Geometry puzzle

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Hertafeld
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Geometry puzzle

Postby Hertafeld » Thu Nov 05, 2009 11:11 pm UTC

My friend gave me a puzzle today in which three circles with radius 1 are lined up, with the middle one intersecting each of the other two at a point, like so: OOO. They are inscribed in a semicircle so that the diameter intersects each circle along the bottom and the curved part intersects each of the end circles. My task is to find the radius of the big circle.

Sorry for the odd description; I attached a picture. It isn't perfect but you get the idea.

I'm sure you all will figure it out pretty quick. I just want to know: Can this be solved after taking High School Algebra 2 and Geometry? It'd be disappointed to find that after giving up the solution involved a math concept I simply wasn't familiar with.

Thanks, and happy geometry'ing!

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captainwalrus
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Re: Geometry puzzle

Postby captainwalrus » Fri Nov 06, 2009 12:03 am UTC

Hmmm, You can do it but I haven't taken geometry since 9th grade, so I don't really remember that well... Why wouldn't it just be 5 though?
It's over 9 x 10^3!

Nitrodon
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Re: Geometry puzzle

Postby Nitrodon » Fri Nov 06, 2009 12:31 am UTC

I don't think it requires anything that you wouldn't know. Also, the answer isn't 5.

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phlip
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Re: Geometry puzzle

Postby phlip » Fri Nov 06, 2009 12:53 am UTC

I'm pretty sure it's
Spoiler:
[imath]2\phi[/imath]


A hint, if you're trying to figure it out:
Spoiler:
If you have two tangential circles, and you draw a line through the two centres, it will go through the tangent point.

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Hertafeld
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Re: Geometry puzzle

Postby Hertafeld » Sat Nov 07, 2009 2:45 am UTC

I just looked at the hint.
I was thinking that it looked that way, but I wasn't sure how to prove it. So there's a postulate stating that? If that's the case, then it'd just be:
Spoiler:
1 + sqrt(5)

, right?

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Cosmologicon
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Re: Geometry puzzle

Postby Cosmologicon » Sat Nov 07, 2009 4:40 am UTC

Hertafeld wrote:I just looked at the hint.
I was thinking that it looked that way, but I wasn't sure how to prove it. So there's a postulate stating that?
It's pretty easy to show, I think
Spoiler:
Consider the straight line segment between the centers. Since the circles don't overlap, this segment is as large as the sum of the radii, plus the distance between the two circles along the line. But the path from one center to the tangent point to the other center is exactly equal to the sum of the radii, so the straight-line distance can be no larger than this by the triangle inequality. So the distance between the two circles along the line is 0.

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skeptical scientist
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Re: Geometry puzzle

Postby skeptical scientist » Sat Nov 07, 2009 6:25 am UTC

Er, cosmo, the smaller circle is tangent to the larger circle on the interior, so that argument doesn't work as given. The same idea works though.
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Poohblah
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Re: Geometry puzzle

Postby Poohblah » Sun Nov 08, 2009 5:59 am UTC

so, if I understand this correctly, the diameter of the large circle is tangential to each of the three smaller circles, and the two outer small circles are both tangential to the large circle?

mr-mitch
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Re: Geometry puzzle

Postby mr-mitch » Sun Nov 08, 2009 6:21 am UTC

1) Draw three circles of radius 1 along a horizontal line just touching each other (well, the two outer ones don't touch each other..)
2) Draw a horizontal line across the bottom
3) complete the semi-circle so that the arc just touches the two outer circles.

That's what it looks like.

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phlip
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Re: Geometry puzzle

Postby phlip » Sun Nov 08, 2009 6:33 am UTC

Poohblah wrote:so, if I understand this correctly, the diameter of the large circle is tangential to each of the three smaller circles, and the two outer small circles are both tangential to the large circle?

Yep:
circles.png
circles.png (6.67 KiB) Viewed 3690 times


Since I'm drawing it anyway, here's the full solution:
Spoiler:
circles_soln.png
circles_soln.png (7.86 KiB) Viewed 3667 times

O is the centre of the big circle, A is the centre of the right-hand small circle, B and T are tangent points.

AB and AT are both radii of the right circle, OT is a radius of the big circle.

As I mentioned before in the hint, OAT is a straight line, so |OT| = |OA| + |AT|. |AT| = 1, since it's a radius. That just leaves |OA|.

∠OBA is a right angle, so we can use Pythagoras. |AB| = 1, since it's a radius, and |OB| = 2, since it's across two circles.

So |OA| = [imath]\sqrt{1^2 + 4^2}[/imath] = [imath]\sqrt5[/imath]

So |OT| = [imath]\sqrt5 + 1[/imath] = [imath]2\phi[/imath]

A rigorous proof would, to get that |OB| = 2, probably show that OBAC, where C is the centre of the middle circle, is a rectangle, and hence |OB| = |AC|. And AC goes through the tangent point of those two small circles, again because it goes through both circles' centres, so its length is just the two radii, which is 2. I'd start with ∠OBA = ∠BOC = a right angle, and |OC| = |AB|... which I'm almost certain is enough to ensure it's a rectangle, but I don't know how to prove that rigorously, it's been too long since I've done geometry.
I think you'd also have to prove that O is the tangent point of the centre circle, but for that, you can simply say that both O, and the tangent point, are on the vertical axis of symmetry, so they must be the same point.

So, required premises to support this proof:
  • The line through the centres of two tangent circles goes through the tangent point.
  • A line tangent to a circle is perpendicular to the radius at the tangent point.
  • If you have three points in a line, you can add the distances up between the two smaller intervals to get the length of the full interval.
  • Pythagoras's theorem.
  • Definition of a circle: radii are all the same length.
I think most of that would be in your average High School geometry class, though I'm not sure about the first point. I can't remember whether I learned that little tidbit in HS geometry or elsewhere...

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Tirian
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Re: Geometry puzzle

Postby Tirian » Sun Nov 08, 2009 7:43 am UTC

Spoiler:
HS Geometry is certainly capable of proving your hint. If you were to draw the tangent at T and call it L, the second factoid demonstrates that OT and AT are both perpendicular to L, so they are parallel to each other, and the three points are collinear since both line segments pass through T.

Hertafeld
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Re: Geometry puzzle

Postby Hertafeld » Fri Nov 13, 2009 11:25 pm UTC

Yep, it certainly makes sense now.
Spoiler:
My friend who told this to me rattled off the names of some of the smarter kids in our school and how long it took them to solve so I was looking for a much less simple solution. My dilemma was simply knowing that you could draw that line through both centers and the tangent point. Thinking about it now, though, it's rather clear it will go through all three. And high school geometry most definitely covered that.
Thanks everyone!

mr-mitch
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Re: Geometry puzzle

Postby mr-mitch » Sat Nov 14, 2009 4:11 am UTC

Enjoy this one,

Find the radius given the following situation with a quadrant of a circle and a square of side length 1,

http://img291.imageshack.us/img291/2664/34353904.jpg

Find the original radius.

I accidentally exported the whole thing, not just the image, so it's very small, but the image is large (that's why I posted a direct link) and I didn't save it.

If you guys/girls want it bigger I'll do it again. The square is tangential to the appropriate sections of the quadrant.

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skeptical scientist
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Re: Geometry puzzle

Postby skeptical scientist » Sat Nov 14, 2009 4:46 am UTC

It's not technically clear from that diagram that vertical reflection is a symmetry of your figure; without that additional assumption, the problem does not have a unique solution.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

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mr-mitch
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Re: Geometry puzzle

Postby mr-mitch » Sat Nov 14, 2009 6:47 am UTC

As far as I know, there's only one way to have a square tangential to all sides, which implies the angles. If you don't accept this (or it isn't true) the triangle made by the square is a 45,45,90 triangle.

But you don't need to know this in order to solve the problem. :)

Nitrodon
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Re: Geometry puzzle

Postby Nitrodon » Sat Nov 14, 2009 6:57 am UTC

Spoiler:
The center of the circle is equidistant from the endpoints of the right side of the square. Thus, the center is on the perpendicular bisector of the side in question. This is also the perpendicular bisector of the left side, so the triangle in the diagram is indeed a right isosceles triangle (that is, 45-45-90). This can be used to show that the distance from the midpoint of the right side of the square to the center of the circle is 3/2.

Hence, the radius of the circle is [imath]\sqrt{10}/2[/imath]

mr-mitch
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Re: Geometry puzzle

Postby mr-mitch » Mon Nov 16, 2009 5:26 am UTC

Correct, but below is a much simpler way..

Spoiler:
Consider a circle made out of the four quadrants. You then have a + made out of the squares in the circle. Draw the diameter connecting two opposite corners squares, this is 2r, creating a 1,3,2r sided triangle. Use Pythagoras to solve for r = sqrt(10)/2

Hertafeld
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Re: Geometry puzzle

Postby Hertafeld » Tue Nov 17, 2009 12:28 am UTC

Is it...
Spoiler:
Sqrt(5/2)? We have a right triangle with lengths of 1/2 and 3/2. Or, sqrt(2)/2 and sqrt(2). Both have the same answer, I think.

mr-mitch
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Re: Geometry puzzle

Postby mr-mitch » Tue Nov 17, 2009 2:25 am UTC

Yes.

Spoiler:
sqrt(5/2) = sqrt(10)/2


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