## Geometry puzzle

A forum for good logic/math puzzles.

Moderators: jestingrabbit, Moderators General, Prelates

Hertafeld
Posts: 10
Joined: Sat Oct 31, 2009 2:02 am UTC

### Geometry puzzle

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!

captainwalrus
Posts: 11
Joined: Tue Nov 03, 2009 2:37 am UTC

### Re: Geometry puzzle

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
Posts: 497
Joined: Wed Dec 19, 2007 5:11 pm UTC

### Re: Geometry puzzle

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

phlip
Restorer of Worlds
Posts: 7554
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia
Contact:

### Re: Geometry puzzle

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.

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

Hertafeld
Posts: 10
Joined: Sat Oct 31, 2009 2:02 am UTC

### Re: Geometry puzzle

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?

Cosmologicon
Posts: 1806
Joined: Sat Nov 25, 2006 9:47 am UTC
Location: Cambridge MA USA
Contact:

### Re: Geometry puzzle

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.

skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

### Re: Geometry puzzle

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.
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.

"With math, all things are possible." —Rebecca Watson

Poohblah
Posts: 53
Joined: Thu Feb 26, 2009 3:54 am UTC

### Re: Geometry puzzle

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
Posts: 477
Joined: Sun Jul 05, 2009 6:56 pm UTC

### Re: Geometry puzzle

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.

phlip
Restorer of Worlds
Posts: 7554
Joined: Sat Sep 23, 2006 3:56 am UTC
Location: Australia
Contact:

### Re: Geometry puzzle

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 (6.67 KiB) Viewed 4144 times

Since I'm drawing it anyway, here's the full solution:
Spoiler:
circles_soln.png (7.86 KiB) Viewed 4121 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...

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

Tirian
Posts: 1891
Joined: Fri Feb 15, 2008 6:03 pm UTC

### Re: Geometry puzzle

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
Posts: 10
Joined: Sat Oct 31, 2009 2:02 am UTC

### Re: Geometry puzzle

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
Posts: 477
Joined: Sun Jul 05, 2009 6:56 pm UTC

### Re: Geometry puzzle

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

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.

skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

### Re: Geometry puzzle

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.

"With math, all things are possible." —Rebecca Watson

mr-mitch
Posts: 477
Joined: Sun Jul 05, 2009 6:56 pm UTC

### Re: Geometry puzzle

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
Posts: 497
Joined: Wed Dec 19, 2007 5:11 pm UTC

### Re: Geometry puzzle

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
Posts: 477
Joined: Sun Jul 05, 2009 6:56 pm UTC

### Re: Geometry puzzle

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
Posts: 10
Joined: Sat Oct 31, 2009 2:02 am UTC

### Re: Geometry puzzle

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
Posts: 477
Joined: Sun Jul 05, 2009 6:56 pm UTC

### Re: Geometry puzzle

Yes.

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