A trigonometric equation

tan(x + y) = tanx · tan y

 

 

 

 

The equation tan(x + y) = tanx * tan y corresponds to the graph in figure 1. It should really consists of closed simple curves, the gaps is due to the computer program. The “bubbles” are symmetric around the line… y = x, they are infinite in number and separated by π in x- and y-direction. This is naturally a direct consequence of the period length π of the tangent function. The second picture is a magnification of the “bubble” going through the origin, in sequel this “bubble” is called B0. The third picture shows tan x versus tan y.

A few challenges:

1. Find some “simple” solution (e g rational multiple of π or maybe a little less simple) corresponding to a point on B0.

2. Solve the equation for y, to get y = f(x) for intervals corresponding to a points on B0.

3. Find the exact coordinates for the point where B0 intersects the line y = x.

Expected projection

Find the mathematical expectation of the area of the projection of a cube with edge of length onto a plane with an isotropically distributed random direction of projection.

(Proposed by Bhargav Gnv/FB/Math)

________________________________________________________

Let the cube be positioned in a 3d coordinate system with centre at the origin and edges parallel to axes,and with edge length 1. Let the direction of projection be given by a unit vector n = (cos α, cos β, cos γ) where α, β, γ are the angles of the vector to the coordinate axes. Also, for simplicity we confine ourselves to the first octant. Because of symmetry that doesn’t affect the result.

Now, from any direction exactly three non-parallel faces of the cube can be seen, including the ”degenerate” cases when some face is seen from its own plane. Therefore the area of the projection is cos α + cos β + cos γ. Isotropically distributed random direction of projection means that every position of the tip of vector n on the unit sphere around the origin is equally probable. Since the area element of the sphere is sin θ dθ dφ in polar coordinates the probability density f(θ) for the tip’s position is proportional to sin θ (no φ-dependence by azimuthal symmetry). By normalization f(θ) = (2/π)*sin θ. Thus the expectation value
<A> = ∫∫ (cos α + cos β + cos γ)(2/(π))*sin θ dθ dφ
Σ
where Σ = [0,π/2]x[0,π/2].

To calculate this we must express the direction cosines cos α, cos β, cos γ in polar angles θ, φ. Let’s say that α, β, γ are the angles with the x-, y-, and x-axis respectively. Then the tip of the vector n is at (sin θ cos φ, sin θ sin φ, cos θ). Using the scalar product of n with the unit vetors e_x, e_y, e_z, these components are equal to the direction cosines. Thus,
<A> = ∫∫ (sin θ cos φ + sin θ sin φ + cos θ)(2/π)*sin θ dθ dφ =
= (2/π)*∫∫ sin²θ (cos φ + sin φ) + cos θ sin θ dθ dφ =
= (2/π) * ∫ [θ=0,π/2] sin²θ dθ * ∫ [θ=0,π/2] cos φ + sin φ dφ
+ (2/π) * (π/2) * ∫ [θ=0,π/2] cos θ sin θ dθ =
= (2/π)*(π/4)*2 + 1*1/2*2 = 1 + 1 = 2.

With edge a we get of course <A> = 2a². Seems resonable, since three faces are seen contracted and each face has area a², so <A> must be something between a² and 3a².

Ooppss! ∫ [θ=0,π/2] cos θ sin θ dθ = 1/4*2 = 1/2 so <A> = 3a²/2.

Also, the estimate can be improved. The greatest projected area occurs in the direction (1/√3, 1/√3, 1/√3) and that area is 3a²/√3 = a²√3. So we must have a² < <A> < a²√3, which satisfied by <A> = 3a²/2. – In general the expaction value of a non-constant random variable must be strictly between the smallest and the greatest value.

Integral distances

Find coordinates of a set of eight non-collinear planar points so that each has an integral distance from others.

(Proposed by Bhargav Gnv/FB/Math)

_________________________________________________

These eight points all lie on a circle with radius 4225, so no three of them are collinear:

A = [1183, 4056]
B = [1183, -4056]
C = [2975, 3000]
D = [2975, -3000]
E = [-2975, 3000]
F = [-2975, -3000]
G = [-1183, 4056]
H = [-1183, -4056]

The 28 connecting segments all have integer length:

AB = 8112 , AC = 2080 , AD = 7280 , AE = 4290 , AF = 8190 , AG = 2366 ,
AH = 8450
BC = 7280 , BD = 2080 , BE = 8190 , BF = 4290 , BG = 8450 , BH = 2366
CD = 6000 , CE = 5950 , CF = 8450 , CG = 4290 , CH = 8190
DE = 8450 , DF = 5950 , DG = 8190 , DH = 4290
EF = 6000 , EG = 2080 , EH = 7280 , FG = 7280 , FH = 2080
GH = 8112

Reinventing the natural logarithm

The natural logarithm ln(x) can be defined as ∫(1 to x) 1/t dt. Using this definition, prove that it is a logarithm, i.e. as follows.

The logarithm of base g, log_g, is defined with the biimplication a=g^x ⇔ log_g(a)=x. Prove that there exists a constant e, such that log_e(a)=∫(1 to a) 1/t dt.

(Because this is a reinvention of the natural logarithm, be careful not to use any fact you know from the natural logarithm, including any fact of the constant e.)

(Proposed by Mauri Ericson Sombowadile/FB/Math)

Sequence of perfect squares

Prove that each term of the following sequence is a perfect square:

49, 4489, 444889, 44448889, 4444488889, 444444888889, …

In general, the nth term has n 4′s and n-1 8′s and one 9.

(Laars Helenius/FB/Math)

______________________________________________

Notation: <n> = 444…48…889 with n digits 4, etc (n>=1).
9·<n> = 10·<n> – <n> = 444…8…8890 – 444…4…8889
Here the first … in the terms mean 4:s, and the second … mean 8:s. The 8 in the first term and the 4 in the second one are in the same position (n digits to the right of them). Thus, 9·<n> = 4·10²ⁿ + 4·10ⁿ + 1 = (2·10ⁿ + 1)² where the last 1 comes from 90 – 89.

Now, 2·10ⁿ + 1 is obviously divisible by 3, i e 2·10ⁿ + 1 =3·m (m integer) so <n> = (3·m)²/9 = m², i e <n> is a square. – qed

Comparing cos(sinθ) and sin(cosθ)

If θ is expressed in radians, which is greater, cos(sinθ) or sin(cosθ)?

(Proposed by Souradeep P./FB/Math)

__________________________________________________

Let f(x) = cos(sin x) – sin(cos x), where 0 ≤ x ≤ 2π. This is enough since the function has period 2π. We consider three subintervals.

(1) 0 ≤ x ≤ π/2:
f(x) = cos(sin x) – sin(cos x) = [cos²(sin x) - sin²(cos x)] / [cos(sin x) + sin(cos x)] =
= (1/2)*[cos(2sin x) + cos(2cos x)] / [cos(sin x) + sin(cos x)] =
= cos(sin x + cos x) * cos (sin x – cos x) / [cos(sin x) + sin(cos x)]

Since |sin x ± cos x| ≤ √2 < π/2 for all x, cos(sin x + cos x) * cos (sin x – cos x) > 0.

When 0 ≤ x ≤ π/2, 0≤ sin x, cos x ≤ 1, so 0 < cos(sin x) ≤ 1, 0 ≤ sin(cos x) ≤ 1.
Thus cos(sin x) + sin(cos x) > 0 and f(x) > 0.

(2) π/2 ≤ x ≤ 3π/2:
-1 ≤ sin x ≤ 1, and -1 ≤ cos x ≤ 0, so cos(sin x) > 0, and sin(cos x) ≤ 0.
Thus f(x) = cos(sin x) – sin(cos x) > 0.

(2) 3π/2 ≤ x ≤ 2π:
Since f(x) = f(2π – x) it follows from case (1) that f(x) > 0.

Cosequently cos(sin x) > sin(cos x) for all x.

Non-elementary functions

This is a big subject. What I have in mind at present is xˣ, xˣ⁻¹ and W(x), the Lambert W-function. (To be used at other places.)

__________________________________________________

The Lambert W-funtion

This has two (real) branches W₀ and W₋₁ defined thus:
(1) If f(x) = x·eˣ, x ≥ -1 then W₀(x) ≡ f⁻¹(x), x ≥ -1/e.
(2) If f(x) = x·eˣ, x ≤ -1 then W₋₁(x) ≡ f⁻¹(x), -1/e ≤ 0.

So W(x·eˣ) = x and W(x)·e^(W(x)) = x with suitable branch.
_________________________________
Example: Find the negative solution of the equation x² = 2ˣ.

Solution: Manipulate, to an equation of the form g(x)·e^g(x) = a.

x² = 2ˣ, x² = e^(ln 2ˣ), x² = e^(x·ln 2), -x = e^(x·(ln 2)/2) (x < 0),
e^(-x·(ln 2)/2) = -1/x, (-x)·e^(-x·(ln 2)/2) = 1, (-x·(ln 2)/2)·e^(-x·(ln 2)/2) = (ln 2)/2,
-x·(ln 2)/2 = W₀((ln 2)/2) ((ln 2)/2 > 0), x = -(2/ln 2)W₀((ln 2)/2)
_____________________
Differentiation

Differentiate W(x)·e^(W(x)) = x and use the chain rule.
W’(x)·e^W(x) + W(x)·e^W(x)·W’(x) = 1, so W’(x) = 1/(e^W(x)(1 + W(x))), or
W’(x) = W(x)/(x(1 + W(x))).

_________________________

More about this function and applications:
http://www.orcca.on.ca/LambertW/
http://en.wikipedia.org/wiki/Lambert_W_function
http://2000clicks.com/MathHelp/BasicSimplifyingLambertWFunction.aspx

________________________________________

The function f(x) = xˣ, x > 0 (or x ≥ 0 if 0⁰ is defined as 1). lim [x→0⁺] xˣ = 1.
_________________________
Differentiation

f’(x) = xˣ(1 + ln x)
The decreases strictly in ]0,1/e] taking all values in [e^(-1/e),1[, increases strictly in [1/e,∞[, and xˣ → ∞ when x → ∞.
__________________________
We will concentrate on the inverse in the intervals ]0,1/e] and and [1/e,∞[.

Let y = xˣ, ln y = x·ln x = e^(ln x)·ln x. Apply W on both side; this gives
W(ln y) = ln x, x = exp(W(ln y)), so f⁻¹(x) = exp(W(ln x)). However this must be made more precise since xˣ is not invertible in the entire interval x > 0 and W has two branches.

If e^(-1/e) ≤ x < 1, -1/e ≤ ln x ≤ 0. Then if 0 < f⁻¹(x) ≤ 1/e, f⁻¹(x) = exp(W₋₁(ln x)).

If x ≥ e^(-1/e) and f⁻¹(x) ≥ 1/e, then ln x ≥ -1/e and f⁻¹(x) = exp(W₀(ln x)).
__________________________
Example 1: The equation xˣ = 2 has the real solution x = f⁻¹(2) = exp(W₀(ln 2)) ≈ 1.559610441.

Example 2: The equation xˣ = ln 2 (> e^(-1/e)) has two real solutions,
x₁ = exp(W₋₁(ln ln 2)) ≈ 0.3366297865 and x₂ = exp(W₀(ln ln 2)) ≈ 0.40004026.

Black curve: y = x^x
Blue curve: y = exp(W₀(lnx))
Red curve: y = exp(W₋₁(ln x))

__________________

Black curve: y = x·e^x
Blue curve: y = W₀(x)
Red curve: W₋₁(x)

_____________________________________

The function f(x) = xˣ⁻¹, x > 0.
---------------------------------------
Differentiation

f'(x) = xˣ⁻¹(1 + ln x - 1/x)
The function decreases strictly in ]0,1], increases strictly in [1,∞[, and xˣ → ∞ when x → ∞ and when x → 0⁺. So xˣ⁻¹ ≥ 1 for all real x with equality iff x = 1. This will be of use in later posts.
---------------------------------
Much like xˣ the function has an inverse in the intervals ]0,1] and and [1,∞[. I will not go into details but to give a closed expression for f⁻¹(x) a ”cousin” of the W-function can be used, defined as the inverse of x·eˣ – x.

______________________________________

(Appendix)

Elliptic functions (and more generally Abelian functions) use integrals as well as inverses. For example the inverse (in suitable interval) of the integral of 1/√((1 – t²)(1 – k²t²)), t = 0 to x, defines the so called sn function, one of the elliptic functions of Jacobi type.

In particular, for parameter value k = 0, this gives first arcsin x then sin x. So the theory of trigonimetric functions can be built from the algebraic function 1/√(1 – x²). For general k the theory of (Jacobi type) elliptic functions are built up.

Group theory problem

Let a, b be elements of a group G. Assume that a has order 5 and that (a³)b = b(a³). Prove that ab = ba.

(Proposed by Joe Rosenthal/FB/Math)

________________________________________________

ab = a⁵ab = a⁶b = a³a³b = a³ba³ = ba³a³ = ba⁶ = baa⁵ = ba

AGM-related limits

Let there be two numbers a and b.
Take the AM and GM of those two call them respectively a_1 and b_1
Then take AM and GM of a_1 and b_1, and call them a_2 and b_2, … .
Continuing this way, if the limit of a_n and b_n as n goes to infinity is x and y
then find these two limit as n goes to infinity:
I) a_1 + a_2 + a_3 + … + a_n – n·x
II) b_1 + b_2 + b_3 + … + b_n – n·y

(Proposed by Sanchar Sharma)

Sachin points out x = y, the AGM of a and b, often denoted M(a,b). In fact, this is a well-known theorem (by Gauss I think).

_________________________________________________

Let s₁(n) = a₁ + a₂ + a₃ +… + a_n – n·x, and s₁(n) = b₁ + b₂ + b₃ +… + b_n – n·y.

Then 2s₁(n) + s₂(n) =

= 2a₁ + 2a₂ + 2a₃ +… + 2a_n – 2n·x + b₁ + b₂ + b₃ +… + b_n – n·y =

= 2a₁ + (a₁ + b₁) + (a₂ + b₂) +… + (a_(n-1) + b_(n-1)) + b₁ + b₂ + b₃ +… b_n – n·(2x + y) =

= 2a₁ + (a₁ + a₂ + … + a_(n-1)) + (b₁ + b₂ +… + b_(n-1)) + b₁ + b₂ + b₃ +… b_n – n·(2x + y) =

= 2a₁ + (s₁(n) – a_n + n·x) + (s₂(n) – b_n + n·y) + s₂(n) + n·y – n·(2x + y) =

= s₁(n) + 2s₂(n) + 2a₁ – a_n – b_n – n·(x – y).

This gives

2s₁(n) + s₁(n) = s₁(n) + 2s₂(n) + 2a₁ – a_n – b_n – n·(x – y),

s₁(n) – s₁(n) = + 2a₁ – a_n – b_n – n·(x – y)

Using x = y = M(a,b) this gives lim[n→∞] (s₁(n) – s₁(n)) = 2a – 2M(a,b).

So at least the difference s₁(n) – s₂(n) has a limit. Notice that x = y is necessary for this to exist! If the limit of one more (linear) combination could be found the problem would be solved. Unfortunately I haven’t been able to find more (yet). – Someone?

Of course we still dont know if the limits of s₁(n) and s₂(n) exist. But either both or none do!

__________________________________

Some numerical results.
Let a = x and b = 1 where x > 1. Then s₁(n) = Σ [k=1,n] (a_k – M(1,x)) and
s₂(n) = Σ [k=1,n] (b_k – M(1,x)) are funtions of x and n. Provided the limits exist,
Σ₁(x) ≡ lim [n→∞] Σ [k=1,n] (a_k – M(1,x)) and Σ₂(x) ≡ lim [n→∞] Σ [k=1,n] (b_k – M(1,x)) are functions of x.

Using Mathematica it seems like the sums for given x converge rapidly and using n = 10 gives almost the same value as larger n. For example (here I use the values for n = 10 and the four numbers are appr values of x, Σ₁(x), Σ₂(x), Σ₁(x)/Σ₂(x), 2a – 2M(a,b)).

{10, 7.0803, -4.41889, 11.4992, 11.4992}
{20, 16.4921, -9.17624, 25.6684, 25.6684}
{50, 47.353, -23.0023, 70.3553, 70.3553}
{70, 68.9654, -32.0087, 100.974, 100.974}
{100, 102.245, -45.3221, 147.567, 147.567}
{150, 159.184, -67.1502, 226.334, 226.334}

Notice that the last two values are equal confirming the earlier result.

_________________________________

Graphs for Σ₁(x) and Σ₂(x), the other the ratio Σ₁(x)/Σ₂(x):

Upper curve: y = Σ₁(x) ≡ lim [n→∞] Σ [k=1,n] (a_k – M(1,x))
Lower curve: y = Σ₂(x) ≡ lim [n→∞] Σ [k=1,n] (b_k – M(1,x))

M(1,x) is known; it’s equal to a certain elliptic integral where x is a parameter.

________________________

y = Σ₁(x)/Σ₂(x)

______________________________________

Convergence proof.

For definiteness we assume that a ≥ b > 0. Then from the theory of the AGM it is known that

(*) a = a₁ ≥ a₂ ≥ … ≥ a_n ≥ a_(n+1) ≥ … ≥ M ≥ … ≥ b_(n+1) ≥ b_n ≥ … ≥ b₂ ≥ b₁ = b > 0 , where M ≡ M(a,b), the AGM of a and b. Also, by induction,

(**) a_(n+1) – b_(n+1) ≤ (a_n + b_n)/2 – b_n = (a_n – b_n)/2 ≤ … ≤ (a – b)/2^n.

From (*) and (**) 0 ≤ a_n – M ≤ a_n – b_n ≤ (a – b)/2^(n-1).

Since Σ [n=1,∞] (a – b)/2^(n-1) is convergent it follows that Σ [n=1,∞] (a_n – M) is also convergent.

Tribonacci numbers

Will there be a formula for a ”n bonacci series’? Like, a ”tri” bonacci is 1, 1, 1, 3, 5, 9, 17, 31, 57, … .  So that an n-bonacci is 1, 1, 1, … ntimes…, n, 2n-1,4 n-3, …?

(Proposed by Souradeep/FB/Math)

_______________________________________

For 1,1,1,3,5,9,17,31,57…. we have a recursion formula
a(n) + a(n+1) + a(n+2) = a(n+3). To solve this first find λ from the characteristic equation 1 + λ + λ² = λ³. Then a(n) = A λ₁ⁿ + B λ₂ⁿ + C λ₃ⁿ where A, B, C are determined by the initial conditions a(1) = a(2) = a(3) = 1.

The result is

a(n) = (λ₂ – 1)·(λ₃ – 1)/(λ₁·(λ₁ – λ₂)·(λ₁ – λ₃))·λ₁ⁿ + (1 – λ₃)·(λ₁ – 1)/(λ₂·(λ₁ – λ₂)·(λ₂ – λ₃))·λ₂ⁿ + (λ₁ – 1)·(λ₂ – 1)/(λ₃·(λ₁ – λ₃)·(λ₂ – λ₃))·λ₃ⁿ

where

λ1 = – (19/216 – √33/72)^1/3 – (√33/72 + 19/216)^1/3 + 1/3 + i·((19·√3/72 + √11/8)^1/3 – (19·√3/72 – √11/8)^1/3)

λ2 = – (19/216 – √33/72)^1/3 – (√33/72 + 19/216)^1/3 + 1/3 + i·((19·√3/72 – √11/8)^1/3 – (19·√3/72 + √11/8)^1/3)

λ3 = (19/27 – √33/9)^1/3 + (√33/9 + 19/27)^1/3 + 1/3

_________________________________

The problem is that the solutions to third degree equations are usual complicated. To get something simpler modify the recursion with suitable coefficients a,b,c so that the third degree eauation a + bλ + cλ² = λ³ gets simple solutions. You can start choosing some numbers λ₁, λ₂, λ₃ and then set a = λ₁ λ₂ λ₃, etc.

Cutting coloured cubes

A carpenter C had a solid wooden cube with whole number dimensions.He painted the entire surface of the cube with red paint.Then he cut the cube into smaller cubes of size 1*1*1 working parallel to the faces of the bigger cube.A certain number of smaller cubes were completely free of paint (x) A certain number of cubes had paint on only one side (y) A certain number number of cubes had paint on 2 sides (z).If (y+z) is 33% of (x) what was the original cube size?

(Proposed by Hrajan Nil)

________________________________________________

Let the original cube have edge length n (integer). Then
x = (n – 2)³, y = 6·(n – 2)², z = 12·(n – 2)
Then there are also 8 small cubes with paint on three sides (those at the original corners).

[ I guess that by 33% you mean 1/3. Then y + z = x/3, which gives
3·[6(n - 2)² + 12(n - 2)] = (n – 2)³. If the binomial powers are expanded we get
3·[6n² - 12n] = n³ – 6n² + 12n – 8, and then n³ – 24n² + 48n – 8 = 0.

This equation can be written (n – 2)(n² – 22·n + 4) = 0.
This equation has just on integer solution, n = 2. – There is one solution very near 22 but not exactly equal.

Aha, there is more to this problem than you may think at first. ]

Suppose 33% mean exactly 33%, i e 33/100. Then we get instead the equation

100·[6(n - 2)² + 12(n - 2)] = 33(n – 2)³. If the binomial powers are expanded we get
100(6n² – 12n) = 33(n³ – 6n² + 12n – 8), and then 33n³ – 798n² + 1596n – 264 = 0.
This equation has the integer solutions n = 2 and n = 22 exactly (and the third solution is n = 2/11).

So the original cube had edge length 2 or 22. However n = 2 gives x = y = z = 0 so I think 22 is meant.

My first attempt with 1/3 instead of 33% is really no part of the solution but I let it be there within [ ] to illustrate how you may or may not think.
_____________________________

Additions:
1. A check of the result. n = 22 gives x = 20³ = 8000, y = 2400, z = 240 so the total number of small cubes is 8000 + 2400 + 240 + 8 = 10648 = 22³. Also (y + z)/x = 2640/8000 = 33/100 so everything is in order.

2. To find the integer solutions (if you don’t take the shortcut asking the computer): An integer solution must divide the constant term 264 = 2³·3·11.

 

Derivative of |x|

Is the derivative of modulus of x possible?If yes,then how?

(Proposed by Avinandan Das)

__________________________________________________

|x| is continuous but not differentiable at x = 0. So it is differentiable for x < 0 where D|x| = D(-x) = -1 and for x > 0 where D|x| = Dx = 1. It is non-differentiable at x = 0 *because* lim [x→0] (|x|/x) does not exist.

However |f(x)| may be differentiable for all x, depeneding on f(x). For example
D|x³| = 3x|x|. Proof:

x < 0: D|x³| = D(-x³) = -3x² = 3(-x)x = 3x|x|
x > 0: D|x³| = D(x³) = 3x² = 3x|x|

Finally (don’t forget this part!), let f(x) = |x³|, so

f’(0) = lim [x→0] [(f(x) - f(0)) / (x - 0)] = lim [x→0] (|x³|/x = lim [x→0] (x²|x|/x) =

= lim [x→0] x|x| = 0 = 3·0·|0|, so f’(x) = 3x|x| for all x ∈ ℝ, or D|x³| = 3x|x|.

 

Number-theoretic minimization problem

If each of the letters A, B and C represents a different digit, what isthe minimum value of  ABC/(A + B + C) ?  In ABC, A isthe hundreds digit, B isthe tens digit and C is the ones digit. A, B and C can be any integer from 0 to 9 both inclusive.

(Proposed by Hrajan Nil/FB/Math)

_______________________________________________

My suggestion how to solve the problem:

When a = 1, b = 8, c = 9 then (100a + 10b + c)/(a + b + c) = 21/2 = 10.5 which it is a reasonable to *guess* is the least value if a ≠ 0. To prove it consider the difference (100a + 10b + c)/(a + b + c) – 21/2.

2·(100a + 10b + c) – 21·(a + b + c) = 179a – (b + 19c) ≥ 179 – (b + 19c)

(if a ≠ 0) ≥ 179 – max(8+19·9,9 + 19·8) = 179 – max(179,161) =

= 179 – 179 = 0, so (100a + 10b + c)/(a + b + c) ≥ 21/2, and we have already seen that 21/2 = 10.5 is a possible value, thus it is the least value if a ≠ 0.

__________________________________

It would interesting to see what happens with other numbers of variables, in particular some general result like ”the quotient of a n-digit number by its digit sum is at least…”.
_____________________
4 variables (a ≠ 0): q ≡ (1000a + 100b + 10c + d)/(a + b + c + d)

Guess: a = 1, b = 0, c = 8, d = 9 gives the (1000 + 80 + 9)/18 = 1089/18 = 121/2 = 60.5

Proof that q cannot be less than this:
2·(1000a + 100b + 10c + d) – 121·(a + b + c + d) = 1879a + 79b – (101c + 119d)

≥ 1879 – (101c + 119d) ≥ 1879 – max(101·8 + 119·9,101·9 + 119·8) =

= 1879 – (101·8 + 119·9) = 0, thus

(1000a + 100b + 10c + d)/(a + b + c + d) ≥ 60.5 so the least value is 60.5.

___________________________

5 variables (a ≠ 0): q ≡ (10000a + 1000b + 100c + 10d + e)/(a + b + c + d + e)

Guess: a = 1, b = 0, c = 7, d = 8, e = 9 gives q = 10789/25=431.56

Proof that q cannot be less than this:
25·(10000a + 1000b + 100c + 10d + e) – 10789·(a + b + c + d + e)

= 239211·a + 14211·b – 8289·c – 10539·d – 10764·e

≥ 239211 – 9·(921·c + 1171·d + 1196·e) ≥ 239211 – 9·max(921·c + 1171·d + 1196·e)

= 239211 – 9·(921·7 + 1171·8 + 1196·9) = 0, thus

q ≥ 10789/25 = 431.56. Since this is a possible value, the least value of q is 431.56.

_________________________________

6 variables (a ≠ 0): (100000·a + 10000·b + 1000·c + 100·d + 10·e + f)/(a + b + c + d + e + f) ≥ 106789/31 and takes this value when a = 1, b = 0, c = 6, d = 7, e = 8, f = 9.

Now it is not too hard to guess the least value for 7-10.

_________________________________

7 variables (a ≠ 0): Least value of (1000000·a + 100000·b + 10000·c + 1000·d + 100·e + 10·f + g)/(a + b + c + d + e + f + g) is 117421/4=29355.25.

 

Bisectors and isosceles triangles

Prove that in a general triangle ABC if two bisectors are equal, this triangle must be isosceles.

(Proposed by Ali Abouzar/FB/Math)

The statement is known as the ”Steiner-Lehmus theorem”. There is much material on it on the web, including proofs. In Coxeter’s book ”Geometry revisited” there is a proof that is elementary as well as easy.

____________________________________________________

Proof by geometry (my own though certainly many people have constructed it or something similar before. After all the problem is a theorem that has attracted interest for well over a century).
_________________________
First we derive a formula for the length of a bisector (I got the idea for this from my friend Tomas Carnstam, Lund University, Sweden). – Let a, b, c be the sides of a triangle and L the length of the bisector from C. Then two expressions for the triangle area gives

bL sin(C/2) + aL sin(C/2) = ba sin C.

This gives together with the formula sin(2v) = 2 sin v cos v, L = 2ab/(a + b)·cos(C/2).
________________________
Using this formula, if the bisectors from C and B are equally long,

2ab/(a + b)·cos(C/2) = 2ac/(a + c)·cos(B/2). Square this to get

b²(a + c)²·cos²(C/2) = c²(a + b)²·cos²(B/2)

b²(a + c)²·(1 + cos C) = c²(a + b)²·(1 + cos B). Then insert (by the cosine theorem)

cos C = (a² + b² – c²)/(2ab) etc, to get

b²(a + c)²·(1 + (a² + b² – c²)/(2ab)) = c²(a + b)²·(1 + (a² + c² – b²)/(2ac))

b(a + c)²·(2ab + a² + b² – c²) = c(a + b)²·(2ac + a² + c² – b²)

b(a + c)²(a + b – c) = c(a + b)²(a + c – b)

Then it is straightforward to expand and then factorize to get

(b – c)(a³ + b²c + bc² + a²b + a²c + 3abc) = 0.

Here the second factor is obviously positive so b – c = 0, i e b = c.

The three doors problem

Triangle problem

(FB/Math/BM – Prposed by Souradeep P.)

Given a triangle ABC, is there any maximum value of Rr, where R=circumradius, r=inradius?

Triangle theorem

(FB/Math/BM – Proposed by Alexander Bogomolny)

Let α + β + γ = 180° so that the three are the angles in a triangle. Then cos²(α) + cos²(β) + cos²(γ) = 1 iff the triangle is right.

_______________________________________________

‎(*) α + β + γ = 180⁰ and
(**) 0 < α, β, γ < 180⁰

From (*) γ = 180⁰ – (α + β) and 1 – cos²γ = sin²γ = sin²(α + β) Then we get the following sequence of equivalent statements

…cos²α + cos²β + cos²γ = 1

cos²α + cos²β = sin²(α + β)

cos²α + cos²β = sin²α cos²β + cos²α sin²β + 2 sin α cos β cos α sin β

cos²α – cos²α sin²β + cos²β – sin²α cos²β= 2 sin α cos β cos α sin β

cos²α(1 – sin²β) + cos²β(1 – sin²α) = 2 sin α cos β cos α sin β

cos²α cos²β + cos²β cos²α = 2 sin α cos β cos α sin β

cos²α cos²β = sin α cos β cos α sin β

cosα cosβ(cosα cosβ – sin α sin β) = 0

cosα = 0 or cosβ = 0 or cos α cos β = sinα sin β

cosα = 0 or cosβ = 0 or cot β = tan α

α = 90⁰ or β = 90⁰ or α + β = 90⁰

α = 90⁰ or β = 90⁰ or γ = 90⁰

The last line is now equivalent to the triangle being right-angled.

Integration using zeta(2)

(FB/Math/BM – Proposed by Noriyasu Tamakoshi)

Evaluate the integral from 0 to 1 of ln(1 + x)/x.

______________________________________________

∫ [0,1] ln(1+x)/x dx = ∫ [0,1] Σ [n=1,∞] (-1)ⁿ⁻¹/n xⁿ/x dx =

= Σ [n=1,∞] (-1)ⁿ⁻¹/n ∫ [0,1] xⁿ⁻¹ dx =

= Σ [n=1,∞] (-1)ⁿ⁻¹/n [xⁿ/n]|[0,1] =

= Σ [n=1,∞] (-1)ⁿ⁻¹/n·1/n =

= Σ [n=1,∞] (-1)ⁿ⁻¹/n² = π²/12
_______________________________
For the last equality, notice that Σ [n=1,∞] 1/n² = π²/6 (=zeta(2)). Then

Σ [n=1,∞] (-1)ⁿ⁻¹/n² =

= Σ [n=1,∞] (-1)²ⁿ⁻¹/(2n)² + Σ [n=1,∞] (-1)²ⁿ⁻²/(2n-1)² =

= -1/4·Σ [n=1,∞] 1/n² + Σ [n=1,∞] 1/(2n-1)² =

= -1/4·Σ [n=1,∞] 1/n² + Σ [n=1,∞] 1/n² – Σ [n=1,∞] 1/(2n)² =

= -1/2·Σ [n=1,∞] 1/n² + Σ [n=1,∞] 1/n² =

= 1/2·Σ [n=1,∞] 1/n² = 1/2·π²/6 = π²/12

Advanced integration

(FB/Math/BM – Proposed by Fawaz Alotaibi)

______________________________________________

Mathematica gives this (and I don’t feel inclined to check it

1/16 (4 ArcCos[3]^2 + ArcCosh[3] ArcCosh[577] +
4 (ArcCoth[Sqrt[2]] Log[4] + Log[2 - Sqrt[2]]^2 -
Log[2 + Sqrt[2]]^2 – 2 PolyLog[2, 3 - 2 Sqrt[2]] +
8 PolyLog[2, -1 + Sqrt[2]]))

______________________________

If the upper integration limit is set to π (instead of π/2) the integral is probably exactly π²/4. (I checked to 16 digits using Mathematica.)

Power factors in Mersenne numbers

(FB/Math/BM – unsolved)

Which of the Mersenne numbers (2ⁿ – 1) are divisible by 5k where k is a positive integer?

Conjecture: