The Maths Site
    
HomeForumPuzzlesAudioVideoToolsDonateHistoryContact
Log in here







Forgotten your password?



Or Click to register

iSquared Magazine

The Daily Rant

Puzzles
Reuleaux triangle

This puzzle isn't too hard, but it will introduce you to a cool geometrical object -- the Reuleaux triangle (pronounced "Roo-low"). This "triangle" is pictured in the diagram at right (the yellow shape). As you can see, it really isn't a triangle at all because its sides aren't straight. To build a Reuleaux triangle, start with the three vertices of an equilateral triangle (with all three sides of length 1). Now draw three circles. Each circle is centered at one of the three vertices, and passes through the other two vertices. The area of overlap between these circular disks is the Reuleaux triangle.

pic
Click to enlarge


This shape has several interesting properties. One of them is that it is the same width no matter which way it is rotated. (That is, it will always just barely fit through an opening 1 unit wide). The circle, of course, has this same property, but not many non-circular shapes do. A square (with side-length 1), for instance, will fit through an opening of width 1 if taken through with sides parallel to the opening. But otherwise, it won't fit!

Have you ever wondered why manhole covers are round instead of square? The answer is found in this very property: a square manhole cover could be picked up, rotated, and dropped through the hole it covers. However, a circular manhole cover won't fit through that hole no matter how it is rotated. Maybe sometime someone will manufacture manhole covers shaped like the Reuleaux triangle!

pic
Click to enlarge


Now for the puzzle: find the exact area of the Reuleaux triangle! Be sure to simplify your answer as much as possible.

Good luck!

 
Social Bookmarking
Add to: Digg Add to: del.icio.us Add to: StumbleUpon Add to: Google
 
Comments
Login or register to add your comments.
terveloc18:36:04, 17 Nov 07
terveloc's picture
Newbie
Group:Members
Points: 0
Posts: 5
Warn level: 0

A = 3*(1/3)*pi*r^2 - 2*(1/2)*r^2*sin(Pi/3)

This simplifies to r^2(pi - sqrt(3)/2), which agrees with cosmo.

 
spotcatbug14:59:43, 11 Oct 07
spotcatbug's picture
Newbie
Group:Members
Points: 0
Posts: 0
Warn level: 0

I get:

r^2(pi/2 - 1)


 
cosmo14513:14:46, 09 Oct 07
cosmo145's picture
Newbie
Group:Members
Points: 0
Posts: 0
Warn level: 0

I got the area be R^2[{pi-root(3)}/2]

I drew in the equilateral triangle, whose sides are length R. The 3 segments that are left over are simply the area of the arc, made by 2 of the triangle sides and a angle of 60, minus the triangle.

 
Lawrence Lee09:37:22, 09 Oct 07
ironic1's picture
Newbie
Group:Members
Points: 0
Posts: 0
Warn level: 0

New to the site. Nice puzzle. I got ((2 pi minus the square root of 3) over 4) times r squared, where r is the radius of the circle.

I got this by figuring that the area of one "wedge" was a sixth of pi r squared and that the Reuleaux must equal this area times 3 minus twice the area of the inner equilateral triangle.

 


© Copyright Andy Drizen 2004-2007 | Website by Think In Colour | Proudly sponsored by JSC Solutions
Valid XHTML 1.0 Transitional Valid CSS!