Selected Problems from Iberoamerican Mathematics Olympiad, 2020

Problem 1

Let $ABC$ be an acute scalene triangle such that $AB <AC$. The midpoints of sides $AB$ and $AC$ are $M$ and $N$, respectively. Let $P$ and $Q$ be points on the line $MN$ such that $\angle CBP = \angle ACB$ and $\angle QCB = \angle CBA$. The circumscribed circle of triangle $ABP$ intersects line $AC$ at $D$ ($D\ne A$) and the circumscribed circle of triangle $AQC$ intersects line $AB$ at $E$ ($E \ne A$). Show that lines $BC$, $DP$ and $EQ$ are concurrent.

Solution

Let us prove that the lines $BC$, $DP$, and $EQ$ intersect at the midpoint of $BC$.

Notice that since $\angle CBP = \angle ACB$ and $MN \parallel BC$, then the quadrilateral $BCNP$ is an isosceles trapezoid and $\angle BPM = \angle CNQ$. Since $\angle CBP = \angle ACB$ and $\angle QCB = \angle CBA$, then $\angle PBM = \angle QCN$. The triangles $BPM$ and $CNQ$ are therefore congruent by the ASA congruence. From here $PQ = PM+MQ =QN+MQ = MN = \frac{BC}{2}$.

 
Let $K_1$ be the intersection of the lines $PD$ and $BC$. Since $BPAD$ is cyclic, then $\angle PBM = \angle PBA = \angle PDA$. However since $\angle PBM = \angle QCN$, then $QC \parallel PD$ and the quadrilateral $PQCK_1$ is a parallelogram. This implies that $PQ=CK_1$ and $K_1$ is the midpoint of $AC$.

Let $K_2$ be the intersection of the lines $EQ$ and $BC$. Since $QEAC$ is cyclic, then $\angle QCN = \angle QCA = \angle BEQ$. However since $\angle PBM = \angle QCN$, then $QE \parallel PB$ and the quadrilateral $BPQK_2$ is a parallelogram. This implies that $PQ=BK_2$ and $K_2$ is the midpoint of $AC$.

Problem 2

Let $T_n$ denotes the least natural such that

$$n\mid 1+2+3+\cdots +T_n=\sum_{i=1}^{T_n} i$$ Find all naturals $m$ such that $m\ge T_m$

Solution

Answer: $m$ is any number that is not a power of $2$.

First, let us show that all powers of $2$ do not work. Let $m=2^a$, where $a \in \mathbb{N}$. Let us assume that $m \geq T_m$, i.e. $2^a \geq T_m$. Therefore

$$ \frac{T_m \cdot \left( T_m + 1 \right)}{2} $$

is divisible by $2^a$ or equivalently $T_m \cdot \left( T_m + 1 \right)$ is divisible by $2^{a+1}$. However since $\text{gcd} \left( T_m , T_m + 1 \right) = 1$ and $T_m \leq 2^a$, then it is divisible by at most $2^{a}$. Contradiction.

 

Now let us show that all numbers that are not powers of $2$ work. Let $m=2^a \cdot s$, where $a,s \in \mathbb{N}$ and $s$ is an odd number greater than $1$. We will show that there exists an integer $k < m$, such that $m$ divides $1+2+…+k$.

Let us consider the numbers $x_i=(2i-1) \cdot s$ and $y_j=(2j-1) \cdot s-1$, where $i,j=1,2,…,2^{a-1}$. We claim there exists $i$ or $j$, such that one of the sums of the form $1+2+…+x_i$ or $1+2+…+y_j$ is divisible by $m$. Indeed

$$ \frac{1+2+…+x_i}{m} = \frac{x_i(x_i+1)}{2m} = \frac{(2i-1) ((2i-1) \cdot s+1)}{2^{a+1}} $$

Since $s$ is relatively prime with $2^{a+1}$, then $s$ is invertible and there exists $i$, such that the congruence is true

$$ (2i-1) \cdot s \equiv -1 \pmod{2^{a+1}} $$

Similarly

$$ \frac{1+2+…+y_j}{m} = \frac{y_j(y_j+1)}{2m} = \frac{((2j-1) \cdot s-1)(2j-1) }{2^{a+1}} $$

Since $s$ is relatively prime with $2^{a+1}$, then $s$ is invertible and there exists $j$, such that the congruence is true

$$ (2j-1) \cdot s \equiv 1 \pmod{2^{a+1}} $$

Notice that one of the indices $(2i-1)$ or $(2j-1)$ will belong to the interval $\left[ 1, 2^{a}-1 \right]$ and the claim is proven.

Problem 4

Show that there exists a set $\mathcal{C}$ of $2020$ distinct, positive integers that satisfy simultaneously the following properties:

$\bullet$ When one computes the greatest common divisor of each pair of elements of $\mathcal{C}$, one gets a list of numbers that are all distinct.

$\bullet$ When one computes the least common multiple of each pair of elements of $\mathcal{C}$, one gets a list of numbers that are all distinct.

Solution

Let us show that there exists a set of $n$ such numbers that satisfy the conditions of the problem.

Let $p_1$, $p_2$, … , $p_{2n}$ be the first $2n$ prime numbers. Consider the numbers

$$ x_1 = p_1 \cdot p_2 \cdot … \cdot p_n $$

$$ x_2 = p_2 \cdot p_3 \cdot … \cdot p_{n+1} $$

… …

$$ x_n = p_{n+1} \cdot p_{n+2} \cdot … \cdot p_{2n} $$

It is not hard to see that the set $\{ x_1, x_2, … , x_n \}$ satisfies the conditions of the problem.

Problem 5

Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$ f(xf(x-y))+yf(x)=x+y+f(x^2) $$

for all real numbers $x$ and $y$.

Solution

Substitute $x=0$, $y=1$:

$$ f(0)+f(0) = 1+f(0) $$

This implies that $f(0)=1$.

Substitute $x=1$, $y=1$:

$$ f(1)+f(1) = 2+f(1) $$

This implies that $f(1)=2$.

Substitute $y=x$:

$$ f(x)+xf(x) = 2x+f(x^2) $$

Substitute $y=x-1$:

$$ f(2x)+xf(x) -f(x) = 2x-1+f(x^2) $$

From the last two equations we have

$$ f(2x) = 2f(x)-1 $$

Substitute $x$ for $2x$ in the equation $ f(x)+xf(x) = 2x+f(x^2) $:

$$ f(2x)+2xf(2x) = 4x+f(4x^2) $$

This implies that

$$ 2f(x) – 1 +2x \left( 2f(x)-1 \right) =4x + f(4x^2) $$

$$ 2f(x) – 1 +2x \left( 2f(x)-1 \right) =4x + 2f(2x^2)-1 $$

$$ 2f(x) – 1 +2x \left( 2f(x)-1 \right) =4x + 2 \left( 2f(x^2)-1\right)-1 $$

$$ 2f(x) – 1 +2x \left( 2f(x)-1 \right) =4x + 4f(x^2)-3 $$

Now we combine the last equation with the equation $ f(x)+xf(x) = 2x+f(x^2) $ multiplies by $4$:

$$ 2f(x) – 1 +2x \left( 2f(x)-1 \right) -4 \left( f(x)+xf(x) \right) = 4x + 4f(x^2)-3 -4 \left(2x+f(x^2) \right) $$

which implies

$$ 2x+2 = 2f(x) $$

$$ f(x) =x+1 $$

It is not hard to see that this function satisfies the original functional equation.

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