1. ## GLaw’s challenge problems

Just an idea of mine. From now on I will try and set a challenge problem on a regular basis. First problem:

Find all solutions $x,y,z\in\mathbb R$ to the equation

$x^2+y^2+z^2\ =\ k(xy+yz+zx)$

where $-1\leq k\leq1$.

2. ## Re: GLaw’s challenge problems

I haven't double checked my calcs, so perhaps I bungled it somewhere, but I think I got it. The final result I got is:
Spoiler:
Without going into the derivation's details, I get the origin only $(x = y = z = 0)$ when $k < 1$, and the line $x(t) = t, y(t) = t, z(t) = t, t \in \mathbb{R}$ when $k = 1$.

(And when $k > 1$, something interesting that I wasted my time calculating - a double cone through the origin).

Nice problem. Thanks.

3. ## Re: GLaw’s challenge problems

You’re welcome.

I’ll post my solution at the end of the week, in case there are others who want to have a go at the problem.

4. ## Re: GLaw’s challenge problems

Thanks for the problem. I have been very busy for the past few weeks so I did not have time to devote to working out the problem in full, but it was interesting to play around with in my brief moments of spare time.

5. ## Re: GLaw’s challenge problems

Solution to Problem #1:
Spoiler:
As $-1\leq k\leq1$, we can use the substitution $k=\dfrac{2t}{1+t^2}$. This gives

$(1+t^2)(x^2+y^2+z^2)\ =\ 2t(xy+yz+zx)$

$\implies\ (x-ty)^2+(y-tz)^2+(z-tx)^2\ =\ 0$

$\implies\ \begin{array}{rcl} x &=& ty \\ y &=& tz \\ z &=& tx \end{array}$

If $t=1$ (i.e. $k=1$) we have $x=y=z$. Otherwise substituting the third equation into the second and the second into the first gives $x=t^3x$ $\implies$ $x=0$, whence $y=z=0$.

Hence the solutions are $x=y=z$ if $k=1$ and $(x,y,z)=(0,0,0)$ if $-1\leq k<1$.

Challenge Problem #2:

Let $a_1,\ldots,a_n$ be positive real numbers such that $a_1+\cdots+a_n=n$. Show that
$$a_1^k+\cdots+a_n^k\ \geqslant\ 2n-\ln\left(a_1^{a_1}\cdots a_n^{a_n}\right)$$
for all positive integers $k$.

6. ## Re: GLaw’s challenge problems

I’ve just discovered a mistake in the second problem – the correct version should be as follows:

Let $a_1,\ldots,a_n$ be positive real numbers such that $a_1+\cdots+a_n=n$. Show that
$$a_1^k+\cdots+a_n^k\ \geqslant\ n-\ln\left(a_1^{a_1}\cdots a_n^{a_n}\right)$$
for all positive integers $k$.

7. ## Re: GLaw’s challenge problems

My solution to the first one

Spoiler:

$x^2+y^2+z^2 = k(xy+yz+zx)$

First, conceptualize the solution set as a set points $(x, y, z) \in \mathbb{R}^3$.

A direct check shows that the origin is always in the solution set for any value of $k$, so will hunt for non-origin solutions only.

The origin is obviously the only point in the solution for $k = 0$. So assume going forward that $k \ne 0$.

Next, the equation is homogeneous... if $(x, y, z)$ is a solution, then so is $(t x, t y, t z)$ for any $t \in \mathbb{R}$.

By homogeneity, given a non-origin point P that's a solution, every point on the line through the origin and P is a solution.

Thus the total solution set is entirely determined by the solutions which are on the unit sphere (and if none, then the solution set is just the origin).

The solution set the is union of the all points on the set of lines going through the origin and any point of the solution on the unit sphere.

Holding off on the unit sphere for a moment, use that $(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx)$ to eliminate the cross terms on the right-hand side.

Do that by adding $\left(\frac{k}{2}\right)(x^2 + y^2 + z^2)$ to both sides, getting

$\left(1 + \frac{k}{2}\right)(x^2 + y^2 + z^2) = \left(\frac{k}{2}\right)(x+y+z)^2$. Dividing both sides by $\frac{k}{2}$ (recall assuming $k>0$) gives:

$(x+y+z)^2 = \left( \frac{1 + \frac{k}{2}}{\frac{k}{2}}\right)(x^2 + y^2 + z^2)$, and so

(*) $(x+y+z)^2 = \left(\frac{2 + k}{k}\right)(x^2 + y^2 + z^2)$

Now by (*), the left hand side must be non-negative, as must right-hand side's factor $(x^2 + y^2 + z^2)$. Thus the only solution will be the origin when $\frac{2 + k}{k} < 0$.

The solution set of $\frac{2 + k}{k} < 0 \ (k \ne 0)$ is $-2 < k < 0$.

Thus the origin is the only solution for $-2 < k \le 0$.

Now, assume $k$ is such that $\frac{2 + k}{k} > 0$ (meaning $k \notin (-2, 0\]$ ), and consider just the points on the unit sphere ( $x^2 + y^2 + z^2 = 1$).

Take the square root of (*) and it reads

$| \ x+y+z \ | = \sqrt{\frac{2 + k}{k}}$.

Letting $c_k = \sqrt{\frac{2 + k}{k}}$, that says that any solution point $(x_0, y_0, z_0)$ on the unit sphere must also be on one of the two planes $x+y+z= c_k$ and $x+y+z = -c_k$.

Now if a solution $(x_0, y_0, z_0)$ on the unit sphere (so $x_0^2 + y_0^2 + z_0^2 = 1$), then by homogenity, its antipodal point $(-x_0, -y_0, -z_0)$ is also a solution on the unit sphere.

So if that point is also on the plane $x+y+z= c_k$ (meaning $x_0+y_0+z_0= c_k$), then $(-x_0, -y_0, -z_0)$ is a solution, and that satisfies $-x_0 - y_0 - z_0 = -c_k$.

Likewise, if a solution $(x_0, y_0, z_0)$ is on the unit sphere and in the plane $x+y+z= -c_k$, then its antipodal point $(-x_0, -y_0, -z_0)$ is a solution on the unit sphere and in the plane $x+y+z= c_k$.

Therefore, to find all non-origin solutions, we only need to find the intersection of unit sphere and the plane $x+y+z= c_k$, since at the end we'll build all solutions by including all the points on all the lines through the origin and a point on that intersection. The solutions on the unit sphere intersect the plane $x+y+z = -c_k$ will be captured in that process.

Note that the case $k = -2$ gives $c_k = 0$ is the case where those planes, $x+y+z = -c_k$ and $x+y+z = -c_k$, are the same plane... they've come together and met. Everything applies in that case as well.

In $\mathbb{R}^3$, a plane intersect the unit sphere is either the empty set (no intersection), a single point (the plane is tangent to the sphere), or a circle (the plane slices through the sphere).

Thus if you draw the all lines through the origin the go through a plane intersect the unit sphere, you'll either get the empty set, or a single line, or a double cone.

Thus the final solution to the problem will either be (depending on $k$): just the origin (since the origin is always a solution!), or a single line through the origin, or a double cone through the origin.

Now work out the precise solutions for a given $k$:

##################################################

Again, consider $k \notin \(-2, 0\]$, and $c_k = \sqrt{\frac{2 + k}{k}}$, and look only for the points on the unit sphere intersect the plane $x+y+z= c_k$.

One of that plane's normal vectors is $\vec{n} = (1, 1, 1)$. Let $\hat{n} = \frac{1}{\sqrt{3}} (1, 1, 1)$. Notice that this doesn't vary with $k$.

Let the plane's point closest point to the origin be defined as $\vec{q}_k$.

Then $\vec{q}_k = R_k \hat{n}$ for some $R_k$. (That's geometrically obvious and has a straightforward proof.)

Whether, and/or how, the plane intersects the unit sphere is determined entirely by $|R_k|$.

If $|R_k| > 1$, then there's no intersection. If $|R_k| = 1$, then there's a single point of intersection (the plane's tangent to the sphere). If $|R_k|<1$, then the intersection is a circle.

To find $R_k$, use that $\vec{q}_k = R_k \hat{n} = R_k \left(\frac{1}{\sqrt{3}} (1, 1, 1)\right) = \frac{R_k}{\sqrt{3}} (1, 1, 1)$ is a point on the plane $x+y+z= c_k$.

Thus $\left( \frac{R_k}{\sqrt{3}} \right) + \left( \frac{R_k}{\sqrt{3}} \right) + \left( \frac{R_k}{\sqrt{3}} \right) = c_k$, so $3\left( \frac{R_k}{\sqrt{3}} \right)= c_k$, so

$\sqrt{3} R_k = c_k$, $\sqrt{3} R_k = \sqrt{\frac{2 + k}{k}}$, so $R_k = \sqrt{\frac{2 + k}{3k}}$. (Note that $R_k \ge 0$, as the geometric picture suggested.)

With the formula $R_k = \sqrt{\frac{2 + k}{3k}} (> 0)$ (remember it's always assumed that $k \notin \(-2, 0\]$), can now classify solutions according to $k$.

############

There's no intersection when $R_k > 1$, hence whenever $\sqrt{\frac{2 + k}{3k}} > 1$. Solving that for k gives $0 < k < 1$, and in that case the grand solution to the problem is just the origin.

That can be rolled into the "origin only" solution when $k \notin (-2, 0\]$, so that: for $-2 < k < 1$, the only solution to the main problem is just the origin ( $x = y = z = 0$).

Thus the grand solution is the origin only in the case where $-2 < k < 1$

$x = 0, y = 0, z = 0$

############

There's only one point of intersection when $R_k = 1$, so when $\sqrt{\frac{2 + k}{3k}} = 1$. That occurs only when $k = 1$.

The single point of intersection between the unti sphere and the plane in that case is the point $\vec{q}_1 = R_1 \hat{n} = (1) \hat{n} = \frac{1}{\sqrt{3}} (1, 1, 1)$.

Therefore the solution set when $k = 1$ is the line through the origin and that point $\vec{q}_1 = \frac{1}{\sqrt{3}} (1, 1, 1)$, so is the line $\left\{ \ \vec{q}_1 \tilde{t} \in \mathbb{R}^3 \ | \tilde{t} \in \mathbb{R} \right\}$.

Letting $t = \frac{\tilde{t}}{\sqrt{3}}$, that can be written parametrically without using the square root.

Thus the grand solution, a line, in the case where $k=1$, is given by

$x(t) = t, y(t) = t, z(t) = t, t \in \mathbb{R}$, or even more concisely written as the line x = y = z

############

The intersection is a circle when $0 \le R_k<1$, so when $\sqrt{\frac{2 + k}{3k}} < 1$ and $k \notin \(-2, 0\]$, so when either $k>1$ or $k \le -2$.

In the case $R_k<1$, the circle in 3-space formed by the plane intersect the unit sphere, will have center $\vec{q}_k$, and a normal vector $\hat{n}$ (since it's on the plane with that normal vector).

Let $r_k$ be the radius of that circle. Then by looking at the triangle from the origin, to $\vec{q}_k = R_k \hat{n}$, to a point on the circle, it's a right triangle with hypotenuses 1 and legs $|R_k|, r_k$.

Therefore $r_k = \sqrt{1 - R_k^2} = \sqrt{1 - \left(\sqrt{\frac{2 + k}{3k}}\right)^2} = \sqrt{1 - \frac{2 + k}{3k}} = \sqrt{\frac{2k - 2}{3k}}$ $= \sqrt{\frac{2}{3}}\sqrt{\frac{k - 1}{k}}$.

To find unit vectors mutually orthogonal to $\hat{n} = \frac{1}{\sqrt{3}} (1, 1, 1)$, find one unit vector orthogonal to it, and then take the cross product to find another (they'll form an orthonormal basis for $\mathbb{R}^3$).

It's easy to find one vector orthogonal to it, $(1, -1, 0)$, so let $\hat{u} = \frac{1}{\sqrt{2}} (1, -1, 0)$.

Then let $\hat{v} = \hat{n} \times \hat{u} = \frac{1}{\sqrt{6}}(1, 1, -2)$.

The points on the circle can be parameterized using sines and cosines, except instead of the classic x and y axes of $\mathbb{R}^2$, it's using the $\hat{u}$ and $\hat{v}$ "axes" in $\mathbb{R}^3$.

To get to a point on that circle, go to $\vec{q}_k = R_k \hat{n}$ and then move radius-times-cosine in the $\hat{u}$ (like classic x-axis) direction, then radius-times-sine in the $\hat{y}$ (like classic y-axis) direction.

The points on that circle will be given by: $R_k \hat{n} + r_k \cos(\theta) \hat{u} + r_k \sin(\theta) \hat{v}$ for all $\theta \in \mathbb{R}$.

In coordinates, that's $\frac{R_k}{\sqrt{3}} (1, 1, 1) + \frac{r_k \cos(\theta)}{\sqrt{2}} (1, -1, 0) + \frac{r_k \sin(\theta)}{\sqrt{6}}(1, 1, -2)$, so

$x = \frac{R_k}{\sqrt{3}} + \frac{r_k \cos(\theta)}{\sqrt{2}} + \frac{r_k \sin(\theta)}{\sqrt{6}}$

$= \sqrt{\frac{2 + k}{k}} + \left(\frac{\cos(\theta)}{\sqrt{2}} + \frac{\sin(\theta)}{\sqrt{6}}\right)\sqrt{\frac{2} {3}}\sqrt{\frac{k - 1}{k}}$

$= \sqrt{\frac{2 + k}{k}} + \left(\frac{\cos(\theta)}{\sqrt{3}} + \frac{\sin(\theta)}{3}\right) \sqrt{\frac{k - 1}{k}}$.

$y = \frac{R_k}{\sqrt{3}} - \frac{r_k \cos(\theta)}{\sqrt{2}} + \frac{r_k \sin(\theta)}{\sqrt{6}}$

$= \sqrt{\frac{2 + k}{k}} + \left( - \frac{\cos(\theta)}{\sqrt{3}} + \frac{\sin(\theta)}{3} \right) \sqrt{\frac{k - 1}{k}}$.

$z = \frac{R_k}{\sqrt{3}} - 2 \frac{r_k \sin(\theta)}{\sqrt{6}}$

$= \sqrt{\frac{2 + k}{k}} - \frac{2 \sin(\theta)}{3}\sqrt{\frac{k - 1}{k}}$.

Thus the grand solution, the double cone, in the case where $k>1$ or $k \le -2$, is given by
$x(t, \theta, k) = \left$\sqrt{\frac{2 + k}{k}} + \left(\frac{\cos(\theta)}{\sqrt{3}} + \frac{\sin(\theta)}{3}\right) \sqrt{\frac{k - 1}{k}}\right$ t$
$y(t, \theta, k) = \left$\sqrt{\frac{2 + k}{k}} + \left( - \frac{\cos(\theta)}{\sqrt{3}} + \frac{\sin(\theta)}{3} \right) \sqrt{\frac{k - 1}{k}}\right$ t$
$z(t, \theta, k) = \left$\sqrt{\frac{2 + k}{k}} - \frac{2 \sin(\theta)}{3}\sqrt{\frac{k - 1}{k}}\right$ t$

My solution to the second one:
Spoiler:

This is just an application of Jensen's Inequality: For a convex function, the image of a weighted average is less than or equal to the weighted average of the images.

$f\left( \frac{\sum_{i = 1}^N c_i x_i}{\sum_{i = 1}^N c_i} \right) \le \frac{\sum_{i = 1}^N c_i f(x_i)}{\sum_{k = 1}^N c_i}$.

If $f(x) = x^k + x \ln(x)$, then $f'(x) = kx^{k-1} + 1 + \ln(x)$ and so $f''(x) = k(k - 1)x^{k-2} + 1/x > 0$.

Thus $f$ is convex on its domain ( $x > 0$), since its second derivative is always positive.

Apply Jemsen's inequality to an even weighting, $c_i = 1$ for $1 \le i \le n$, of the points $\{a_i\}_{i = 1}^n$, to get:

$f\left( \frac{\sum_{i = 1}^n a_i}{n} \right) \le \frac{\sum_{i = 1}^n f(a_i)}{n}$.

But since we're told that $\sum_{i = 1}^n a_i = n$, have that $f\left( \frac{\sum_{i = 1}^n a_i}{n} \right) = f(1) = (1)^k + (1)\ln(1) = 1$.

Thus $1 \le \frac{\sum_{i = 1}^n f(a_i)}{n}$, so $\sum_{i = 1}^n f(a_i) \ge n$.

Thus $\left( \sum_{i = 1}^n a_i^k \right) + \left( \sum_{i = 1}^n a_i \ln(a_i) \right) = \sum_{i = 1}^n \left( a_i^k + a_i \ln(a_i) \right) = \sum_{i = 1}^n f(a_i)$ $\ge n$, so

$\sum_{i = 1}^n a_i^k \ge n - \sum_{i = 1}^n a_i \ln(a_i)$.

8. ## Re: GLaw’s challenge problems

This one frustrated me. I was running down the wrong road completely.

I suggest GLaw's next challenge be in a new thread. I suspect few saw this challenge.

9. ## Re: GLaw’s challenge problems

Solution to Problem #2:
Spoiler:
Apply Jensen’s inequality to the convex function $\mathbb R^+\to\mathbb R;\ x\mapsto x^k+x\ln x$.

Challenge Problem #3:

Show that the following groups are isomorphic:

• $\left(\mathbb Z[x],+\right)$, polynomials with integer coefficients under addition
• $\left(\mathbb Q^+,\times\right)$, positive rational numbers under multiplication

10. ## Re: GLaw’s challenge problems

Here's a simple solution to the group isomorphism problem:
Spoiler:
Each of the two groups is free abelian with a countable base. (For the multiplicative group of positive rationals, a base is the set of primes; the group of polynomials has a base $\{x^i:i\geq0\}$.) So the two are isomorphic.

11. ## Re: GLaw’s challenge problems

I’m back!

Solution to Problem #3:
Spoiler:
An isomorphism is given by $\phi:\mathbb Z[x] \to \mathbb Q^+$; $\phi(0)=1$ and if $f = a_1 + a_2x + \cdots + a_nx^{n-1} \not\equiv 0$, $\phi(f) = 2^{a_1}3^{a_2}\cdots p_n^{a_n}$ where $p_n$ is the $n$th prime.

Challenge Problem #4:

Find all integers $a$ such that

$\frac{2 \cdot 1000^k - a}{27}$

is an integer for all integers $k\geqslant0$. Similarly find all integers $b$ and $c$ such that

$\frac{2 \cdot 10^{3k-1} - b}{27}\ \text{and}\ \frac{2 \cdot 10^{3k-2} - c}{27}$

are integers for all integers $k\geqslant1$.

12. ## Re: GLaw’s challenge problems

A solution using congruence mod 27:
Spoiler:
For any integer $x$, ${x\over27}$ an integer is the same as saying $x\equiv 0\pmod {27}$ . Easily, the multiplicative order of 10 mod 27 is 3. So
1. $2\cdot 1000^k-a\equiv 0\pmod {27}$ if and only if $2(10^{3})^k-a\equiv 0\pmod {27}$ if and only if $a\equiv 2\pmod {27}$ or $a=2+27q\text{ for some integer }q$
2. $2\cdot 10^{3k-1}\equiv b\pmod {27}$ iff $2\cdot 10^{-1}\equiv b\pmod {27}$. Here $10^{-1}=19$ is the multiplicative inverse of 10. So $b\equiv 11\pmod {27}$
3. $2\cdot 10^{3k-2}\equiv c\pmod {27}$ iff $2\cdot 10^{-2}\equiv c\pmod {27}$ iff $c\equiv 20 \pmod {27}$

13. ## Re: GLaw’s challenge problems

Originally Posted by GLaw
I’m back!

Solution to Problem #3:
Spoiler:
An isomorphism is given by $\phi:\mathbb Z[x] \to \mathbb Q^+$; $\phi(0)=1$ and if $f = a_1 + a_2x + \cdots + a_nx^{n-1} \not\equiv 0$, $\phi(f) = 2^{a_1}3^{a_2}\cdots p_n^{a_n}$ where $p_n$ is the $n$th prime.

Challenge Problem #4:

Find all integers $a$ such that

$\frac{2 \cdot 1000^k - a}{27}$

is an integer for all integers $k\geqslant0$. Similarly find all integers $b$ and $c$ such that

$\frac{2 \cdot 10^{3k-1} - b}{27}\ \text{and}\ \frac{2 \cdot 10^{3k-2} - c}{27}$

are integers for all integers $k\geqslant1$.
An answer equivalent to johng's using induction.

Spoiler:
$\dfrac{2 * 1000^0 + 25}{27} = \dfrac{2 + 25}{27} = \dfrac{27}{27} = 1 \in \mathbb Z.$

$Let\ k\ be\ any\ integer \ge 0\ such\ that\ \dfrac{2 * 1000^k + 25}{27} \in \mathbb Z.$

There is at least one such integer, namely 0.

$Let\ u = \dfrac{2 * 1000^k + 25}{27} \implies u \in \mathbb Z.$

$Let\ v = 74 * 1000^k,\ which\ obviously \in \mathbb Z.$

$\dfrac{2 * 1000^{(k+1)}}{27} - u = \dfrac{2 * 1000^{(k+1)} + 25 - (2 * 1000^k + 25)}{27} = \dfrac{2 * 1000^{(k+1)} - 2 * 1000^k}{27} \implies$

$\dfrac{2 * 1000^k * (1000 - 1)}{27} = 1000^k * \dfrac{2 * 999}{27} = v.$

$So\ \dfrac{2 * 1000^{(k+1)}}{27} - u = v \implies \dfrac{2 * 1000^{(k+1)}}{27} = u + v \in \mathbb Z.$

$therefore, n \in \mathbb\ Z\ and\ n \ge 0 \implies \dfrac{2 * 1000^n + 25}{27} \in \mathbb Z.$

$Let\ n = any\ non-negative\ integer\ and\ x = \dfrac{2 * 1000^n + 25}{27}.$

$x \in \mathbb Z.$

$Let\ m\ be\ any\ integer\ and\ a = 27m - 25 \implies a \in \mathbb Z.$

$\dfrac{2 * 1000^n - a}{27} = \dfrac{2 * 1000^n + 25 - 27m}{27} = \dfrac{2 * 1000^n + 25}{27} - \dfrac{27m}{27} = x - m \in \mathbb Z.$

So the answer is that a must be of the form a = 27m - 25, where m is an integer.

I don't have time to do the follow ons right now. But thanks: this was fun.

14. ## Re: GLaw’s challenge problems

Solution to Problem #4:
Spoiler:
Define

$\begin{array}{rcl}s_1 &=& 1 \\ s_2 &=& 11 \\ s_3 &=& 111 \\ &\vdots& \end{array}$

$\therefore\ s_n\ \equiv\ 1\pmod5\ \forall n=1,2,\ldots$

$\Rightarrow\ s_n\ \equiv\ 1,6,11\pmod{15}$

$\text{Also}\ s_n\ \equiv\ \text{sum of digits}\ =\ n\pmod3$

$\therefore\ s_n\ \equiv\ 1,11,6\pmod{15}\ \text{if}\ n\equiv1,2,3\pmod3\ \text{respectively}$

$\Rightarrow\ 10^n\ =\ 9s_n+1\ \equiv\ 10,100,55\pmod{135}\ \text{if}\ n\equiv1,2,3\pmod3$

$\Rightarrow\ 2\cdot10^{n-1}\ \equiv\ 2,20,11\pmod{27}\ \text{if}\ n\equiv1,2,3\pmod3$

$\Rightarrow\ a\equiv2\pmod{27},\ b\equiv11\pmod{27},\ c\equiv20\pmod{27}$

Challenge Problem #5:

Let $a,b,c$ be non-negative real numbers. Prove that

$(a+b)\sqrt{ab}+(b+c)\sqrt{bc}+(c+a)\sqrt{ca}\ \leqslant\ \frac{a^2+b^2+c^2+3(ab+bc+ca)}2$

15. ## Re: GLaw’s challenge problems

Hopefully your last one this year

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