MHB A Few Interesting Problems I Found

[eddybob123]

1. What is the smallest degree of a non-constant polynomial $$f(x)$$, such that all roots of $$f(x)$$ are in the set $${0,2,3}$$, and the derivative of $$f(x)$$ is divisible by $$8x^2-24x+7$$?

2. Let x, y, and z be positive real numbers such that

$$xyz=945$$

and

$$x(y+1)+y(z+1)+z(x+1)=385$$

What is the minimum possible value of $$z+\frac{y}{2}+\frac{x}{4}$$?

3. Let $$ A$$ be a 6x6 matrix such that $$A^k$$ is the identity matrix for some positive integer $$ k$$. The smallest $$k$$ is called the order of $$A$$. What is the largest possible order of $$A$$?

4. What is the most number of non-parallel lines in 7 dimensional space such that the angle between any two of them is the same?

Enjoy! (Giggle)

 

[caffeinemachine]

1. What is the smallest degree of a non-constant polynomial $$f(x)$$, such that all roots of $$f(x)$$ are in the set $${0,2,3}$$, and the derivative of $$f(x)$$ is divisible by $$8x^2-24x+7$$?

2. Let x, y, and z be positive real numbers such that

$$xyz=945$$

and

$$x(y+1)+y(z+1)+z(x+1)=385$$

What is the minimum possible value of $$z+\frac{y}{2}+\frac{x}{4}$$?

3. Let $$ A$$ be a 6x6 matrix such that $$A^k$$ is the identity matrix for some positive integer $$ k$$. The smallest $$k$$ is called the order of $$A$$. What is the largest possible order of $$A$$?

4. What is the most number of non-parallel lines in 7 dimensional space such that the angle between any two of them is the same?

Enjoy! (Giggle)

Discussion on the 3rd question.

If $A$ is allowed to have complex entries then I believe there are matrices which would have arbitrarily large orders.

Let $p_1,\ldots,p_6$ be distinct primes and let $\xi_i$ be the (complex) primitive root of $X^{p_i}-1$ over $\mathbb C$. Then the diagonal matrix $A$ having $\xi_1,\ldots,\xi_6$ on its diagonal has order $\prod_{i=1}^6p_i$. By choosing $p_i$'s judiciously, we can make this order arbitrarily large.

I am working on the case when $A$ can have only real entries. Here block diagonal matrices having blocks of size at most $2\times 2$ will help I guess. Thoughts?

 

[eddybob123]

Actually, $$A$$ can only have positive integer entries. Let me clarify a bit:
Let $$A$$ be a positive integer matrix such that $$A^k$$ is the identity matrix for some (finite) positive integer $k$. The smallest $k$ is the order of $A$, which must exist by definition. What is the largest possible order of such a matrix?

 

[I like Serena]

Actually, $$A$$ can only have positive integer entries. Let me clarify a bit:
Let $$A$$ be a positive integer matrix such that $$A^k$$ is the identity matrix for some (finite) positive integer $k$. The smallest $k$ is the order of $A$, which must exist by definition. What is the largest possible order of such a matrix?

I'm confused.
Are we still talking about 6x6 matrices?

Anyway, if all entries have to be positive integers, there is no solution at all (for matrices of at least 2x2).
Not for the smallest order nor for the largest order.
Even the identity matrix itself does not fit the criteria.

Proof
The image of a unit vector is a vector with positive integer entries, meaning it has length greater than 1 (assuming at least dimension 2).
Therefore repeated application of the matrix will never yield the original unit vector.

Discussion on the 3rd question.

If $A$ is allowed to have complex entries then I believe there are matrices which would have arbitrarily large orders.

No need for complex entries.
Pick a rotation matrix around an arbitrary small angle.
It has an order that can become as large as you want.

 

[caffeinemachine]

No need for complex entries.
Pick a rotation matrix around an arbitrary small angle.
It has an order that can become as large as you want.

Thanks. :)

 

[eddybob123]

Anyone have thoughts on the other questions? :D

 

[I like Serena]

1. What is the smallest degree of a non-constant polynomial $$f(x)$$, such that all roots of $$f(x)$$ are in the set $${0,2,3}$$, and the derivative of $$f(x)$$ is divisible by $$8x^2-24x+7$$?

Spoiler

Since all roots of f(x) are in the set {0,2,3}, f(x) must be of the form:
$$f(x)=nx^k(x-2)^l(x-3)^m \qquad (1)$$
Its derivative is:
$$f'(x)=n\Big(kx^{k-1}(x-2)^l(x-3)^m + x^k l (x-2)^{l-1}(x-3)^m + x^k(x-2)^l m(x-3)^{m-1}\Big)$$
$$f'(x)=n\Big(k(x-2)(x-3) + l x(x-3) + m x(x-2)\Big)x^{k-1}(x-2)^{l-1}(x-3)^{m-1}$$
$$f'(x)=n\Big((k+l+m)x^2 - (5k+3l+2m)x+6k\Big)x^{k-1}(x-2)^{l-1}(x-3)^{m-1}$$

Since none of the roots of $$8x^2-24x+7$$ is either 0, 2, or 3, it follows that
$$n\Big((k+l+m)x^2 - (5k+3l+2m)x+6k\Big)$$
must be divisible by $$8x^2-24x+7$$.
Since the degrees are the same, we might as well set them equal to each other.
Solving the system, where we pick n such that the solution has the smallest integers, yields $k=7, l=27, m=14$ with $n=1/6$.

From (1) we can see that the smallest degree of f(x) is $k+l+m=7+27+14=48$. $\qquad \blacksquare$

 

[anemone]

2. Let x, y, and z be positive real numbers such that

$$xyz=945$$

and

$$x(y+1)+y(z+1)+z(x+1)=385$$

What is the minimum possible value of $$z+\frac{y}{2}+\frac{x}{4}$$?

Enjoy! (Giggle)

I'll tackle the easiest one out of these headache challenging problems... (Giggle)

My solution:

Spoiler

We're asked to find the minimum value of $$z+\frac{y}{2}+\frac{x}{4}$$.(*)

But notice that we're also given the expression $x(y+1)+y(z+1)+z(x+1)=385$, so it's a justifiable move to add one to each of the term in (*) and hence we get:

$$z+\frac{y}{2}+\frac{x}{4}=(z+1)+ \left( \frac{y+1}{2} \right)+\left( \frac{x+1}{4} \right)-1-\frac{1}{2}-\frac{1}{4}=(z+1)+\left( \frac{y+1}{2} \right)+\left( \frac{x+1}{4} \right)-\frac{7}{4}$$

By applying the AM-GM inequality to the following three terms, namely:

$$(z+1), \;\;\left( \frac{y+1}{2} \right), \;\;\left( \frac{x+1}{4} \right)$$, we get

$$(z+1)+\left( \frac{y+1}{2} \right)+\left( \frac{x+1}{4} \right) \ge 3\sqrt[3]{(z+1)\left( \frac{y+1}{2} \right)\left( \frac{x+1}{4} \right)}$$

$$(z+1)+\left( \frac{y+1}{2} \right)+\left( \frac{x+1}{4} \right) \ge \frac{3}{2}\sqrt[3]{(z+1)(y+1)(x+1)}$$

The remaining effort from now on is to cudgel the brain to find for the value for $$\sqrt[3]{(z+1)(y+1)(x+1)}$$, then we're almost done!

I see that from $$xyz=945$$, we can perform the same algebraic manipulation like what we did above, that is to add one and then minus another one in each of the variable $x$, $y$ and $z$ and by doing that, we obtain:

$$xyz=945$$

$$(x-1+1)(y-1+1)(z-1+1)=945$$

$$(x+1)(y+1)(z+1)-((xy+xz+yz)+x+y+z))-1=945$$

but we know that from the given equation $x(y+1)+y(z+1)+z(x+1)=385$, this also implies $xy+xz+yz+x+y+z=385$, thus the equation above becomes

$$(x+1)(y+1)(z+1)-(385)-1=945$$

$$(x+1)(y+1)(z+1)=945+385+1=1331=11^3$$

Hence, this gives

$$(z+1)+\left( \frac{y+1}{2} \right)+\left( \frac{x+1}{4} \right) \ge \frac{3}{2}\sqrt[3]{11^3}$$

$$(z+1)+\left( \frac{y+1}{2} \right)+\left( \frac{x+1}{4} \right) \ge \frac{33}{2}$$

and therefore

$$z+\frac{y}{2}+\frac{x}{4} +\frac{7}{4} \ge \frac{33}{2}$$

$$z+\frac{y}{2}+\frac{x}{4} \ge \frac{59}{4}$$

that is, the minimum possible value of $$z+\frac{y}{2}+\frac{x}{4}$$ is $\dfrac{59}{4}$, if we use the AM-GM inequality to crack this problem.