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# Geometry

Geometry is the gate to Science.  This gate is
so small that one can only enter it as a child
.
William Clifford   (1845-1879)

### Related Links (Outside this Site)

The Geometry Center (University of Minesota).
The Geometry Junkyard by David Eppstein (UC Irvine).
Geometry from the Land of the Incas  by  Antonio Gutierez.

## Elementary Geometry and Topology

(E. C. of Pasadena, CA. 2000-10-10)
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. 2001-02-12)
[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. 2001-02-10)
What is the formula to determine the area of a circle?
[What about] a triangle, a trapezoid, a sphere, [a cylinder, a cone] ?

• A circle of radius R has area pR2 (p=3.14159265...).
• The area of a trapezoid is the arithmetic average (i.e., the half-sum) 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 4pR2 (its volume is 4pR/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 4pR2, 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 (R2=H2+r2) comes in handy to give you the lateral area of the conical surface as pr Ö(H2+r2) Note that the case R=r (or H=0) corresponds to a "flat" circular cone, which is simply a circle of area pR2... Back to our first formula!

(2015-06-15)   Simson line of a point on the circumcircle of a triangle
The line containing its three projections along the sides of the triangle.

By definition,  the  pedal triangle  of a point P with respect of a triangle ABC is the triangle formed by the orthogonal projections of P along the three sides of ABC.

That pedal triangle is flat  (i.e., its vertices are collinear)  if and only if P is on the  circle circumscribed to ABC.  (In that case, the line on which the three projections are located is called the  pedal line  or  Simson line  of P with respect to ABC.)

This result is due to Robert Simson (1687-1768).  It was first published by  William Wallace (1768-1843)  who was born 8 days before Simson died!

Arguably, the most remarkable point on the circumcircle is the Steiner point (1826).  The Simson line of the Steiner point is parallel to the line joining the circumcenter  O  to the symmedian point  K  (or  Lemoine point, 1873 ).

Simson line   |   Steiner deltoid   |   Jakob Steiner (1796-1863)   |   Emile Lemoine (1840-1912; X1860)

(2015-06-16)   Isogonal Conjugation
One of the crown jewels of modern geometry.

For a given triangle  ABC,  where A, B and C are not collinear, let's consider a point  P  which is not a vertex.  For any vertex  (say, A)  we build the line which is symmetrical to  AP  with respect to the  (internal)  angular bissector at  A.  The amazing fact is that the three lines so constructed  (one per vertex)  are concurrent!  Their common point of intersection  P*  is called the  isogonal conjugate  of P.

The conjugate of a vertex isn't well defined  (one of the three lines is undefined and the other two are identical).  Likewise, for any point on the circumcircle, the three lines are parallel  (they intersect "at infinity").

Otherwise, isogonal conjugation is an  involution  (which is to say that the isogonal conjugate of the isogonal conjugate of any point is itself).  This involution has four fixed foints  (the incenter and the three excenters).

• The isogonal conjugate of the  incenter  is itself.
• The isogonal conjugate of any  excenter  is itself.
• The  circumcenter  and  orthocenter  are isogonal conjugates.
• The isogonal conjugate of the  centroid  is the  symmedian point.

Isogonal conjugate   |   Isotomic conjugate

(K. D. of USA. 2000-11-13)   Euler's line  (Euler, 1765)
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 stand is often called  Euler's line  (it's undefined for an equilateral triangle).

The centroid  G  is between the orthocenter  H  and the circumcenter  O.  The distance  HG  is twice the distance  GO.  Recall that:

• The  centroid  G  is where the three medians intersect.
• The  orthocenter  H  is where the three altitudes intersect.
• The  circumcenter  O  is where the 3 perpendicular bisectors meet.
• The  incenter  I  is where the 3 interior angle bisectors meet.
• An  excententer  is where an interior bissector meets two outer ones.

Except for isoceles triangles, the  incenter  is  not  on the Euler line.  Neither are any of the three  excenters.

The recently-discovered  Exeter point  (1986) is on the Euler line.

Triangles have a Magic Highway (9:59)  by  Zvezdelina Stankova   (Numberphile, 2016-02-01)

Triangle Centres and the Euler Line (15:08)  by  Zvezdelina Stankova   (Proofs, 2016-02-02)

(2011-05-08)   Euler's Circle  (9-point circle)
The  nine-point circle  named after Euler wasn't discovered by him.

The  incenter  of a scalene triangle is not on its  Euler line,  unlike another  remarkable  point  E  which is located exactly halfway between  H  and  O:  That point  E  is the center of the so-called  Euler circle  (or  9-point circle)  which goes though 9 special points of the triangle: the 3 midpoints of the sides, the feet of the 3 altitudes and the 3 midpoints from the orthocenter H to the vertices.  The  9-point circle  has half the radius of the circumcircle.

 Karl Feuerbach

### Euler's Circle and Feuerbach's Theorem :

Feuerbach's theorem  (1822)  states that the  9-point circle is tangent externally to the three  excircles  and internally to the  incircle  (at a point called  Feuerbach's point).

The existence of the 9-point circle was unknown to Euler.  The basic fact that the middles of the sides and the feet of the altitudes all belong to the same circle was discovered independently by Charles Brianchon (1783-1864; X1803), Jean-Victor Poncelet (1788-1867; X1807) and Karl Wilhelm von Feuerbach (1800-1834).  The remark that the same circle also goes through the midpoints between the orthocenter and the vertices was first made by Olry Terquem (1782-1862; X1801)  who coined the term  9-point circle,  which stuck...

That might be unfortunate in view of the fact that there are many more than  9  special points on  Euler's circle :  In 1996,  Jingchen Tong  &  Sidney Kung  named  24  of those!

The  Poncelet point  of a quadrilateral  ABCD  is the point of intersection of the four respective  Euler circles  of  ABC, ABD, ACD and BCD.

Twelve New Points on the Nine-Point Circle   by  Jingchen Tong & Sidney Kung   (1996)

(2012-11-05)   Power of a point with respect to a circle.
The  intersecting chords  theorem  (Jakob Steiner, 1826).

Consider a point  P  at a distance  d  from the center of a circle of radius  r.

If a line through  P  intersects that circle at points  A  and  B  (A = B  if the line is tangent to the circle)  then the following quantity is called the  power of P with respect to the circle.  It doesn't depend on the intersecting line.

d2 - r2   =   PA . PB

MN  denotes the  linear abscissa  whose magnitude is the Euclidean distance between  M  and  N.  Its sign depends on the orientation of the line.

UV  +  VW   =   UW         (Chasles relation)

With respect to a circle, a point has negative power if it's inside the circle, positive power if it's outside and zero power if it's on the circle itself.

Cut-the-Knot  by  Alexander Bogomolny :   Power of a Point   |   Intersecting Chords Theorem

(2011-05-09)   Trilinear  &  Barycentric Coordinates
Two dual types of  homogeneous coordinates  in the Euclidean plane.

Trilinear coordinates  (trilinears)  and barycentric coordinates are examples of  homogeneous coordinates.  This is to say that, in either system, two proportional triplets represent the same point of the Euclidean plane.

In both systems, by convention, the triplet  (0,0,0)  represents the point at infinity  (spanning the entire  horizon  of the plane, in all directions).

A finite point  M  of barycentric coordinates  (x,y,z)  is defined in terms of the three base points  A,B,C  by the following relation:

(x+y+z) M   =   x A  +  y B  +  z C

The trilinear coordinates of a point of barycentric coordinates  (x,y,z)  are  (x/a, y/b, z/c).  Under the usual geometric interpretation, a, b and c are the pairwise distances between the three base points and, therefore, must satisfy the triangular inequality.  However, the above correspondence can be investigated abstractly without that requirement...

Actually, barycentric coordinates describe a general vector space without resorting to any metric concept, whereas the mapping from barycentric to trilinear coordinates is a way to endow the plane with a definite metric  (as is a linear mapping from the plane to its dual).  Barycentric coordinates are to contravariant cartesian coordinates what trilinears are to covariant coordinates.  When the triangular inequality fails for a, b and c, the metric so defined is  Lorentzian,  not  Euclidean.

Trilinear Coordinates   |   Barycentric Coordinates

(M. Muz Zviman, Ph.D. 2001-11-28; e-mail)
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  x2/a2 + y2/b2=1 is:  x = a sin(q)   and   y = b cos(q) . We assume  a>b and define e=1-b2/a2. 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).

The above parameterization is used conventionally, because it turns out to be numerically superior to the complementary one which would make  q  small near the  sharper  apex.  That's 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(1-e2sin2q) 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 old-fashioned quadrature )  known as the  incomplete elliptic integral of the second kind :

a E(q,e)   =   a ó
õ
q
0
 Ö 1 - e2 sin2 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 single-argument 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)C(½,n) e2n sin2na, for n=0, 1, 2, 3... Each such term may then be integrated individually using the formula:

ó
õ
q
0
sin2n a da  =
 q 22n
ì
î
2n
n
ü
þ
+
 1 22n

 n å k = 1

 (-1)k k
ì
î
2n
n-k
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 2001-11-29, Muz Zviman wrote: Thank you for the quick answer. The website is great.   Best regards, Muz

On 2002-12-31, David W. Cantrell proposed:
A 0.56% approximation to the above (first posted to the sci.math newsgroup).

(Jaleigh. B. of Minonk, IL. 2000-11-26 twice)
What is the formula for the perimeter of an ellipse?
(S. H. of United Kingdom. 2001-01-25)
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 x2/a2 + y2/b2=1 with a>b, a is called the major radius, and b is the minor radius. The quantity e = Ö(1-b2/a2) 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)/(2n-1) [(2n)!/(2n n!)2]2 e2n. The first such term (for n=0) is equal to 1 whereas all the others are negative correction terms :

P/2pa = 1 - [1/4]e2 - [3/64]e4 - [5/256]e6 - [175/16384]e8 - [441/65536]e10 ...

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:

 (1)
 P  » pÖ 2(a2+b2) - (a-b)2/2

A 1914 formula due to Srinivasa Ramanujan (1887-1920) is

 (2)
 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 = (a-b)2/(a+b)2 :

 (3)
 P  » p (a+b) [ 1 + 3h / ( 10+Ö 4-3h ) ]

The relative error of this formula for ellipses of low eccentricities is fabulous:

(-3/237 ) e20 [ 1 + 5 e2 + 11107/768 e4 + 4067/128 e6 + 3860169/65536 e8 + ... ]

In 1917, Hudson came up with a formula without square roots, which is traditionally expressed in terms of the quantity L = h/4 = (a-b)2/[2(a+b)]2 :

 (4)

P » p (a+b)/4 [ 3(1+L) + 1/(1-L) ]

In 2000, Roger Maertens proposed the following so-called "YNOT formula":

 (5)

P » 4 (ay+by) 1/y   or   P » 4a (1 + (1-e2)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.

A popular upper bound formula is due to Euler (1773):

 (6)
 P  » pÖ 2(a2+b2)

The following simple lower bound formula is due to Johannes Kepler (1609):

 (7)
 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 -3e4/64 [1+e+ ...] -84.61 m
(6) Euler 1773   +11.072 e4/64 [1+e+ ...] +28.20 m
(5) Maertens 2000     +0.3619 (2y-3)e4/64 [1+e+ ...]   +1.97 m
(1)       -3.809 -3e8/214 [1+2e+ ...]   -1.49 10-5
(2) Ramanujan I     -0.416 -e12/221 [1+3e+ ...]   -1.75 10-12
(4) Hudson 1917     -0.189 -9e16/230 [1+4e+ ...]   -1.39 10-18
(3) Ramanujan II     -0.0402 -3e20/237 [1+5e+ ...]   -1.63 10-25

(Sherry of Murray, KY. 2000-10-19)
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:

x2/a2 + y2/b2 = 1

The area of such an ellipse is simply   S = pab.

(B. K. of Honolulu, HI. 2000-10-12)
How can I determine the volume of an oval object? [egg-shaped solid]

The volume of an ellipsoid of equation   x2/a2 + y2/b2 + z2/c2 = 1   is

V   =   ( 4p/3 ) a b c

This is a good approximation for other egg-shaped 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.

(Scott of Emeryville, CA. 2000-10-10)
How do I calculate the surface area of a cylinder
and an oblate sphere (flying saucer shaped) ?
(M. P. of Williamsport, PA. 2000-10-16)
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 = 2pa2 [ 1 + (1-e2) atanh(e)/e ] ,     or
S = 2pa2 [ 1 + (b/a)2 atanh(e)/e ]     [ See proof. ]
In this, e is Ö(1-b2/a2), where b<a is the "polar radius" (the distance from either pole to the center) and atanh(e) is   ½ ln((1+e)/(1-e))   [also denoted argth(e) ].

The surface area S of a prolate ellipsoid ("cigar-like") 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 = 2pb2 [ 1 + (a/b) arcsin(e)/e ]
This shows that a very elongated ellipsoid has an area of p2ab (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 noted by Archimedes of Syracuse (c.287-212BC).

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.

William Van Drent, Ph.D. (2001-08-16; e-mail)
Staff Scientist / New Product Development Manager. Digital Measurement Division. ADE Technologies, Inc. Newton, MA.
[...]  For an ellipse of equation
A x2 + B y2 + C xy + D x + E y + F = 0,
express  (in terms of A, B, C, D, E, and F ) ... We'll assume  A+B  is positive  (after changing all signs, if needed).

Background : The above general quadratic equation describes planar curves known as conic sections  (because they can be 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. It may also be a so-called degenerate conic which consists of a pair of lines (intersecting, parallel or equal) in the case of a quadratic polynomial that is a product of two first-degree 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: x2+y2+4=0. For completeness, a 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: x2+y2 = 0, which is to say (x+iy)(x-iy) = 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+y-i) = 0].

As the  question  is only about real ellipses, so is the following discussion:

First, we 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 gives:

(B-A) 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 cos2q + B sin2q + C cos q sin q ] X2 + [A sin2q + B cos2q - C cos q sin q ] Y2 + [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 X2 and Y2. We find the former equal to (1/2)[A(1+1/cos 2q) + B(1-1/cos 2q)], whereas the latter equals (1/2)[A(1-1/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 X-axis to be the major one, the coefficient of X2 is inversely proportional to the square of the major radius and is thus smaller than the coefficient of Y2 (which is is inversely proportional to the square of the minor radius). In other words, (A-B)/cos 2q is negative. With this in mind, we can 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  (we're assuming  A+B > 0):

 cos 2q = (B-A)/Q    and    sin 2q = -C/Q ,    where    Q = Ö (A-B) 2 + C 2

This determination of the inclination isn't 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 X2 and Y2 respectively boil down to (A+B-Q)/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 > Q2 , 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 = (CE-2BD)/(4AB-C 2 )   and   yo = (CD-2AE)/(4AB-C 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.

To determine the principal radii of the ellipse,  we first need the value of the equation's constant term  (call it 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, which yields:

K   =   F + (CDE - AE 2 - BD 2 ) / (4AB-C 2 )

In the properly tilted frame centered at (xo,yo), the equation of the ellipse is thus: (A+B-Q) X2 + (A+B+Q) Y2 + 2K   =   0 , which we just need to identify with the standardized equation X2/a2 + Y2/b2 = 1 in order to obtain the values of the principal radii, and/or their squares:

a2   =   -2K / (A+B-Q)
b2   =   -2K / (A+B+Q)

Thus, a real ellipse is described only when   (A+B)K < 0   and   4AB > C 2.

Ujjwal Rane of Mumbai, India  (2010-09-28; e-mail)   Parabola
Parabolic arc of given extremities and prescribed apex between them.

By definition, a  parabola  is the set of points, in the Euclidean plane, that are equally distant from a given point  (the parabola's  focus  F)  and a prescribed straight line  (the parabola's  directrix).  The  axis  of a parabola is the perpendicular to the  directrix  trough the  focus.  The point at the intersection of the parabola and its  axis  is the parabola's  apex  (O).

More generally, we call  apex  of a planar curve any point where its curvature  is extremal.  There are four such points in a proper ellipse.  An hyperbola has two apices  (plural of apex).  A parabola has only one apex, which can be characterized as above.

If the apex  O  of a parabola is between two of its points  A  and  B,  we want a construction of the focal point  F  based on  A,  O  and  B.

Let's first determine the  locus  of the foci of all the parabolas through point  A  whose apex is at  O.

(Danny of Lincoln, RI. 2000-10-19)
Let's say I have a parabola, f(x) = x2/50.  Where is the focal point?

In a parabola of equation y=x2/(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=x2/50, 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. 2000-11-12)
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 d1 be the distance from the chord to the point of incidence (x,y) on the parabola and let d2 be the distance from (x, y) to the focus.  Show that the sum of the distance d1+d2 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=x2/4f (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 x2+(y-f)2 = 4fy+(y-f)2 = (y+f)2. In other words, the distance d2 is (y+f). On the other hand, d1 is equal to A-y (where A is some constant which depends on how far you drew the "chord" described in the question).  Therefore, d1+d2 = f+A = constant.

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. 2000-10-01)
How do I find the centroid of a circular segment?

Use Guldin's theorem (named after Paul Guldin 1577-1643), which is also called Pappus theorem in the English-speaking 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. 2000-10-22)
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 (3-Ö3)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, Ö6-Ö2 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.

Wikipedia :   Prince Rupert's cube

Prince Rupert's paradox (7:18)  by  Burkard Polster  (2017-07-20)

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 is 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 (2-Ö2).

The same result can be obtained with standard trigonometric functions:  The ratio  a/d  is the sine of a  p/8  angle  (22.5°; a full turn divided by 16)  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!

(2001-02-14)   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 n-sided 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  22n + 1   There are probably only five of these, namely:  3, 5, 17, 257 and 65537.

An explicit construction of a  65537th  root of unity with straightedge and compass was given in 1894 by Johann Gustav Hermes (1846-1912)  after spending  12 years  on the project...  His 200-page manuscript is now preserved in Göttingen.
nn-gon Perimeter/diameter ratio = n sin(p/n)
2digon22
3triangle2.598 076 211+ (3/2)Ö3
4square2.828 427 125- 2 Ö2
5pentagon2.938 926 261+ (5/2)Ö((5-Ö5)/2)
6hexagon33
8octagon3.061 467 459- 4 Ö(2-Ö2)
10decagon3.090 169 944- 5 (Ö5-1)/2
12dodecagon3.105 828 541+ 3 (Ö3-1)Ö2
15pentadecagon3.118 675 363- (15/8)[Ö(10+2Ö5 ) - Ö3 (Ö5-1)]
16hexadecagon3.121 445 152+ 8 Ö(2 - Ö(2+Ö2))
17heptadecagon3.123 741 803- 17 Ö((1-c)/2)    [ c=cos(2p/17) ]
c = { 2Ö[ 17+3Ö17-Ö(2(17-Ö17))-2Ö(2(17+Ö17)) ] + Ö(2(17-Ö17)) - 1 + Ö17 } / 16
20icosagon3.128 689 301- 5 Ö( 8 - 2Ö(10+2Ö5))
24tetracosagon3.132 628 613+ 6 Ö( 8 - 2Ö2 - 2Ö6)
30triacontagon3.135 853 898+ (15/4) [ Ö(30-6Ö5) - Ö5 - 1 ]
¥circle3.141 592 654- p

If a is the side of an n-gon of diameter d, the side b of the 2n-gon 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/2-c, 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 b2 =  1/2 [ 1 - Ö(1-a2) ]. In other words, if x is the square of the side of the n-gon of unit diameter, the square y of the side of the 2n-gon of unit diameter is given by y =  1/2 [ 1 - Ö(1-x) ] (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 fixed-precision computations are used blindly with the formula "as is"). Starting with the trivial case of the hexagon, Archimedes of Syracuse (c.287-212BC) iterated this 4 times to compute the ratio of the circumference to the diameter in a 96-sided 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 (1539-1610): 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 St 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. 2002-02-19)
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 n-gon with just square roots only when the n-gon is constructible with straightedge and compass.  A beautiful result of Gauss (1796) says that an n-gon 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 so-called 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 Carl Friedrich Gauss (1777-1855).  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 (2001-03-28)
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 n-sided polygon of diameter D has a perimeter P = nD sin(p/n) and an area S = n D2/4 sin(p/n) cos(p/n), which boils down to S=P2/[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 P2/4p (or pR2 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. 2000-10-08)
A circle is a two dimensional object that has constant width; its height remains constant regardless of the orientation. Can you construct another two-dimensional figure that has constant width, but is not a circle?

There are plenty of examples. The simplest is the so-called Reuleaux triangle, pictured at right and named after the German engineer Franz Reuleaux (1829-1905): 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 (1839-1889) states that the perimeter of any curve of constant width is p times its width.

 On 2000-10-09, 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 scaled-down 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, à la  Martin Gardner?  Any curve of constant width might be called a  roller  and those based on polygons could be dubbed  polygroller :  triangroller, pentagroller, heptagroller, etc.

You may have noticed that using a light-duty 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. 2000-12-03)
One can construct figures of  constant "diameter" 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 --universally understood as 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).

1. Draw the arc of the circle (of radius AC) centered on C going from A to the intersection B' with the line BC.
2. Draw the arc of the circle (of radius AC-BC) centered on B going from B' to the intersection B" with the line AB.
3. Draw the arc of the circle (of radius AB+AC-BC) centered on A from B" to the intersection C" with the line AC.
4. Draw the arc of the circle (of radius AB-BC) centered on C from C" to the intersection C' with the line BC.
5. 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+AC-BC). 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).

To get 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 W-R. 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. 2000-10-09)
What are some 3d shapes of constant width, besides spheres?

Surprisingly enough, an obvious three-dimensional generalization of the Reuleaux triangle  doesn't  work:  Consider the Reuleaux tetrahedron  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].

In 1911,  Ernst Meissner  and  Friedrich Schilling  turned the above idea into an actual solution by "rounding" three of the six edges of the above solid.  The resulting solid of constant width is now called a Meissner tetrahedron  (there are two distinct types, as the unrounded edges may either form a triangle or meet at a vertex).  The original Meissner tetrahedra do not possess tetrahedral symmetry, but there's a unique way to round all edges the same way to preserve that symmetry.

A simpler way to generate a 3D solid of constant width is to rotate 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 cross-section 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 width-preserving 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...

Actual Meissner tetrahedra in action  captured on video by  Brady Haran  (2013-11-11)
Matt Parker  &  Steve Mould  (the man in the previous video)  sell  classical Meissner tetrahedra.
The more modern version  (featuring tetrahedral symmetry)  doesn't seem to be commercially available.

ciderspider (Mark Barnes, UK. 2000-10-09)
A 4D hypershpere is a 4D object with  constant width.  What are other 4D shapes with constant width?  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. 2000-10-19)
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's no great 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. 2000-05-03)
What is the formula for the hyper-volume of a four-dimensional sphere?

If R is the radius of a 4-D hypersphere, its hyper-volume is simply pR/2 .

More generally, in n dimensions, a sphere of radius R has a volume equal to:

V   =   Rn pn/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 Rn in the above is:

• pk/k! with k=n/2 when n is even, or
• 2n pk k!/n! with k=(n-1)/2 when n is odd.
A formula valid in both cases  (using the  double-factorial  notation)  is given below. In other words, the "hypervolume" of an n-dimensional sphere of unit radius is:
• 1 for n=0 (the "0-volume" 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),
• p2/2 for n=4 (the original question), and 8p2/15 for n=5,
• p3/6 for n=6, and 16p3/105 for n=7,
• p4/24 for n=8, and 32p4/945 for n=9,
• p5/120 for n=10, and 64p5/10395 for n=11,
• p6/720 for n=12, and 128p6/135135 for n=13,
• ... ...
• (pk/k!) for n=2k, and (2k+1pk/n!!) for n=2k+1.

In the above, we used the (standard)  double-factorial  notation  n!!  as a shorthand for n(n-2)(n-4)... which is the product of all positive integers up to n which have the same parity as n:

0!!=1,  1!!=1,  2!!=2,  3!!=3,  4!!=8,  5!!=15,  6!!=48,  7!!=105 ...

To retain the relation  n!! = (n-2)!! n   for all positive integers, including  n = 1,  the convention is made that  (-1)!! = 1.  For completeness, a less useful extension is available for all  odd  negative integers:

(-1)!! = 1,  (-3)!! = -1,  (-5)!! = 1/3,  (-7)!! = -1/15,  ...

Using the double-factorial 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 aRn is the hypervolume of an n-dimensional ball of radius R, then naRn-1 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 (n-1) dimensions] has the following values: 2 for n=1, 2p for n=2, 4p for n=3, 2p2 for n=4, 8p2/3 for n=5, p3 for n=6... We may replace n/n!! by 1/(n-2)!! in the following formula [retaining the case n=1 with the convention (-1)!! =1]:

S  =   2 Rn-1 pn/2 / G(n/2)   =   2n-k pk Rn-1 n/n!!    [where n is 2k or 2k+1]

Of particular interest is the so-called Einstein-Eddington universe, which is defined as the 3-dimensional boundary of the 4-dimensional hypersphere of radius R. The above shows that the volume of the Einstein-Eddington universe is 2pR3. If this is meant to be a model of the Universe we live in [capital "U"], the distance R to the "center" of the 4-D 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 2D3/p.

(Jerry of Nashville, TN. 2000-11-18)
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 so-called 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. 2000-11-19)   Descartes-Euler
How many edges (lines) are in a cylinder?

We'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.

At first glance, this may look like a  "counterexample"  to the  Descartes-Euler formula, which states that "in a polyhedron" the numbers of faces (F), edges (E) and vertices (V) obey the following relation:

F - E + V   =   2

Our cylinder has 3 faces (top, bottom, lateral), 2 edges (top and bottom circles) and no vertices, so that F-E+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 Descartes-Euler formula has 3 restrictions, namely:

1. 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).
2. 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.
3. 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 the two new vertices.  Now that all 3 conditions are met, what do we have? 3 faces, 3 edges and 2 vertices.  Since 3-3+2 is indeed 2, the Descartes-Euler formula does hold.

The better way to fix the formula does not involve introducing unnecessary edges or vertices.  It involves the so-called Euler characteristic  introduced by Leonhard Euler in 1752 and often denoted c (chi):

### The Euler Characteristic  c  ( chi )

The fundamental properties of c (chi) may be summarized as follows :

1. Any set with a single element has a c of 1 :   "x,  c ( {x} )  =  1
2. c is additive:  For two disjoint sets E and F,  c(EÈF) = c(E) + c(F)
3. If E is homeomorphic to F, then   c(E) = c(F)
("Homeomorphic" is the precise term for topologically "like".)

Using those three properties as axioms, we could show by induction that, if it's defined at all, the c of n-dimensional space can only be equal to (-1)n.  (HINT:  A plane divides space into 3 disjoint parts; itself and 2 others...)

The  c  of shapes dissected into parts of known  c  can then be derived...  For example, a circle has zero  c  because it's formed by gluing to a single point  (c = 1)  both extremities of an  open  line segment  (whose  c  is  -1  because it's homeomorphic to an infinite straight line).

In particular,  the ordinary Descartes-Euler formula is valid 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 our cylinder (whose c is also 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.

• 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 n-dimensions) = (-1)n
• c (circle, or semi-open segment) = 0
• c (surface of a sphere) = 2
• c (surface of an infinite cylinder) = 0
• c (surface of torus) = 0
• etc.

Note (2000-11-19) :   The orthodox definition of the Euler-Poincaré 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...