p1. For a positive integer, let's call $<a>$ the number obtained by multiplying each figure of a by $2$ and writing the numbers thus obtained. For example $<126> = 2412$ and $<809> = 16018$. Prove that it is not possible find two different positive integers $a$ and $b$ such that $<a> = <b>$. p2. In how many ways is it possible to cut a graph paper of $6\times 6$ starting at the bottom of the paper and working to the top if it can only be cut on the lines of the grid, the two pieces in which it is divided must be equal and it can't be cut down (see figure)? Note: Two pieces are considered the same if you can place one over the other and they fit perfectly. p3. In an equilateral triangle $ABC$ with side $2$, side $AB$ is extended to a point $D$ so that $B$ is the midpoint of $AD$. Let $E$ be the point on $AC$ such that $\angle ADE = 15^o$ and take a point $F$ on $AB$ so that $|EF| = |EC|$. Determine the area of the triangle $AFE$. p4. For each positive integer $n$ we consider $S (n)$ as the sum of its digits. For example $S (1234) = 1 + 2 + 3 + 4 = 10$. Calculate $S (1)- S (2) + S (3)- S (4) +...- S (2012) + S (2013)- S (2014) $ p5. Given $102$ points on a circle, next to one of them is written a $ 1$ and next to each other a $0$. The allowed operation consists of choosing a point that has a $1$ and change the number of that point, and also the number of its two neighbors, the one on the left and on the right (where there is a $ 1$, write $0$ and where there is a $0$ is written $ 1$). Show that it is impossible, with permitted operations, to achieve that all the points have a $0$. p6. Consider a convex figure $P$ in the plane. For a point $Z$ of the plane outside the figure we denote by $|Z, P|$ the smallest length of the segments that join $Z$ with some point of $P$. Consider a line $L$ that does not intersect the figure $P$, two points $X, Y$ over $L$, and $M$ the midpoint between $X$ and $Y$. Prove that $\frac{|X, P| + |Y, P|}{2} \ge |M, P|$. PS. Seniors P3, P4 were also proposed as Juniors P3, easier P4.
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Tags: algebra, geometry, combinatorics, number theory, chilean NMO
15.10.2021 19:16
parmenides51 wrote: p1. For a positive integer, let's call $<a>$ the number obtained by multiplying each figure of a by $2$ and writing the numbers thus obtained. For example $<126> = 2412$ and $<809> = 16018$. Prove that it is not possible find two different positive integers $a$ and $b$ such that $<a> = <b>$. p2. In how many ways is it possible to cut a graph paper of $6\times 6$ starting at the bottom of the paper and working to the top if it can only be cut on the lines of the grid, the two pieces in which it is divided must be equal and it can't be cut down (see figure)? Note: Two pieces are considered the same if you can place one over the other and they fit perfectly. p3. In an equilateral triangle $ABC$ with side $2$, side $AB$ is extended to a point $D$ so that $B$ is the midpoint of $AD$. Let $E$ be the point on $AC$ such that $\angle ADE = 15^o$ and take a point $F$ on $AB$ so that $|EF| = |EC|$. Determine the area of the triangle $AFE$. p4. For each positive integer $n$ we consider $S (n)$ as the sum of its digits. For example $S (1234) = 1 + 2 + 3 + 4 = 10$. Calculate $S (1)- S (2) + S (3)- S (4) +...- S (2012) + S (2013)- S (2014) $ p5. Given $102$ points on a circle, next to one of them is written a $ 1$ and next to each other a $0$. The allowed operation consists of choosing a point that has a $1$ and change the number of that point, and also the number of its two neighbors, the one on the left and on the right (where there is a $ 1$, write $0$ and where there is a $0$ is written $ 1$). Show that it is impossible, with permitted operations, to achieve that all the points have a $0$. p6. Consider a convex figure $P$ in the plane. For a point $Z$ of the plane outside the figure we denote by $|Z, P|$ the smallest length of the segments that join $Z$ with some point of $P$. Consider a line $L$ that does not intersect the figure $P$, two points $X, Y$ over $L$, and $M$ the midpoint between $X$ and $Y$. Prove that $\frac{|X, P| + |Y, P|}{2} \ge |M, P|$. PS. Seniors P3, P4 were also proposed as Juniors P3, easier P4. is p3) \(9\sqrt(2)-5\sqrt(6)\)
16.10.2021 01:02
1) Suppose $a\in N^+$ and $<a>$ is given. The first digit of $<a>$ is 1 or even. If it 1 and $<a>$ starts with $10,12,14,18$ then the first digit of $a$ is known. In that case we can delete the first two digits from $<a>$ and start from scratch. If the first digit of $<a>$ is $2,4,6,8$ then $a$ starts with $1,2,3,4$ and we can delete the first digit from $<a>$. Continuing it follows that $a$ is unique. 4) Since $2k$ does not have final digit 9, we have $-s(2k)+s(2k+1)=1$ for all $k$. Hence $1+(-s(2)+s(3))+\cdots+ (-s(2012)+s(2013))-7=-6+1006=1000$.