Math Quiz Thread

[ryan]

Is this thread here to make me feel stupid?

No, it's free lessons.

 

[beero1000]

Some classical geometry:

1. Show how to generate an infinite number of primitive Pythagorean triples (i.e. relatively prime triples of integers such that \(a^2 + b^2 = c^2\))
2. Given a parabola drawn arbitrarily in the plane, a compass, and a straightedge, show how to double the cube (i.e. take a segment of length 1 and construct a segment of length \(\sqrt[3]{2}\)).
3. Show that the existence of a rectangle (a quadrangle with four right angles) is equivalent to Euclid's fifth postulate.

 

[Bomb#20]

Some classical geometry:

1. Show how to generate an infinite number of primitive Pythagorean triples (i.e. relatively prime triples of integers such that \(a^2 + b^2 = c^2\))

For any natural number k, let a = 2k+1, b = (a² - 1)/2, c = b + 1.

The hard question is to show how to generate all of them.

 

[lpetrich]

Some classical geometry:
Show that the existence of a rectangle (a quadrangle with four right angles) is equivalent to Euclid's fifth postulate.

Counterexample: a surface of a sphere with a slice in it. The surface on each side of the slice is continuous with the slice's surface. The slice can have rectangles on it, but the overall surface violates Euclid's fifth postulate. Should one try to prove something like (rectangles possible everywhere) == (Euclid's fifth postulate)?

Some classical geometry:

1. Show how to generate an infinite number of primitive Pythagorean triples (i.e. relatively prime triples of integers such that \(a^2 + b^2 = c^2\))

For any natural number k, let a = 2k+1, b = (a² - 1)/2, c = b + 1.

Or

• a = 2k+1
• b = 2k² + 2k
• c = 2k² + 2k + 1

It is easy to show that a, b, and c are relatively prime.

 

[beero1000]

Counterexample: a surface of a sphere with a slice in it. The surface on each side of the slice is continuous with the slice's surface. The slice can have rectangles on it, but the overall surface violates Euclid's fifth postulate. Should one try to prove something like (rectangles possible everywhere) == (Euclid's fifth postulate)?

To clarify: the implication is "equivalent in Neutral geometry", i.e. including all the Euclidean postulates except the fifth. It turns out that Elliptic geometry is inconsistent with Neutral geometry, which is surprising to most people as they think of the three geometries as built on the same base. The axiom "There exists a rectangle" is called Clairaut's axiom.

 

[lpetrich]

I checked on the postulates of neutral geometry, in this page, for instance, and it contains the SAS Postulate. It states that if two triangles have the same values of side-angle-side, then the rest of their values are the same, meaning that the triangles are congruent.

On a curved surface, to a first approximation, the sum of the angles in a triangle is 180d + (curvature)*(triangle area) where the curvature is averaged over the triangle region. That's exact for spherical triangles and maybe also for pseudospherical ones. So for the SAS Postulate to hold, the surface must have constant curvature.

I'll prove Euclid's fifth postulate -> all triangles have angle sum = 180d -> rectangles from right triangles

Euclid's fifth. From two arbitrary points P1 and P2, make a segment S. Make a line L1 go through P1 without including S, and likewise for line L2 for P2. If the angles inside L1 and L2 on one side of S add up to less than 180d, then L1 and L2 will meet on that side and only on that side.

This yields Playfair's postulate, that if L1 does not intersect L2, then the angles inside L1 and L2 on each side of S add up to 180d, and there is thus only one possible line L2 for P1, P2, and L1.

It's fairly easy to get the sum of a triangle's angles from that. Take triangle ABC. Extend AB to infinity making L1. Make a line L2 through C that does not intersect L1. Then exterior angle AC-L2 has the size of angle BAC, and the exterior angle BC-L2 has the size of angle ABC. Those two angles' sizes add to the size of ACB to give 180d.

Now construct a rectangle from a right triangle ABC with the right angle at A. Add a point D, making second triangle CDB. Set the size of CBD equal to the size of ACB. Its size adds to the size of ABC to make 90d. Make BD as long as AC. By SAS, CDB is thus congruent to ABC, meaning that the sizes of ACB and BCD add to 90d and CDB is a right angle. Thus, four right angles and opposite sides equally long -- a rectangle.

Going the other way is more difficult.

 

[lpetrich]

With the SAS axiom, it's possible, however.

Take that single rectangle, with vertices ABCD around it. It can be split into two right triangles, ABC and CDA, which are congruent to within reflection. Construct a line like DACB with angles DAC and ACB the same. By the SAS postulate, the result will be a duplicate of the original rectangle. One can use such constructions to make a larger rectangle by tiling.

Likewise, one can split a rectangle by taking a zigzag path across it, where the zigzag vertices are evenly spaced in the zigzag direction. By the SAS axiom, one gets congruent rectangles that fill the original rectangle. So one can construct a rectangular tiling that fills the original rectangle.

By doing this tiling, one can construct rectangles with arbitrary dimensions from the original rectangle. From such rectangles, one can construct right triangles with arbitrary dimensions, and from them, general triangles with arbitrary dimensions. All of them will have angle sums equaling 180d or 2 right angles.

I'm not sure how to get from there to Euclid's 5th postulate.

 

[beero1000]

With the SAS axiom, it's possible, however.

Take that single rectangle, with vertices ABCD around it. It can be split into two right triangles, ABC and CDA, which are congruent to within reflection. Construct a line like DACB with angles DAC and ACB the same. By the SAS postulate, the result will be a duplicate of the original rectangle. One can use such constructions to make a larger rectangle by tiling.

Likewise, one can split a rectangle by taking a zigzag path across it, where the zigzag vertices are evenly spaced in the zigzag direction. By the SAS axiom, one gets congruent rectangles that fill the original rectangle. So one can construct a rectangular tiling that fills the original rectangle.

By doing this tiling, one can construct rectangles with arbitrary dimensions from the original rectangle. From such rectangles, one can construct right triangles with arbitrary dimensions, and from them, general triangles with arbitrary dimensions. All of them will have angle sums equaling 180d or 2 right angles.

I'm not sure how to get from there to Euclid's 5th postulate.

This, combined with your constant curvature observation is enough. The key idea is the notion of the defect of a triangle, which is how much the sum of the angles of the triangle deviate from 180 degrees. You can prove that if a triangle has zero defect, then all triangles have zero defect, if a triangle has positive defect, then all triangles have positive defect, and if a triangle has negative defect, then all triangles have negative defect. From that, knowing that two triangles have angle sum of 360 means that their defects sum to zero...

 

[lpetrich]

]Given a parabola drawn arbitrarily in the plane, a compass, and a straightedge, show how to double the cube (i.e. take a segment of length 1 and construct a segment of length \(\sqrt[3]{2}\)).

Straightedge / ruler = lines
Compass = circles
Circle + parabola = quartic equation (power 4), which is more than enough for the duplication of the cube.

I'll base the coordinates on the parabola's size scale, position, and orientation. Coordinates (x,y) where the parabola is y = x². Place a circle with radius sqrt(5)/2 at x = 1 and y = 1/2. It will intersect at a point with x satisfying x*(x³-2) = 0. Two points, one at x = 0, and the other at x = 2^1/3. Thus duplicating the cube.

I'll now take on the problem of trisecting a general angle. Like duplicating the cube, it also requires solving a cubic equation: y = 4x³ - 3x for x, where x = cos(a) and y = cos(3a).

More generally, for a circle at (x0,y0) with radius sqrt(x0² + y0²), we get x*(x³ + (1-2y0)*x - 2x0). We get a trisection solution for x0 = cos(3a) and y0 = 2: x = 2*cos(a).

Positioning the circle's center only needs ruler-and-compass constructions relative to the parabola, and its radius is easy: make the circle go through the parabola's center point.

 

[Bomb#20]

Positioning the circle's center only needs ruler-and-compass constructions relative to the parabola, and its radius is easy: make the circle go through the parabola's center point.

What's the construction to find a parabola's center point?

 

[beero1000]

Positioning the circle's center only needs ruler-and-compass constructions relative to the parabola, and its radius is easy: make the circle go through the parabola's center point.

What's the construction to find a parabola's center point?

Perfect lead-in to followup questions:

1. Show how to construct, with compass and straightedge, the appropriate foci and directrices of a conic drawn arbitrarily in the plane.
2. Show how to construct, with compass and straightedge, any number of distinct points lying on a conic, given foci and directrices drawn arbitrarily in the plane.

If you're tired of compass and straightedge geometry:

1. If 3 mutually orthogonal n-dimensional vectors each have entries +1 or -1, show that n is divisible by 4.
2. Find a (simple) formula for \(\frac{d^n}{dx^n} [\cos(x)^3]\).
3. Are there any Fibonacci numbers that are multiples of 314159?

 

[lpetrich]

Show how to construct, with compass and straightedge, the appropriate foci and directrices of a conic drawn arbitrarily in the plane.

The foci and directrices one can construct from the center of the conic section, its major-axis length a and direction, and its eccentricity e. Each focus is distance +- a*e from the center along the major-axis direction, and each directrix is perpendicular to the major-axis direction, intersecting at +- a/e along that direction. That is true for the ellipse and the hyperbola, but there are several degenerate conic sections: a circle (e = 0), a parabola (e = 1, a*(1-e^2) finite), a line segment (e = 1-, a finite), an interrupted line segment (e = 1+, a finite), and a line (e infinite, a*e finite).

Off to algebra to get a solution. A conic section is given by a quadratic equation in the two rectangular coordinates, and it thus has 5 parameters. Overall location: 2, size: 1, shape (eccentricity): 1, orientation: 1.
\(a_{20} x^2 + 2a_{11} xy + a_{02} y^2 + 2a_{10} x + 2a_{01} y + a_{00} = 0\)

Or more generally:
\(X \cdot A_2 \cdot X + 2A_1 \cdot X + A_0 = 0\)

A line can have 2 intersections with a general conic section, and a circle can have 2, 3, or 4 intersections. It's easiest to work with lines' intersection, though even those have some square roots in them. However, the point halfway in between two intersection points is much more convenient. It does not have square roots. From two parallel lines, the directions between the intersections yield a constraint on a02, a11, and a20. Repeating with a pair of parallel lines in a different direction yields another constraint, and thus a complete fix on the quadratic terms to within an overall factor. For convenience of solution, we can impose a constraint like a02²+2a11²+a20² = 1 and get all three values. These values with the center points' locations give us a01 and a10, and an intersection with the curve gives us a00.

We now have everything we need to find the size, shape, orientation, and location of this curve. The coefficients a02, a11, and a20 form matrix A, and we get the shape and orientation with A's eigenvalues and eigenvectors. Fortunately, it only requires solving a quadratic equation. Likewise, finding the center only requires solving some linear equations. So it's feasible by ruler and compass.

Show how to construct, with compass and straightedge, any number of distinct points lying on a conic, given foci and directrices drawn arbitrarily in the plane.

With a directrix on the x axis and a focus at (0,f), the equation for the curve is
\(\sqrt{x^2 + (y-f)^2} = e y\)
or
\(x^2 + (1-e^2)y^2 - 2fy + f^2 = 0\)
or
\(x^2 + (1-e^2)\left(y - \frac{f}{1-e^2}\right)^2 - \frac{(ef)^2}{1-e^2} = 0\)

giving a quadratic equation of x and y in terms of the other one. Ruler and compass again.

If 3 mutually orthogonal n-dimensional vectors each have entries +1 or -1, show that n is divisible by 4.

First, multiply all three by the first one. The first one will have all 1's, and one concludes that the other two have half 1's and half -1's. This, n must be a multiple of 2. Let's see what happens when we multiply the 2nd and 3rd ones.

Both +: npp, both -: nmm, 2nd one + 3rd one -: npm, 2nd one - 3rd one +: nmp
n = npp + npm + nmp + nmm

From orthogonality with the first one,
npp + npm = nmp + nmm
npp + nmp = npm + nmm
Form orthogonality with each other:
npp + nmm = npm + nmp

Or, using a convenient shorthand:

+ + + + : n
+ + - - : 1st, 2nd orthogonal
+ - + - : 1st, 3rd orthogonal
+ - - + : 2nd, 3rd orthogonal
Add the four together: 4*npp = n
Add two and subtract two: 4*npm = 4*nmp = 4*nmm = n
Thus, n is a multiple of 4.

It seems that with v orthogonal +1/-1 vectors, n must be a multiple of 2^v-1.

Find a (simple) formula for \(\frac{d^n}{dx^n} [\cos(x)^3]\).

One can use a trigonometric identity:
cos(3x) = 4*cos(x)³ - 3*cos(x)
or
cos(x)³ = (cos(3x) + 3*cos(x))/4
One gets
\(\frac{d^n}{dx^n} [\cos(x)^3] = \frac{3^n (\cos (3x))^{(n)} + 3 (\cos (x))^{(n)}}{4}\)
where the superscript is the nth derivative.

Are there any Fibonacci numbers that are multiples of 314159?

A way of addressing that is to consider the question of which numbers can be factors of some Fibonacci number. If any number can be a factor of some Fibonacci number, then that proves this result. If not, then one might be able to check if 314159 is one of those numbers.

 

[lpetrich]

Back to Pythagorean triples: a² + b² = c², where a, b, and c are integers
If there is no k such that {a,b,c} = k*{a',b',c'} where a', b', and c' are another Pythagorean triple, then the triple is primitive. All the elements of one are relatively prime with respect to each of the other ones, because if two of them aren't, then the third one must share their common factor.

One of a and b must be even, and the other two are odd, meaning that c is always odd.

Let a be the even one. Then a² = c² - b². Set c = (s + t) and b = (s - t). Since b and c are both odd, s and t must therefore be integers. Also, s and t must be relatively prime, or else b and c would share their common factor. We find
a² = 4*s*t

Since s and t are relatively prime, s and t must both be squares: s = n² and t = m². Thus,
a = 2*m*n
b = n² - m²
c = n² + m²
where m and n are relatively prime. One of them must be odd and the other one must be even.

-

Returning to that ruler-and-compass finding of parameters of a conic section, I've made a discovery. With a line, I find a halfway point between its two intersections. With a parallel line I find another halfway point. I draw a line through the two points. I then repeat this operation with two lines with a different direction. The two halfway-point lines intersect at the conic's center. That gives us 2 of the conic's 5 parameters.

One can find the remaining three by drawing three lines through the center and finding where they meet the curve. From the reciprocals of the squares of the distances to the curve, one can find a20/a00, a11/a00, and a02/a00, and from there, the remaining 3 parameters.

 

[Bomb#20]

Are there any Fibonacci numbers that are multiples of 314159?

A way of addressing that is to consider the question of which numbers can be factors of some Fibonacci number. If any number can be a factor of some Fibonacci number, then that proves this result. If not, then one might be able to check if 314159 is one of those numbers.

Any number can be a factor of some Fibonacci number. No, I don't have a proof; it's intuitively obvious.

But one can check any particular number without a general proof, and yes, infinitely many Fibonacci numbers are multiples of 314159. The smallest one, curiously, is

Show

F₃₁₄₁₅₈

 

[Bomb#20]

No, I don't have a proof; ...

Okay, now I have a proof; it's easy to show that for all N there must exist a Fibonnaci number F_M divisible by N, where M < N². But it looks like we can do much better. Can anyone prove a tighter bound?

 

[beero1000]

No, I don't have a proof; ...

Okay, now I have a proof; it's easy to show that for all N there must exist a Fibonnaci number F_M divisible by N, where M < N². But it looks like we can do much better. Can anyone prove a tighter bound?

There's lots of interesting stuff on the Fibonacci numbers. Take a look at  Pisano_period.

In particular the bound can be improved to M ≤ 6N in general. For N = 314159, N is prime and -1 mod 5, so we get the even nicer result that M must divide N - 1.

 

[beero1000]

Returning to that ruler-and-compass finding of parameters of a conic section, I've made a discovery. With a line, I find a halfway point between its two intersections. With a parallel line I find another halfway point. I draw a line through the two points. I then repeat this operation with two lines with a different direction. The two halfway-point lines intersect at the conic's center. That gives us 2 of the conic's 5 parameters.

One can find the remaining three by drawing three lines through the center and finding where they meet the curve. From the reciprocals of the squares of the distances to the curve, one can find a20/a00, a11/a00, and a02/a00, and from there, the remaining 3 parameters.

There's a really nice property of the auxiliary circle of a conic. Draw a tangent to the conic, and draw perpendiculars to the tangent where it intersects the auxiliary circle. Each line then goes through a focus.

 

[Kharakov]

Are there any Fibonacci numbers that are multiples of 314159?

A way of addressing that is to consider the question of which numbers can be factors of some Fibonacci number. If any number can be a factor of some Fibonacci number, then that proves this result. If not, then one might be able to check if 314159 is one of those numbers.

Any number can be a factor of some Fibonacci number. No, I don't have a proof; it's intuitively obvious.

But one can check any particular number without a general proof, and yes, infinitely many Fibonacci numbers are multiples of 314159. The smallest one, curiously, is

Show

F₃₁₄₁₅₈

Reminded me of this (which I glanced at years ago).

 

[lpetrich]

Now some stuff on the loxodrome / rhumb line. The angles on a sphere made by latitude and longitude are Δχ = Δβ, cos(β)*Δλ where β is the latitude and λ is the longitude. For bearing slope m,
(Δβ)/(cos(β)*Δλ) = m

This has solution
\(\beta = \arcsin(\tanh(m\lambda)) ;\ \lambda = \frac{1}{m} \text{arctanh}(\sin(\beta))\)
This is related to the Gudermannian function: gd(x) = arctan(sinh(x)) = arcsin(tanh(x))

The distance is (Δχ)² = (Δβ)² + (cos(β)*Δλ)² and it has the solution
\(\chi = \frac{\sqrt{m^2+1}}{m} \beta\)

One can thus find the bearing from the ratio of intervals of the longitude and the inverse Gudermannian of the latitude, and that's what I did earlier. One also gets the distance to travel, using a simple formula.

 

[lpetrich]

Here's one. Consider a family of polynomials y_n(x) with degree n in x: constant, linear, quadratic, cubic, quartic, quintic, ...

They satisfy the differential equation p2(x)*y'' + p1(x)*y' + (p0(x) + q)*y = 0

From the y's being those polynomials, what constraints can one find for p0, p1, p2, and q?