💡Okay, so I have this problem where there's a rectangle on graph paper that's 200 cells wide and 100 cells tall. We start coloring from the top left corner and move in a spiral pattern, turning right when we hit an edge or a already colored cell. I need to figure out which cell will be colored last. Hmm, okay.First, let me visualize this. Imagine a grid that's wider than it is tall—200 columns and 100 rows. Starting at the top left, which would be cell (1,1). Then we move right along the top row, coloring each cell until we reach the end, which is cell (1,200). Then, we turn right, which would take us down along the rightmost column, from (1,200) to (100,200). Then, we turn right again, moving left along the bottom row from (100,200) to (100,1). Then, we turn right again, moving up along the leftmost column from (100,1) back to (1,1). But wait, (1,1) is already colored, so we can't go there. So, does that mean we stop? Or do we continue inward?I think we continue inward, creating a spiral. So after completing the outermost rectangle, we move inward by one cell on each side, creating a new smaller rectangle of 198 columns and 98 rows. Then, we repeat the same process: move right, down, left, up, and so on.So, each time we complete a full loop around the rectangle, we reduce the width and height by 2 (since we've gone in by one cell on each side). This process continues until we can't form a complete loop anymore.Let me try to figure out how many such loops we can make before we're left with a single row or column. The initial dimensions are 200x100. Each loop reduces the width and height by 2. So, the number of loops would be determined by how many times we can subtract 2 from both dimensions before one of them becomes less than or equal to 1.Wait, but 200 is even and 100 is even, so we can subtract 2 fifty times from 100 to get to 0, but since we can't have a rectangle with 0 height or width, we need to stop when one of the dimensions becomes 1.Actually, let's think differently. Each loop reduces the width and height by 2, so the number of complete loops would be floor((min(width, height))/2). In this case, min(200,100) is 100, so floor(100/2) is 50. So, we can make 50 complete loops.But wait, after 50 loops, the width would be 200 - 2*50 = 100, and the height would be 100 - 2*50 = 0. Hmm, that doesn't make sense because height can't be 0. Maybe I need to adjust my thinking.Perhaps the number of loops is such that after each loop, both width and height are reduced by 2, but we stop when either width or height becomes 1. So, starting from 200x100:Loop 1: 198x98Loop 2: 196x96...Loop 49: 200 - 2*49 = 200 - 98 = 102 columns, and 100 - 2*49 = 100 - 98 = 2 rows.So, after 49 loops, we're left with a 102x2 rectangle. Now, can we make another loop? Let's see:Loop 50: 102 - 2 = 100 columns, 2 - 2 = 0 rows. Again, 0 rows doesn't make sense. So, perhaps we can't complete the 50th loop because the height would become 0.Therefore, after 49 loops, we're left with a 102x2 rectangle. Now, how do we color this?Starting from the top left of this 102x2 rectangle, which would be cell (50,50) in the original grid. Wait, let me think.Each loop reduces the starting position by 1 in both row and column. So, after 49 loops, the starting position would be (50,50). The rectangle is 102 columns wide and 2 rows tall.So, starting at (50,50), we move right along the top row to (50,151). Then, we turn right and move down to (51,151). Then, we turn right and move left along the bottom row from (51,151) to (51,50). Then, we turn right and move up, but we can't go back to (50,50) because it's already colored. So, does that mean we stop?Wait, but we have to color all cells. So, after moving left from (51,151) to (51,50), the next move would be up to (50,50), but that's already colored. So, we can't proceed further. Therefore, the last cell to be colored is (51,50).Wait, let me verify. After coloring the top row from (50,50) to (50,151), then down to (51,151), then left to (51,50). The next move would be up, but (50,50) is already colored, so we can't go there. Therefore, the last cell colored is (51,50).But let me think again. When we start the 50th loop, we have a 102x2 rectangle. Starting at (50,50), moving right to (50,151), then down to (51,151), then left to (51,50). Now, the next move would be up, but (50,50) is already colored, so we can't proceed. Therefore, the last cell is indeed (51,50).Alternatively, maybe the last cell is (50,50) because after moving left to (51,50), the next move is up to (50,50), but since it's already colored, we stop. So, the last cell is (51,50).Yes, I think that's correct. So, the last cell to be colored is (51,50).

💡Okay, so I need to figure out how many integers ( n ) satisfy the inequality ( 100 < n^2 < 400 ). Hmm, let's break this down step by step. First, I know that ( n^2 ) means ( n ) multiplied by itself. So, I'm looking for integers whose squares are greater than 100 but less than 400. That means ( n ) has to be a number such that when I square it, it doesn't just equal 100 or 400, but it's somewhere in between.Let me start by finding the square roots of 100 and 400 because that might help me determine the range of ( n ). The square root of 100 is 10, and the square root of 400 is 20. So, if ( n^2 ) is greater than 100, then ( n ) must be greater than 10. Similarly, if ( n^2 ) is less than 400, then ( n ) must be less than 20.Wait, but ( n ) could also be negative because squaring a negative number gives a positive result. So, I need to consider both positive and negative integers here. That means ( n ) could be between -20 and -10 as well as between 10 and 20.Let me write this out to make it clearer:- For positive integers: ( 10 < n < 20 )- For negative integers: ( -20 < n < -10 )Now, I need to list out all the integers that fall into these ranges. Starting with the positive integers:- The smallest integer greater than 10 is 11.- The largest integer less than 20 is 19.So, the positive integers that satisfy the condition are 11, 12, 13, 14, 15, 16, 17, 18, and 19. That's 9 numbers.Now, for the negative integers:- The largest integer greater than -20 is -19.- The smallest integer less than -10 is -11.So, the negative integers that satisfy the condition are -11, -12, -13, -14, -15, -16, -17, -18, and -19. That's also 9 numbers.Adding both sets together, I have 9 positive integers and 9 negative integers, making a total of 18 integers.Wait a second, did I count zero? No, because zero squared is 0, which is less than 100, so it doesn't satisfy the condition. Okay, so zero isn't included.Let me double-check to make sure I didn't miss any numbers. For the positive side, starting at 11 and going up to 19, that's 9 numbers. For the negative side, starting at -11 and going down to -19, that's also 9 numbers. So, 9 plus 9 equals 18.Is there any other integer that I might have missed? Let's see, numbers like 10 or -10: 10 squared is exactly 100, which doesn't satisfy ( n^2 > 100 ), so they're excluded. Similarly, 20 squared is 400, which doesn't satisfy ( n^2 < 400 ), so they're also excluded.Alright, I think I've covered all the possible integers that satisfy the condition. So, the total number of integers ( n ) that satisfy ( 100 < n^2 < 400 ) is 18.

💡Alright, so I'm trying to solve this problem about a stingy knight who has gold coins in chests. There are two parts: part (a) with six chests and part (b) with eight chests. The knight notices that he can evenly distribute the coins among any number of chests from 2 up to 5 (in part (a)) or up to 7 (in part (b)). The question is whether it's possible to evenly distribute all the coins across all the chests without looking inside.Starting with part (a), let's see. If the knight can evenly distribute coins among any two chests, that means the number of coins in each pair of chests must be equal when combined. So, if I take any two chests, say chest A and chest B, the total number of coins in A and B must be even because you can split them equally. This suggests that the total number of coins in all six chests must be even because there are three pairs, and each pair has an even number of coins.Next, the knight can also distribute coins equally among any three, four, or five chests. For three chests, the total number of coins in any three chests must be divisible by three. Similarly, for four chests, the total must be divisible by four, and for five chests, divisible by five. So, the total number of coins must be divisible by 2, 3, 4, and 5. The least common multiple (LCM) of these numbers is 60, so the total number of coins must be a multiple of 60.Wait, but 60 is divisible by 6, which is the number of chests. So, if the total number of coins is a multiple of 60, it's also a multiple of 6. That means we can evenly distribute the coins across all six chests. So, for part (a), the answer should be yes, it's possible.Now, moving on to part (b) with eight chests. The knight can distribute coins equally among any two, three, four, five, six, or seven chests. Similar to part (a), the total number of coins must be divisible by 2, 3, 4, 5, 6, and 7. The LCM of these numbers is 420. So, the total number of coins must be a multiple of 420.But 420 divided by 8 is 52.5, which isn't an integer. That means we can't evenly distribute 420 coins into eight chests because each chest would need to have 52.5 coins, which isn't possible since we can't have half coins. Therefore, it's not possible to evenly distribute all the coins across all eight chests.Wait, but the problem says "if there were eight chests, and the stingy knight could evenly distribute the coins in any 2, 3, 4, 5, 6, or 7 chests." So, does that mean the total number of coins is a multiple of 420, but 420 isn't divisible by 8? So, does that mean it's impossible?Alternatively, maybe there's a different way to think about it. Maybe the total number of coins doesn't have to be exactly 420, but just a multiple of 420. But even then, any multiple of 420 divided by 8 would still result in a non-integer because 420 and 8 are co-prime? Wait, no, 420 and 8 share a common factor of 4. So, 420 divided by 4 is 105, and 8 divided by 4 is 2. So, 105 and 2 are co-prime. So, 420k divided by 8 would be (105k)/2, which is still not an integer unless k is even. But if k is even, say k=2, then total coins would be 840, and 840 divided by 8 is 105, which is an integer. Wait, so maybe it is possible?Hmm, this is confusing. Let me think again. If the total number of coins is a multiple of 420, and if we can choose k such that 420k is divisible by 8, then it's possible. Since 420 and 8 have a common factor of 4, we can write 420k = 4*105k, and 8 = 4*2. So, 420k divided by 8 is (105k)/2. For this to be an integer, 105k must be even. Since 105 is odd, k must be even. So, if k is even, say k=2, then 420*2=840, and 840 divided by 8 is 105, which is an integer. Therefore, it is possible to evenly distribute all the coins across all eight chests if the total number of coins is 840.But wait, the problem says "if there were eight chests, and the stingy knight could evenly distribute the coins in any 2, 3, 4, 5, 6, or 7 chests." It doesn't specify that the total number of coins is exactly 420, just that it's a multiple of 420. So, if the total number of coins is 840, which is a multiple of 420, then it's possible to distribute evenly across all eight chests. Therefore, the answer might be yes for part (b) as well.But I'm not sure. Maybe I'm missing something. Let me check. If the total number of coins is 420, then 420 divided by 8 is 52.5, which isn't an integer. But if the total number of coins is 840, which is 420*2, then 840 divided by 8 is 105, which is an integer. So, as long as the total number of coins is a multiple of 420 and also a multiple of 8, it's possible. Since 420 and 8 have a least common multiple of 840, the total number of coins must be a multiple of 840. Therefore, it is possible to evenly distribute all the coins across all eight chests if the total number of coins is a multiple of 840.But the problem doesn't specify the total number of coins, just that the knight can distribute them equally among any 2, 3, 4, 5, 6, or 7 chests. So, the total number of coins must be a multiple of 420, but not necessarily 840. Therefore, it's not guaranteed that the total number of coins is a multiple of 8, so it's not necessarily possible to distribute them evenly across all eight chests.Wait, but if the total number of coins is a multiple of 420, and we can choose any number of chests from 2 to 7, then the total number of coins must be a multiple of all those numbers, which is 420. But 420 isn't a multiple of 8, so the total number of coins isn't necessarily a multiple of 8. Therefore, it's not possible to guarantee that the coins can be evenly distributed across all eight chests.I think that's the correct reasoning. So, for part (a), it's possible, and for part (b), it's not possible.

💡Okay, so I have this function f(x) = sin x (sin x + √3 cos x), and I need to find its maximum value. The options are A: 2, B: 1 + √3, C: 3/2, and D: 1. Hmm, let me think about how to approach this.First, I should probably expand the function to make it easier to work with. So, let's distribute sin x:f(x) = sin x * sin x + sin x * √3 cos xThat simplifies to:f(x) = sin²x + √3 sin x cos xAlright, so now I have sin squared x and sin x cos x terms. I remember that there are some trigonometric identities that can help simplify expressions like this. Maybe I can use the double-angle formulas or something similar.Let me recall the double-angle identities:sin²x = (1 - cos 2x)/2andsin x cos x = (sin 2x)/2So, substituting these into the function:f(x) = (1 - cos 2x)/2 + √3*(sin 2x)/2Let me write that out:f(x) = (1/2) - (cos 2x)/2 + (√3 sin 2x)/2Hmm, so now I have:f(x) = 1/2 + (√3 sin 2x - cos 2x)/2I think I can combine the sine and cosine terms into a single trigonometric function using the amplitude-phase form. The general formula is A sin θ + B cos θ = C sin(θ + φ), where C = √(A² + B²) and tan φ = B/A.In this case, A is √3 and B is -1 because the expression is √3 sin 2x - cos 2x. So, let's compute C:C = √[(√3)² + (-1)²] = √(3 + 1) = √4 = 2Okay, so the amplitude is 2. Now, let's find the phase shift φ:tan φ = B/A = (-1)/√3So, φ = arctan(-1/√3). I know that tan(π/6) = 1/√3, so arctan(-1/√3) would be -π/6. Alternatively, since tangent is periodic with period π, we can also represent this as 11π/6, but I think -π/6 is simpler here.So, putting it all together:√3 sin 2x - cos 2x = 2 sin(2x - π/6)Therefore, the function f(x) becomes:f(x) = 1/2 + (2 sin(2x - π/6))/2Simplify that:f(x) = 1/2 + sin(2x - π/6)Now, to find the maximum value of f(x), I need to find the maximum value of sin(2x - π/6). The sine function oscillates between -1 and 1, so the maximum value of sin(2x - π/6) is 1.Therefore, the maximum value of f(x) is:1/2 + 1 = 3/2So, the maximum value is 3/2, which corresponds to option C.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expanded f(x) correctly: sin²x + √3 sin x cos x.2. Applied double-angle identities correctly: (1 - cos 2x)/2 + (√3 sin 2x)/2.3. Combined the terms: 1/2 + (√3 sin 2x - cos 2x)/2.4. Recognized the need to express √3 sin 2x - cos 2x as a single sine function.5. Calculated amplitude C correctly: √(3 + 1) = 2.6. Found phase shift φ correctly: arctan(-1/√3) = -π/6.7. Rewrote the expression as 2 sin(2x - π/6).8. Substituted back into f(x): 1/2 + sin(2x - π/6).9. Noted that sin(2x - π/6) has a maximum of 1, so maximum f(x) is 3/2.Everything seems to check out. So, I'm confident that the maximum value is 3/2, which is option C.**Final Answer**boxed{C}

💡Okay, so I have this geometry problem here about an acute triangle ABC. The problem talks about inscribing squares in the triangle with two vertices on each side, and then defining points A1, B1, and C1 as the centers of these squares. The goal is to prove that the lines AA1, BB1, and CC1 are concurrent, meaning they all meet at a single point.First, I need to visualize what's going on. Let me try to draw a rough sketch in my mind. Triangle ABC is acute, so all its angles are less than 90 degrees. Now, they're talking about inscribing a square in this triangle. The square has two vertices on side BC, one on AB, and one on AC. The center of this square is A1. Similarly, we have squares inscribed with two vertices on AC and AB, giving us centers B1 and C1 respectively.So, the first thing I need to understand is how exactly these squares are inscribed in the triangle. Let me focus on the square with two vertices on BC. Let's denote the square as KLMN, where K and N are on BC, L is on AB, and M is on AC. The center of this square is A1. Since it's a square, all sides are equal, and all angles are right angles.Now, since A1 is the center, it should be equidistant from all four vertices of the square. That means the distance from A1 to K, L, M, and N is the same. Also, the lines connecting A1 to these vertices should form 45-degree angles because the diagonals of a square bisect the angles.Hmm, maybe I can use coordinate geometry here. Let me assign coordinates to the triangle ABC. Let's place point B at (0,0), point C at (c,0), and point A somewhere in the first quadrant at (a,b). Since the triangle is acute, all coordinates should be positive, and the slopes of the sides should be such that all angles are less than 90 degrees.Now, the square KLMN has two vertices on BC. Let me denote K as (k,0) and N as (n,0). Since it's a square, the side length is |n - k|. The other two vertices L and M are on AB and AC respectively. Let me find the coordinates of L and M.Since KL is a side of the square, it should be perpendicular to KN. The slope of KN is 0 because it's along the x-axis. Therefore, the slope of KL should be undefined or infinite, meaning it's a vertical line. Wait, but that can't be because L is on AB, which is a line from (0,0) to (a,b). So, if KL is vertical, then L would have the same x-coordinate as K, which is (k, y). But since L is on AB, its coordinates must satisfy the equation of AB.The equation of AB can be written as y = (b/a)x. So, if L is at (k, y), then y = (b/a)k. Similarly, since KL is a side of the square, the length from K to L should be equal to the length from K to N. The length KN is |n - k|, and the length KL is the vertical distance from K to L, which is |y - 0| = |(b/a)k|. Since it's a square, these lengths should be equal:|n - k| = |(b/a)k|Assuming everything is positive because we're dealing with lengths, we can drop the absolute value:n - k = (b/a)kSo, n = k + (b/a)k = k(1 + b/a) = k(a + b)/aOkay, so n is expressed in terms of k. Now, let's find the coordinates of M. Since it's a square, the side LM should be equal and perpendicular to KL. Since KL is vertical, LM should be horizontal. So, from point L at (k, (b/a)k), moving horizontally to the right by the same length as KL, which is (b/a)k, we reach point M.Wait, but M is on AC. The equation of AC is from (a,b) to (c,0). Let me find the equation of AC. The slope is (0 - b)/(c - a) = -b/(c - a). So, the equation is y - b = (-b/(c - a))(x - a). Simplifying, y = (-b/(c - a))(x - a) + b.So, point M is at (k + (b/a)k, (b/a)k). Let me write that as (k(1 + b/a), (b/a)k). Let me denote this as (k(a + b)/a, (b/a)k). Now, this point M must lie on AC, so plugging into the equation of AC:(b/a)k = (-b/(c - a))(k(a + b)/a - a) + bLet me simplify the right-hand side:First, compute the term inside the parenthesis:k(a + b)/a - a = (k(a + b) - a^2)/aSo, the equation becomes:(b/a)k = (-b/(c - a)) * (k(a + b) - a^2)/a + bMultiply both sides by a to eliminate denominators:b k = (-b/(c - a)) * (k(a + b) - a^2) + a bLet me distribute the -b/(c - a):b k = [ -b k(a + b)/(c - a) + b a^2/(c - a) ] + a bBring all terms to one side:b k + b k(a + b)/(c - a) - b a^2/(c - a) - a b = 0Factor out b:b [ k + k(a + b)/(c - a) - a^2/(c - a) - a ] = 0Since b ≠ 0, we can divide both sides by b:k + k(a + b)/(c - a) - a^2/(c - a) - a = 0Combine like terms:k [1 + (a + b)/(c - a)] - [a^2/(c - a) + a] = 0Let me compute 1 + (a + b)/(c - a):1 + (a + b)/(c - a) = (c - a + a + b)/(c - a) = (c + b)/(c - a)Similarly, a^2/(c - a) + a = a^2/(c - a) + a(c - a)/(c - a) = [a^2 + a(c - a)]/(c - a) = [a^2 + a c - a^2]/(c - a) = (a c)/(c - a)So, plugging back in:k (c + b)/(c - a) - (a c)/(c - a) = 0Multiply both sides by (c - a):k (c + b) - a c = 0So, k = (a c)/(c + b)Okay, so we found k in terms of a, b, c. Then, n = k(a + b)/a = (a c)/(c + b) * (a + b)/a = c(a + b)/(c + b)So, n = c(a + b)/(c + b)So, the coordinates of K are (k, 0) = ( (a c)/(c + b), 0 )Coordinates of N are (n, 0) = ( c(a + b)/(c + b), 0 )Coordinates of L are (k, (b/a)k) = ( (a c)/(c + b), (b/a)(a c)/(c + b) ) = ( (a c)/(c + b), (b c)/(c + b) )Coordinates of M are (k(a + b)/a, (b/a)k ) = ( (a c)/(c + b) * (a + b)/a, (b c)/(c + b) ) = ( c(a + b)/(c + b), (b c)/(c + b) )Wait, interesting. So, point M is at ( c(a + b)/(c + b), (b c)/(c + b) )But point N is at ( c(a + b)/(c + b), 0 ). So, the vertical distance from N to M is (b c)/(c + b). Since the square has side length equal to the distance from K to L, which is (b c)/(c + b). So, that makes sense.Now, the center A1 of the square is the midpoint of the diagonals. So, the diagonal from K to M. The midpoint would be average of coordinates.Coordinates of K: ( (a c)/(c + b), 0 )Coordinates of M: ( c(a + b)/(c + b), (b c)/(c + b) )Midpoint A1:x-coordinate: [ (a c)/(c + b) + c(a + b)/(c + b) ] / 2 = [ c(a + a + b) / (c + b) ] / 2 = [ c(2a + b) / (c + b) ] / 2 = c(2a + b)/(2(c + b))Wait, hold on:Wait, (a c)/(c + b) + c(a + b)/(c + b) = c/(c + b) [ a + a + b ] = c/(c + b)(2a + b). So, divided by 2, it's c(2a + b)/(2(c + b))Similarly, y-coordinate: [0 + (b c)/(c + b)] / 2 = (b c)/(2(c + b))So, A1 is at ( c(2a + b)/(2(c + b)), (b c)/(2(c + b)) )Hmm, that's the center. Now, I need to find the equation of line AA1.Point A is at (a, b). Point A1 is at ( c(2a + b)/(2(c + b)), (b c)/(2(c + b)) )Let me compute the slope of AA1.Slope m_AA1 = [ (b c)/(2(c + b)) - b ] / [ c(2a + b)/(2(c + b)) - a ]Simplify numerator:( b c / (2(c + b)) ) - b = b [ c / (2(c + b)) - 1 ] = b [ (c - 2(c + b)) / (2(c + b)) ] = b [ (c - 2c - 2b) / (2(c + b)) ] = b [ (-c - 2b) / (2(c + b)) ] = -b(c + 2b)/(2(c + b))Denominator:c(2a + b)/(2(c + b)) - a = [ c(2a + b) - 2a(c + b) ] / (2(c + b)) = [ 2a c + b c - 2a c - 2a b ] / (2(c + b)) = [ b c - 2a b ] / (2(c + b)) = b(c - 2a)/(2(c + b))So, slope m_AA1 = [ -b(c + 2b)/(2(c + b)) ] / [ b(c - 2a)/(2(c + b)) ] = [ - (c + 2b) ] / (c - 2a )So, m_AA1 = -(c + 2b)/(c - 2a)Hmm, that's the slope of AA1.Similarly, I can find the equations for BB1 and CC1, but this seems quite involved. Maybe there's a better approach.Wait, the problem is to show that AA1, BB1, CC1 are concurrent. That is, they meet at a single point. In triangle geometry, concurrency often relates to Ceva's Theorem. Maybe I can apply Ceva's Theorem here.Ceva's Theorem states that for concurrent lines from vertices of a triangle to the opposite sides, the product of certain ratios equals 1. Specifically, if lines AA1, BB1, CC1 are concurrent, then:[ (BL / LC) * (CM / MA) * (AN / NB) ] = 1Where BL, LC, etc., are lengths on the sides.But in this case, the lines are not necessarily intersecting the opposite sides, but rather going to the centers of squares. So, maybe I need a different version of Ceva's Theorem or another approach.Alternatively, maybe using vectors or barycentric coordinates could help. But since I already started with coordinate geometry, perhaps I can continue in that direction.But this seems getting too messy. Maybe I should try to find a property or a symmetry that can help.Wait, all the squares are similar in their construction. Each is inscribed with two vertices on a side, and the other two on the other sides. Their centers are A1, B1, C1. So, perhaps there is a homothety or similarity transformation that relates these centers.Alternatively, maybe the lines AA1, BB1, CC1 are related to some known centers of the triangle, like centroid, orthocenter, etc. But I'm not sure.Wait, another idea: since each square is inscribed with two vertices on a side, maybe the center A1 lies on some specific line or has some specific ratio related to the triangle's sides.Alternatively, maybe using trigonometric Ceva's Theorem. Since we're dealing with angles, perhaps we can express the concurrency condition in terms of sines of angles.Trigonometric Ceva's Theorem states that lines from the vertices of a triangle are concurrent if and only if:[ (sin(angle BAA1) / sin(angle CAA1)) * (sin(angle CBB1) / sin(angle ABB1)) * (sin(angle ACC1) / sin(angle BCC1)) ] = 1So, if I can compute these sine ratios, maybe I can show their product is 1.To do that, I need to find the angles at each vertex. Let's consider angle BAA1 and angle CAA1.Looking back at the square inscribed with two vertices on BC, the center A1. The lines from A to A1 make angles with AB and AC.Given that A1 is the center of the square, and the square has sides equal and at 45 degrees to the sides of the triangle.Wait, perhaps the angles BAA1 and CAA1 can be related to the angles of the triangle and 45 degrees.Wait, in the square, the sides from A1 to the square's vertices are at 45 degrees to the sides of the square. So, maybe the angles between AA1 and AB or AC can be expressed in terms of the triangle's angles plus or minus 45 degrees.But I need to think more carefully.Let me consider triangle AKA1, where K is one of the square's vertices on BC. Since A1 is the center, A1K is equal to half the diagonal of the square. The angle at A1 is 45 degrees because it's a square.Wait, maybe using the Law of Sines in triangles involving A1.In triangle AKA1, we have sides A1K, A1A, and AK. Similarly, in triangle ANA1, we have sides A1N, A1A, and AN.Since A1K = A1N, as A1 is the center, and angles at K and N are 45 degrees, perhaps the Law of Sines can relate the angles at A.Wait, let me try that.In triangle AKA1:sin(angle KAA1) / A1K = sin(angle A1KA) / A1ASimilarly, in triangle ANA1:sin(angle NAA1) / A1N = sin(angle A1NA) / A1ABut A1K = A1N, so:sin(angle KAA1) / sin(angle A1KA) = sin(angle NAA1) / sin(angle A1NA)But angles A1KA and A1NA are related to the angles of the triangle.Wait, angle A1KA is equal to angle ABC + 45 degrees, because the square is inscribed such that side KN is parallel to BC, so the angle at K is the same as angle B, but offset by 45 degrees due to the square.Similarly, angle A1NA is equal to angle ACB + 45 degrees.Therefore, we have:sin(angle KAA1) / sin(angle NAA1) = sin(angle B + 45°) / sin(angle C + 45°)Similarly, for the other centers B1 and C1, we can derive similar ratios.Therefore, the product of these ratios would be:[sin(B + 45°)/sin(C + 45°)] * [sin(C + 45°)/sin(A + 45°)] * [sin(A + 45°)/sin(B + 45°)] = 1Thus, by trigonometric Ceva's Theorem, the lines AA1, BB1, CC1 are concurrent.Wait, that seems to work out. So, the key idea was to use the Law of Sines in the triangles formed by the centers and the vertices, relate the angles using the properties of the square, and then apply Ceva's Theorem in its trigonometric form.I think that's the solution. It was a bit involved, but breaking it down step by step helped me see how the concurrency condition is satisfied.

💡Alright, so I've got this problem about an ( n times n ) grid with some cells marked by asterisks. The condition given is that if you cross out any rows (but not all of them), there will still be at least one column that has exactly one asterisk left. And from that, I need to prove that if you cross out any columns (again, not all of them), there will be at least one row with exactly one asterisk left.Hmm, okay. Let me try to unpack this step by step.First, let's restate the given condition to make sure I understand it. If I remove any subset of the rows (as long as I don't remove all of them), there will still be at least one column that has exactly one asterisk remaining. So, even if I remove a bunch of rows, there's always some column that has precisely one star left in it.Now, I need to show that the same kind of property holds for columns. That is, if I remove any subset of the columns (again, not all), there will still be at least one row with exactly one asterisk remaining.I think the key here is to somehow use the given condition about the rows to deduce something about the columns. Maybe there's some symmetry or duality here that I can exploit.Let me consider the structure of the grid. If every time I remove some rows, there's a column with exactly one asterisk, that suggests that the asterisks are arranged in such a way that no column has too many asterisks to begin with. Otherwise, removing some rows might leave multiple asterisks in a column, which would violate the condition.Wait, actually, the condition is that after removing any rows (but not all), there's still a column with exactly one asterisk. So, it's not that every column has exactly one asterisk, but rather that no matter how you remove rows, you can't eliminate all columns with exactly one asterisk.This makes me think that the grid must have a certain kind of structure where each column has at least one asterisk, and perhaps the asterisks are arranged in such a way that they don't overlap too much.Maybe I can think about the dual problem. If I transpose the grid, turning rows into columns and vice versa, then the original condition about rows would translate into a condition about columns. But I'm not sure if that helps directly.Alternatively, perhaps I can use induction on the size of the grid. Let's see. For ( n = 1 ), it's trivial because there's only one cell, and it must have an asterisk. So, removing the only row or column would leave nothing, but since we can't remove all rows or columns, the condition holds trivially.For ( n = 2 ), let's consider a 2x2 grid. Suppose we have some asterisks. The condition says that if we remove one row, there's still a column with exactly one asterisk. So, let's say we have two rows, and if we remove one, the remaining row must have a column with exactly one asterisk.Wait, but in a 2x2 grid, if we remove one row, we're left with one row. So, for there to be a column with exactly one asterisk, that row must have exactly one asterisk. So, each row must have exactly one asterisk. Otherwise, if a row had two asterisks, removing the other row would leave a row with two asterisks, which doesn't satisfy the condition.So, in the 2x2 case, each row must have exactly one asterisk. Therefore, each column must also have exactly one asterisk because it's a Latin square kind of situation. So, in this case, if we remove any column, we're left with one column, which has exactly one asterisk. So, the condition holds.Hmm, interesting. So, for ( n = 2 ), the grid must be such that each row and each column has exactly one asterisk. That is, it's a permutation matrix.Wait, is that the case for larger ( n )? If each row has exactly one asterisk, then each column must also have exactly one asterisk, right? Because if a column had more than one, then removing the other rows would leave multiple asterisks in that column, which would violate the condition.Wait, no. Let me think again. If each column has exactly one asterisk, then removing any subset of rows would leave some columns with zero asterisks and some with one. But the condition says that after removing any rows (but not all), there is still at least one column with exactly one asterisk.So, if each column has exactly one asterisk, then removing any subset of rows would leave some columns with one asterisk (those whose single asterisk wasn't in the removed rows) and others with zero. So, as long as not all rows are removed, there will be at least one column with exactly one asterisk.Therefore, if the grid is such that each column has exactly one asterisk, then the given condition is satisfied. Similarly, if each row has exactly one asterisk, then the dual condition (for columns) would also be satisfied.But wait, in the 2x2 case, we saw that both conditions lead to a permutation matrix. So, perhaps in general, the grid must be a permutation matrix? That is, exactly one asterisk in each row and each column.If that's the case, then the conclusion follows trivially because if you remove any subset of columns, you're left with some columns, each of which has exactly one asterisk in their respective rows. So, each remaining column has exactly one asterisk, and thus, each remaining row has exactly one asterisk.But wait, no. If you remove columns, the remaining columns might have their asterisks in different rows. So, actually, each remaining row could have multiple asterisks if the columns you removed had asterisks in different rows.Wait, no. If it's a permutation matrix, each row has exactly one asterisk. So, if you remove some columns, you're effectively removing some asterisks from the rows. So, each row could have zero or one asterisk remaining, depending on whether the column containing its asterisk was removed.Therefore, if you remove some columns, each row that still has its asterisk in one of the remaining columns will have exactly one asterisk, and the others will have zero. So, as long as not all columns are removed, there will be at least one row with exactly one asterisk.Yes, that makes sense. So, if the grid is a permutation matrix, then both the given condition and the desired conclusion hold.But is the grid necessarily a permutation matrix? Or is there a more general structure?Wait, let's think about it. Suppose the grid is not a permutation matrix. For example, suppose there's a column with two asterisks. Then, if we remove the other ( n-2 ) rows, we'd be left with two rows, each having an asterisk in that column. So, that column would have two asterisks, which would violate the given condition because after removing those rows, there should still be a column with exactly one asterisk.Wait, no. The given condition says that after removing any subset of rows (but not all), there exists at least one column with exactly one asterisk. So, even if some columns have multiple asterisks, as long as there's at least one column with exactly one, it's okay.But if a column has two asterisks, then removing the other ( n-2 ) rows would leave two asterisks in that column, but maybe there's another column with exactly one asterisk.Hmm, this is getting complicated. Maybe I need to think about the structure more carefully.Let me try to formalize the given condition. For any subset ( S ) of the rows (where ( S ) is not empty and not all rows), there exists a column ( j ) such that the number of asterisks in column ( j ) not in ( S ) is exactly one.Similarly, we need to show that for any subset ( T ) of the columns (where ( T ) is not empty and not all columns), there exists a row ( i ) such that the number of asterisks in row ( i ) not in ( T ) is exactly one.So, perhaps I can use the given condition to deduce something about the columns, and then use that to show the desired property for the rows.Alternatively, maybe I can use a proof by contradiction. Suppose that there exists a subset of columns ( T ) such that every row has either zero or at least two asterisks not in ( T ). Then, I need to show that this leads to a contradiction with the given condition.Wait, let's try that. Suppose, for contradiction, that there exists a subset of columns ( T ) (not all columns) such that every row has either zero or at least two asterisks not in ( T ). Then, consider the complement of ( T ), which is a subset of columns ( T' ) (also not all columns, since ( T ) is not empty). For each row, the number of asterisks in ( T' ) is either zero or at least two.Now, consider the set of rows that have at least two asterisks in ( T' ). Let's call this set ( R ). If ( R ) is not all rows, then there exists at least one row not in ( R ), which has zero asterisks in ( T' ). But then, considering the given condition, if we remove all rows not in ( R ), we should still have a column with exactly one asterisk. However, in ( T' ), every row in ( R ) has at least two asterisks, so removing the other rows doesn't affect ( T' ). Therefore, in ( T' ), every column has at least two asterisks, which contradicts the given condition that there must be a column with exactly one asterisk.Wait, that seems a bit convoluted, but I think the idea is that if every row has at least two asterisks in ( T' ), then removing the other rows would leave ( T' ) with columns that have at least two asterisks, which violates the given condition.Therefore, our assumption must be wrong, meaning that there must exist at least one row with exactly one asterisk not in ( T ). Hence, the desired conclusion holds.Hmm, that seems to work. So, by assuming the opposite and reaching a contradiction with the given condition, we've shown that the desired property must hold.But let me double-check this reasoning. Suppose we have a subset of columns ( T ) such that every row has either zero or at least two asterisks in ( T' ) (the complement of ( T )). Then, if we remove all rows not in ( R ) (where ( R ) is the set of rows with at least two asterisks in ( T' )), we're left with ( R ) and ( T' ). In ( T' ), every row in ( R ) has at least two asterisks, so every column in ( T' ) must have at least two asterisks (since each column is covered by multiple rows). But this contradicts the given condition that after removing some rows, there must be a column with exactly one asterisk.Therefore, our initial assumption is false, and there must be at least one row with exactly one asterisk not in ( T ).Okay, that makes sense. So, the key idea is to use the given condition about rows to show that the columns cannot all have too many asterisks, and thus, when we remove some columns, there must still be a row with exactly one asterisk left.I think this approach works. It's a bit abstract, but the contradiction seems valid. So, in summary, by assuming that after removing some columns, every row has either zero or at least two asterisks, we derive a contradiction with the given condition about rows. Therefore, there must be at least one row with exactly one asterisk remaining after removing any subset of columns (not all).Yeah, I think that's the way to go. It's a bit tricky, but I think the contradiction method is the right approach here.