p1. Inside a rectangle a point $P$ is marked. Segments are drawn that join the vertices with $P$. In an alternate way, the sectors that are formed are colored. Show that the sum of the areas of the painted sectors is equal to the sum of the unpainted sectors. p2. Find the maximum number of different paths that can be built on a pool table to join two balls on the $n$-cushion table. No ball is touching a band. A band is defined when a ball bounces off one side of the table, with the angle of incidence equal to the angle of departure. p3. Using only two different digits $2$ and $d$, the following $90$-digit number is formed: $m= 2d22d222d...$ If $m$ is a multiple of $9$, determine all possible values of the digit $d$. p4. Calculate all the solutions $x,y,z$ in the positive real numbers of the following system: $$x(6- y) = 9\,\, ,\,\,y(6-z) = 9\,\, ,\,\ z(6-x) = 9$$ PS. Seniors p1 was also Juniors p1.
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Tags: algebra, number theory, geometry, combinatorics, chilean NMO
kentok
09.02.2024 05:56
Solution p4. Note that \[3=\frac{x+y+z+(6-x)+(6-y)+(6-z)}{6}\ge \sqrt[6]{xyz(6-x)(6-y)(6-z)}=\sqrt[6]{9^3}=3\]Hence equality holds that is $x=y=z=6-x=6-y=6-z$. Hence the only solution is $(x,y,z)=(3,3,3)$.
whatdohumanitarianseat
09.02.2024 10:54
Notice that the amount of $2$'s increase every time they are written in front of a $d$, which means that before reaching 90 digits, there must be $x$ sections of $2$'s written, which also means there are $x$ number of $d$'s written down. Notice that since the amount of $2$'s increase linearly with 1 as the common difference, we can implement the Arithmetic Sequence Sum Formula to estimate our $x$. We try out numbers for $x$ until we stumble upon $x=12$, in which
$\frac{(1+12)\dot(12)}{2} = 78$
This means that there are in total, $12$ complete groups of $2$'s, since if $x=13$, then the sum would be
$\frac{(1+13)\dot(13)}{2}=91$
Which is greater than $90$, even though we didn't include the amount of $d$'s yet.
Anyways, after we get $x=12$, we verify it further by adding $12$ (amount of $d$'s) to the total number of $2$'s to get the total number of digits, which is
$78+12=90$
which verifies that $x=12$.
Recall that for a number to be divisible by $9$, the digits of this number need to add up to a multiple of 9.
$(2\cdot78+12d)\equiv 0\mod{9}$
Since we already know that $2 \cdot 78 \equiv 3 \mod{9}$
That means that $12d \equiv 6 \mod{9}$
and since $12 \equiv 3 \mod{9}$
We see that $3d \equiv 6 \mod{9}$
and we see that $d\in{2,5,8}$
$2$ is not solution for $d$ as the problem states that $d$ has to be different from 2, so in total, the solutions for $d$ are:
$5,8$