Wednesday, November 30th

Problem 
Find all naturals $a, b$ such that $\frac{a+1}{b}$ and $\frac{b+1}{a}$ are naturals.

Solution
The only possibilities are $a=b$ and $b=a+1$ (or $a=b+1$). Thus the only solutions are $(1,1)$, $(1,2)$ and $(2,3)$.

Tuesday, November 29th

Problem
Find all the possible areas on an hexagon with equal angles and whose sides are 1,2,3,4,5, and 6 in some order.

Solution
Up to cyclic permutations and reflexions, the only possibilities for the sides are 1,6,2,4,3,5 and 1,6,3,2,5,4. Thus the possible values of areas are $\frac{67\sqrt{3}}{4}$ and $\frac{65\sqrt{3}}{4}$.

Monday, November 28th

Problem
Let $E$ be an ellipse and $E'$ be its reflection along one of its tangents. Find the locus of the foci of $E'$ as the tangent line to $E$ varies.

Solution
Let $A$ and $B$ be the foci of $E$. Then by considering the reflexions of $A$ and $B$ over the tangent lines to $E$ one can see that the locus of $A$ is homothetic to the original ellipse with center at $B$ and a factor of 2. Likewise the locus of $B$ is homothetic to $E$ with center at $A$ and a factor of 2.

Tuesday, November 22nd

Problem
Let ABCD be a square and F be a point on BC; draw the perpendicular to DF through B and let it cut DC at Q. Find the angle FQC.

Solution
Let P be intersection BQ with DF. Since the quadrilaterals CQPF and CDBQ are both cyclic, we have that the angles FQC, FPC and DBC are equal we have that FQC is 45.

Monday, November 21st

Problem
Let $f(x), g(x)$ be two continuous real functions such that $\int_0^1f^2(x)dx=\int_0^1 g^2(x)dx=1$. Prove that there is a real number $c$ such that $f(c)+g(c)\le 2$.

Solution
Suppose that $f(c)+g(c) > 2$ for all $c\in[0,1]$, then $\int_0^1 f(x)g(x)dx>1$, but by Cauchy-Schwarz $\int_0^1 f(x)g(x)dx\le1$ which gives a contradiction, hence such $c$ exists.

Friday, November 18th

Problem
Let $p(x)$ be a polynomial with real coefficients such that $p(x)=x$ doesn't have real roots. Prove that $p(p(x))=x$ doesn't have real roots either.

Solution
Without loss of generality suppose that $p(x)>x$ for all $x$. Then $p(p(x))>p(x)>x$ and thus $p(p(x))=x$ does not have any real roots.

Thursday, November 17th

Problem
Find all the functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x )-y) f(x+f(y ))=x^2-y^2$ for all $x,y \in \mathbb{R}$.

Solution 
Let $c=f(0)$, then $f(c)=0$. Also, $f(c-y)f(f(y))=-y^2$, and with $y=c$   $f(f(c))=-c=f(0)=c$, thus $c=0$. Let $y=f(x)$, then $x^2-f(x)^2=(x-f(x))(x+f(x))=0$ for all $x$. Since $f(x)=-x$ does not satisfy the equation, the only solution is $f(x)=x$.

Wednesday, November 16th

Problem
Let $I_m=\int_0^{2\pi}\cos(x )\cos(2s)\dots\cos(mx)dx$. For which integers &1\le m\le 10$ is $I_m\neq 0$?

Solution
Since $\cos(kx)$ is a $k$ degree even or odd polynomial in $\cos(x)$ whether $k$ is even or odd respectively. Since the integral of an odd power of $\cos(x)$ is zero and the integral of an even power is nonzero, we need the integral in $I_m$ to be an even polynomial. Therefore the values that satisfy $I_m\neq0$ are $m=3,4,7,8$.

Tuesday, November 15th

Problem
Find the largest positive integer $b$ such that there exists an integer a that satisfies $3\cdot2^a+1=b^2$. 


Solution
We have that $3\cdot 2^a=b^2-1=(b-1)(b+1)$, thus one of the factors $b\pm1$ has to be a power of 2. After considering the two options, we have the possible values for b to be 5 and 7. Thus $b=7$ is the largest possible value for $b$.

Monday, November 14th

Problem
Let A and B be different nxn matrices with real entries. If $A^3=B^3$ and $A^2B=AB^2$, can $A^2+B^2$ be invertible? 


Solution
Since $(A-B)(A^2+B^2)=A^3+AB^2-BA^2-B^3=A^3-B^3=0$ and $A-B\neq 0$, we have that $A^2+B^2=0$ and hence, it is not invertible. 

Friday, November 11th

Problem
Find the smallest natural with all its digits equal to 4 that is a multiple of 169.




Solution
A number whose only digits are 4's can be written as $4\frac{10^n-1}{9}$ for some $n$. Then we need to find the smallest $n$ such $13^2|10^n-1$, then by Lagrange's Theorem $n=\phi(13)$ or $n=\phi(13^2)$. Since $n=12$ doesn't work, we have that the smallest $n$ that works is $n=156$ and hence the number we are looking for is

$4\frac{10^{156}-1}{9}$.

Thursday, November 10th

Problem
A number is said decreasing if its digits are non-increasing from left to right. Are there integers n such that 16^n is decreasing? 


Solution 
Since the sum of the digits of $16^n$ is $6n+1$ and $16^n$ has $\lfloor n\log(16)\rfloor$ digits, in order to have a decreasing power of 16, its digit sum has to be at least $6\lfloor n\log(16)\rfloor \ge 6.2n$, hence it is not possible to have such number.

Wednesday, November 9th

Problem
Let $x_1, x_2, \dots, x_n$ be positive reals and let $S=x_1+x_2+...+x_n.$ Prove that 

$(1+x_1)(1+x_2)...(1+x_n)\leq 1+\frac{S}{1!}+\frac{S^2}{2!}+...+\frac{S^n}{n!}$.

Solution
By the McLaurin inequalities we have that

$S_1^{j}\geq S_j$

where $S_i$ is the $i$th elementary symmetric average. Thus 

$S^j\frac{\binom{n}{j}}{n^j}\geq S_j\binom{n}{j}$

and is it easily seen that 

$\frac{1}{j!}\geq \frac{\binom{n}{j}}{n^j}$

therefore

$\frac{S^j}{j!}\geq S_j\binom{n}{j}=\sigma_j$

where $\sigma_j$ is the $j$th elementary symmetric function. Thus expanding the LHS of the inequality gives the result.

Tuesday, November 8th

Problem 
Inscribe a circle of radius $r$ in a 30,60,90 triangle. Find the distances from the center of the circle to the vertices as a function of $r$.


Solution
Let $ABC$ be the triangle where $A, B, C$ are the vertices with angles of 30, 90 and 60 degrees respectively. Let $O$ be the center of the circle. Then $OA=r/\sin (15)$, $OB=r/\sin(45)$ and $OC=r/\sin(30)$. It is well known the values of the two later ones, so just remains to calculate the value of $\sin(15)$.


$\sin(15)=\sqrt{\frac{1-\cos(30)}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}$.

Monday, November 7th

Problem
Given any five points on a sphere, show that some four of them must lie on a closed hemisphere.


Solution
Take two different points and construct the great circle that passes throughout them. By the pigeon-hole principle there is a hemisphere (closed) that contains 2 of the 3 remaining points. Hence this hemisphere will contain 4 of the 5 initial points. 

Friday, November 4th

Problem 
Consider $p=1010..101$ where there are $n$ 1's and $(n-1)$ 0's. Find all possible $n$ such that $p$ is prime.


Solution 

Notice that $p=\frac{10^{2n}-1}{99}$. Thus $p=\frac{1}{9\times 11}(10^n-1)(10^n+1)$. For $n>2$ and $n$ even, $99| 10^n-1$ and $10^n-1>99$, hence $p$ cannot be prime. For $n$ odd, $9|10^n-1$ and $11|10^n+1$ and $10^n-1>9$ and $10^n+1>11$. Hence the only solution is $n=2$ and $p=101$.

Thursday, November 3rd

Problem
Can an arc of a parabola inside a circle of radius 1 have a length greater than 4? 




Solution
No. Without loss of generality let $y=ax^2$ be the parabola. After a bit, one can be convinced that in order to maximize the arc length inside the circle, it has to have its center at $(0,1)$. 


By finding the arc length as a function of $a$ and applying the fundamental theorem of calculus, one can see that the arc length of the parabola inside the circle is an increasing function of $a$. Since for $a=0$ the arc length is 0 and for $a=\infty$ the arc length is 4, there is no such solution for the problem.

Wednesday, November 2nd

Problem 
Find all possible naturals $x,y,z$ such that $xyz=4104$ and $x+y+z=77$


Solution 
First we notice that $4104=2^3 3^3 19$. There are a total of $\binom{5}{2}\binom{5}{2}\binom{3}{2}$ such triples that satisfy the first condition. For the second condition, we notice that one of the numbers $x,y,z$ has to be a multiple of 19. Going through the possibilities (19, 38, 57, etc) we can easily find that the only triples that work are $(4,54,19)$ and $(3,36,38)$ and their permutations. 

Tuesday, Novermber 1st

Problem
Find the minimum $k >2$, such that there are $k$ consecutive numbers such that their sum of squares is a square.


Solution
Let $k=2m+1$ and let the $k$ numbers be $(n-m),(n-(m-1)), \dots, n, \dots, (n+m)$. Then after adding the squares, we have

$(2m+1)\left(n^2+\frac{m(m+1)}{3}\right)$

has to be a square, which working case by case, has the smallest solution for $m=5$ and $n=23$ (or $n=1$ for any integers).

Letting $k=2m$, then adding the squares gives

$m\left(2n^2+2n+\frac{2m^2+1}{3}\right)$

which analyzing the first cases gives no solution for $m<6$ and hence $k=11$ is the minimum.

Monday, October 31st

Problem


A strip of width w is the set of all points which lie on, or between, two parallel lines distance w apart. Let $S$ be a set of $n \geq 3$ points on the plane such that any three different points of $S$ can be covered by a strip of width 1. Prove that S can be covered by a strip of width 2. 


Solution  

Since we have a finite set, there is a pair of points in $S$ which are the furthest apart from each other. Then every point in $S$ is in a strip  of width 1 with one of its sides being the line passing through the maximal pair. Hence $S$ lies inside the union of two such possible strips.