π with arbitrary precision, using the Law of Sines.
Yahoo Answers. Shouldn’t Pi be equal to 4? (2011). Yahoo Answers. Referenced March 27, 2013.
Things of Interest. Troll Pi Explained. (2010). Referenced March 27, 2013.
Weisstein, Eric W. Law of Sines. MathWorld-A Wolfram Web Resource. Referenced March 27, 2013.
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