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I have the following problem:

In my class, we did a majorly complicated method to figure this out but I think there is a better way to do this... Here is the exact problem:

A belt fits snugly around the two circular pulleys shown.

Find the distance between the centers of the pulleys. Round to the nearest hundredth.

The method we used required that I create this huge triangle across the page. It didn't sound like the best way to do it. Does anybody have a "none-hacked" way to do it (as in, a more straight forward procedure)?

EDIT

I'm sorry about this, yes, the diagram is misleading, I had to recreate the image in an annoying program. Yes, lines RS and line QP are both tangent to both circles. Sorry about that, I never explained the context of the question.

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Does snugly means the lines through $QP$ and $RS$ are tangent to both circles. And hence $SNP$ and $RMQ$ are not straight lines? Diagram is misleading in this respect. –  Eric Naslund May 24 '11 at 16:37
    
@Eric Naslund Sorry, I just updated the details. Yes, thanks for pointing that out. –  Xan May 24 '11 at 16:43
    
Good programs do draw things like this are GeoGebra and CarMetal, where you can define the RELATIONSHIPS between geometrical elements, add labels, etc, and take a screenshot. –  heltonbiker Mar 22 '12 at 17:50

1 Answer 1

up vote 6 down vote accepted

Assuming that "fits snugly" means that PQ is tangent to both circles, at Q and P respectively: Note that the radii MQ and NP are both perpendicular to PQ (as it is tangent). This means that MQ and NP are parallel. Draw a line through N parallel to PQ, meeting MP at Q', say.

Gears: Direct common tangent

So QPNQ' is a rectangle, and NQ'=PQ=14 (in length). The part of MQ "above" Q', namely MQ', has length MQ - Q'Q = MQ - NP = 5 - 4 = 1.

This means that MN, the distance between the centres, is the hypotenuse of a right-angled triangle MQ'N with MQ'=MQ-NP and Q'N=QP, so $$\text{MN} = \sqrt{(\text{MQ}-\text{NP})^2 + \text{QP}^2} = \sqrt{(5-4)^2 + 14^2} = \sqrt{197} \approx 14.04.$$

In general, if the length of the PQ-like part (distance between points of tangency) is $l$, and the circles have radii $r_1$ and $r_2$, then the distance between the centres is $\sqrt{l^2 + (r_1-r_2)^2}$.

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Our resulting answer was 14.03 so you may be on to something (we may have rounded weird). It's going to take me a minute to decipher your explanation though :) –  Xan May 24 '11 at 16:51
    
Nope, you got it! :D –  El'endia Starman May 24 '11 at 16:52
    
Well, taking a look though, how can we assume RQ and SP is a straight line? They can't because one is smaller (the circles) so, for them all to be tangent, SNP appears to be almost a straight angle or RMQ could be almost a straight angle or even both. I'm confused as to how you figured them parallel (MQ and NP). I'll take another look at it. –  Xan May 24 '11 at 16:57
    
Hmmm... OK, you are right, thanks so much! I must just be a little bit slow, math class ended 3 hours ago and so must my brain :) –  Xan May 24 '11 at 16:59
    
@DalexL: I did not assume RMQ and SNP are straight lines; in fact we don't need to care about R and S (about which we directly know nothing useful, anyway). MQ and NP are parallel because they are both perpendicular to PQ. Let me try drawing a figure, though I'm very bad at such tools... –  ShreevatsaR May 24 '11 at 17:01

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