p1. For each unit square on a $5 \times 9$ board write the number $1$ or $0$. Then calculate the sum of all the numbers in each column and also in each row so are obtained $14$ numbers. Suppose H is a set containing those numbers. Determine the maximum number of members of $H$. p2. The natural number $k> 2$ is said to be beautiful if for every natural number $n \ge 4$ with $5n + 1$ a perfect square number, can be found real numbers $a_1, a_2,..., a_k$ such that $$n+1= a_1^2+ a_2^2+...+a_k^2.$$Find the smallest beautiful number p3. Given triangle $ABC$, the three altitudes intersect at point $H$. Determine all points $X$ on the side $BC$ so that the symmetric of $H$ wrt point $X$ lies on the circumcircle of triangle $ABC$. p4. Let $a$, $b$, and $c$ be real numbers whose absolute value is not greater than $ 1$. Prove that $$\sqrt{|a-b|}+\sqrt{|b-c|}+\sqrt{|c-a|}\le 2+\sqrt2$$ p5. On a $2017\times n$ chessboard , Ani and Banu play a game. First player choose a square and then color it red. Next player selects a square from an area that has not been colored red and then color it in red. The selected square can be any size but must accurately cover a number of square units on a chessboard. Then the two players take turns doing the same thing. One player is said to have won, if the next player can no longer continue the game. If Ani gets the first turn, determine all values of $n\ge 2017$ until Ani has a strategy to win the game.
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Tags: algebra, geometry, number theory, combinatorics, Indonesia Regional MO
27.05.2023 18:23
P1 Case 1: there is no column that has sum of 0 For each of 5 row, there can be 10 different sums (0,9). Because there are only 5 rows, we can make 5 different sum for each row, and because we want this value to be different from the column, we set the sum are $\{5,6,7,8,9\}$, but this implies each column cannot have sum of 0 For each of 9 column, there can be 6 different sum (0,5) Because any column cannot have sum of 0, so there are only 5 different sum $\{1,2,3,4,5\}$ Case 2: there is one column that has sum of 0 This implies the maximum sum of a row is 8 (otherwise there is no 0) we set the sum are $\{4,5,6,7,8\}$ Because one column has 0 and we want the sums are different from the row, we set $\{0,1,2,3,4\}$ So the maximum member of $H$ is 9 One of the configuration is shown below
28.05.2023 08:44
P5 If $n=2017$ then Ani just need to place a $2017\times 2017$ square then it is ended and Ani won the game Now we want to know if $n>2017$ Case 1: $2017\times (2k+1)$ ($n$ is odd) Ani just need to place a $2017\times 2017$ in the middle, then it will divide the the chessboard with same area, for the next move, Ani just need to mirror Banu's move Case 1: $2017\times (2k)$ ($n$ is even) Ani just need to place a $2016\times 2016$ in the middle, then it will divide the the chessboard with same area, also, there will be a $1\times 2016$ stripes, it can also be divided to 2 stripes with same area. For the next move, Ani just need to mirror Banu's move So Ani always has strategy to win the game
23.11.2024 19:01
P4 Hope this isn't a fakesolve Notice that the ineq is symmetric, so we can assume a≥b≥c. Moreover, because 1≥a≥b≥c≥-1, so a-c≤2. So, the problem is to prove : sqrt(a-b) + sqrt(b-c) + sqrt (a-c) ≤ 2 + sqrt(2). By CS, sqrt(a-b) + sqrt(b-c) ≤ sqrt((1+1)(a-b+b-c)) = sqrt(2(a-c)). Then, sqrt(a-b)+sqrt(b-c)+sqrt(a-c) ≤ sqrt(2(a-c)) + sqrt(a-c) ≤ 2+√2, where the last inequality follows from the fact that a-c≤2. Equality cases happen when a-c=2 and a-b=b-c, that is (a, b, c)=(1, 0,-1) and it's permutations. So we're done.
08.12.2024 22:13
parmenides51 wrote: p3. Given triangle $ABC$, the three altitudes intersect at point $H$. Determine all points $X$ on the side $BC$ so that the symmetric of $H$ wrt point $X$ lies on the circumcircle of triangle $ABC$. There are only $2$ such points namely, the feet of the altitudes from $A$ upon $ BC$ and the midpoint of $BC.$
11.12.2024 01:55
P1) The sums must be in the set of {$0, 1, \dots 9$}. Therefore, the maximum number of elements in $H$ is $10$, where every sum occurs once. Let's say there are sums of both $0$ and $9$. The sums of $0$ and $9$ would have to be from two rows, or else this would not be possible. The rest of the rows must have sums of $6, 7, 8$ as the columns cannot have sums greater than $5$. The columns will never have a sum of $5$ as there is at least one $0$ in each column from the row with sum $0$. Therefore, the answer is $10-1=\boxed {9}$