IMO shortlist, 1998

G7. $ABC$ is a triangle with $\angle ACB = 2 \angle ABC$. $D$ is a point on the side $BC$ such that $DC = 2 BD$. $E$ is a point on the line $AD$ such that $D$ is the midpoint of $AE$. Show that $\angle ECB + 180 = 2 ∠EBC$.

 

Solution.  Put $D$ in the center. We can rewrite everything in terms of $a$ - the number corresponding to $A$ and $b$ - the number corresponding to $B$. Thus $C$ is $-2b$, E is $-a$.
The relation between the angles at $B$ and $C$ is easily reduced to the following equation:
$$\frac{(a-b)^2}{|a-b|^2}\frac{a+2b}{|a+2b|}=\frac{b^3}{|b|^3}.$$
Squaring and representing modules with conjugates, we obtain
$$\frac{(a-b)^2}{(\bar a-\bar b)^2}\frac{a+2b}{\bar a+2\bar b}=\frac{b^3}{\bar b^3}.$$
Clearing the denominator and setting everything to the left we obtain
$$a^3\bar b^3-3a\bar b^3 b^3-\bar a^3b^3+3\bar a\bar b^2 b^3=0\ (*)$$
Similarly, the equality we want to proof is equivalent to
$$\left(\frac{\frac{-b}{|-b|}}{\frac{-a-b}{|-a-b|}}\right)^2=(-1)
\left(\frac{\frac{-a+2b}{|-a+2b|}}{\frac{b}{|b|}}\right)$$
Both of the arguments of the complex numbers on both sides are between 180 and 360, so squaring is equivalent transformation here. As above we obtain
$$\frac{(a+b)^2}{(\bar a+\bar b)^2}\frac{a-2b}{\bar a-2\bar b}=\frac{b^3}{\bar b^3}.$$
Everything on the left and expand - we obtain
$$-a^3\bar b^3+3a\bar b^3 b^3+\bar a^3b^3-3\bar a\bar b^2 b^3=0.$$
But this is just the same as $(*)$.

A problem told by Zhivko Petrov

 The following integral was proposed for homework to Applied Math, by Zhivko Petrov. 

$\textbf{Problem}.$ Evaluate

$$\int_{0}^1\frac{\ln(1-x+x^2)}{x^2-x}\text{d} x.$$

Later I communicated the problem to prof. Gadjev and he proposed a neat solution (to be presented later).

Now we elaborate a solution, on an idea proposed by prof. Babev. I would like to thank the afformentioned people, as well as David Petrov for pointing me to that idea.

$\textbf{Solution}.$ Introduce $$I(y)=\int_0^1\frac{\ln(1-y(x-x^2))}{x^2-x}\text{d}x.$$

We need to find $I(1)$. Clearly $I(0)=0$. Using differentiation under the integral sign we obtain 

$$I'(y)=\int_0^1\frac{1}{1-y(x-x^2)}\text{d}x.$$

Assuming that $y\in [0,1]$, one can easily integrate the last to obtain 

$$I'(y)=\frac{4 \arcsin\left(\frac{\sqrt{y}}{2}\right)}{\sqrt{y(4-y) }}.$$

The  latter is very easy to integrate (for example making the change $y=4t^2$) in order to obtain

$$I'(y)=\left(4 \arcsin\left(\frac{\sqrt{y}}{2}\right)^2\right)'.$$

Thus 

$$I(1)=\int_0^1I'(y)\text{d}y+I(0)= 4 \arcsin\left(\frac{\sqrt{1}}{2}\right)^2-4 \arcsin\left(\frac{\sqrt{0}}{2}\right)^2+0=4 \arcsin\left(\frac{1}{2}\right)^2=\frac{\pi^2}{9}.$$

A problem proposed by prof. Gadjev

Evaluate 

$$\int_0^{\infty}\frac{\ln x}{x^2+2x+5}\, \text{d}x$$

$\textbf{Solution.}$

The indefinite integral is not expressible in elementary functions. Denote the integral by $I$. First make the change of variables $\displaystyle x\to \frac{1}{x}$ to obtain 

$$I=-\int_0^{\infty}\frac{\ln x}{5x^2+2x+1}\, \text{d}x\ \ $$

In the original integral make the change of variables $x\to 5x$ to obtain 

$$ I=\int_0^{\infty}\frac{\ln (5x)}{25x^2+10x+5}5\, \text{d}x=\int_0^{\infty}\frac{\ln x}{5x^2+2x+1}\, \text{d}x+\int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x=$$

$$-I+\int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x$$

whence 

$$I=\frac{1}{2} \int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x=\frac{1}{8}\arctan(2)\ln(5)$$

Geometric-analytic problem

Let $f:\mathbb{R}^2\to\mathbb{R}$ be strictly positive Lipschitz function with constant $1/2$. Let $A$ be a nonempty subset of $\mathbb{R}^2$, such that if $x\in A$ and $y\in\mathbb{R}^2$ with $\|x-y\|=f(x)$, then $y\in A$. Prove that $A=\mathbb{R}^2$.

The problem was proposed to Bulgarian TST, 2009. Do you know earlier source or some context of the problem?

A problem proposed by prof. Babev

Consider the sequence
$$a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t.$$
It could be proven that
$$\lim_{n\to \infty}n a_n=\frac{1}{2}$$
 Now we are going to prove that
$$\lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} $$
and show how to derive all such limits.
$\textbf{Proof.}$ Since $\displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2}$ (for any $n\in \mathbb{N}$) we can rewrite the limit in question as
$$\lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right)$$
$$=\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right)$$
Changing the variables $t\to n+s$ and using periodicity of sine, the last simplifies to
$$ \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s$$
Since for $s\in[0,1]$, $\displaystyle \frac{n}{n+s}$ is bounded between $1$ and $ \displaystyle \frac{n}{n+1}$, the last limit is bounded between the limits
$$ \lim_{n\to \infty}\int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s\ \ \mbox{and}\ \ \lim_{n\to \infty}\frac{n}{n+1}\int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s$$
which both are equal to $\displaystyle  \int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s=-\frac{1}{4}$, hence the result follows.

$\textbf{Comment.}$ In a similar vein it could be proven that
$$\lim_{n\to\infty}n\left(n\left(na_n-\frac{1}{2}\right)+\frac{1}{4}\right)=\int_0^1\sin^2(\pi s)s^2\text{d}s$$
and we could extend this further.
This could be viewed as a "Taylor series of the sequence at infinity", namely
$$a_n=\sum_{j=1}^\infty b_j\left(\frac{1}{n}\right)^j$$
where $$b_j= \int_{0}^{1}\sin^2(\pi s)(-s)^{j-1}\text{d}s$$
Having this perspective leads to general and more insightful solution. Consider the series
$$\sum_{j=1}^\infty b^j \left(\frac{1}{n}\right)^j=\sum_{j=1}^\infty\int_{0}^{1}\sin^2(\pi s)(-s)^{j-1}\left(\frac{1}{n}\right)^j\text{d}s=\int_{0}^{1}\sin^2(\pi s)\frac{1}{n+s}\text{d}s.$$
(where we have exchanged summation and integration and then calculated the sum of the geometric progression)
After changing the variables $s\to t-n$ we arrive at the integral
$$\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t$$
which is exactly $a_n$.

Basel problem type sum, proposed by prof. Skordev

Evaluate $$\sum_{n=1}^{\infty}\frac{1}{2^n n^2}.$$

Proof. The function
$$S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2}$$
is well defined for $x\in[-1,1]$ and thus we need to find $S(1/2)$. Using differentiation, we observe that
$$S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t.$$
If we make change of variables in the latter integral $t\to 1-t$ we obtain
$$S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1)$$
On the other hand, using integration by parts, we obtain
$$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t.$$
Notice that $\displaystyle \lim_{t\to 0}\log t \log(1-t)=0$ ($\log (1-t)\sim t$ and then L'Hopital), hence
$$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2)$$
Plugging $x=1/2$ and equating $(1)$ to $(2)$ we obtain
$$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2.$$
On the other hand  $$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t+\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t=\int_{0}^1\frac{\log(1-t)}{t}\text{d} t=-\frac{\pi^2}{6}$$
Substracting the last two equalities we obtain  $$\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t=-\frac{\pi^2}{12}-\frac{1}{2}\log\left(\frac{1}{2}\right)^2$$
and thus $\displaystyle S\left(\frac{1}{2}\right)=\frac{\pi^2}{12}+\frac{1}{2}\log\left(\frac{1}{2}\right)^2$.

Some math jokes (which turn out to be correct)

Prove that $$\int_{0}^{\infty}\frac{1}{(x^2+1)(x^{\pi}+1)}\text{d}x=\int_{0}^{\infty}\frac{1}{(x^2+1)(x^{e}+1)}\text{d}x$$
Prove that $$\int_{0}^{\pi/2}\frac{1}{1+\tan^{\pi}(x)}\text{d}x=\int_{0}^{\pi/2}\frac{1}{1+\tan^{e}(x)}\text{d}x$$

$\textbf{Proof.}$ Both of the integrals are being solved, when the peculiar powers are being replaced by a parameter "a" and then differentiation with respect to "a"  is performed. Actually the integrals are related vie the change of variables $x\to\arctan(x)$.
For the first integral, denote
$$f(x,a)= \frac{1}{(x^2+1)(x^{a}+1)}.$$
Then $$\frac{\partial }{\partial a}f(x,a)=\frac{x^a \log (x)}{\left(x^a+1\right) \left(x^a+1\right)^2}$$
Consider the integral
$$\int_{0}^{1}\frac{\partial }{\partial a}f(x,a)\text{d}x.$$ Making the change $x\to 1/x$ this transforms to the integral
$$\int_{\infty}^{1}\frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{-1}{x^2}\text{d}x=\int_{1}^{\infty}\frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{1}{x^2}\text{d}x$$
Now it remains to observe that
$$ \frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{1}{x^2}=-\frac{\partial }{\partial a}f(x,a)$$
so that
$$\int_{0}^{1}\frac{\partial }{\partial a}f(x,a)\text{d}x=-\int_{1}^{\infty}\frac{\partial }{\partial a}f(x,a)\text{d}x$$
and hence
$$\int_{0}^{\infty}\frac{\partial }{\partial a}f(x,a)\text{d}x=0.$$
Thus  $$\int_{0}^{\infty}f(x,a)\text{d}x$$
is a constant independent of $a$, which proves the equality. Plugging $a=0$ we see moreover that the value of the integrals is $\displaystyle \frac{\pi}{4}$.

$\textbf{Addendum}$. Another solution proceeds as directly making the change of variable $x\to 1/x$. Thus

$$\int_0^\infty\frac{1}{(x^2+1)(x^{a}+1)}\text{d}x=\int_0^\infty\frac{x^a}{(x^2+1)(x^{a}+1)}\text{d}x.$$
Summing both integrals yields the solution.

The second problem could be done in a similar manner.