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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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"$s$ will always be a little longer than $r$, so I eliminate A and C" - I don't think this is a good argument since for the sequence of numbers $a_{n}=\frac{1}{n}$ every element is always greater than $0$ but the limit is $0$. so $s-r<1$ does not imply it also holds in the limit – Belgi Aug 31 '12 at 20:33
@Belgi Right, I'm not fond of the argument either. My idea was that by taking an auxiliary line from the lower left vertex perpendicular to $r$ will always leave some portion of $s$ to the left of it. – Dedede Aug 31 '12 at 20:38
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. – thisfeller Aug 31 '12 at 20:52
up vote 7 down vote accepted

From the cosine law,


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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Thanks Ross, I follow this now. – Dedede Aug 31 '12 at 20:55
@Dedede: I made the limit explicit. – Ross Millikan Aug 31 '12 at 22:35

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


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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. – marty cohen Sep 1 '12 at 3:06

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