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This is a question from an old released exam.

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By the triangle inequality, $s-r<1$, so I eliminate answers D and E. Intuitively, since the lower angle between $1$ and $r$ is fixed at $110^\circ$, $s$ will always be a little longer than $r$, so I eliminate A and C to find B as the correct answer.

This is pretty informal, is there a more rigorous way one could prove the limit?

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    Hint: suppose the given angle were 90 degrees, and the angle between$r$and$s$were "a". Then you would have "r = scos(a)" As$r$and$s$increase without bound, a tends to 0, so cos(a) tends to 1, and you have r = s, so "s - r" tends to 0. Now drag out the law of cosines and see what happens given one angle at 110 and one side fixed at 1.2012-08-31

3 Answers 3

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From the cosine law,

$s^2=1+r^2-2r\cos\alpha,$

where $\alpha=110^\circ$.
We can rewrite this as

$s-r=\frac{1-2r \cos\alpha}{s+r}$

and remember that $\cos\alpha \lt 0$.
As you say, $0 \lt s-r \lt 1$, so

$\frac{1-2r \cos\alpha}{2r+1} \lt s-r \lt \frac{1-2r \cos\alpha}{2r}.$

The left inequality is bounded away from $0$ and the right from $1$, so the answer is B. In fact, we can say

$\frac{1/r-2 \cos\alpha}{2+1/r} \lt s-r \lt \frac{1/r-2 \cos\alpha}{2}$

so the limit is $-\cos\alpha=-\cos 110^\circ$.

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    @Dedede: I made the limit explicit.2012-08-31
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A qualitative answer.

When $s$ and $r$ tend both to infinity, keeping the said elements fixed, the sides $s$ and $r$ tend to be parallel, so their difference is the projection of the third side on their common direction, given by

$s-r\to|1\cdot\cos\alpha|$

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By the cosine law, $c = \cos 110 = \frac{1+r^2-s^2}{2r}$, or $2 r c = 1 + r^2 - s^2$.

Let $d = |c|$, so $d > 0$. Since $s^2 = r^2 +2rd+1$, $s = r\sqrt{1 + 2d/r + 1/r^2} =r (1+d/r + O(1/r^2)) = r + d + O(1/r)$, so $ s-r \to d = -\cos 110$.

A slightly modified look:

$s^2 = r^2+2rd+d^2 + 1-d^2 = (r+d)^2+1-d^2$, so $s = \sqrt{(r+d)^2+1-d^2} =(r+d)\sqrt{1 + (1-d^2)/(r+d)^2} = (r+d)(1 + O(1/(r+d)^2)) = r+d + O(1/(r+d)) $.

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    Why was I downvoted? My answer was independent and correct.2012-09-01