Problem 1
Let $N=2021^2k+2021$, where $k$ is a positive integer. Alfredo calculates the sum of all the positive divisors of $N$ that are multiples of $43$ and obtains $S_A$ as a result. Bianca calculates the sum of all the positive divisors of $N$ that are multiples of $47$ and obtains $S_B$ as a result. Find the value of $S_A/S_B$ and show that it does not depend on $k$.
Solution
Notice that $2021 = 43 \cdot 47$. Let us rewrite $N$ as
$$ N=2021 \cdot (2021k+1) = 43 \cdot 47 \cdot (2021k+1) $$
The number $(2021k+1)$ is relatively prime with both $43$ and $47$. Let $a_1$, $a_2$, … , $a_n$ be all positive divisors of the number $(2021k+1)$.
The sum $S_A$ of positive divisors of $N$ that are multiples of $43$ is
$$ 43 \cdot a_1+ 43 \cdot a_2+…+43 \cdot a_n+43 \cdot 47 \cdot a_1+ 43 \cdot 47 \cdot a_2+…+43 \cdot 47 \cdot a_n = $$
$$ \left(a_1+a_2+…+a_n\right) \cdot \left( 43 +43\cdot47 \right) = \left(a_1+a_2+…+a_n\right) \cdot 2064 $$
The sum $S_B$ of positive divisors of $N$ that are multiples of $47$ is
$$ 47 \cdot a_1+ 47 \cdot a_2+…+47 \cdot a_n+47 \cdot 43 \cdot a_1+ 47 \cdot 43 \cdot a_2+…+47 \cdot 43 \cdot a_n = $$
$$ \left(a_1+a_2+…+a_n\right) \cdot \left( 47 +47\cdot43 \right) = \left(a_1+a_2+…+a_n\right) \cdot 2068 $$
From here we have that
$$ \frac{S_A}{S_B} = \frac{ \left(a_1+a_2+…+a_n\right) \cdot 2064}{\left(a_1+a_2+…+a_n\right) \cdot 2068}= \boxed{ \frac{516}{517} } $$
Problem 2
In a regular pentagon, segments are drawn each connecting one end of each side with $n$ different points on the adjacent side clockwise. Into how many non-overlapping regions is the pentagon divided?
Solution
Let us pick the first vertex and draw the first $n$ segments. They will divide the pentagon into $n+1$ regions.
Let us pick the second vertex and draw the next $n$ segments. They will divide the pentagon into $n(n+1)$ regions. The same will be true for the third and fourth vertices.
Check out our courses!
Learn more
Let us pick the last vertex and draw the last $n$ segments. They will divide the pentagon into $n(2n+1)$ regions.
From here the number of all regions is
$$ (n+1) + 3n(n+1) + n(2n+1) = \boxed{ 5 n^2 + 5 n + 1 } $$
Problem 3
A circular game board is divided into $n$ sectors ($n \geq 3$) each being empty or having one token. One move consists in selecting a sector with a token, removing the token, and “reversing the polarity” of the neighboring sectors (i.e. emptying an occupied sector and occupying an empty sector). Initially, there is only one occupied sector and you want to make all sectors empty. Find all values of $n$, for which this is possible.
Solution
Let us first show that it is impossible to make all sectors empty for $n$ being a multiple of $3$, $n \geq 3$. Let us color the sectors into three colors $A$, $B$ and $C$ in the following pattern: $ABCABC…ABC$. Let $n_A$, $n_B$ and $n_C$ be the numbers of tokens on the colors $A$, $B$ and $C$ respectively. Let us consider the following triple
$$ \left( \left| n_A-n_B\right|, \left| n_B-n_C\right|, \left| n_C-n_A\right| \right) $$
One move changed the parity of the numbers $n_A$, $n_B$ and $n_C$ to the opposite. However, it keeps the triple invariant modulo $2$. It is not hard to see that in the end the triple becomes $(0,0,0)$, while in the beginning it is not so. Contradiction.
Let us now show that it is always possible to make all sectors empty for $n$ not being a multiple of $3$, $n \geq 4$. Let $n$ be either $1$ or $2$ modulo $3$. It would be enough to show that when $n$ is $1$ modulo $3$ we can obtain a configuration with only one empty sector and when $n$ is $2$ modulo $3$ we can obtain a configuration with only two consecutive empty sectors.
Let $0$ represent an empty sector and $1$ represent an occupied sector. Let some part of the sector be given as $01000…$. We can obtain three consecutive sectors with tokens in the following three steps:
$$ 01000… \rightarrow 10100… \rightarrow 11010… \rightarrow 11101… $$
It is clear that we can continue these operations until we get either $4$ or $5$ sectors left.
For the case when there are $4$ sectors left we can leave one empty sector as follows:
$$ 0100… \rightarrow 1010… \rightarrow 1101… $$
For the case when there are five sectors left we can leave two empty sectors as follows:
$$ 01000… \rightarrow 10100… \rightarrow 11010… \rightarrow 11101 … \rightarrow 111100 $$
Problem 4
Let $\Gamma$ be the circumcircle of the triangle $ABC$. The bisectors of the angles $B$ and $C$ intersect $\Gamma$ at the points $B’$ and $C’$ respectively. Let $D$ be any point on the arc $BC$ that does not contain the point $A$. Lines $AB$ and $DC’$ intersect at $P$. Lines $AC$ and $DB’$ intersect at $Q$. Line $DB$ intersects the circumcircle of the triangle $BPC’$ at $X$. Line $DC$ intersects the circumcircle of the triangle $CQB’$ at $Y$. Prove that $DX=DY$.
Solution
Let us show that the triangles $ADB’$ and $YDB’$ are congruent. Indeed,
$ \angle CDB’ = \angle CBB’ = \angle ABB’ = \angle ADB’ $
Also
$ \angle AB’D = \angle ACD = \angle YB’Q $
Therefore, the triangles $ADB’$ and $YDB’$ are congruent by ASA-congruence and $DA=DY$. Similarly, the triangles $ADC’$ and $XDC’$ are congruent by ASA-congruence and $DA=DX$. From here $DX=DY$ as needed.
