# Tag Info

### Knowing $x^2-x-1$ is a factor of $p(x)=ax^5 + bx^4 + 1$ , find a,b.

A simple long division method yields $p(x)=ax^5+bx^4+1=(x^2-x-1)((ax^3)+(a+b)x^2+(2a+b)x+(3a+2b))+\ (5a+3b)x+(3a+2b+1)$ as we know $(x^2-x-1)$ is factor of $p(x)$, The remainder $(5a+3b)x+(3a+2b+1)$ ...
Accepted

### Knowing $x^2-x-1$ is a factor of $p(x)=ax^5 + bx^4 + 1$ , find a,b.

Your method is on-target. You just need to reduce the quintic until both the equations are of same degree to compare coefficients. I will show an example with highest degree of $2$ (reduction to ...
1 vote

Accepted

### If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$

Some thoughts. By Holder inequality, we have \begin{align*} &\left(\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a^2 + 4bc}} \right)^2\cdot \sum_{\mathrm{cyc}} (a^2 + 4bc)(4b + 4c - bc + 4ab + 4ac)^3 \\ \...
Accepted

### IMO 1987/P1 - Combinatoric approach

Your claim that $p_n(k)=\binom nk(n-k-1)!$ is not correct. In the case where $k=0$, your formula would imply that there are $(n-1)!$ permutations with zero fixed points, so $(n-1)!$ derangements of $n$...
1 vote
Accepted

1 vote

### Diagonals in an inscribed quadrilateral can simultaneously be angle bisectors of triangles involving their midpoints

Let $\angle CDE=\alpha, ~\angle CDM=\beta$ and $\angle MDB=\gamma$. Simple angle chasing (using $MD=ME$ and $AC||DE$) leads to $\angle ADB=\beta$. Hence, $DB$ is the $D$-symmedian of $\triangle ADC$. ...

### Show that for all $n$ there exist some $n$-digit number with no $0$ in it whose digit sum divides it.

It's a classical induction problem to show that Lemma. Given $m\ge 1$, there is an $m-$digit multiple of $2^m$ whose digits are all $1$ or $2$. Let $2^{m-1}\le n<2^m$ and $x$ be the $(m+1)-$digit ...
Accepted

### how to solve $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{x^2}{\sin(x)} \ln\left(2^{\sin^3(x)}+ 5^{\cos^3(x)} \right)dx$?

I am a bit skeptical about a possible closed form for $$I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{x^2}{\sin(x)} \log\left(2^{\sin^3(x)}+ 5^{\cos^3(x)} \right)\,dx$$ It can write (I suppose that ...
Accepted

### Closed form of $\prod_{n=0}^{\infty}\frac1{1+x^{2^n}}$

This is a telescoping product. $$(1-x^{2^n})(1+x^{2^n}) = 1 - x^{2^{n+1}}.$$ Therefore, $$(1-x) \prod_{n=0}^N (1+x^{2^n}) = 1 - x^{2^{N+1}}.$$ The rest of the details I leave to you as an exercise.
1 vote

### Bases of three $n$-dimensional subspaces of $\mathbb R^{2n}$ which have pairwise trivial intersection.

Pushing through OP's approach OP starts with a (any) basis $\{ c_i \}$ of $C$. OP shows that there is a unique $a_i \in A, b_i \in B$ such that $a_i + b_i = c_i$. As pointed out in the comments, we ...

### Given that $a,b,c>0$ and $abc=1$, prove that $a+b+c+\frac{3}{ab+bc+ca} \geq 4$

Elaborating on my hint, show that $(a+b+c)^2 \geq 3(ab+bc+ca)$ $a+b+c \geq 3$ let $t = a+b+c$, then $LHS \geq t + \frac{ 9}{t^2}$. Finally, show that for $t \geq 3$, $t + \frac{9}{t^2 } \geq 4$. ...

### Why does the following condition holds in the geometric optimisation problem?

$n_1+n_2>9$ would mean that the circles around $A_1,A_2$ contain at least 3 common points -- they cannot possibly host 10 distinct points because the total number of points is 7. But if 3 distinct ...
1 vote
Accepted

### BMO2 2011/12 question on cyclic quadrilateral and showing two circumcircles have same radius

According to the diagram below, you can follow these steps: Step $1$: $\triangle EBC$ and $\triangle EAD$ are similar. So, $\angle QEC= \angle SED.$ Step $2$: Similarly, you should be able to show ...
Accepted

### Minimizing $\sum_{cyc}\frac{\sqrt{5a+8bc}}{8a+5bc}$ with $\sum_{cyc}ab=1$
Some thoughts. Let $$x := 5a + 8bc, \quad y := 5b + 8ca, \quad z := 5c + 8ab,$$ $$u := 8a + 5bc, \quad v := 8b + 5ca, \quad w := 8c + 5ab.$$ Let A := \frac{2\sqrt{2} - \sqrt{5}}{3}, \quad B := \frac{...