Math Quiz Thread

[Jokodo]

I recently had to find a general formula for the area of an n-sided regular polygon with a circumcircle of radius r. I refused to google the solution and came up with a working formula myself. That's about as much trigonometry as I can take (and that's all the more reason to work it out myself).

Did have a practical application too, at least I thought so. I needed to adjust the radius of polygons with varying n, created by drawing them inside a circle, such as to give them the same area. At least I thought so. I ended up using only triangles and squares, so I might as well have hardcoded the ratios.

Feel free to laugh at me.

Show

Consider that an n-sided polygon is made up of n triangles, each covering the area between two adjacent corners and the centre.

The angle at the center is \(\frac{2\pi}{n}\)
Each of those triangles is in turn made up of two rectangular triangles, cutting the far side in half.

The angle at the center for those is \(\frac{\pi}{n}\)

The area of each of those rectangular triangles is thus \(r^2\frac{\cos{\frac{\pi}{n}}\sin{\frac{\pi}{n}}}{2}\)

Given that there are 2n of those triangles in the polygon, the area of the polygon is \(r^2n\cos{\frac{\pi}{n}}\sin{\frac{\pi}{n}}\)

The ratio radius_of_a_circle/radius_of_a_larger_circle_such_that_n-sided_polygon_drawn_into_it_has_same_area is thus \(\sqrt{\frac{\pi}{n\cos{\frac{\pi}{n}}\sin{\frac{\pi}{n}}}}\)

 

[fast]

Math quiz question for those who have no clue what half the symbols in this thread represent

(3+2*4)-(3*2-4)-2= ?

 

[lpetrich]

Jokodo,

Show

That is correct.

For circumscribed-circle radius r_c,
Inscribed-circle radius = \(r_c \cos \frac{\pi}{n}\)
Circumference = \(2n r_c \sin \frac{\pi}{n}\)
Area = \(n r_c^2 \cos \frac{\pi}{n} \sin \frac{\pi}{n}\)

For inscribed-circle radius r_i,
Circumscribed-circle radius = \(r_i \sec \frac{\pi}{n}\)
Circumference = \(2n r_i \tan \frac{\pi}{n}\)
Area = \(n r_i^2 \tan \frac{\pi}{n}\)

Archimedes had famously approximated pi using an circle, an inscribed polygon, and a circumscribed one, and comparing their circumferences. One gets
\( n \sin \frac{\pi}{n} < \pi < n \sin \frac{\pi}{n} \)

He started out with some value of n that has simple formulas for these angles, like n = 3 or 4, and he used half-angle trigonometric identities to improve the result, doubling n at each step. I remember once implementing this algorithm in Mathematica.

 

[Loren Pechtel]

Math quiz question for those who have no clue what half the symbols in this thread represent

(3+2*4)-(3*2-4)-2= ?

Show

(11)-(2)-2 = 7

 

[fast]

Math quiz question for those who have no clue what half the symbols in this thread represent

(3+2*4)-(3*2-4)-2= ?

Show

(11)-(2)-2 = 7

Easy as pie.

 

[beero1000]

Jokodo,

Show

That is correct.

For circumscribed-circle radius r_c,
Inscribed-circle radius = \(r_c \cos \frac{\pi}{n}\)
Circumference = \(2n r_c \sin \frac{\pi}{n}\)
Area = \(n r_c^2 \cos \frac{\pi}{n} \sin \frac{\pi}{n}\)

For inscribed-circle radius r_i,
Circumscribed-circle radius = \(r_i \sec \frac{\pi}{n}\)
Circumference = \(2n r_i \tan \frac{\pi}{n}\)
Area = \(n r_i^2 \tan \frac{\pi}{n}\)

Archimedes had famously approximated pi using an circle, an inscribed polygon, and a circumscribed one, and comparing their circumferences. One gets
\( n \sin \frac{\pi}{n} < \pi < n \sin \frac{\pi}{n} \)

He started out with some value of n that has simple formulas for these angles, like n = 3 or 4, and he used half-angle trigonometric identities to improve the result, doubling n at each step. I remember once implementing this algorithm in Mathematica.

Speaking of Archimedes - he famously approximated the square root of 3 as lying between 265/153 and 1351/780 (the upper bound is surprisingly good, on par with an 8th order Taylor approximation).

If you had to compute root 3 (by hand), how would you do it?

 

[lpetrich]

With Mathematica:

N[{265/153,Sqrt[3], 1351/780},16]
{1.732026143790850,1.732050807568877,1.732051282051282}

%-%[[2]]
{-0.000024663778028,0.*10^-16,4.74482405*10^-7}

By hand, a simple way of doing is by interpolation and by searching by changing the last digit. Since 1² = 1 and 2² = 4, sqrt(3) must be between 1 and 2.

1: 1
2: 4

1.7: 2.89
1.8: 3.24

1.73: 2.9929
1.74: 3.0276

1.732: 2.99982
...

Naive but general.

There is, however, a much faster way: Newton's method. To find x = sqrt(a), one does
x(new) = (1/2)*(x + a/x)

When close to a solution, it converges quadratically.

Mathematica again:

sqrtnm[a_, x_, n_] := NestList[(1/2)*(# + a/#) &, x, n]
sqrtnm[N[3, 16], N[2, 16], 5] // TableForm
2.000000000000000
1.750000000000000
1.732142857142857
1.732050810014728
1.732050807568877
1.732050807568877

One can extend this algorithm to any power. For x = a^1/p, we do
x(new) = (1/p)*((p-1)*x + (a/x^p-1))

For a reciprocal square root, we get
x(new) = (1/2)*x*(3 - a*x²)

For a reciprocal, we get
x(new) = x*(2 - a*x)

So one can use Newton's method to do division.

 

[lpetrich]

Archimedes and pi

archpi[nsinit_,niter_] := Module[{ns,tana,seca,sina,initres,res},
ns = nsinit;
tana = Switch[ns,3,Sqrt[3],4,1,5,Sqrt[5-2Sqrt[5]],6,1/Sqrt[3],_,Return[]];
tana = N[tana,16];
seca = Sqrt[1 + tana^2];
sina = tana/seca;
initres = ns*{1,sina,tana};
res = Table[
ns *= 2;
tana = tana/(1+seca);
seca = Sqrt[1+tana^2];
sina = tana/seca;
ns*{1,sina,tana},
{niter}];
Prepend[res,initres]
]

archpi[3,26] // TableForm

(number of sides, pi lower limit, pi upper limit)

3	2.598076211353316	5.196152422706632
6	3.000000000000000	3.464101615137755
12	3.105828541230249	3.215390309173472
24	3.132628613281238	3.159659942097500
48	3.139350203046867	3.146086215131435
96	3.141031950890510	3.142714599645368
192	3.141452472285462	3.141873049979824
384	3.141557607911858	3.141662747056849
768	3.141583892148318	3.141610176604690
1536	3.141590463228050	3.141597034321526
3072	3.141592105999272	3.141593748771352
6144	3.141592516692157	3.141592927385097
12288	3.141592619365384	3.141592722038614
24576	3.141592645033691	3.141592670701998
49152	3.141592651450768	3.141592657867844
98304	3.141592653055037	3.141592654659306
196608	3.141592653456104	3.141592653857171
393216	3.141592653556371	3.141592653656638
786432	3.141592653581438	3.141592653606504
1572864	3.141592653587704	3.141592653593971
3145728	3.141592653589271	3.141592653590838
6291456	3.141592653589663	3.141592653590054
12582912	3.141592653589761	3.141592653589859
25165824	3.141592653589785	3.141592653589810
50331648	3.141592653589791	3.141592653589797
100663296	3.141592653589793	3.141592653589794
201326592	3.141592653589793	3.141592653589793

N[Pi,16]
3.141592653589793

 

[Jokodo]

Jokodo,

Show

That is correct.

For circumscribed-circle radius r_c,
Inscribed-circle radius = \(r_c \cos \frac{\pi}{n}\)
Circumference = \(2n r_c \sin \frac{\pi}{n}\)
Area = \(n r_c^2 \cos \frac{\pi}{n} \sin \frac{\pi}{n}\)

For inscribed-circle radius r_i,
Circumscribed-circle radius = \(r_i \sec \frac{\pi}{n}\)
Circumference = \(2n r_i \tan \frac{\pi}{n}\)
Area = \(n r_i^2 \tan \frac{\pi}{n}\)

Archimedes had famously approximated pi using an circle, an inscribed polygon, and a circumscribed one, and comparing their circumferences. One gets
\( n \sin \frac{\pi}{n} < \pi < n \sin \frac{\pi}{n} \)

He started out with some value of n that has simple formulas for these angles, like n = 3 or 4, and he used half-angle trigonometric identities to improve the result, doubling n at each step. I remember once implementing this algorithm in Mathematica.

I know it's correct. I didn't expect you to laugh at me because of the solution, but because of the fact that this represents about the most trig I can do myself.

 

[lpetrich]

Some more.

Traditional geometry is based on ruler-and-compass constructions. This can be shown to be equivalent to these operations on numbers: elementary-arithmetic operations (addition, subtraction, multiplication, division) and square roots. Proof?

Show that a regular pentagon can be constructed by ruler and compass. You don't need to provide the details of a construction, just to show that it is feasible.

 

[lpetrich]

To clarify the first one a bit, what operations are needed to find the intersections of

• A line and a line
• A line and a circle
• A circle and a circle

?

This is in terms of the coordinates and other parameters for specifying these geometric objects.

 

[beero1000]

To clarify the first one a bit, what operations are needed to find the intersections of

• A line and a line
• A line and a circle
• A circle and a circle

?

This is in terms of the coordinates and other parameters for specifying these geometric objects.

Interestingly enough, these observations are what Descartes is famous for, not coordinate systems (those date back to the ancients). The notion that one could take a geometric problem, describe it algebraically, solve the algebra, and have the algebraic solution correspond to the geometric one is so deeply ingrained that people don't realize how big of a conceptual leap it was.

For the second question, my favorite constructions use  Carlyle circles. They are a really cool mix of the algebraic and the geometric.

 

[lpetrich]

Some posts earlier, I'd discussed getting the axioms of Euclidean geometry out of metric-space differential geometry. One can get most of them locally, even if not necessarily globally. The SAS postulate of neutral geometry gives constant curvature, and Euclid's famous fifth postulate gives zero curvature.

Is it fair to say that that's an extension of Descartes's work?


As to the second part, I'll give a hint. For a regular n-gon, the angle (a vertex) - (center) - (center of a side that contains the vertex) is (π/n), and to find vertex positions, side lengths, and other such quantities, one must find trigonometric functions of (π/n). So for a regular pentagon, one must find sin(π/5), cos(π/5), etc.

So I will rephrase my challenge. Is it possible to find cos(π/5) = cos(36d) by starting with rational numbers and doing a finite number of arithmetic operations and square roots?

 

[lpetrich]

Here's another hint. Find the formula for cos(5*a) in terms of cos(a). Set a = π/5, and you will get an equation for cos(π/5).

More generally, can anyone here show the following:

For n >= 0,
cos(n*arccos(x)) = polynomial with degree n in x

For n > 0,
sin(n*arccos(x)) = sqrt(1-x²) * polynomial with degree (n-1) in x

You can do that without finding those polynomials' general formulas, and all you will need is trig identities and mathematical induction.

 

[beero1000]

Some posts earlier, I'd discussed getting the axioms of Euclidean geometry out of metric-space differential geometry. One can get most of them locally, even if not necessarily globally. The SAS postulate of neutral geometry gives constant curvature, and Euclid's famous fifth postulate gives zero curvature.

Is it fair to say that that's an extension of Descartes's work?

In a sense, yes. The Greeks had all of the pieces they needed to anticipate Descartes (and Newton/Leibniz), they just didn't manage to put them together. In particular, Archimedes knew calculus (essentially), Menaechmus dealt with coordinate systems, Apollonius talked about tangent planes to surfaces, etc. However, without the power of a universal algebraic description of the operations, each situation needed to be individually described in order to work out the computation, and only the nicest geometric objects were tractable, even to the best mathematicians on the planet.

It stayed that way until Descartes (and others, in particular also Fermat) realized that algebra was exactly the necessary tool to describe general geometry (and in a weaker sense, geometry was a useful tool to describe algebraic equations). Mathematicians then immediately began using algebra to describe geometric objects, and determining their properties.

One of the main open questions of the 17th century was a procedure for finding tangents to arbitrary plane curves, extending Apollonius' work on tangents of conics. In particular, Apollonius had shown (geometrically) how to find tangents to conics at arbitrary points (in modern notation he showed that the derivative of x² was 2x). That was question that led to differential calculus. Similarly, Archimedes' results on finding areas and volumes of shapes led to integral calculus.

At that point, differential geometry was a hop-skip-and-jump away. Geometers now had tools powerful enough to work with geometries that had algebraic descriptions without nice geometric ones, they could go to dimensions higher than 3, they could talk about the 'geometry' of functions or equations, etc...

 

[lpetrich]

Now for the operations needed to find the coordinates of points in the intersections of lines and circles.

Specify a line by x = p + n * t where p is a point, n is a direction vector, and t is a parameter, an independent variable. Specify a circle with ((x - c)²) = r² where c is the center point and r is the radius.

Two lines intersecting give a system of linear equations:

p1 + n1 * t1 = p2 + n2 * t2

2 equations in 2 unknowns: t1 and t2, and one gets the solution by basic arithmetic. Let ε = {{0,1},{-1,0}} be the 2D antisymmetric symbol. Then
t1 = - (n2 . ε . (p1 - p2)) / (n2 . ε . n1)
t2 = - (n1 . ε . (p2 - p1)) / (n1 . ε . n2)

One finds the point's coordinates with more arithmetic.

A line and a circle intersecting:

((p - c + n*t)²) = r²
(n.n) * t² + 2((p - c).n) * t + ((p - c)²) = r²

Clearly a quadratic equation in t (degree 2), with solution
t = - ((p - c).n)/(n.n) +- sqrt( r²*(n.n) - ((p - c)²)*(n.n) +((p - c).n)² ) / (n.n)

A circle and a circle intersecting:

((x - c1)²) = r1²
((x - c2)²) = r2²

We can express x = c1 + xa*c12 + xb*(ε.c12) where c12 = c2 - c1 and xa and xb are to be found.

Going in reverse, xa = ((x - c1).c12)/(c12.c12), xb = ((x - c1).ε.c12)/(c12.c12) so every possible x can be specified by its xa and xb values.


The circle equations become
(xa² + xb²)*(c12.c12) = r1²
((1-xa)² + xb²)*(c12.c12) = r2²

giving
xa = (1/2)*( (r1² - r2²)/(c12.c12) + 1)

This in turn yields
xb = +- (1/2) * sqrt(- (r1² - r2²)²/(c12.c12)² + 2*(r1² + r2²)/(c12.c12) - 1)


Lines and circles are degenerate 2D conic sections. However, the intersection of two general 2D conics requires solving a quartic equation (degree 4), giving as many as 4 intersection points.

Some ancient mathematicians departed from ruler-and-compass orthodoxy, like Archimedes. He often used a "neusis", a marked ruler that pivots around some point, with each end being on some curve. One can show that a neusis with two lines yields a quartic in the neusis's position. Solving a general quartic equation requires solving a cubic (degree 3) equation. Thus, if you allow a neusis, you can duplicate the cube and trisect a general angle.

 

[Kharakov]

Here's another hint. Find the formula for cos(5*a) in terms of cos(a).

Show

z is complex.
\(z=\cos(a) + i \sqrt {1- \cos^2{a}}\)

take \(z^n\).... the real part is \(\cos(n \, \times \, a)\).

 

[lpetrich]

Kharakov,

Show

that is indeed the general solution, but I'm asking about the n = 5 solution. Could you please expand zⁿ for n = 5 and take the real part?

 

[lpetrich]

Here are some challenges related to prime numbers.

Consider numbers P(p,n,-) = pⁿ - 1 and P(p,n,+) = pⁿ + 1 where p is a prime and n a positive integer.

If P(p,n,-) is a prime, what constraints can you find for p and n?
Likewise for P(p,n,+). They need not be the same constraints.

 

[Kharakov]

Kharakov,

Show

that is indeed the general solution, but I'm asking about the n = 5 solution. Could you please expand zⁿ for n = 5 and take the real part?

Show

I'm a bit lazy... so looked up the binomial expansion, which I should have memorized. Anyway-
\(x^5 - 10x^3 y^2 + 5 xy^4\)

which is:

\(cos^5(a) -10 \, cos^3(a) [1-cos^2(a)] + 5 cos(a) [1-cos^2(a)]^2 \)

\(cos^5(a) -10 \, cos^3(a) + 10 \, cos^5(a) + 5 cos(a) [1-2cos^2(a)+ cos^4(a)] \)

\(cos^5(a) -10 \, cos^3(a) + 10 \, cos^5(a) + 5 cos(a) -10cos^3(a)+ 5 cos^5(a) \)

\(16 \, cos^5(a) -20 \, cos^3(a) + 5 cos(a) \)