Geometry and Topology
(E. C. of Pasadena, CA.
20001010)
How do I determine the radius of an arc drawn on paper?
I have the drawing scale but not the radius of the arc. I also have a compass.
(R. R. of Utica, IL.
20010212)
[How do I] find the exact center of a circle?

Take a pair of points on the arc. Open your compass wide enough so two (equal) circles
drawn with these point as centers will intersect.
Draw the two circles (or at least part of them) to determine their two points of intersection.
Draw a line through the two points of intersection.
Take another pair of points on the arc and do the above construction again to draw
another line the same way.
The point where the two lines you've drawn intersect is the center of the circle your original
arc belongs to.
(And, of course, the distance between that center and any point on the arc is the radius you
were after.)
(Bobby of Diboll, TX.
20010210)
What is the formula to determine the area of a circle?
[What about] a triangle, a trapezoid, a sphere, [a cylinder, a cone] ?

Let's start with simple planar surfaces:

A circle of radius R has area
pR^{2}
(p=3.14159265...).

The area of a trapezoid is the arithmetic average (i.e., the halfsum)
of its two parallel bases multiplied by its height
(the height is the distance between the bases).
The area of a rectangle (width multiplied by height) may be seen as
a special case of this...

A triangle may be also considered a special type of trapezoid
(with one base of zero length) and its area is [therefore]
half the product of a side by the corresponding height.
Let's proceed with the simplest curved surfaces:

The surface area of a sphere of radius R is
4pR^{2}
(its volume is
4pR^{3 }/3 ).

More generally, we may consider the surface
(sometimes called a spherical frustum or "frustrum")
which consists of the part of the surface of a sphere between two parallel planes that
intersect it.
The surface area of such a frustum is
2pRH, if H is the distance between the two planes.
When H=2R, the frustum consists of the entire sphere and the above formula does
give an area of
4pR^{2}, as expected.

An ordinary right cylinder of height H is the surface generated by a straight
segment of length H perpendicular to a plane containing the trajectory
of one of its extremities (a simple curve of length L).
The surface area of such a cylinder is simply LH.
In particular, if the above "trajectory" is a circle, we have a circular cylinder
of radius R, whose surface area is 2pRH.
(Note that a spherical frustum has the same area as the right cylinder
circumscribed to it, a remarkable fact first discovered by Archimedes of Syracuse.)

We may also consider the lateral surface area of the conical surface
generated by a segment of length R with one fixed extremity,
when the other extremity has a trajectory of length L (this spherical trajectory is
not planar unless it happens to be a circle, which is true only for an ordinary
circular cone).
The area of such a surface is simply RL/2. In particular, the lateral
surface area of an ordinary circular cone is pRr,
if r is the radius of its base and
R is the distance from the circumference of the base to the apex.
If you're given the height H of the cone instead of R, the Pythagorean theorem
(R^{2}=H^{2}+r^{2}) comes in handy to give you the
lateral area of the conical surface as
pr
Ö(H^{2}+r^{2})
Note that the case R=r (or H=0) corresponds to a "flat" circular cone,
which is simply a circle
of area pR^{2}...
Back to our first formula!
Historically, ancient mathematicians did convince themselves that
LH/2 was, as stated above, the area of the surface
generated by a segment of length R when one of its extremities (the "apex")
is fixed and the other extremity has a trajectory of length L.
The record shows that they did this for planar geometry (in which case the trajectory is
a circle centered on the apex), but the same reasoning would apply
to nonplanar trajectories as well.
They reasoned that the trajectory (the circle) could be approximated
with a polygonal line with many small sides.
The surface could then be seen as consisting of many "thin" triangles whose heights
were very nearly equal to R, whereas the base was very nearly a portion of the
trajectory. As the area of each triangle is R/2 times such a portion, the area of the
whole surface is R/2 times the length of the entire trajectory QED.
Of course, this type of reasoning was made fully rigorous only with the advent of
infinitesimal calculus, but it convinced everyone well before that of the
existence of a single number p which would
give both the circumference (2pR) and the
surface area (pR^{2})
of a circle of radius R... (The symbol p itself appeared only
in the early 18th century, but the existence of the constant it represents was clear
thousands of years ago.)
(K. D. of USA.
20001113)
[In a triangle,] what is the relationship between the centroid,
the circumcenter and the orthocenter?

For any triangle, these 3 points are collinear.
The straight line on which they are is sometimes called "Euler's line"
(it's undefined for an equilateral triangle).
The centroid G is between the orthocenter H and the circumcenter I.
The distance HG is twice the distance GI.
You may also want to know that there is another remarkable point E on this line,
exactly halfway between H and I:
The point E is the center of the socalled "Euler's circle" or "9point circle"
which goes though 9 special points of the triangle: the 3 midpoints of the sides,
the feet of the 3 perpendiculars to the sides from a vertex,
and the 3 midpoints from the orthocenter H to each vertex.

(M. Muz Zviman, Ph.D. 20011128; email)
How do I calculate the length of an elliptic arc between two chosen points?

For the length (perimeter) of the entire circumference, see our
(unabridged) answer to the
next question.
A parametric equation for an ellipse of cartesian equation
x^{2}/a^{2}
+ y^{2}/b^{2}=1
is:
x = a sin(q) and
y = b cos(q) .
We assume a>b and define
e^{2 }=1b^{2}/a^{2}.
The above figure shows how q may be determined using
an auxiliary circle whose radius is the ellipse's major radius a.
It suffices to calculate the elliptic arc, shown as a red line
in the picture, from the flat apex at
q=0) to an arbitrary point (conventionally, no more than a quarter
of a perimeter away).
The length of the arc between two points is obtained by adding or subtracting two such
quantities (possibly adding a multiple of a quarter of the
perimeter as well).
The above parameterization is used conventionally, because it turns out to be numerically
superior to the complementary one which would make the parameter
q small near the pointed part of the ellipse.
This is especially so in the case of very elongated ellipses.
The length of an elementary arc is obtained as the square root of
(dx)^{2}+(dy)^{2}, which boils down to
aÖ(1e^{2}sin^{2}q) dq.
[This is simply an infinitesimal expression of the Pythagorean theorem:
At infinitesimal scales, every ordinary curve looks straight, and a small piece of
it appears as the hypotenuse of a tiny right triangle of sides dx and dy.]
The length of the elliptic arc corresponding to the angle q
may thus be expressed as a simple
integral (an oldfashioned quadrature)
known as the incomplete elliptic integral of the second kind:
a E(q,e) = a 
ó^{q} õ_{0} 
Ö 


1  e^{2 }sin^{2 }a 
da 

This function (E) was introduced because the integral has no expression in terms of more
elementary functions.
(The function E also comes in a singleargument version known as the
complete elliptic integral of the second kind, namely
E(e) = E(p/2,e), which is
a quarter of the perimeter of an ellipse of eccentricity e
and unit major radius.)
To compute the integral when e is not too close to 1 and/or
q is not too close to p/2
[in which case other efficient approaches exist, see
elsewhere on this site],
we may expand the square root in the integrand as a sum of
infinitely many terms of the form
(1)^{n }C(½,n) e^{2n }sin^{2n}a,
for n=0, 1, 2, 3...
Each such term may then be integrated individually using the formula:
ó^{q} õ_{0} 
_{ }sin^{2n }a
da = 
^{ }q^{ } 

2^{2n} 

ì2n î n 
ü þ 
+ 
^{ }1^{ } 

2^{2n} 




(1)^{k} 

k 

ì 2n înk 
ü þ 
sin 2kq_{ } 
When q = p/2, all the sines vanish and only the first
term remains. This translates into the simple series
given in the next article. Otherwise, what we are left with is
2q/p times that complete integral plus
the Fourier series of some odd periodic function
of q
(whose period is p)...
On 20011129, Muz Zviman wrote:
Thank you for the quick answer. The website is great.
Best regards, Muz

On 20021231,
David W. Cantrell
proposed:
A 0.56% approximation
to the above (first posted to the sci.math newsgroup).

(Jaleigh. B. of Minonk, IL.
20001126
twice)
What is the formula for the perimeter of an ellipse?
(S. H. of United Kingdom.
20010125
What is the formula for the circumference of an ellipse?

The following is a summary. For more details, see our
unabridged discussion.
There is no simple exact formula: There are simple formulas but they are not exact
and there are exact formulas but they are not simple.
If the ellipse is of equation
x^{2}/a^{2}
+ y^{2}/b^{2}=1
with a>b, a is called the major radius,
and b is the minor radius.
The quantity e = Ö(1b^{2}/a^{2})
is the eccentricity of the ellipse.
An exact expression for the ellipse perimeter P involves the sum of infinitely
many terms of the form (1)/(2n1) [(2n)!/(2^{n} n!)^{2}]^{2} e^{2n}.
The first such term (for n=0) is equal to 1 whereas all the others are negative
correction terms :
P/2pa = 1
 [1/4]e^{2}
 [3/64]e^{4}
 [5/256]e^{6}
 [175/16384]e^{8}
 [441/65536]e^{10} ...
Note that for a circle (e=0) of radius a, the above does give the
circumference as 2p times the radius.
Among the many approximative formulas for the perimeter of an ellipse, we have:
 
P » pÖ  2(a^{2}+b^{2})
 (ab)^{2}/2 
A 1914 formula due to the Indian mathematician
S.
Ramanujan (18871920) is
 
P » p [ 3(a+b) 
Ö  (3a+b)
(a+3b)  ] 
A second 1914 formula, also due to Ramanujan, is expressed in terms of the quantity
h = (ab)^{2}/(a+b)^{2} :
 
P » p (a+b)
[ 1 + 3h / ( 10+Ö  43h 
) ] 
The relative error of this formula for ellipses of low eccentricities
is fabulous:
(3/2^{37 }) e^{20}
[ 1 + 5 e^{2} +
11107/768 e^{4} +
4067/128 e^{6} +
3860169/65536 e^{8} +
... ]
In 1917, Hudson came up with a formula without square roots, which is traditionally
expressed in terms of the quantity
L = h/4 = (ab)^{2}/[2(a+b)]^{2} :
P » p (a+b)/4 [ 3(1+L) + 1/(1L) ]
In 2000, Roger Maertens proposed the following socalled
"YNOT formula":
P »
4 (a^{y}+b^{y}) ^{1/y}
or
P »
4a (1 + (1e^{2})^{y/2} )^{1/y}
with y = ln(2)/ln(p/2)
The special value of y (the "YNOT constant") makes the formula
exact for circles, whereas it is clearly also exact for flat ellipses
(b=0 and P = 4a).
The relative error of the
YNOT formula never exceeds 0.3619%.
It is highest for the perimeter of an ellipse whose eccentricity is about 0.979811
[ pictured at right ] with an
aspect ratio a/b slightly above 5.
We sould also mention the popular upper bound formula due to Euler (1773):
 
P » pÖ  2(a^{2}+b^{2}) 
The following simple lower bound formula is due to Johannes Kepler (1609):
 
P » 2p Ö  ab 
The precision of all of the above formulas is summarized in the table below.
The last column shows the absolute error (in meters) of each formula when it is
used to compute the circumference of an ellipse with the same eccentricity and the same
size as the Earth Meridian. Note that even the humble #1 formula is accurate to
15 mm, or about one tenth
of the width of a human hair!
(For Ramanujan's first formula,
this would be one sixtieth of the diameter of a hydrogen atom.
We lack a physical yardstick for the more precise formulas...)
Except for Maertens' YNOT formula, the modest precision
shown for the "worst case" corresponds to a completely flat ellipse
(of perimeter 4a).
Perimeter Formula 
Relative Error 
D for Earth Meridian (m) 
Worst (%)  Low Eccentricity 
(7) 
Kepler 1609 
100 
3e^{4}/64 [1+e^{2 }+ ...]

84.61 m 
(6) 
Euler 1773 
+11.072 
e^{4}/64 [1+e^{2 }+ ...]

+28.20 m 
(5) 
Maertens 2000 
+0.3619 
(2y3)e^{4}/64 [1+e^{2 }+ ...]

+1.97 m 
(1) 

3.809 
3e^{8}/2^{14}
[1+2e^{2 }+ ...]

1.49 10^{5} 
(2) 
Ramanujan I 
0.416 
e^{12}/2^{21}
[1+3e^{2 }+ ...]

1.75 10^{12} 
(4) 
Hudson 1917 
0.189 
9e^{16}/2^{30}
[1+4e^{2 }+ ...]

1.39 10^{18} 
(3) 
Ramanujan II 
0.0402 
3e^{20}/2^{37}
[1+5e^{2 }+ ...]

1.63 10^{25} 
For more information, see our
unabridged discussion.
(Scott of Emeryville, CA.
20001010)
How do I calculate the surface area of a cylinder
and an oblate sphere (flying saucer shaped)_{ }?
(M. P. of Williamsport, PA.
20001016)
What is the surface area of an ellipsoid?
The (lateral) surface area of a circular cylinder of radius R and height H is
2pRH.
The surface area S of an oblate ellipsoid
(generated by an ellipse rotating around its minor axis)
of equatorial radius a and eccentricity e is given by:
S = 2pa^{2 }[ 1 +
(1e^{2}) atanh(e)/e ] ,
or
S = 2pa^{2 }[ 1 +
(b/a)^{2 }atanh(e)/e ]
In this, e is
Ö(1b^{2}/a^{2}),
where b<a is the "polar radius"
(the distance from either pole to the center)
and atanh(e) is ½ ln((1+e)/(1e))
[also denoted argth(e) ].
The surface area S of a prolate ellipsoid ("cigarlike") generated by
an ellipse rotating around its major axis
(so that the equatorial radius b is smaller than the polar radius a) is given by:
S = 2pb^{2 }[ 1 +
(a/b) arcsin(e)/e ]
This shows that a very elongated ellipsoid has an area of
p^{2}ab
(e is close to 1 and b is much less than a),
which is about 21.46% less than the lateral area of the circumscribed cylinder
(4pab),
whereas these two areas are equal in the case of a sphere,
as first noted by Archimedes of Syracuse (c.287212BC).
Now, it's not nearly as easy to work out the surface area of a general ellipsoid of cartesian
equation (x/a)^{2}+(y/b)^{2}+(z/c)^{2}=1.
No elementary formula for this one!
The general formula
involves elliptic functions, which "disappear" only for solids of revolution.
(Sherry of Murray, KY.
20001019)
What's the formula for the area of an oval?
If your "oval" is an ellipse of major radius a and minor radius b,
its cartesian equation with the proper choice of coordinates is:
x^{2}/a^{2}
+ y^{2}/b^{2} = 1
and its area is simply S = pab.
William Van Drent, Ph.D.
(20010816; email)
Staff Scientist / New Product Development Manager.
Digital Measurement Division.
ADE Technologies, Inc.
Newton, MA.
[...] In an ellipse of equation
A x^{2} + B y^{2} + C xy +
D x + E y + F = 0,
what are (in terms of A, B, C, D, E, and F )
 the inclination q of the major axis,
 the coordinates (xo,yo) of the center,
 the major radius (a) and the minor radius (b)?
Note: Somewhere in the discussion below, we shall assume that
A+B is positive.
If necessary, change the sign of all coefficients in the equation to make it so.
Background :
The socalled general quadratic equation you are giving describes planar curves
which are known as conic sections, for historical reasons (the "ordinary" ones
are obtained as the the intersection of a plane and a full cone,
defined as the surface generated by a straight line rotating around an axis that
intersects it).
A conic section may be an ellipse (possibly a circle), a parabola, or a hyperbola.
There are also socalled degenerate conics: intersecting lines,
parallel lines, or double line
(a line in which every point is a "double point" of the curve).
Such degenerate cases occur when our quadratic polynomial happens to be the product of
two firstdegree polynomials.
It's also possible for such an equation to describe what's called an
imaginary ellipse, which is an empty set in the real plane
(but would not be if imaginary coordinates were allowed).
For example, the equation of an imaginary circle could be something like:
x^{2}+y^{2}+4=0.
For completeness, your general quadratic equation with real coefficients could
also describe a pair of imaginary lines. Such lines correspond to a single
solution point if they intersect
[example: x^{2}+y^{2} = 0, which is to say
(x+iy)(xiy) = 0],
or an empty set in the real plane when they don't
[example: (x+y)^{2}+1 = 0, which is to say
(x+y+i)(x+yi) = 0].
Your question is limited to the case of real ellipses and we will restrict our discussion
to this.
However the extension to other cases is similar and should be fairly clear
from what follows.
First, we shall notice that we may get rid of any existing cross term
("xy" with a nonzero C coefficient) by tilting
the coordinate axes. If we do so by an angle q (see figure),
the new coordinates X and Y (note capitalization)
are best obtained as the scalar products of the unit
vectors of the new tilted axes.
These vectors are
(cos q,sin q)
and
(sin q,cos q):
X = x cos q + y sin q
Y = x sin q + y cos q

conversely Þ
(change q to q ) 
x = X cos q  Y sin q
y = Y sin q + Y cos q

The above expressions of x and y in terms of X and Y give us the curve's equation in the
tilted frame. Equating to zero the coefficient of XY in this gives:
(BA) sin 2q
+ C cos 2q = 0
We could thus obtain q within an integral multiple
of p/2 as half an arctangent, but let's not
rush things! What we really want is the inclination of the major axis,
which is determined within an integral multiple of p...
When the above relation is satisfied, the rest of the equation reads:
[A cos^{2}q
+ B sin^{2}q
+ C cos q sin q ]
X^{2} +
[A sin^{2}q
+ B cos^{2}q
 C cos q sin q ]
Y^{2} +
[D cos q +
E sin q ]
X +
[D sin q +
E cos q ]
Y + F = 0
Using the previous relation, we may reduce the above coefficients of X^{2}
and Y^{2}. We find the former equal to
(1/2)[A(1+1/cos 2q) +
B(11/cos 2q)],
whereas the latter equals
(1/2)[A(11/cos 2q) +
B(1+1/cos 2q)].
We are only interested in the elliptic case, so these two are of the same sign,
which is also the sign of their sum A+B.
As stated in our preliminary note, we shall assume that sum to be positive
(without loss of generality, since an equivalent equation is clearly obtained by
changing the sign of all coefficients).
Now, if we want the Xaxis to be the major one, the coefficient of X^{2}
is inversely proportional to the square of the major radius
and is thus smaller than the coefficient of Y^{2}
(which is is inversely proportional to the square of the minor radius).
In other words,
(AB)/cos 2q is negative.
With this in mind, we may now fully specify the inclination of the major axis
(within a multiple of p, of course) by giving the
sine and cosine of the angle 2q
(again, we are assuming: A+B > 0):
 
cos 2q = (BA)/Q
and
sin 2q = C/Q ,
where
Q = Ö  (AB)^{ 2} + C^{ 2} 
For obvious reasons, this determination of the inclination is not valid when Q=0.
(Q=0 implies A=B and C=0, which corresponds to the trivial
case where the ellipse is, in fact, a circle for which any direction may be considered
"major".)
The above coefficients of X^{2} and Y^{2} respectively boil down to
(A+BQ)/2 and (A+B+Q)/2.
We shall need these simple expressions below.
We may also remark that
the curve described is indeed an ellipse real or imaginary
when (A+B)^{2} > Q^{2} ,
so these two coefficients do have the same sign.
This relation translates into 4AB > C^{ 2}.
The coordinates of the ellipse center are fairly easy to compute directly in the original
frame of reference: We are simply looking for
(xo,yo)
such that the transforms x=xo+u and y=yo+v
yield an equation where the coefficients of u and v are zero (so that the origin will be
a center of symmetry)... This translates into the two simultaneous equations:
0 = 2 A xo + C yo + D
0 = C xo + 2 B yo + E
Therefore:
xo = (CE2BD)/(4ABC^{ 2} )
and
yo = (CD2AE)/(4ABC^{ 2} ).
This argument may be used to show that any conic section has a center,
except in the case of the parabola,
when 4AB = C^{ 2}.
Finally, we come to the determination of the principal radii of the ellipse.
For this, we first need the value of the equation's constant term K,
in a frame of reference centered
at the above point (xo,yo).
Knowing that the tilt of the axes is irrelevant to this constant K,
we may as well compute it at zero tilt. It boils down to:
K =
F + (CDE  AE^{ 2}  BD^{ 2} )
/ (4ABC^{ 2} )
In the properly tilted frame centered at (xo,yo),
the equation of the ellipse is thus:
(A+BQ) X^{2} + (A+B+Q) Y^{2}
+ 2K = 0 , which we just need to identify with the standardized
equation X^{2}/a^{2} + Y^{2}/b^{2} = 1
in order to obtain the values of the principal radii, and/or their squares:
a^{2} =
2K / (A+BQ)
b^{2} =
2K / (A+B+Q)
The equation thus describes a real ellipse only when
(A+B)K < 0
and 4AB > C^{ 2}.
(B. K. of Honolulu, HI.
20001012)
How can I determine the volume of an oval object? [eggshaped solid]
The volume of an ellipsoid of equation
x^{2}/a^{2}
+ y^{2}/b^{2}
+ z^{2}/c^{2} = 1
is 4pabc/3.
This formula is a good approximation for other eggshaped ovals which
are nearly elliptical: 2a is the diameter
(i.e. the largest width), 2b is the largest width for a direction perpendicular
to the diameter and 2c is the width in the direction perpendicular to both
previous directions.
Each such width is measured between two parallel planes perpendicular to
the direction being considered.
(Danny of Lincoln, RI.
20001019)
Let's say I have a parabola, f(x) = x^{2}/50. Where is the focal point?
In a parabola of equation y=x^{2}/(2p), the "parameter" p is twice
the distance from the focal
point to the apex (both points being on the parabola's axis of symmetry).
In the parabola y=x^{2}/50, the parameter is 25 and the focal distance is 12.5.
Since the apex is at x=0 and y=0, the focal point is at x=0 and y=12.5.
(K. P. of Clarksville, MD.
20001112)
The path followed by a ray of light from a star to the focus of [a parabolic] mirror
has [a] special property.
Draw a chord of the parabola that is above the focus and parallel to the directrix.
Consider a ray of light parallel to the axis as it crosses the chord,
hits the parabola and is reflected to the focus.
Let d_{1} be the distance from the chord to the point
of incidence (x,y) on the parabola
and let d_{2} be the distance from (x, y) to the focus.
Show that the sum of the distance
d_{1}+d_{2}
is constant, independent of the particular point of incidence.
This particular property is true of any optical system:
The optical length from the object to the image is a constant regardless of the path taken
(the optical length is proportional to the time it takes light to travel in a given medium,
so you have to take into account the index of refraction in the case of lenses, where glass
is involved).
There's no glass in a reflector so the optical length and actual length are the same thing,
hence the result.
The only complication is that when the object is at infinity, you should count distances
from a plane perpendicular to the rays (that's what the "chord" in the question
is all about) instead of dealing with infinite distances:
The reasoning is that all points of such a plane are "at the same distance" from the object;
a small portion of such a plane can be seen as a portion of the sphere which is centered on
the object at a great distance.
If you prefer a purely geometrical approach,
you may consider that a parabola is what an ellipse becomes when you send one of its foci
"to infinity".
The fact that the sum of the distances to the foci is constant on the ellipse translates
into the property you are asked to prove for the parabola.
If neither of the above convinces you (or your teacher),
you may use a more elementary approach,
starting with the equation of the parabola y=x^{2}/4f
(where f is the focal distance).
The square of the distance from a point (x,y) on the parabola to the focal point (0,f) is
x^{2}+(yf)^{2} = 4fy+(yf)^{2} = (y+f)^{2}.
In other words, the distance d_{2} is (y+f).
On the other hand, d_{1} is equal to Ay
(where A is some constant which depends on how far you drew the "chord"
described in the question).
Therefore d_{1}+d_{2} =
f+A = constant. QED.
(This, by the way is one way to actually prove that a parabolic mirror is an
optical system which correctly "focuses" a point at infinity.)
(D. F. of Bozeman, MT.
20001001)
How do I find the centroid of a circular segment?
Use Guldin's theorem (named after Paul Guldin 15771643),
which is also called Pappus theorem in the Englishspeaking world.
The theorem states that the area of a surface of
revolution is equal to the product of the length of the meridian by the length of the circular
trajectory of the meridian's centroid.
(The volume of a solid of revolution is also obtained
as the area of the meridian surface
by the length of the circular trajectory of the centroid of that surface.)
Make the segment rotate around the diameter of the circle which is parallel to the segment's
chord and apply the theorem:
Your meridian is the circular segment of radius R, length L and
chord H=2R´sin(L/2R).
The surface is a spherical segment of area 2pRH.
If D is the distance of the centroid to the center of the circle,
its trajectory has a length 2pD and
Guldin's theorem tells us that:
2pRH=2pDL.
Therefore: D=RH/L, and that gives you the position of the centroid.
johnrp (John P. of Middletown, NJ.
20001022)
Suppose you have two wooden cubes, one just slightly larger than the other.
How can you cut a hole through the smaller cube so that the larger cube will fit through?
Make the axis of the hole a line that goes through two opposite corners of the cube.
Viewed in the direction of that axis, the cube appears as a regular hexagon.
If the side of the cube is 1, the side of this hexagon is
Ö6/3 (approximately 0.8165).
Now, in a regular hexagon of side A, we may
inscribe a square of side
(3Ö3)A
or about 1.268A (one of the sides of the square is parallel to
one of the sides of the hexagon).
When A is Ö6/3, this means that a square of side
Ö6 Ö2
will fit.
Well, Ö6Ö2
is about 1.03527618... so we may cut in a cube a square hole with a
side 3.5% larger than the side of the cube.
A cube "just slightly larger" will easily go through such a hole.
(Adrian Brancato. 20010214; email)
Thanks for the wonderfully informative website, Dr. Michon [...]
I am still unable to solve the problem for which I originally
went to your site. Can you provide a formula for me:
For an octagon, given the "diameter" (i.e. the distance between two
opposite vertices) I need to determine the length of each side.
This is not an academic endeavor; we have to build a large display of
strawberries (the diameter is 6' at the base, with ever decreasing diameters as
the conical structure rises).
Thanks for the kind words, Adrian...
In a regular octagon of side a, the diameter d can be seen as the
hypotenuse of a right triangle whose sides are a and a+2b
(see figure), where b is the side of a square of diagonal a, so that we have
2b=a Ö2 and
d^{ 2} =
a^{ 2} [1 +
(1+Ö2)^{2}]
or
d^{ 2} =
a^{ 2} [4 +
2 Ö2 ]. Take the square root of that, and
you have the desired relation between the side a and the diameter d,
which boils down numerically to
d/a = 2.61312592975...
or, if you prefer,
a/d = 0.382683432365...,
which is half the square root of (2Ö2).
Another way to obtain the same result
is by using standard trigonometric functions. The ratio a/d is the sine
of a p/8 angle
(22.5° or one 16^{th}
of a full turn), which does equal 0.382683432365... according to my trusty scientific
calculator.
All told, your 6' diameter display should have a side almost exactly equal to 2.2961'
(within 0.18 mm or about 1/700 of the width of a human
hair), which is roughly
2' 3^{ 9}/16".
Hope the display will look good!
(20010214)
Constructible Regular Polygons, Constructible Angles
An ancient problem solved by Carl Friedrich Gauss in 1796 (at age 19).
In the previous article, we could have noticed that
8 times the side of the octagon is, of course, its perimeter.
For an nsided polygon, the ratio P/d of the perimeter P
to the diameter d is
n sin(p/n) , which tends to
p as n tends to infinity.
Listed below are the first values of this ratio which may be expressed by radicals.
Gauss showed that this is the case if [and only if] n is the product of a
power of 2 by (zero or more) distinct Fermat primes
(A003401).
Fermat primes are prime numbers of the form
2^{2n} + 1
There are probably only five of these, namely: 3, 5, 17, 257 and 65537.
n  ngon 
Perimeter/diameter ratio = n sin(p/n) 

2  digon  2  2 
3  triangle  2.598 076 211+ 
(3/2)Ö3 
4  square  2.828 427 125 
2 Ö2 
5  pentagon  2.938 926 261+ 
(5/2)Ö((5Ö5)/2) 
6  hexagon  3  3 
8  octagon  3.061 467 459 
4 Ö(2Ö2) 
10  decagon  3.090 169 944 
5 (Ö51)/2 
12  dodecagon  3.105 828 541+ 
3 (Ö31)Ö2 
15  pentadecagon  3.118 675 363 
(15/8)[Ö(10+2Ö5 )
 Ö3 (Ö51)] 
16  hexadecagon  3.121 445 152+ 
8 Ö(2 
Ö(2+Ö2)) 
17  heptadecagon  3.123 741 803 
17 Ö((1c)/2)
[ c=cos(2p/17) ]

c =
{ 2Ö[ 17+3Ö17Ö(2(17Ö17))2Ö(2(17+Ö17)) ] + Ö(2(17Ö17))  1 + Ö17 } / 16

20  icosagon  3.128 689 301 
5 Ö( 8 
2Ö(10+2Ö5)) 
24  tetracosagon  3.132 628 613+ 
6 Ö( 8 
2Ö2  2Ö6) 
30  triacontagon  3.135 853 898+ 
(15/4) [ Ö(306Ö5) 
Ö5  1 ] 

¥  circle  3.141 592 654 
p 
If a is the side of an ngon of diameter d, the side b of
the 2ngon of the same diameter may be obtained simply with the pythagorean
theorem as the hypotenuse of a right triangle whose sides are a/2
and d/2c, where c is the third side of a right triangle
with hypotenuse d/2 and side a/2. All told, for a unit diameter, we have
b^{2} =
^{ 1}/2
[ 1  Ö(1a^{2})
].
In other words, if x is the square of the side of the ngon of unit diameter, the
square y of the side of the 2ngon of unit diameter is given by
y = ^{ 1}/2
[ 1  Ö(1x) ] (there's just one caveat
which is not a problem with hand computation and that's about the difference of
nearly equal quantities in the square bracket, which may cause a crippling
loss of precision when fixedprecision computations are used blindly with the formula
"as is"). Starting with the trivial case of the hexagon,
Archimedes of Syracuse (c.287212BC)
iterated this 4 times to compute the ratio of the circumference to
the diameter in a 96sided polygon (namely 3.141 031 95... which is
about 178.5 ppm below the value of p).
Using a complementary estimate of the circumscribed polygon, Archimedes could then
produce the first rigorous bracketing of what we now call "p".
Until better methods where found at the dawn of calculus,
this was essentially the basic method used
to compute more and more decimals of p...
The last person in history who used Archimedes' method to compute
p with record precision was the Dutchman
Ludolph van Ceulen (15391610): A professor of mathematics at the
University of Leyden, he published 20 decimals in 1596 and
32 decimals in a posthumous 1615 paper. It is said that, at the end
of his life, he worked out 3 more decimals which were engraved on
his tombstone in the S^{t}
Peter Church at Leyden. To this day,
p is still sometimes called
Ludolph's Number or the Ludolphine Number,
especially by the Germans ("die Ludolphsche Zahl").
WiteoutKing
(Lowell, MA. 20020219)
What are the areas of regular polygons with sides of length one?

A regular polygon with n sides of length 1 consists of n congruent triangles of
base 1 and height ½ / tan(p/n).
Its area is therefore equal to:
¼ n / tan(p/n)
This happens to be equal to
Ö3/4 for a triangle,
1 for a square,
Ö(25+10Ö5)/4
for a regular pentagon,
3Ö3/2 for an hexagon,
2+2Ö2 for an octagon,
etc.
The surface area of the regular heptagon of unit side
cannot be expressed using just square roots, sorry!
In general, you can express the area of an ngon with just square roots
only when the ngon is constructible with straightedge and compass.
A beautiful result of Gauss (1796) says that an ngon is so constructible
if and only if n is equal to a power of two (1, 2, 4, 8, 16, ...)
possibly multiplied by a product of distinct socalled Fermat primes.
Only 5 such primes are known (3, 5, 17, 257 and 65537) and
there are most probably no unknown ones...
Ruling out n=1 and n=2, the only acceptable values of n are therefore
3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, ... (A003401).
For other values of n
(namely 7, 9, 11, 13, 14, 18, 19, 21, 22, 23, 25,
26, 27, 28, 29, 31, 33, 35, 36, 37, 38, 39, ... A004169),
you will have to be satisfied with the simple trigonometric formula given above.
The above result is the first entry (dated March 30, 1796)
in the mathematical diary of Karl Friedrich Gauss (17771855).
It was the solution to a problem that had been open for nearly 2000 years,
and Gauss had solved it as a teenager!
This discovery was decisive in helping Gauss choose a career in mathematics
(he was also considering philology at the time). We should all be glad he did...
cem72 (20010328)
If you have a triangle and a pentagon both with a perimeter of 10,
which has the greater area?

Among triangles of a given perimeter,
the equilateral triangle is the one with the largest area.
Similarly, among pentagons of a given perimeter,
the regular pentagon is the one with the largest area.
If a regular pentagon and an equilateral triangle have the same perimeter,
the pentagon has a larger area than the triangle
(see below for the exact expressions of those areas).
On the other hand, for a given perimeter,
you can build scalene triangles or irregular pentagons with as small an area as you wish
(including a zero area for flat or "degenerate" polygons).
Therefore, for irregular polygons, there is no definite answer to your question.
For the record, a regular nsided polygon of diameter D has a perimeter
P = nD sin(p/n) and an area
S = n D^{2}/4 sin(p/n)
cos(p/n), which boils down to
S=P^{2}/[4 n tan(p/n)].
If you know that tan(x)/x is an increasing function of x when x is between 0 and
p/2,
you can easily deduce that S is an increasing function of n when P is held constant...
The more sides in a regular polygon of given perimeter, the larger the area.
For example an equilateral triangle with a perimeter of 10 has a surface of
25/[3 tan(p/3)] or about 4.811257,
whereas the regular pentagon of the same perimeter has a surface
25/[3 tan(p/3)], which is about 6.8819096.
The limiting case is, of course, the circle:
As n tends to infinity S tends to P^{2}/4p
(or pR^{2}
with P=2pR, if you prefer).
A circle with a perimeter of 10 has an area of 25/p,
which is about 7.957747.
johnrp (John P. of Middletown, NJ.
20001008)
A circle is a two dimensional object that has constant width;
its height remains constant regardless of the orientation.
Can you construct another twodimensional figure that has constant width,
but is not a circle?

There are plenty of examples. The simplest is the socalled
Reuleaux triangle,
pictured at right and named after the German engineer Franz Reuleaux (18291905):
Just take the three vertices of an equilateral triangle and connect each pair of vertices
with an arc of a circle centered on the third vertex. An interesting
theorem
due to Joseph Emile Barbier (18391889) states that the perimeter of
any curve of constant width is p times its width.
On 20001009, Mark Barnes (UK)
wrote:
You can do the same thing with any regular polygon [having an odd number of sides].
An example of a shape of constant diameter [constant width] in England is
the fifty pence coin (also the 20p which is the same shape but smaller).
You can form this shape by drawing a regular heptagon,
then using a compass to construct arcs between adjacent corners,
the centre for the arc being the corner three corners along.
The shape was chosen when the fifty pence coin was introduced at the time of decimalisation
in 1971.
It was chosen because it was easily identified by feel  all other coins were circles 
but the constant diameter [constant width] allows it to roll like circular coins,
meaning it can be used in slot machines.

Note that with any shape of constant width you can construct infinitely many new ones:
The (convex hull of the) envelope of the circles of radius R centered on a curve of
constant width is also a curve of constant width. (If R is small enough,
the envelope includes another shape inside the original one which may only be a scaleddown
version of it.)
The rounded shapes are also constructed with arcs of circles centered on the vertices
of the original polygon.
The radius of each such arc is either R or R+D, where D is the (constant) diameter of
the original shape with sharp corners...
Why not coin a word for these shapes.
How about modifying the names of the polygons into nice macaronic terms:
triangroller, pentagroller, heptagroller, polygroller...
Call any curve of constant width a roller:
Martin Gardner would probably have loved to adocate this
in his former Scientific American column...
You may have noticed that using a lightduty handheld drill on
a thin piece of metal often results in a hole which is not round, but instead in the shape
of a rounded "triangroller" (I am sure that other shapes of constant width
do occur, but they may be less frequent and/or less noticeable).
This is because, in 2 dimensions, a drill bit cuts at two points a fixed distance apart,
but the axis of the drill bit may vibrate...
Such weird holes do not occur either with a thick piece of metal or when using a drill
press (unless the bit happens to be very flexible).
It is fairly easy to figure out why...
Think about it.
johnrp (John P. of Middletown, NJ.
20001203)
One can construct figures of constant diameter [constant width]
from a regular polygon (with an odd number of vertices)
by drawing small circles of radii R around each vertex and then drawing arcs
from each vertex as to connect the two opposite circles at a tangent.
Is there a way to do this with an arbitrary polygon?
For example, if I have a triangle that is not equilateral?
Sure. There are plenty of such irregular curves of constant width...
Call them irregrollers!
Note: It's probably better to use the accepted term "width" in this context.
Although little confusion is possible with "diameter" here,
the standard meaning is different and you may have a need to mention the diameter
the largest width in related discussions.
You may build such a shape around a scalene triangle ABC as follows.
In this description, we assume, AC is the longest side and BC the shortest
(AC>AB>BC).

Draw the arc of the circle (of radius AC) centered on C going from A to
the intersection B' with the line BC.

Draw the arc of the circle (of radius ACBC) centered on B going from B' to
the intersection B" with the line AB.

Draw the arc of the circle (of radius AB+ACBC) centered on A from B" to
the intersection C" with the line AC.

Draw the arc of the circle (of radius ABBC) centered on C from C" to
the intersection C' with the line BC.

Finally draw the arc of the circle (of radius AB) centered on B from C' back to A.
The five arcs you've drawn make up the perimeter of a shape of constant width (W=AB+ACBC).
It has at least one sharp corner
(2 in case of an isoceles triangle with a base AC larger than the other sides,
and 3 sharp corners in case of an equilateral triangle).
If you want a smooth curve you may increase all of the above radii by the same quantity R.
The construction is trivially modified by introducing only two points A' and A"
at a distance R from A on AB and AC respectively.
The construction starts with A" and ends with an arc of radius R from A' to A",
to close the curve.
Alternatively, you may describe the new "rounded" shape as the set of all points
at a distance R from the (inside of) the previous shape...
You may want to notice that these curves need not involve any circular arcs at all...
What you want is to have conjugate arcs (not necessarily circular)
on opposite sides of your shape of constant width W so that
if the radius of curvature at one point of an arc is R,
the radius of curvature at the corresponding point on the other arc is WR.
This means the two arcs have the same evolute.
More precisely, if you roll a segment of length W on any curve you care to choose
(without inflexion points) you obtain a pair of conjugate arcs as the trajectories
of the segment's endpoints!
(Conjugate circular arcs correspond to the degenerate case,
where the above "base" curve is reduced to a single point,
so the segment just rotates instead of rolling.)
When using such building blocks to make an actual shape of constant width,
you only have to make sure the perimeter closes up into a convex shape
(you could easily end up with some kind of double spiral).
Remarkably, a few symmetry remarks allow you to find immediately entire families
of curves with constant width.
Take the deltoid, for example
(it does not have to be an exact deltoid;
any curve with the same general features and symmetries will do):
All its (closed) convex involutes are curves of constant width!
They look very much like the rounded equilateral triangles you mentioned in your question,
but without any circular arcs on the perimeters...
ciderspider (Mark Barnes, UK.
20001009)
A sphere is a 3d object with constant diameter [constant width].
What are other 3d shapes with constant diameter?
At first, we thought a threedimensional generalization of
the Reuleaux triangle could work.
Consider the shape pictured at right (image courtesy of
FastGeometry).
This 3D solid is obtained as the intersection of the four balls of radius R
centered on the vertices of a regular tetrahedron of side R.
If the solid is on an horizontal table, its highest point will indeed be at a height
R over the surface of the table, provided the point of contact [with the table]
is either one of the 4 vertices or is somewhere
in the midst of one of the spherical faces. So far so good.
However, the point of contact could also be on one of the edges, in which case
the highest point does move on the opposite edge if we
rotate the solid around the tangent to the edge at the point of contact
(as we may).
This opposite edge is an arc of a circle whose axis of symmetry goes through
the two extremities of the edge of contact.
As [part of] this arc rotates around a different axis, the height of its highest point
varies, which shows that this solid does not have constant width;
in fact, its width varies between R and
(Ö3  ½Ö2) R
[» 1.024944 R].
On the other hand, we do generate a 3D solid of constant width
by rotating any 2D shape of constant width around an axis of symmetry,
if it has one (the Reuleaux triangle has 3).
This works because, any rotation of such a solid is a combination of three independent
rotations which all preserve the width between two given parallel planes,
namely:
a rotation around the solid's axis of symmetry (obviously), a rotation around an
axis perpendicular to the two planes (think about it) and, finally,
a rotation around an axis parallel to the planes and perpendicular to
the axis of symmetry (which is seen "sideways" as a 2D rotation of a crosssection of
constant width).
[Notice that the first two rotations may coincide, but only when
there are 2 independent rotations of the last type, so we always have 3 independent
widthpreserving elementary rotations.]
Once you have a solid of constant width, you may build infinitely many others,
since, for any D>0, the set of all points
within a distance D of some given solid of constant width
is also a solid of constant width...
ciderspider (Mark Barnes, UK.
20001009)
A 4D hypershpere is a 4D object with constant diameter [constant width].
What are other 4D shapes with constant diameter? 5D? 6D?
The construction(s) outlined at the end of the previous article seem to remain valid to
obtain a symmetrical shape of constant width in N+1 dimensions from one in N dimensions.
(Michael of Reston, VA.
20001019)
Can you have a Cartesian coordinate system where the axes are mutually perpendicular
and the number of axes is greater than 3?
Yes, absolutely!
That's what happens in a Euclidean space with 4 dimensions or more.
It may be difficult (or impossible) to
visualize a space with more than 3 dimensions,
but there is no great mathematical difficulty in considering the set of all quadruplets
of real numbers (x,y,z,t), which is what 4D space really is.
(B. N. of Auburn, AL.
20000503)
What is the formula for the hypervolume of a fourdimensional sphere?
If R is the radius of a 4D hypersphere,
its hypervolume is simply
p^{2 }R^{4 }/2 .
More generally, in n dimensions, a sphere of radius R has a volume equal to:
V = R^{n}
p^{n/2} / G(1+n/2)
Using the definition of the
Gamma function (G)
in terms of factorials (the notation being
k! = 1´2´3´ ... ´k ),
the coefficient of R^{n} in the above is:
 p^{k}/k! with k=n/2 when n is even, or
 2^{n} p^{k} k!/n!
with k=(n1)/2 when n is odd.
A formula valid in both cases (using the doublefactorial notation) is given below.
In other words, the "hypervolume" of an ndimensional sphere of unit radius is:
 1 for n=0 (the "0volume" of a point must be so defined for consistency!),
 2 for n=1 (length of a segment of "radius" 1),
 p for n=2 (area of a unit disc),
and 4p/3 for n=3 (volume of a sphere),
 p^{2}/2 for n=4 (the original question),
and 8p^{2}/15 for n=5,
 p^{3}/6 for n=6,
and 16p^{3}/105 for n=7,
 p^{4}/24 for n=8,
and 32p^{4}/945 for n=9,
 p^{5}/120 for n=10,
and 64p^{5}/10395 for n=11,
 p^{6}/720 for n=12,
and 128p^{6}/135135 for n=13,
 ... ...
 (p^{k}/k!) for n=2k,
and (2^{k+1}p^{k}/n!!) for n=2k+1.
In the above, we used the (standard) "doublefactorial" notation n!! as a shorthand for
n(n2)(n4)...
which is the product of all positive integers up to n which have
the same parity as n [note that n!! = (n2)!! n ]:
0!!=1, 1!!=1, 2!!=2, 3!!=3,
4!!=8, 5!!=15, 6!!=48, 7!!=105 ...
Using this doublefactorial notation, it's possible to give a cute formula
valid in n dimensions, whether n is even (n=2k) or odd (n=2k+1), namely:
V =
(p/2)^{k }(2R)^{n} / n!!^{ }
 What about the hyperarea of [the boundary of] such a hypersphere?
The above hypervolume could be obtained by integrating the hyperarea of a
shell from 0 to R. Conversely, if aR^{n} is the hypervolume of an
ndimensional ball of radius R, then naR^{n1}
must be its "hypersurface area" (S).
(Except for n=0, which we rule out as meaningless.)
For a hypersphere of unit radius in n dimensions, this means that the hypersurface
"area" [i.e., the measure in (n1) dimensions] has the following values:
2 for n=1,
2p for n=2,
4p for n=3,
2p^{2} for n=4,
8p^{2}/3 for n=5,
p^{3} for n=6...
We may replace n/n!! by 1/(n2)!!
in the following formula [retaining the case n=1 with the convention
(1)!! =1]:
S = 2 R^{n1}
p^{n/2} / G(n/2)
=
2^{nk}
p^{k}
R^{n1}
n/n!!^{ }
[where n is 2k or 2k+1]
Of particular interest is the socalled EinsteinEddington universe,
which is defined as the 3dimensional boundary of the 4dimensional hypersphere of radius R.
The above shows that the volume of the EinsteinEddington universe
is 2p^{2 }R^{3}.
If this is meant to be a model of the Universe we live in [capital "U"],
the distance R to the "center" of the 4D sphere is quite literally
out of this world and it may be better to consider the
maximal possible distance D between two points in the Universe.
As D is simply pR,
the volume of the Universe is 2D^{3}/p.
(Jerry of Nashville, TN.
20001118)
What [polyhedron] has six faces?

A polyhedron with 6 faces is called a hexahedron.
The cube
is an hexahedron, but that's certainly not the only one:
The socalled triangular dipyramid
is another possibility
(with 5 vertices and 9 edges, this solid may be obtained by "adding" one vertex to a
tetrahedron to make it look like two tetrahedra "glued" on a common face).
A third hexahedron is the pentagonal pyramid
(6 vertices, 10 edges; a pyramid whose base is a pentagon).
The above three are the only hexahedra which exist in a version where all 6 faces
are regular polygons.
The least symmetrical of all hexahedra is the
tetragonal antiwedge (it has only one possible symmetry,
a 180° rotation). This skewed hexahedron has
the same number of edges and vertices as the pentagonal pyramid.
Its faces consist of 4 triangles and 2 quadrilaterals.
Such a solid may be obtained by considering two quadrilaterals
that share an edge but do not form a triangular prism. First, join with an edge the
two pairs of vertices closest to the edge shared by the quadrilaterals. To complete the
polyhedron, you must join two opposite vertices of the nonplanar quadrilateral
that you're left with.
This can be done in one of two ways (only one of which gives a convex polyhedron).
Loosely speaking, there are two types of
tetragonal antiwedges which are mirror images of each other;
each is called an enantiomer, or enantiomorph of the other.
The tetragonal antiwedge is thus the simplest example of a chiral
polyhedron (in particular, any other hexahedron can be distorted into a shape
which is its own mirror image). Because of this unique property among hexahedra,
the tetragonal antiwedge
may also be referred to as the chiral hexahedron.
The other types of hexahedra are more symmetrical and simpler to visualize.
One of them may be constructed by cutting off
one of the 4 base corners of a square pyramid to create a new triangular face.
This hexahedron has 7 vertices and 11 edges.
Its faces include 3 triangles, 2 quadrilaterals and 1 pentagon.
It could also be obtained by cutting an elongated square pyramid
(the technical name for an obelisk)
along a bisecting plane through the apex of the pyramid and the
diagonal of the base prism, as pictured at left.
For lack of a better term, we may therefore call this hexahedron an
hemiobelisk.
Also with 7 vertices and 11 edges, there's a solid which we may call
a hemicube (or square hemiprism), obtained
by cutting a cube in half using a plane going through two opposite corners and
the midpoints of two edges. Its 6 faces include 2 triangles and
4 quadrilaterals.
With 8 vertices and 12 edges, the cube
(possibly distorted into some kind of irregular prism or truncated tetragonal pyramid)
is not the only solution:
Consider a tetrahedron, truncate two of its corners and you have a
pentagonal wedge.
It has as many vertices, edges and faces
as a cube, but its faces consist of 2 triangles,
2 quadrilaterals and 2 pentagons.
The above 7 types (8 if you counts both chiralities of tetragonal antiwedges)
include all possible hexahedra. By contrast, there's only one
tetrahedron. There are two types of pentahedra
(exemplified by the square
pyramid and the triangular prism). There are 7 types of hexahedra, as we've just seen.
34 heptahedra, 257 octahedra, 2606 enneahedra, 32300 decahedra, 440564 hendecahedra, etc.
(see our detailed table of the enumeration, elsewhere
on this site).
For an unabridged discussion of hexahedra
and more general information about polyhedra,
see our dedicated Polyhedra Page...
(J. T. of Summerville, SC.
20001119)
How many edges (lines) are in a cylinder?
I assume you're talking about a finite cylinder; the "ordinary kind"
with two parallel bases, which are usually circular (as opposed, say,
to an infinite cylinder with an infinite lateral surface and no bases).
The answer is, of course, that there are two edges, the two circles.
I think you figured this out by yourself and did not need anybody to tell you,
so I suppose your real concern is elsewhere...
Because you used the term "edges" I suspect you think you've found an
exception to the DescartesEuler formula, which states that "in a polyhedron"
the numbers of faces (F), edges (E) and vertices (V)
are related by the formula: FE+V=2.
In a way, you have such a "counterexample": In a cylinder, there are 3 faces
(top, bottom, lateral), 2 edges (top and bottom circles) and no vertices,
so that FE+V is 1, not 2! What could be wrong?
Nothing is wrong if things are precisely stated.
Edges and faces are allowed to be curved, but the DescartesEuler formula
has 3 restrictions, namely:

It only applies to a (polyhedral) surface which is topologically "like" a sphere
(imagine making the polyhedron out of flexible plastic and blowing air into it,
and you'll see what I mean). Your cylinder does qualify (a torus would not).

It only applies if all faces are "like" an open disk.
The top and bottom faces of your cylinder do qualify, but the lateral face
does not.

It only applies if all edges are "like" an open line segment.
Neither of your circular edges qualifies.
There are two ways to fix the situation.
The first one is to introduce new edges and vertices artificially to
meet the above 3 conditions.
For example, put a new vertex on the top edge and on the bottom edge.
This satisfies condition (3),
since a circle minus a point is "like" an open line segment.
The remaining problem is condition (2); the lateral face is not "like"
an open disk (or square, same thing).
To make it so, "cut" it by introducing a regular edge between
your two new vertices.
Now that all 3 conditions are met, what do we have?
3 faces, 3 edges and 2 vertices.
Since 33+2 is indeed 2, the DescartesEuler formula does hold.
The better way to fix the formula does not involve introducing
unnecessary edges or vertices.
It involves the socalled Euler characteristic,
often denoted c (chi):
The Euler Characteristic c
( chi )
The fundamental properties of c (chi)
may be summarized as follows_{ }:
 Any set with a single element has a c of 1 :
"x,
c ( {x} ) = 1
 c is additive:
For two disjoint sets E and F,
c(EÈF)
= c(E) + c(F)
 If E is homeomorphic to F, then
c(E) = c(F)
("Homeomorphic" is the precise term for topologically "like".)
Using the above 3 properties as axioms, it's not difficult to show by induction
that, if it's defined at all, the c of ndimensional
space can only be equal to (1)^{n}.
(Hint: A plane divides space into 3 disjoint parts; itself and 2 others...)
 c (point) = 1
 c (entire straight line, or open segment) = 1
 c (plane or open disc) = 1
 c (space or open ball) = 1
 c (space with ndimensions) =
(1)^{n}
 c (surface of a sphere) = 2
 c (surface of an infinite cylinder) = 0
 c (surface of torus) = 0
 c (circle, or semiopen segment) = 0
 etc.
Now, back to our problem:
Why is the DescartesEuler formula valid to begin with?
Well, that's because the c of a sphere's surface is 2
and it's "made from" disjoint faces, edges and vertices, each respectively with a
c of 1, 1 and 1.
In the "natural" breakdown of your cylinder (whose c
is indeed 2), you have no vertices, two ordinary faces (whose
c is 1) and one face whose c is 0
(the lateral face), whereas the c of both edges is 0.
The total count does match.
Note
(20001119) :
The orthodox definition of the EulerPoincaré characteristic does not
use the above 3 fundamental properties as "axioms" but instead is closer to the
historical origins of the concept (generalized polyhedral surfaces).
It would seem natural to extend the definition of c
to as many objects as the axioms would allow.
This question does not seem to have been tackled by anyone yet..._{ }
Consider, for example, the union A of all the intervals
[2n,2n+1[
from an even integer (included) to the next integer (excluded).
The union of two disjoint sets homeomorphic to A can
be arranged to be either the whole number line or another set homeomorphic to A.
So, if c(A) was defined to be x, we would
simultaneously have x = x+x and
1 = x+x.
Thus, x cannot possibly be any ordinary number,
and the latter equation says x is nothing like a signed infinity either [as
(+¥)+(+¥) ¹ 1].
At best, x could be defined as an unsigned infinity
(¥) like the
"infinite circle" at the horizon of the complex plane
(¥+¥ is undetermined).
This could be a hint that a proper extension of c would
have complex values...
