Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a question from an old released exam.

enter image description here

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?

share|improve this question
    
"$s$ 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

3 Answers 3

up vote 7 down vote accepted

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 $-\cos\alpha=-\cos 110^\circ$.

share|improve this answer
    
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

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

share|improve this answer

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)) $.

share|improve this answer
    
Why was I downvoted? My answer was independent and correct. –  marty cohen Sep 1 '12 at 3:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.