(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) circlesdrawn 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.
Takeanother pair of points on the arc and do the above construction again to drawanother line the same way.
The point where the two lines you've drawn intersect is the center of the circle your originalarc belongs to.(And, of course, the distance between that center and any point on the arc is the radius youwere 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] ?
Let's start with simple planar surfaces:
Acircle of radius R has areaR(=3.14159265...).
The area of atrapezoid 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 arectangle (width multiplied by height) may be seen asa special case of this...
Atriangle 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 asphere of radius R is4R (its volume is4R/3 ).
More generally, we may consider the surface(sometimes called a sphericalfrustum or "frustrum")which consists of the part of the surface of a sphere between two parallel planes thatintersect it.The surface area of such afrustum is2RH, if H is the distance between the two planes.When H=2R, the frustum consists of the entire sphere and the above formula doesgive an area of 4R, as expected.
An ordinaryrightcylinder of height H is the surface generated by a straightsegment of length H perpendicular to a plane containing the trajectoryof 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 acircular cylinderof radius R, whose surface area is 2RH.(Note that a spherical frustum has the same area as the right cylindercircumscribed to it, a remarkable fact first discovered byArchimedes of Syracuse.)
We may also consider the lateral surface area of theconical surfacegenerated by a segment of length R with one fixed extremity,when the other extremity has a trajectory of length L (this spherical trajectory isnot planar unless it happens to be a circle, which is true only for an ordinarycircular cone).The area of such a surface is simply RL/2. In particular, the lateralsurface area of an ordinarycircular cone is Rr,if r is the radius of its base andR is the distance from thecircumference 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 thelateral area of the conical surface asr(H2+r2)Note that the case R=r (or H=0) corresponds to a "flat" circular cone,which is simply a circleof area R...Back to our first formula!
(2015-06-15) The line containing its three projections along the sides of the triangle.
By definition, the pedal triangle of a point P with respectof a triangle ABC is the triangle formed by the orthogonal projections of P along thethree sides of ABC.
That pedal triangle is flat (i.e., its vertices are collinear) if and onlyif P is on the circle circumscribed to ABC. (In that case, theline 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 toRobert 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) whose Simson line is parallel to the line joining the circumcenter O and theLemoine point K (1873) where the three symmedians intersect.
(2015-06-16) 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 symmetricalto AP with respect to the (internal) angular bissectorat 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 linesis 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-points (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.
(K. D. of USA.2000-11-13) 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 thecircumcenter O. The distance HG is twice the distance GO. Recall that:
The centroid G is where the threemedians intersect.
The orthocenter H is where the threealtitudes intersect.
An excenter is where an interior bissector meets two outer ones.
Except for isosceles 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.
(2011-05-08) 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 locatedexactly 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 altitudesand the 3 midpoints from the orthocenter H to the vertices. The 9-point circle has half the radius of thecircumcircle.
Euler's Circle and Feuerbach's Theorem :
Feuerbach's theorem (1822) states that the 9-point circleis tangent externally to the three excircles and internally tothe 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 altitudesall belong to the same circle was discovered independently byCharles Brianchon(1783-1864;X1803),Jean-Victor Poncelet(1788-1867;X1807) andKarl Wilhelm von Feuerbach(1800-1834). The remark that the same circle also goes through the midpoints between the orthocenter and thevertices was first made byOlry Terquem(1782-1862;X1801) who coined the term 9-point circle, which stuck...
The Poncelet point of a quadrilateral ABCD is the point of intersection of the four respective Euler circles of ABC, ABD, ACD and BCD.
(2012-11-05) . The intersecting chords theorem (Jakob Steiner, 1826).
Consider a point P at a distance d from the centerof 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 = .
denotes the linear abscissa whose magnitude is the Euclidean distance between M and N. Its sign depends on the orientation of the line.
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.
(2020-02-09) 'stheorem for cyclic quadrilaterals. The products of the opposing sides add up to the product of the diagonals.
(2011-05-09) Two dual types of homogeneous coordinates in the Euclidean plane.
Trilinear coordinates (trilinears) and barycentric coordinates areexamples of homogeneous coordinates. This is to say that, in either system, two proportional triplets represent thesame 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 ofthe 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 andcare the pairwise distances between the three base points and, therefore, must satisfy thetriangular inequality. However, the above correspondence can be investigated abstractly without that requirement...
Actually, barycentric coordinates describe a generalvector space without resorting to any metric concept, whereasthe mapping from barycentric to trilinear coordinates is a way toendow 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 trilinearsare to covariant coordinates. When the triangular inequality fails fora,b andc, the metric sodefined is Lorentzian, not Euclidean.
(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 thenext question.
Aparametric equation for an ellipse ofcartesian equation x/a+ y/b=1is: x =a sin() and y =b cos().We assume a>b and definee=1-b/a.The above figure shows how may be determined usingan auxiliary circle whose radius is the ellipse's major radius a.
It suffices to calculate the elliptic arc (shown as a in the picture) from theflat apex (at=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 suchquantities (possibly adding a multiple of aquarter of theperimeter).
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 a(1-esin) d. [This is simply an infinitesimal expression of the Pythagorean theorem:At infinitesimal scales, every ordinary curve looks straight, and a small piece ofit appears as the hypotenuse of a tiny right triangle of sides dx and dy.] The length of the elliptic arc corresponding to the angle may thus be expressed as a simple integral (an old-fashionedquadrature ) known as the incomplete elliptic integral of the second kind :
a E(,e) = a
0
1 -esin
d
This function (E) was introduced because the integral has no expression in terms of moreelementary functions. (The function E also comes in a single-argument version known as thecomplete elliptic integral of the second kind, namelyE(e) = E(/2,e), which isa quarter of the perimeter of an ellipse of eccentricityeand unitmajor radius.) To compute the integral whene is not too close to 1 and/or is not too close to /2[in which case other efficient approaches exist, seeelsewhere on this site],we may expand the square root in the integrand as a sum ofinfinitely many terms of the form(-1)C(½,n) esin,for n=0, 1, 2, 3...Each such term may then be integrated individually using the formula:
0
sin2n d =
22n
2n n
+
1
22n
(-1)k
k
2n n-k
sin 2k
When /2, all the sines vanish and only the firstterm remains. This translates into thesimple seriesgiven in the next article. Otherwise, what we are left with is2 times thatcomplete integral plustheFourier series of some odd periodic functionof (whose period is )...
(Jaleigh. B. of Minonk, IL.2000-11-26twice) 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?
There is no simple exact formula: There are simple formulas but they are not exactand there are exact formulas but they are not simple.
If the ellipse is of equationx/a+ y/b=1witha>b,a is called themajor radius,andb is theminor radius.The quantitye = (1-b/a)is theeccentricity of the ellipse.
Anexact expression for the ellipse perimeter P involves the sum of infinitelymany terms of the form (-1)/(2n-1) [(2n)!/(2 n!)]e.The first such term (for n=0) is equal to 1 whereas all the others are negativecorrection terms :
A second 1914 formula, also due to Ramanujan, is expressed in terms of the quantityh = (a-b)/(a+b) :
(3)
P (a+b)[ 1 + 3h / ( 10+
4-3h
) ]
Therelative error of this formula for ellipses of low eccentricitiesisfabulous:
(-3/)e[ 1 + 5e +11107/e +4067/e +3860169/e +... ]
In 1917, Hudson came up with a formula without square roots, which is traditionallyexpressed in terms of the quantityL =h/4 = (a-b)/[2(a+b)] :
(4)
P (a+b)/4 [ 3(1+L) + 1/(1-L) ]
In 2000, Roger Maertens proposed the following so-called" formula":
(5)
P 4 (ay+by)1/y or P 4a (1 + (1-e)y/2 )1/y with y = ln(2)/ln(/2)
The special value of y (the " constant") makes the formulaexact for circles, whereas it is clearly also exact for flat ellipses(b=0 and P = 4a). Therelative error of the formula never exceeds 0.3619%. It's highest for the perimeter of an ellipse whose eccentricity is about 0.979811 (pictured at right) with anaspect ratioa/b slightly above 5.
A popularupper bound formula is due toEuler (1773):
(6)
P
2(a+b)
The following simplelower bound formula is due toJohannes Kepler (1609):
(7)
P
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 isused to compute the circumference of an ellipse with the same eccentricity and the samesize as the Earth Meridian. Note that even the humble #1 formula is accurate to15 m, or aboutone tenthof 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' formula, the modest precisionshown for the "worst case" corresponds to acompletely flat ellipse(of perimeter 4a).
(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 radiusa and minor radiusb,its cartesian equation (with the proper choice of coordinates) is:
x/a+ y/b = 1
The area of such an ellipse is simply S = ab.
(B. K. of Honolulu, HI.2000-10-12) How can I determine the volume of an oval object? [egg-shaped solid]
The volume of anellipsoid of equation x/a+ y/b+ z/c = 1 is
V = ( 4/3)a b c
This is a goodapproximation for other egg-shaped ovals whichare nearly elliptical: 2a is thediameter(i.e. the largest width), 2b is the largest width for a direction perpendicularto the diameter and 2c is the width in the direction perpendicular to bothprevious directions.Each suchwidth is measured between two parallel planes perpendicular tothe 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 is2RH.
The surface area S of anoblate ellipsoid(generated by an ellipse rotating around its minor axis)of equatorial radiusa and eccentricitye is given by:
S = 2a2[ 1 +(1-e2) atanh(e)/e ] , or S = 2a2[ 1 +(b/a)2atanh(e)/e ] [ Seeproof. ]
In this,e is(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 aprolate ellipsoid ("cigar-like") generated byan ellipse rotating around its major axis(so that the equatorial radiusb is smaller than the polar radiusa) is given by:
S = 2b2[ 1 +(a/b) arcsin(e)/e ]
This shows that a very elongated ellipsoid has an area of2ab(e is close to 1 andb is much less than a),which is about 21.46% less than the lateral area of the circumscribed cylinder(4ab),whereas these two areas are equal in the case of a sphere,as noted byArchimedes of Syracuse (c.287-212BC).
Now, it's not nearly as easy to work out the surface area of a general ellipsoid of cartesianequation (x/a)2+(y/b)2+(z/c)2=1.No elementary formula for this one!Thegeneral formulainvolves 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 ofA, B, C, D, E, andF ) ...We'll assume A+B is positive (after changing all signs, if needed).
As the question is only about real ellipses, so is the following discussion:
First, we notice that we may get rid of any existingcross term("xy" with a nonzeroC coefficient) by tiltingthe coordinate axes. If we do so by an angle (see figure),the new coordinates X and Y (note capitalization)are best obtained as the scalar products of the unitvectors of the new tilted axes. These vectors are(cos ,sin )and(-sin ,cos ):
X = x cos + y sin Y = -x sin + y cos
conversely (change to - )
x = X cos - Y sin y = Y sin + Y cos
The above expressions of x and y in terms of X and Y give us the curve's equation in thetilted frame. Equating to zero the coefficient of XY gives:
(B-A) sin + C cos = 0
We could thus obtain within an integral multipleof as half an arctangent, but let's notrush things! What wereally want is the inclination of themajor axis,which is determined within an integral multiple of ...When the above relation is satisfied, the rest of the equation reads:
[A cos2+B sin2+C cos sin ]X2 +[A sin2+B cos2-C cos sin ]Y2 +[D cos +E sin ]X +[-D sin +E cos ]Y +F = 0
Using the previous relation, we may reduce the above coefficients of X2and Y2. We find the former equal to(1/2)[A(1+1/cos 2) +B(1-1/cos 2)],whereas the latter equals(1/2)[A(1-1/cos 2) +B(1+1/cos 2)].We are only interested in the elliptic case, so these two are of the same sign,which is also the sign of their sumA+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 bychanging the sign of all coefficients).Now, if we want the X-axis to be the major one, the coefficient of X2isinversely proportional to the square of themajor radiusand is thussmaller than the coefficient of Y2(which is isinversely proportional to the square of theminor radius).In other words,(A-B)/cos 2 isnegative.With this in mind, we can fully specify the inclination of the major axis(within a multiple of , of course) by giving thesine and cosine of the angle (we're assumingA+B > 0):
cos = (B-A)/Q and sin = -C/Q , where Q =
(A-B) +C
This determination of the inclination isn't valid when Q=0. (Q=0 impliesA=B andC=0, which corresponds to the trivialcase 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 expressionsbelow.
The coordinates of the ellipse center are fairly easy to compute directly in the originalframe of reference: We are simply looking for (x,y)such that the transforms x=x+u and y=y+vyield an equation where the coefficients of u and v are zero (so that the origin will bea center of symmetry)... This translates into the two simultaneous equations:
0 = 2A x +C y +D 0 =C x + 2B y +E
Therefore: x = (CE-2BD)/(4AB-C 2 ) and y = (CD-2AE)/(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 centeredat the above point (x,y). 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 (x,y),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 standardizedequationX2/a2 + Y2/b2 = 1in 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 Raneof Mumbai, India (2010-09-28; e-mail) Parabolic arc of given extremities and prescribed apex between them.
By definition, a parabola isthe 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).
If the apex O of a parabola is between two of its points A and B, we want a construction of the focalpoint 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 istwicethe distance from the focalpoint 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] mirrorhas [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 be the distance from the chord to the pointof incidence (x,y) on the parabolaand let d be the distance from (x, y) to the focus. Show that the sum of the distanced+dis 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 glassis 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 distancesfrom a plane perpendicular to the rays (that's what the "chord" in the questionis 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 onthe 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 translatesinto 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) isx2+(y-f)2 = 4fy+(y-f)2 = (y+f)2.In other words, the distance d is (y+f).On the other hand, d is equal to A-y(where A is some constant which depends on how far you drew the "chord"described in the question). Therefore, d+d =f+A = constant.
This, by the way, is one way to actuallyprove that a parabolic mirror is anoptical 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 ofrevolution is equal to the product of the length of the meridian by the length of the circulartrajectory of the meridian's centroid. (Thevolume of a solid of revolution is also obtainedas the area of the meridian surfaceby the length of the circular trajectory of the centroid of thatsurface.)
Make the segment rotate around the diameter of the circle which is parallel to the segment'schord and apply the theorem: Your meridian is the circular segment of radius R, length L andchord H=2Rsin(L/2R). The surface is a spherical segment of area 2RH. If D is the distance of the centroid to the center of the circle,its trajectory has a length 2D andGuldin's theorem tells us that: 2RH=2DL. 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 is6/3 (approximately 0.8165).
Now, in a regular hexagon of side A, we mayinscribe a square of side(3-3)Aor about 1.268A (one of the sides of the square is parallel toone of the sides of the hexagon).When A is 6/3, this means that a square of side6 -2will fit.
Well, 6-2is about 1.03527618... so we may cut in a cube a square hole with aside 3.5% larger than the side of the cube.A cube "just slightly larger" will easily go through such a hole.
(Adrian Brancato. 2001-02-14; e-mail) Thanks for the wonderfully informative website, Dr. Michon [...]I am still unable to solve the problem for which I originallywent to your site. Can you provide a formula for me: For an octagon, given the "diameter" (i.e. the distance between twoopposite vertices) I need to determine the length of each side. This is not an academic endeavor; we have to build a large display ofstrawberries (the diameter is 6' at the base, with ever decreasing diameters asthe conical structure rises).
Thanks for the kind words, Adrian... In a regular octagon of sidea, the diameterd is thehypotenuse of a right triangle whose sides area anda+2b(see figure), whereb is the side of a square of diagonala, so that we have2b=a 2 andd =a [1 +(1+2)]ord =a [4 +2 2 ]. Take the square root of that, andyou have the desired relation between the sidea and the diameterd,which boils down numerically tod/a = 2.61312592975...or, if you prefer,a/d = 0.382683432365...,which is half the square root of (2-2).
The same result can be obtained with standard trigonometric functions: The ratio a/d is the sine of a /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 m or about 1/700 of the width of a humanhair) which is roughly2' 3/". Hope the display will look good!
(2001-02-14) An ancient problem solved by Carl Friedrich Gauss in 1796 (at age 19).
In theprevious article, we could have noticed that 8 times the side of the octagon is, of course, itsperimeter. For ann-sided polygon, the ratio P/d of the perimeter P to the diameter d is n sin(/n) , which tends to 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 apower 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 (A019434).
=/[ 1 - (1-a)].In other words, ifx is the square of the side of the n-gon of unit diameter, thesquarey of the side of the 2n-gon of unit diameter is given byy = /[ 1 - (1-x) ] (there's just one caveat--which is not a problem with hand computation-- and that's about the difference ofnearly equal quantities in the square bracket, which may cause a cripplingloss 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 tothe diameter in a 96-sided polygon (namely 3.141 031 95... which isabout 178.5 ppm below the value of ).Using a complementary estimate of the circumscribed polygon, Archimedes could thenproduce the first rigorous bracketing of what we now call "".Until better methods where found at the dawn of calculus,this was essentially the basic method usedto compute more and more decimals of ...The last person in history who used Archimedes' method to compute with record precision was the DutchmanLudolph van Ceulen (1539-1610): A professor of mathematics at theUniversity of Leyden, he published 20 decimals in 1596 and32 decimals in a posthumous 1615 paper. It is said that, at the endof his life, he worked out 3 more decimals which were engraved onhis tombstone in the SPeter Church at Leyden. To this day, is still sometimes calledLudolph's Number or theLudolphine 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 ofbase 1 and height½ / tan(/n). Its area is therefore equal to:
¼ n / tan(/n)
This happens to be equal to3/4 for a triangle,1 for a square,(25+105)/4for a regular pentagon,33/2 for an hexagon,2+22 for an octagon,etc.
The surface area of the regular heptagon of unit sidecannot be expressed using just square roots, sorry! In general, you can express the area of an n-gon with just square rootsonly when the n-gon isconstructible with straightedge and compass. A beautiful result of Gauss (1796) says that an n-gon is soconstructibleif and only if n is equal to a power of two (1, 2, 4, 8, 16, ...)possibly multiplied by a product ofdistinct so-calledFermat primes. Only 5 such primes are known (3, 5, 17, 257 and 65537) andthere aremost probably no unknown ones... Ruling out n=1 and n=2, the only acceptable values of n are therefore3, 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 ofCarl 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 aregular pentagon and anequilateral 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 perimeterP = nD sin(/n) and an areaS = n D2/4 sin(/n)cos(/n), which boils down toS=P2/[4 n tan(/n)].If you know that tan(x)/x is an increasing function of x when x is between 0 and/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 of25/[3 tan(/3)] or about 4.811257,whereas the regular pentagon of the same perimeter has a surface25/[3 tan(/3)], which is about 6.8819096.
The limiting case is, of course, the circle:As n tends to infinity S tends to P2/4(or R2with P=2R, if you prefer).A circle with a perimeter of 10 has an area of 25/,which is about 7.957747.
johnrp (John P. of Middletown, NJ.2000-10-08) A circle is a two dimensional object that hasconstant width;its height remains constant regardless of the orientation.Can you construct another two-dimensional figure that hasconstant width,but is not a circle?
There are plenty of examples. The simplest is the so-calledReuleaux 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 verticeswith an arc of a circle centered on the third vertex. A theorem with a surprising proof is due to Joseph Emile Barbier (1839-1889): Barbier's theorem states that the perimeter of any curve of constant widthis times its width.
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 isthe 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 decimalizationin 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 ofconstant 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-downversion of it.)The rounded shapes are also constructed with arcs of circles centered on the verticesof the original polygon.The radius of each such arc is either R or R+D, where D is the (constant) diameter ofthe original shape with sharp corners...
Why notcoin a word for these shapes,à laMartin Gardner? Any curve of constant width might be called a roller and those based onpolygonscould be dubbed polygroller : triangroller,pentagroller,heptagroller, etc.
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 arcsfrom 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 suchirregular curves of constant width... Call themirregrollers!
Such a shape can be built around a scalene triangle ABC as follows, assuming WLG that BC < AB < AC.
Draw the arc of the circle (of radius AC) centered on C going from A tothe intersection B' with the line BC.
Draw the arc of the circle (of radius AC-BC) centered on B going from B' tothe intersection B" with the line AB.
Draw the arc of the circle (of radius AB+AC-BC) centered on A from B" tothe intersection C" with the line AC.
Draw the arc of the circle (of radius AB-BC) centered on C from C" tothe 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+AC-BC).It has at least one sharp corner(2 in case of an isosceles 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 pointsat a distance R from the (inside of) the previous shape...
ciderspider (Mark Barnes, UK.2000-10-09) What are some 3d shapes of constant width, besides spheres?
Surprisingly enough, an obvious three-dimensional generalization ofthe Reuleaux triangle doesn't work: Consider theReuleaux tetrahedron pictured at right (image courtesy ofFastGeometry). This 3D solid is obtained as the intersection of the four balls of radius Rcentered 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 heightR over the surface of the table, provided the point of contact [with the table]iseither one of the 4 verticesor is somewherein the midst of one of the spherical faces. So far so good. However, the point of contact could also be on one of theedges, in which casethe highest point doesmove on the opposite edge if werotate 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 whoseaxis of symmetry goes throughthe two extremities of the edge of contact. As [part of] this arc rotates around a different axis, the height of its highest pointvaries, which shows thatthis solid does not have constant width:
Its width varies between R and ½ R = 1.024944... R.
In 1911, Ernst Meissner and Friedrich Schilling turned the above idea into a solution by rounding three of the six edgesof the above solid. The resulting solids of constant width are calledMeissner tetrahedra (there aretwo types, as the unrounded edges may either form a triangle or meet at a vertex). Meissner tetrahedra are conjectured to be the solids of prescribed constant width with the least volume.
The original Meissner tetrahedra don't possess tetrahedral symmetry, but there's a unique way to preserve tetrahedral symmetry by rounding all six edges of a Reuleaux tetrahedron using the envelope of spheres tangentto the two curved faces and to the tetrahedron's corresponding straight edge.
A simpler way to generate a 3D solid of constant width is torotate any 2D shape of constant width around an axis of symmetry, if it has one (theReuleaux triangle has 3). This works because, any rotation of such a solid is a combination of three independentrotations which allpreserve the width between two given parallel planes,namely: a rotation around the solid's axis of symmetry (obviously), a rotation around anaxis perpendicular to the two planes (think about it) and, finally,a rotation around an axis parallel to the planes and perpendicular tothe axis of symmetry (which is seen "sideways" as a 2D rotation of a cross-section ofconstant width). [Notice that the first two rotations may coincide, but only whenthere are 2 independent rotations of the last type, so we always have 3 independentwidth-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 pointswithin a distance D of some given solid of constant widthisalso a solid of constant width...
ciderspider (Mark Barnes, UK.2000-10-09) A 4D hypersphere 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 toobtain 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 perpendicularand 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) tovisualize a space with more than 3 dimensions,but there's no great difficulty in considering the set of all quadrupletsof 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 simply2 R4 /2 .
More generally, in n dimensions, a sphere of radius R has a volume equal to:
V = Rnn/2 / (1+n/2)
Using the definition of theGamma function ()in terms of factorials (the notation beingk! = k),the coefficient of Rn in the above is the hypervolumeof an n-dimensional ball of unit radius, given by the table below, whose first two lines correspond to the following "volumes":
1 for n = 0 ("0-volume" of a point, so defined for consistency).
2 for n = 1 (length of a segment of "radius" 1).
for n = 2 (area of a disk of unit radius).
4/ 3 for n = 3 (volume of a sphere of unit radius).
Hypervolume of an n-dimensional ball of unit radius :
Even dimension
Odd dimension
n = 2k
k/ k!
n = 2k+1
2nk k!/ n! = 2k+1k/ n!!
0
1
1
2
2
3
4/ 3
4
2/ 2
5
82/ 15
6
3/ 6
7
163/ 105
8
4/ 24
9
324/ 945
10
5/ 120
11
645/ 10395
12
6/ 720
13
1286/ 135135
In the above, we used the (standard) double-factorial notation n!! as a shorthand forn(n-2)(n-4)...which is the product of all positive integers up to n which havethe same parity as n (A006882):
To retain the relation n!! = (n-2)!! n for all positiveintegers, including n = 1, the convention is made that (-1)!! = 1. For completeness, a less useful extension isavailable for all odd negative integers:
Using the double-factorial notation, it's possible to give a cute formulavalid in n dimensions, whether n is even (n=2k) or odd (n=2k+1), namely:
V = (/2)k(2R)n / n!!
Hyperarea of [the boundary of] a hypersphere :
The above hypervolume could be obtained by integrating the hyperarea of ashell from 0 to R. Conversely, if aRn is the hypervolume of ann-dimensional ball of radius R, then naRn-1must 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=1,2 for n=2,4 for n=3,22 for n=4,82/3 for n=5,3 for n=6...We may replace n/n!! by1/(n-2)!!in the following formula [retaining the case n=1 with the convention(-1)!! =1]:
S = 2 Rn-1n/2 / (n/2) = 2n-kkRn-1n/n!! [where n is 2k or 2k+1]
Of particular interest is the so-calledEinstein-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 theEinstein-Eddington universeis 22 R3.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 literallyout of this world and it may be better to consider themaximal possible distance D between two points in the Universe.As D is simply R,the volume of the Universe is 2D3/.
(5:37) by Zach (MajorPrep, 2019-05-24).
(Jerry of Nashville, TN.2000-11-18) What [polyhedron] has six faces?
A polyhedron with 6 faces is called ahexahedron.The is an hexahedron, but that's certainly not the only one:
The so-called is another possibility(with 5 vertices and 9 edges, this solid may be obtained by "adding" one vertex to atetrahedron to make it look like two tetrahedra "glued" on a common face).
A third hexahedron is the (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 facesare regular polygons.
Theleast symmetrical of all hexahedra is thetetragonal antiwedge (it has only one possible symmetry,a 180° rotation). This skewed hexahedron hasthe 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 quadrilateralsthat share an edge but donot form a triangular prism. First, join with an edge thetwo pairs of vertices closest to the edge shared by the quadrilaterals. To complete thepolyhedron, you must join two opposite vertices of the nonplanar quadrilateralthat you're left with.This can be done in one of two ways (only one of which gives a convex polyhedron).Loosely speaking, there aretwo types oftetragonal antiwedges which are mirror images of each other;each is called anenantiomer, orenantiomorph of the other.Thetetragonal antiwedge is thus the simplest example of achiralpolyhedron (in particular, any other hexahedron can be distorted into a shapewhich is its own mirror image). Because of this unique property among hexahedra,thetetragonal antiwedgemay also be referred to as thechiral hexahedron.
The other types of hexahedra are more symmetrical and simpler to visualize.One of them may be constructed by cutting offone 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 anelongated square pyramid(the technical name for anobelisk)along a bisecting plane through the apex of the pyramid and thediagonal of the base prism, as pictured at left.For lack of a better term, we may therefore call this hexahedron an.
Also with 7 vertices and 11 edges, there's a solid which we may callahemicube (orsquare hemiprism), obtainedby cutting a cube in half using a plane going through two opposite corners andthe midpoints of two edges. Its 6 faces include 2 triangles and4 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.It has as many vertices, edges and facesas a cube, but its faces consist of 2 triangles,2 quadrilaterals and 2 pentagons.
The above 7 types (8 if you counts both chiralities oftetragonal antiwedges)include all possible hexahedra. By contrast, there's only onetetrahedron. There are two types of pentahedra(exemplified by the squarepyramid 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 ourdetailed table of the enumeration, elsewhereon this site).For anunabridged discussion of hexahedraand more general information about polyhedra,see our dedicatedPolyhedra Page...
(J. T. of Summerville, SC.2000-11-19) 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 formulahas 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 cylinderdoes 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 facedoes 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 verticesartificially tomeet 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 betweenthe 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 introducingunnecessary edges or vertices. It involves the so-calledEuler characteristic introduced byLeonhard Euler in 1752and often denoted (chi):
The Euler Characteristic (chi )
The fundamental properties of (chi)may be summarized as follows:
Any set with a single element has a of 1 : x, ( {x} ) = 1
is additive: For twodisjoint sets E and F, (EF)= (E) + (F)
If E ishomeomorphic to F, then (E) = (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 of n-dimensionalspace can only be equal to (-1). (: A plane divides space into 3 disjoint parts; itself and 2 others...)
The of shapes dissected into partsof known can then be derived... For example, a circle has zero because it's formed by gluing to a single point ( = 1) both extremitiesof an open line segment (whose is -1 because it's homeomorphic to an infinite straight line).
In particular, the ordinary Descartes-Euler formula is valid because the of a sphere's surface is 2and it's "made from" disjoint faces, edges and vertices, each respectively with a of 1, -1 and 1.
In the "natural" breakdown of our cylinder (whose is also 2), you have no vertices, two ordinary faces (whose is 1) and one face whose is 0(the lateral face), whereas the of both edges is 0. The total count does match.
(point) = 1
(entire straight line, oropen segment) = -1
(plane or open disc) = 1
(space or open ball) = -1
(space with n-dimensions) =(-1)
(circle, orsemi-open segment) = 0
(surface of a sphere) = 2
(surface of an infinite cylinder) = 0
(surface of torus) = 0
etc.
Noteto 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 canbe arranged to be either the whole number line or another set homeomorphic to A. So, if (A) was defined to be x, we wouldsimultaneously have x = x+x and1 = x+x. Thus, x cannot possibly be any ordinary number,and the latter equation says x is nothing like asigned infinity either [as1].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 wouldhave complex values...
(2021) of the Euclidean plane. Their relationship with classical Euclidean axioms.
At the deepest level, all we find are symmetries and responses tosymmetries. StevenWeinberg [1986 Dirac Memorial Lecture]
The plane has at least one axis of symmetry. It's also symmetrical by translation and by any rotation.
Geometry is hard (10:28) by Steve Chow (blackpenredpen, 2020-06-03).
(
If the ellipse is of equationx/a+ y/b=1witha>b,a is called themajor radius,andb is theminor radius.The quantity
with an
(Sherry of Murray, KY.
A x2 +B y2 +C xy +D x +E y +F = 0,
In other words,
Make the axis of the hole a line that goes through two opposite corners of the cube.
cem72 (
(Jerry of Nashville, TN.
Theleast symmetrical of all hexahedra is thetetragonal antiwedge (it has only one possible symmetry,a 180° rotation). This skewed hexahedron hasthe 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 quadrilateralsthat share an edge but donot form a triangular prism. First, join with an edge thetwo pairs of vertices closest to the edge shared by the quadrilaterals. To complete thepolyhedron, 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 aretwo types oftetragonal antiwedges which are mirror images of each other;each is called anenantiomer, orenantiomorph of the other.Thetetragonal antiwedge is thus the simplest example of achiralpolyhedron (in particular, any other hexahedron can be distorted into a shapewhich is its own mirror image). Because of this unique property among hexahedra,thetetragonal antiwedgemay also be referred to as thechiral hexahedron.
The other types of hexahedra are more symmetrical and simpler to visualize.One of them may be constructed by cutting offone 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 anelongated square pyramid(the technical name for anobelisk)along a bisecting plane through the apex of the pyramid and thediagonal of the base prism, as pictured at left.For lack of a better term, we may therefore call this hexahedron an.
Also with 7 vertices and 11 edges, there's a solid which we may callahemicube (orsquare hemiprism), obtainedby cutting a cube in half using a plane going through two opposite corners andthe midpoints of two edges. Its 6 faces include 2 triangles and4 quadrilaterals.
(J. T. of Summerville, SC.