This article is from the Puzzles FAQ,
by Chris Cole firstname.lastname@example.org and Matthew Daly
email@example.com with numerous contributions by others.
Arrange pentominos in 3x20, 4x15, 5x12, 6x10, 2x3x10, 2x5x6 and 3x4x5 forms.
I've seen several different naming schemes used for pentominoes. This is
the system I'm using (I think only F & N require a bit of imagination):
FF I L N PP TTT U U V V W W W X X Y ZZ
FF I L NN PP T UUU V V W W X YY Z
F I L N P T V X X Y ZZ
I LL N Y
A 3x20 solution (the other solution is easily obtained by a rotation of
the section from the Z to the L inclusive):
A 4x15 solution:
2 5x6 rectangles. Joined side-to-side, end-to-end, or stacked, these
enable construction of the 6x10 & 5x12 rectangles, and the 2x5x6 prism:
The 2x3x10 and 3x4x5 solutions are tricky to show - I hope these diagrams
A 2x3x10 solution (shown as 2 layers; Y and L are shared between the
A 3x4x5 solution (3 layers, V F W & L shared between 2 or more layers):
VUUXF VZFFF VNYFW
VUXXX TZZZW NNYPW
VUUXW TTTZW NYYPP
IIIII TLLLL NLYPP
+------------------- firstname.lastname@example.org -------------------+
| The effort to understand the universe is one of the very few things |
| that lifts human life above the level of farce, and gives it some |
| of the grace of tragedy - Steven Weinberg |
What is the expected distance between two random points inside a sphere?
Assume the points are uniformly and independently distributed.
Use spherical polar coordinates, and w.l.o.g. choose the polar axis
through one of the points. Now the distance between the two points is
sqrt ( r1^2 + r2^2 - 2 r1 r2 cos(theta))
and cos(theta) is (conveniently) uniformly distributed between -1 and
+1, while r1 and r2 have densities 3 r1^2 d(r1) and 3 r2^2 d(r2). Split
the total integral into two (equal) parts with r1 < r2 and r1 > r2, and
it all comes down to integrating polynomials.
More generally, the expectation of the n'th power of the distance
between the two points is
2^n . 72 / ((n+3)(n+4)(n+6))
So the various means are:
the (arithmetic) mean distance is 36/35 = 1.028571...
the root mean square distance is sqrt(6/5) = 1.095445...
the geometric mean distance is 2exp(-3/4) = 0.944733...
the harmonic mean distance is 5/6 = 0.833333...
the inverse root mean inverse square distance is
2/3 = 0.666666...
What are the odds that n random points on a sphere lie in the same hemisphere?
1 - [1-(1/2)^(n-2)]^n
where n is the # of points on the sphere.
The question will become a lot easier if you restate it as the following:
What is the probability in finding at least one point such that all the other
points on the sphere are on one side of the great circle going through this
When n=2, the probability= 1 ,
when n=infinity, it becomes 0.
In his Scientific American column which was titled "Curious Maps",
Martin Gardner ponders the fact that most of the land mass of the Earth
is in one hemisphere and refers to a paper which models continents
by small circular caps. He gives the above result.
See "The Probability of Covering a Sphere With N Circular Caps" by
E. N. Gilbert in Biometrika 52, 1965, p323.