Integral Inequality

Prove that:

$$\left( \int_\pi^\infty\frac{\cos x}{x}\ dx\right)^{2} < \frac{1}{{\pi}^{2}} $$

Solution

Integrate by parts:

$$\int \frac{\cos x}{x} dx = \frac{\sin x}{x} + \int \frac{\sin x}{x^2} dx$$

So
$$\int_\pi^\infty\frac{\cos x}{x}\ dx = \int_\pi^\infty \frac{\sin x}{x^2} dx$$

And
$$| \int_\pi^\infty\frac{\cos x}{x}\ dx| = |\int_\pi^\infty \frac{\sin x}{x^2} dx|$$
$$\leq \int_\pi^\infty \frac{| \sin x |}{x^2} dx$$
$$\leq \int_\pi^\infty \frac{1}{x^2} dx $$
$$= \frac{1}{\pi}$$

Number of solutions to modular equation

Prove that there are infinitely many prime numbers $p$ such that there are exactly $p^2$ integer triples $(x,y,z)$ such that $0 \leq x,y,z < p$ and $x^2+y^2 - 2010z^3$ is divisible by $p$.

Solution

We will first show that any prime that has form $p = 6k-1$ and does not divide 2010 satisfies the given condition.

First, let $p$ be such prime. Suppose there are $a,b \in \{ 1, \dots, p-1 \}$ such that $a^3 \equiv b^3 \mod p$.

Since $b^{p-1} \equiv 1 \mod p$ then let $c \equiv ab^{p-2} \mod p$, so that we have $bc \equiv a \mod p$.

Then $b^3c^3 \equiv a^3 \equiv b^3 \mod p$ which means $b^3c^3 \equiv b^3 \mod p$ which then means $c^3 \equiv 1 \mod p$ (since $(b,p) = 1$).

Because $c^{p-1} = c^{6k-2} \equiv 1 \mod p$ and $c^{6k-3} = c^{3(2k-1)} \equiv 1 \mod p$, then we must have $c \equiv 1 \mod p$, which means $a \equiv b \mod p$.

In summary, we've shown that $a^3 \equiv b^3 \Rightarrow a \equiv b \mod p$. This means that the set $\{1^3, \dots, p^3 \} \mod p$ is the same as set $\{ 1, \dots, p \} \mod p$. So given any arbitrary $x$ and $y$, we can find exactly one $z$ such that $z^3 \equiv 2010^{-1} (x^2+y^2) \mod p$. Since there are $p^2$ possible pairs for $(x,y)$, then there are also $p^2$ possible triples for $(x,y,z)$.

Now we show that there are infinitely many primes of the form $6k-1$ which would mean that there must be infinitely many primes of the form $6k-1$ that do not divide 2010.

Note that any prime above 3 must have form $6k+1$ or $6k-1$. If there are only finite number of primes of the form $6k-1$, say $p_1, p_2, \dots, p_n$, consider the number $N = 6p_1p_2 \dots p_n - 1$.
For each $i$, $p_i$ does not divide $N$ because otherwise $p_i$ would also have to divide 1, a contradiction. So $N$ is not divisible by any of the $p_i$s, which means all prime factors of $N$ must be of the form $6k+1$. That means, $N \equiv 1 \mod 6$, a contradiction.

a,b,c integers and cubic number

Suppose $a,b,c$ are positive integers such that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is an integer. Prove that $abc$ is a cubic number.

Solution

We have $ab^2 + bc^2 + ca^2 = kabc$ for some k.

Let $d = \gcd(a,b,c)$. We can replace $a,b,c$ by $a/d, b/d, c/d$ respectively and the problem does not change. Thus, without loss of generality, we may assume that $d = 1$.

Let $p$ be a prime that divides $abc$, which means $p$ divides at least one of $a,b,c$. We also know that $p$ cannot divide all three, since $d = 1$.

If $p$ divides exactly one of $a,b,c$, for example $a$, then $ab^2, ca^2, kabc$ are all divisible by $p$, but not $bc^2$. Impossible. Thus, $p$ must divide exactly two of $a,b,c$.

Suppose $p$ divides $a$ and $b$. Furthermore, let $x$ be the largest integer such that $p^x$ divides $a$. Likewise, let $y$ be the largest integer such that $p^y$ divides $b$.

Since $ab^2, bc^2, kabc$ are all divisible by $b$, then so is $ca^2$. Thus $y \leq 2x$.

Since $ab^2, ca^2, kabc$ are all divisible by $a$, then so is $bc^2$, Thus $y \geq x$, which means $x \leq y \leq 2x$.

Now, since $ab^2, ca^2, kabc$ are all divisible by $p^{2x}$, then so is $bc^2$, which means $y \geq 2x$.

Therefore, $y = 2x$, which means that the degree of $p$ in the factorization of $abc$ is $x+y = 3x$.

For each prime $p$ that divides $abc$, it must occur as a cubic number in its prime factorization. Thus $abc$ is a cubic number.

A non-trivial example is $a=1,b=2,c=4$.

Minimum Reciprocal Length

Given a triangle $ABC$, and point $P$ in its interior. Find points $X,Y$ on $AB$ and $AC$ (or their extensions) such that $XY$ passes through $P$ and
$$\frac{1}{PX} + \frac{1}{PY}$$ is maximized.

Circumcircle and Incircle Tangency

Given a triangle $ABC$, its circumcircle $C_O$ and incircle $C_I$. Suppose $X,Y,Z$ are points on $C_O$ such that $XY$ and $XZ$ are tangent to $C_I$, prove that $YZ$ is also tangent to $C_I$.

Solution

Let us denote $R$ to be the radius of $C_O$ and $r$ to be the radius of $C_I$. Also let $O$ and $I$ to be the centers of the circle. Let also Suppose $AB$ is tangent to $C_I$ at $L$. And suppose $AI$ meets $C_O$ at $M$ and $MO$ meets $C_O$ at $N$.

Now suppose $a = \angle MAZ = \angle MAB = \angle MNC = \angle MCB$. Because $\angle ALI = \angle MCN = 90^o$ then by similarity of $\triangle ALI$ and $\triangle MNC$ we have: $$AI.MC = MN.LI = 2Rr$$

Now $\angle MIC = \angle IAC + \angle ACI = x + \angle ACI = \angle MCB + \angle ICB = \angle MCI$ so that we know $MC = MI$. Thus: $$AI.MI = 2Rr$$

Now, we draw similar figures to the triangle $X,Y,Z$. That is: $XY$ is tangent to $C_I$ at $L'$. And suppose $XI$ meets $C_O$ at $M'$ and $M'O$ meets $C_O$ at $N'$.

In our analysis above, we showed that $AI.MC = 2Rr$ but we didn't use the fact that $BC$ is tangent to $C_I$, so we can repeat the argument to claim that $$XI.M'Z = 2Rr$$.

On the other hand, $XI.IM' = AI.IM = 2Rr$ so we have $IM' = M'Z$ which means $IM'Z$ is an isosceles. $$\angle M'IZ = \angle M'ZI$$ $$\angle M'XZ + \angle IZX = \angle M'ZY + \angle YZI$$ Because $XY$ and $XZ$ are both tangent to $C_I$ then $XM$ is a bisector of $\angle YXZ$ so that $\angle M'XZ = \angle M'XY = \angle M'ZY$. So our equation above becomes: $$\angle IZX = \angle YZI$$ which means $ZI$ is also internal angle bisector, which means $YZ$ is tangent to $C_I$.

Three independent random variables

3 independent random variables $A,B,C$ are drawn from the same distribution. Determine the correlation between $A-B$ and $B-C$.

Polynomial and divisibility

A sequence of polynomials $P_i(x)$ are defined as follows:
$P_1(x) = 1$
$P_2(x) = 1$
$P_{n+2}(x) = (x+2)P_{n+1}(x) - P_n(x), n=1,2,\dots$

Prove that for all $n > 1$, $P_n(x)^2 + x$ is divisible by $P_{n-1}(x)$