Help with geometry problem required

[Harald]

Since we have had a few other off topics here recently, I thought I would have a go at that, too.

Hopefully my pigeon English is good enough to describe the problem:

Imagine a circle of the radius R is divided into four equal sectors. What is the radius r of the biggest possible circle that can be placed inside each of the four sectors in such a way, that the outline of the sector and the outline of the inner circle meet at positions equally spaced (ie 120°) on the circumference of the inner circle.

Cheers,
Harald

[Walter B]

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Hopefully my pigeon English is good enough to describe the problem:

Imagine a circle of the radius R is divided into four equal sectors. What is the radius r of the biggest possible circle that can be placed inside each of the four sectors in such a way, that the outline of the sector and the outline of the inner circle meet at positions equally spaced (ie 120°) on the circumference of the inner circle.

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Can you send the pigeon for another lesson, please?

d;-)

[fhub]

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Imagine a circle of the radius R is divided into four equal sectors. What is the radius r of the biggest possible circle that can be placed inside each of the four sectors in such a way, that the outline of the sector and the outline of the inner circle meet at positions equally spaced (ie 120°) on the circumference of the inner circle.

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Harald, I guess you should post an image of your problem, because my understanding of your text is the following:

And of course this can't be what you mean, because there's only one such inner circle possible, and the points A and B would always build an angle of 90° !?

Franz

[Maximilian Hohmann]

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And of course this can't be what you mean...

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Why not just put the originial text (in german language I suppose) here?

[Harald]

Thanks for the drawing Franz. How did you draw that? I haven't managed with the tools I have to hand.

The problem is not just the pigeons poor English, it has also mad a mistake ;)

Here comes the correction:

Imagine a circle of the radius R is divided into four equal sectors. What is the radius r of the biggest possible circle that can be placed inside each of the four sectors in such a way, that lines from the center of the inner circle, equally spaced at 120°, meet the outline of the sector at identical distances from the center of the inner circle ie at the same radius (that radius is bigger then r of course).

[Walter B]

The pigeon must be the problem still. Seems it can't set points ;-)

d:-)

[Walter B]

Apparently I was too polite in expressing my request: I concur with Maximilian. (Bitte erst mal auf Deutsch die Aufgabe stellen - ich vermute, dass sie schon da nicht verständlich und eindeutig formuliert ist. Wie soll sie dann aber nach der Übersetzung besser werden? Übrigens kannst du der Taube mal beibringen, dass man i.e. - falls d.h. gemeint sein sollte - mit Punkten schreibt.)

d:-I

[Harald]

Sorry Walter, I was at work and *thought* I had a few minutes to spare to post the problem. But as always something came up and it was only 2 minutes ;)

There is no original question in German for this, and you are probably right in assuming, that given the short space of time, I would also have messed this up writing in German.

And yes, the pigeon knows that it is i.e., it is just the spell check at work that doesn't. It's not a good idea when computers start "thinking".

[fhub]

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How did you draw that?

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Well, I've just used the simple 'Paint' program included in Windows.

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Here comes the correction:

Imagine a circle of the radius R is divided into four equal sectors. What is the radius r of the biggest possible circle that can be placed inside each of the four sectors in such a way, that lines from the center of the inner circle, equally spaced at 120°, meet the outline of the sector at identical distances from the center of the inner circle ie at the same radius (that radius is bigger then r of course).

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Oh my god, I still can't imagine how this should look like. :-(

It's indeed better if you post a picture ...

Franz

[Harald]

Ok. I'll give it a try when I get home.

[Harald]

Here we go. Not very pretty, but hopefully good enough:

R is the radius of the big circle, r is the radius of the small circle. x is the length of the 3 vectors pointing from the centre of the small circle to the outline of the sector. The vectors are all 120° apart and have the same length.
Unknown is the position of the centre and the biggest possible radius r under the constraint that the small circle is entirely within the sector of the big circle.

Sorry for the earlier attempts at describing the problem without a drawing and not enough time to do it...

Edited: 21 Jan 2013, 2:01 p.m.

[Mark Storkamp]

(R*cos(15))/(sqrt(2)*cos(15)+1) is the radius I got for the smaller circle.

[Harald]

Could you explain how you got there?

[fhub]

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(R*cos(15))/(sqrt(2)*cos(15)+1) is the radius I got for the smaller circle.

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Yes, that's what I also got, but it's the length x in his picture (i.e. the length of the 3 vectors). If we assume that r would be x, then of course this 'inner' circle is not really within the quarter-circle, it crosses the radius of the outer circle.

I'm still not sure what this r should have to do with x!? In other words: should we now find the biggest circle (with _this_ center of the 3 vectors) which fits in the quarter-circle?

Franz

[Mark Storkamp]

The inner circle does not touch the outer circle, but it touches the two perpendicular lines. I solved for the center of the circle first by giving the 3x120 degree lines a length of 1, then drawing lines to the smaller center, perpendicular to the sides of the wedge. Solve for those, then multiply by sqrt(2) the get diagonal to the center point. Now that plus 1 is R. Now you know the length of the 3x120 lines in terms of R, then multiply by cos(15) so the circle touches the perpendicular lines. (haven't figured out how to include drawings yet).

[fhub]

Ok, I've overlooked that you've multiplied R with cos(15) in the numerator. I have calculated x and my formula looked similar to yours (despite of the *cos(15)), so I first thought your result would also be x, but everything is clear now.

Nevertheless I can't believe that this is really what the original problem was intended for!?

Franz

[Harald]

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Should we now find the biggest circle (with _this_ center of the 3 vectors) which fits in the quarter-circle?

Franz

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Yes, that is exactly what the question is.

Harald

[fhub]

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Yes, that is exactly what the question is.

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Well, then Mark's result is what you've looked for. This radius is just r=x*cos(15), and x is easy to calculate from the triangle with 2 sides R-x and x, and the height r.

Franz

[fhub]

Here's the way I solved it:

(not an exact construction, just a freehand drawing)

(R-x)*sin(45)=x*cos(15) (=r)

So first calculate x=... and then r=x*cos(15)

Franz

Edited: 21 Jan 2013, 4:14 p.m.

[Harald]

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Here's the way I solved it:

(not an exact construction, just a freehand drawing)

(R-x)*sin(45)=x*cos(15) (=r)

So first calculate x=... and then r=x*cos(15)

Franz

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Ok, thanks for that. Now that you have shown the solution, I don't know why I didn't get there myself. I just didn't spot that triangle there....

[Eric Smith]

The inner circle touches the x and y axes but not the outer circle.

If I've done all of my trig correctly, the radius of the inner circle as well as the x and y coordinates of its center are sqrt(2)(3+sqrt(3)) / 6 (1+sqrt(3)).

[Walter B]

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The inner circle touches the x and y axes but not the outer circle.

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If true then the angle between those two touches is 90° - in contradiction to the required 120°.

d:-)

[Eric Smith]

My interpretation of the problem may be wrong, but it wasn't that the three vectors were radii of the small circle, just that they defined the point that was the center of the small circle. I thought that they couldn't all be radii of that circle, because the circle cannot be tangent to both axes and the outer circle.

[Harald]

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My interpretation of the problem may be wrong, but it wasn't that the three vectors were radii of the small circle, just that they defined the point that was the center of the small circle. I thought that they couldn't all be radii of that circle, because the circle cannot be tangent to both axes and the outer circle.

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Your interpretation is correct.
And the inner circle touches both axes and not the outer circle.

[fhub]

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If I've done all of my trig correctly, the radius of the inner circle as well as the x and y coordinates of its center are sqrt(2)(3+sqrt(3)) / 6 (1+sqrt(3)).

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Hmmm? Shouldn't this depend on R? Or what value for R did you assume?

My calculation (shown above) gives x=R*(1-sqrt(3)/3) and the inner radius r=R*sqrt(6)/6=0.4082483*R after simplifying sin(45°) and cos(15°).

Franz

[Eric Smith]

I assumed R=1.

[fhub]

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I assumed R=1.

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Ok, I see now what the problem was in your formula: you should have surrounded the denominator 6 (1+sqrt(3)) with (...). I've simply copied your expression and got a complete nonsense result, so I thought it would be wrong - but in fact only the () was missing. :-)

Franz

[Jim Horn]

Unless I'm mistaken, the problem as stated is self contradictory.

If a smaller circle tangentally touches the larger circle and its two normal diameters (as it has to be if contained within the sector), the contact points *have to be* 135°, 135° and 90° apart. If it touches the normal diameters with points 120° apart, it will have to touch them at two points each and protrude into two of the other sectors. Changing the size of the smaller circle won't change this last situation.

If it *doesn't* touch the two normal diameters, then the 120° angle between contact points won't apply as they won't exist.

So there aren't contact points, they are at 90°, or the circle protrudes into other sectors - all of which are prohibited by the problem. So there is no answer...

(Can God create problems too difficult for Him to solve? Godel: Yes...)

[Kiyoshi Akima]

There's no contradiction here.

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If a smaller circle tangentally touches the larger circle and its two normal diameters...

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Who or what says it touches the larger circle AND its two normal diameters? It is possible for a circle to fit inside another geometric figure without touching its perimeter. The fact that we're looking for the largest circle that fits implies that there is at least one point of contact, but not necessarily three.

[Harald]

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Unless I'm mistaken, the problem as stated is self contradictory.

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No, it doesn't. At least not if I had managed to explain the problem in an understandable way.

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If a smaller circle tangentally touches the larger circle and its two normal diameters (as it has to be if contained within the sector), the contact points *have to be* 135°, 135° and 90° apart. If it touches the normal diameters with points 120° apart, it will have to touch them at two points each and protrude into two of the other sectors. Changing the size of the smaller circle won't change this last situation.

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Correct.

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If it *doesn't* touch the two normal diameters, then the 120° angle between contact points won't apply as they won't exist.

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Also correct. The way this is meant to be, there are 2 contact points. But in order to get the drawing to correctly show this, I would have had to a) have dividers to draw the circles properly and b) have solved the problem to know how to set them first.

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So there aren't contact points, they are at 90°, or the circle protrudes into other sectors - all of which are prohibited by the problem. So there is no answer...

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Not correct. the solution is to reduce the radius of the inner circle until there are only two contact points left.

[Walter B]

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Not correct. the solution is to reduce the radius of the inner circle until there are only two contact points left.

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Two?!?

d:-?

[Harald]

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Two?!?

d:-?

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Or one, depending on where the centre of the circle is. The inner circle is either touching the outer one, or it is touching the two perpendicular lines of the sector.
I assumed the latter.

But since the answer has been posted, I'll try and understand how to get it now.

Edit: And as Franz's drawing shows, there actually are two contact points

Btw, the forum is loading painfully slow for me today. Is anybody else experiencing that too? Took me several minutes to get this edit done.

Edited: 21 Jan 2013, 4:31 p.m.

[Jean-Michel]

CAD software gives me : radius of great cirle being equal to 1, corresponding radius of the inner circle is approx. 0.423
Sorry, I didn't find out how to post an image (copy & paste is no help).

[Jean-Michel]

Trigs gives me: (assuming R = 1)

r (inner circle) = R (great circle) / (1 + cos15/sin45) = 0.4226...

Edited: 22 Jan 2013, 4:34 a.m.

[fhub]

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Trigs gives me: (assuming R = 1)

r (inner circle) = R (great circle) / (1 + cos15/sin45) = 0.4226...

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That's wrong, I'm afraid. Your result is x but not r, a factor *cos(15°) is missing.

It should be r=R*cos(15°)/(sqrt(2)*cos(15°)+1)=R*sqrt(6)/6=0.4082483*R

Franz

Edited: 22 Jan 2013, 4:56 a.m.

[Walter B]

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Sorry, I didn't find out how to post an image (copy & paste is no help).

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There are instructions right on the page were you write your post.

d:-)

[Jean-Michel]

Güten Tag Walter

sorry, but http//... does not speak to me ;-) when it matters to cpy and paste. That's what I get when pushing the "Image" button

[Walter B]

Bonjour Jean-Michel,

Put your drawing at a nice free space in the web and you can refer to it. Or get a guest directory by asking Dave (the curator) - I did it this way, for example. Copy & paste don't work, as you found out yourself. HTH

d:-)

[C.Ret]

Hi,

My best investigation and computation of radius r and vector length x give me the following figure.

As can be seen, the three vectors length is about x= 0.4226 (based on unit great circle R = 1 ) and the inner circle is limited by orthogonal axis of the quadrant r = 0.4082 (when great circle is R = 1).

This show that actually to contact points are present respectively I_x and I_y.

The (OA) segment is one radius of the great (red) circle; OA = R leading to eq.1

As explain in this threat, due to symmetry, the angle between O'C vector (respectively O'B) and vertical axe O'I_x (respectively horizontal axe O'B_x) is 15‹ leading to eq.2

Solving this system as a function of R is trivial:
(eq.3)

Value of r is deduce from (eq. 2).

One may also use this code to get numerical result on his ageless HP classic:

01 LBL A

02 1

03 DEG

04 2 

05 sqrt x

06 GSB 9

07 +

08 ÷

08 ENTER^

09 LBL 9

10 1

11 5

12 COS

13 *

14 RTN

Enter radius R of great circle.Execute code. Radius of inner circle is displayed, lenght of vector is store in stack register y:

5 [ A ] 2.0412 [ x<>y ] 2.1132

Edited: 22 Jan 2013, 11:50 a.m. after one or more responses were posted

[Walter B]

Thanks! Very nice drawing!

d:-)