A Curious Geometric Limit

Published by patrick honner on

Inspired by Alexander Bogomolny (@CutTheKnotMath) and his curious geometric limit, I offer this trigonometric approach.

Consider the following diagram in which we have circle X of radius 3 centered at (3,0) and circle O of radius k centered at the origin.  Call the intersection of the these circles in the first quadrant B.  Let A be the intersection of circle O with the y-axis, and extend line AB until it intersects the x-axis at E.  Our goal is to show that \lim_{kto}OE=12.

Consider the diagram below.


Let \angle{OXB} = \theta.  Some simple angle chasing, using properties of isosceles triangles and right triangles, gives us that \angle{OEA} = \frac{\theta}{4}.

Since AO = k by definition, we have tan(\frac{\theta}{4}) = \frac{k}{OE}, and so OE = \frac{k}{tan(\frac{\theta}{4}) }.

Since OB = k is the base of isosceles triangle OXB, we can drop an altitude from X to OB and find that sin(\frac{\theta}{2}) = \frac{\frac{k}{2}}{BX} = \frac{\frac{k}{2}}{3}.    This gives us k = 6sin(\frac{\theta}{2}) .

So OE =  \frac{k}{tan(\frac{\theta}{4}) } =  \frac{6sin(\frac{\theta}{2})}{tan(\frac{\theta}{4}) }.  By the double angle formula for sine, we have

sin(\frac{\theta}{2}) = 2sin(\frac{\theta}{4})cos(\frac{\theta}{4}),

Making this substitution, and by writing tangent as the quotient of sine and cosine, we get

OE = \frac{6sin(\frac{\theta}{2})}{tan(\frac{\theta}{4}) } = \frac{12sin(\frac{\theta}{4}) cos((\frac{\theta}{4}))}{\frac {sin(\frac{\theta}{4})}{cos(\frac{\theta}{4})} } = 12 {(cos(\frac{\theta}{4}))}^2.

Finally, since \theta\to 0 as k \to 0, we have

\lim_{k\to 0}OE = \lim_{\theta \to 0}12 {(cos(\frac{\theta}{4}))}^2 = 12\lim_{\theta \to 0} {(cos(\frac{\theta}{4}))}^2=12.

patrick honner

Math teacher in Brooklyn, New York


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