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December 31st, 2009, 07:27 AM   #1
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Suspended compass geometry problem

Hi

A friend of mine showed this problem he translated from a olympiad but I can't solve it:
"Suspended compass
Consider a perfectly symmetrical compass. It is constituted by two rigorously equal connecting rods, that join in a vertex the one that we call “pivot”. The opening angle is regulable. Imagine that we suspend the compass for the tip of one of the connecting rods, attaching it to a wire whose other end is attached at the ceiling.
a) Sketch the position of the bar at various angles of opening.
b) What is the opening angle of the compass, so that the pivot is the highest possible?
Hint: The center of mass of the bar, assuming that the rods have densities
uniform is the point where the bisector of the angle of the bar crosses the line
passing through the midpoint of both rods. The mass center has to be, for any angle of opening, in the vertical line of the point of suspension of the wire."

Sorry for the english in the translation. My question is how do you solve this geometrically (I know it seems like physics but the problem is all about geometry)
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December 31st, 2009, 11:06 AM   #2
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Which contest is the problem from? Please post what you have done so far for parts (a) and (b).
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December 31st, 2009, 11:11 AM   #3
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Re: Suspended compass geometry problem

The Problem I believe is from a Portuguese Physics (although this problem is merely geometrical) Olympiads and as for part a) I could complete it. Regarding part b) however I can't solve it.

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December 31st, 2009, 11:19 AM   #4
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Can you post an annotated sketch (i.e., with a letter assigned to each relevant point), so that there is something to refer to?
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December 31st, 2009, 11:52 AM   #5
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Re: Suspended compass geometry problem

I can't post one because I have no scanner and I can't draw it on paint, but please explain me how to do it and assign your letter in process.

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December 31st, 2009, 10:31 PM   #6
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[attachment=0:33cjide6]compass.jpg[/attachment:33cjide6]
Above is a rough diagram (created using paint). P is the pivot and the points M and L are the midpoints of PR and PQ respectively. The center of mass, C, is the midpoint of ML. CN is constructed to be perpendicular to PR.

Let x and y denote angles MPC and MRC respectively. You need to find 2x such that y is maximized. You can take PM = MR = 1.

Show that CN = sin(x)cos(x) = sin(2x)/2.
Show that RN = 1 + CN*tan(x) = (3 - cos(2x))/2.

When tan(y) = CN/RN = sin(2x)/(3 - cos(2x)) is maximized, so is y.

I proceeded from there using calculus, but there may well be an alternative geometrical method.
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