48
votes
Accepted
An interesting trigonometric integral
\begin{align*}
I&= \int_0^\pi \cos^2\cos x+\sin^2\sin x \, dx\\
&= 2\int_0^{\pi/2} \cos^2\cos x+\sin^2\sin x \, dx\\
&= 2\int_0^{\pi/2} \cos^2\sin x+\sin^2\cos x \, dx \quad(x\mapsto\pi/2-...
- 2,729
36
votes
Accepted
What is the least number of concerts needed to be scheduled in order that each musician may listen, as part of the audience, to every other musician?
There are a total of $6\cdot5=30$ (listener, performer)-pairs.
If at a concert we have $n$ performers, then in that concert there are $(6-n)n$ such (listener, performer)-pairs. The maximum value of $(...
- 131k
26
votes
Accepted
Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .
Write
$$x^4-6x+6=(x^2-1)^2+2\left(x-\frac32\right)^2+\frac12$$
and then it is clear that $x^4-6x+6>0$ for all $x\in\mathbb{R}$.
- 10.9k
21
votes
Accepted
Symmetry in Probability (AMC 12A 2023)
The idea in Solution 1, although not explicitly stated, is that if the frog has not reached or passed $10$, irrespective of how much further Flora needs to go, the probability that the next jump ...
- 129k
20
votes
Accepted
Integral from MIT Integration Bee 2023 Finals - $\int_{-1/2}^{1/2} \sqrt{x^2+1+\sqrt{x^4+x^2+1}}\,\textrm{d}x$
They pulled off a sneaky move.
Given a radical having the form
$\sqrt{a+\sqrt{b}},$
we may render
$\sqrt{a+\sqrt{b}}=\sqrt{u}+\sqrt{v}$
$\sqrt{a-\sqrt{b}}=\sqrt{u}-\sqrt{v}$
Multiplying these together ...
- 37.5k
20
votes
How can I get faster at doing math?
In my view, paradoxically you have to be slower in order to be faster. To be fast at maths you have to be less focused on getting the job in front of you right now done, and more focused during the ...
- 9,115
19
votes
Accepted
Math competition question about ways to spell BANANA in a square
$\stackrel{\text{B}}{\begin{array}{|c|c|c|c|c|}\hline
1 & \ &\ \\ \hline
& & \\\hline
& & \\\hline
\end{array}}$
$\to$
$\stackrel{\text{BA}}{\begin{array}{|c|c|c|c|c|}\hline
\ ...
- 13.3k
19
votes
Accepted
Any different way to solve an integral involving a trigonometric ratio?
Decompose the integrand as follows
\begin{align}
& \int_0^{2\pi}\frac{\sin x+1}{\cos x +2}\ dx\\
=& \int_ 0^{2\pi}\left( \frac{\sin x}{\cos x +2}- \frac{\cos x+2-\sqrt3}{\sqrt3(\cos x +2)}+\...
- 92k
17
votes
Evaluate $\lim_{n \to \infty} n \prod_{m = 1} ^ n \left(1 - \frac1m + \frac5{4m ^ 2}\right)$
We have the Weierstrass product for $\cos$:
$$\cos(\pi z) = \prod_{m=1}^\infty\left(1-\frac{4z^2}{(2m-1)^2}\right)$$ valid for all $z\in \Bbb C$.
Setting $z=i$ we get $$\cosh(\pi) = \prod_{m=1}^\infty\...
- 17.1k
17
votes
An interesting trigonometric integral
A slightly different approach :
\begin{align*}
I&= \int_0^\pi \cos^2\cos x+\sin^2\sin x \, dx\\
& = \int_0^\pi \dfrac{1+\cos(2\cos(x))}{2} + \dfrac{1-\cos(2\sin(x))}{2} dx \\
& = \pi + \...
- 29.3k
16
votes
Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .
Still another way :
If $x≤0$, then $x^4-6x≥0\thinspace .$ Therefore, $x>0\thinspace .$ Thus, using the AM-GM inequality you have :
$$x^3+\frac 2x+\frac 2x+\frac 2x=6≥4\sqrt [4]{8}$$
A contradiction ...
- 14.6k
15
votes
Accepted
For what $n$ can we find a degree $\leq n-2$ polynomial such that $P(i) \in \{0 , 1 \}$ for $i \in [n]$, but not all identical.
The conjecture that this is possible for all composite $n$ is false, the smallest composite $n$ that fails this condition is $n=65$.
This is done by computer search. I've used the equivalent form by ...
- 16k
14
votes
Accepted
Prove that $\prod_{1\leq i,j\leq n}\frac{1+a_ia_j}{1-a_ia_j}\geq1$ for $n$ real numbers $a_i\in(-1,1)$
Found here on AoPS:
For $-1 < x < 1$ we have the Taylor series
$$
\ln (1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} x^k \, ,
$$
which implies
$$
\ln(1+x) - \ln(1-x) = 2\sum_{k=1}^\infty \frac{...
- 108k
14
votes
Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .
We can use for example
$$x^4-6x+6 =(x^2-1)^2+2x^2-6x+5 >0$$
indeed $(x^2-1)^2\ge 0$ and
$$36-4\cdot 2\cdot 5 =-4<0$$
- 152k
14
votes
Accepted
Australian Mathematics Competition Year 7-8 Question 21
The answer should be 2. Place a "1" in the center, then a "2" in one of the corners. The black numbers are given while the red numbers are deduced.
It is also easy to see that ...
- 1,469
14
votes
An interesting trigonometric integral
Let's take a completely different approach. Set
$$f(x)=\cos^2\cos x+\sin^2\sin x-1.$$
We will show that $f$ integrates to $0$ over $[0,\pi]$. Observe that
\begin{align*}
f\left(x+\frac{\pi}{2}\right)
&...
- 8,703
14
votes
What is the least number of concerts needed to be scheduled in order that each musician may listen, as part of the audience, to every other musician?
Given a schedule with $n$ concerts, define $S_1,\dots,S_6$ to be subsets of $\{1,\dots,n\}$, where $S_i$ is corresponds to the set of concerts that musician number $i$ played in. For example, if $n=6$,...
- 72.1k
14
votes
Accepted
Show that $x<\dfrac{\sqrt [8]{128}}{2}$
collect $u = x^2 + 1 $ as $ \; \; \; x^{15} u^2 - u + x = 0$ so
$$ x^2 + 1 = \frac{1 \pm \sqrt{1 - 4 x^{16} \;}}{2 x^{15} } $$
and you cannot have $4 x^{16} > 1.$
Also $4 x^{16} = 1$ ...
- 138k
13
votes
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $f(3x)-f(x)=x.$ If $f(8)=7$, then $f(14)$ is equal to
$$f(x)-f\big(\frac{x}{3}\big)=\frac{x}{3}.$$
Similarly, $$f\big(\frac{x}{3^n}\big)-f\big(\frac{x}{3^{n+1}}\big)=\frac{x}{3^{n+1}}.$$
Let us add this up over all $n$.
This gives $f(x)=f(0)+\frac{x}{2}$ ...
- 2,777
13
votes
Accepted
Number of parallelograms in a hexagon of equilateral triangles
You can convince yourself that:
The two points at the acute angles of a parallelogram uniquely determine that parallelogram;
Any two points that don't lie on a common line in the grid can be the two ...
- 137k
13
votes
Accepted
How many $3$-element subsets of $\{1,2,3,...,19,20\}$ have product divisible by $4$?
We use complementary counting. For a set to have the product of its elements not divisible by $4$, there are two cases:
All elements are odd. There are $10$ odd numbers, so there are $\binom{10}{3}$ ...
- 2,769
12
votes
Accepted
Prove that $A^3\equiv I\mod p$.
Here is how I would approach the problem.
The form of $\binom{i + j - 2}{i - 1}$ looks really suspicious. It reminds me of the Vandermonde formula
$$\binom{i + j - 2}{i - 1} = \sum_{k = 1}^{p} \binom{...
- 7,135
12
votes
Accepted
Proving $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2$ If $a,b,c,d \in \mathbb{N}$
$$E>\frac a{a+b+c+d}+\frac b{a+b+c+d}+\frac c{a+b+c+d}+\frac d{a+b+c+d}=1
$$
Similarly, $F=\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}>1$.
Since $E+F=(\frac a{a+b}+\frac b{a+b})+(\...
- 13.3k
12
votes
Accepted
Nigerian Math Olympiad Question
Observations towards a solution.
Prove the following. If you're stuck, state what you've tried.
The total number of blocks in the structure is $n$.
Every layer must be built one at a time (since we ...
- 66.7k
12
votes
Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .
Hint: $\,P(x+1) = x^4 + 4 x^3 + 6 x^2 - 2 x + 1 = x^2(x+2)^2 + x^2 + (x-1)^2\,$.
- 76.2k
12
votes
Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .
This question attracted a huge echo, so let us write down a further decomposition of $f$ into a sum of squares:
$$
x^4-6x+6=\left(x^2 -\frac 32\right)^2 + 3(x-1)^2 + \frac 34\ge \frac 34=0.75\ .
$$
$\...
- 31.5k
11
votes
Accepted
Maximum number of negative coefficients $p(x)^2$ can have
Note: I assume that we are restricting ourselves to polynomials with real coefficients.
We claim that the answer is $2n - 2$, which can be achieved with the polynomial $$p^*(x) = nx^n - x^{n-1} - x^{n-...
- 2,624
11
votes
Show that $\forall (a, b, c) \ \in \mathbf{R}: \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} \ge \frac{b}{a} + \frac{c}{b} + \frac{a}{c}$
Here's one way to solve the problem while avoiding algebra. The key idea is that $\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} = 1$. Thus, we will substitute $x = \frac{a}{b}, y = \frac{b}{c},$ and ...
- 1,292
10
votes
Accepted
Can we arrange $\{1,...,16\}$ in $4\times 4$-grid so {products of rows} = {products of columns}?
RobPratt has answered the question for all $n\le 13$. I shall prove in this answer that when $n\ge 11$, there is no solution.
Let $\pi$ be the prime counting function. I claim is impossible to form a ...
- 72.1k
Only top scored, non community-wiki answers of a minimum length are eligible