💡Okay, so I have this inequality to solve: ( x^2 - 48x + 576 leq 16 ). Hmm, I need to find all the values of ( x ) that make this true. Let me think about how to approach this.First, I remember that solving quadratic inequalities usually involves getting everything on one side so that I can factor or find roots. So, maybe I should subtract 16 from both sides to get the inequality in a standard form. Let me try that:( x^2 - 48x + 576 - 16 leq 0 )Simplifying that, I get:( x^2 - 48x + 560 leq 0 )Alright, now I have ( x^2 - 48x + 560 leq 0 ). I need to solve this quadratic inequality. I think the next step is to factor the quadratic expression if possible. Let me see if I can factor ( x^2 - 48x + 560 ).Factoring quadratics usually involves finding two numbers that multiply to give the constant term (which is 560 here) and add up to the coefficient of the middle term (which is -48 here). So, I'm looking for two numbers that multiply to 560 and add up to -48.Let me list the factors of 560 and see which pair adds up to -48. 560 is a bit large, but let's try:1 and 560: 1 + 560 = 561 (way too big)2 and 280: 2 + 280 = 282 (still too big)4 and 140: 4 + 140 = 144 (closer, but not -48)5 and 112: 5 + 112 = 1177 and 80: 7 + 80 = 878 and 70: 8 + 70 = 7810 and 56: 10 + 56 = 6614 and 40: 14 + 40 = 5416 and 35: 16 + 35 = 5120 and 28: 20 + 28 = 48Oh, wait! 20 and 28 multiply to 560 and add up to 48. But in the quadratic, the middle term is -48x, so both numbers should be negative to add up to -48 and multiply to 560. So, the factors would be -20 and -28.So, the quadratic can be factored as:( (x - 20)(x - 28) leq 0 )Let me check that:( (x - 20)(x - 28) = x^2 - 28x - 20x + 560 = x^2 - 48x + 560 )Yes, that's correct.Now, I have the inequality ( (x - 20)(x - 28) leq 0 ). To solve this, I need to find the intervals where the product of these two factors is less than or equal to zero.First, let's find the critical points where each factor is zero:( x - 20 = 0 ) => ( x = 20 )( x - 28 = 0 ) => ( x = 28 )These points divide the real number line into three intervals:1. ( (-infty, 20) )2. ( (20, 28) )3. ( (28, infty) )Now, I need to test each interval to see where the product ( (x - 20)(x - 28) ) is negative or zero.Let's pick a test point from each interval.1. For ( (-infty, 20) ), let's choose ( x = 10 ): ( (10 - 20)(10 - 28) = (-10)(-18) = 180 ) which is positive.2. For ( (20, 28) ), let's choose ( x = 24 ): ( (24 - 20)(24 - 28) = (4)(-4) = -16 ) which is negative.3. For ( (28, infty) ), let's choose ( x = 30 ): ( (30 - 20)(30 - 28) = (10)(2) = 20 ) which is positive.So, the product ( (x - 20)(x - 28) ) is negative only in the interval ( (20, 28) ). At the critical points ( x = 20 ) and ( x = 28 ), the product is zero, which satisfies the inequality ( leq 0 ).Therefore, the solution to the inequality is ( x ) in the interval ( [20, 28] ).Wait, let me double-check my work to make sure I didn't make any mistakes.Starting from the original inequality:( x^2 - 48x + 576 leq 16 )Subtracting 16:( x^2 - 48x + 560 leq 0 )Factored correctly as ( (x - 20)(x - 28) leq 0 ). Test intervals:- Left of 20: positive- Between 20 and 28: negative- Right of 28: positiveSo, the inequality holds where the product is negative or zero, which is between 20 and 28, including the endpoints.Yes, that seems correct. So, the solution is ( [20, 28] ).**Final Answer**The solution in interval notation is boxed{[20, 28]}.

💡Okay, so I have this piecewise function f(x) defined as:[f(x) =begin{cases}x^2 + 3x + 1 & text{if } x < m, 3x + 6 & text{if } x geq m.end{cases}]And I need to find the sum of all possible values of m that make the graph of y = f(x) continuous. Hmm, continuity at a point means that the left-hand limit and the right-hand limit at that point are equal, and they both equal the function's value at that point. Since the function is defined differently on either side of x = m, I need to ensure that the two pieces meet smoothly at x = m.So, for f(x) to be continuous at x = m, the limit as x approaches m from the left (using the quadratic part) should be equal to the limit as x approaches m from the right (using the linear part), and both should equal f(m). Let me write that out:1. The left-hand limit as x approaches m: (lim_{x to m^-} f(x) = m^2 + 3m + 1).2. The right-hand limit as x approaches m: (lim_{x to m^+} f(x) = 3m + 6).3. The function value at m: (f(m) = 3m + 6) because for x ≥ m, we use the linear part.For continuity, all three should be equal. So, setting the left-hand limit equal to the right-hand limit:[m^2 + 3m + 1 = 3m + 6]Now, let's simplify this equation step by step. First, subtract 3m from both sides:[m^2 + 3m + 1 - 3m = 3m + 6 - 3m][m^2 + 1 = 6]Then, subtract 6 from both sides:[m^2 + 1 - 6 = 0][m^2 - 5 = 0]So, we have a quadratic equation: (m^2 - 5 = 0). To solve for m, we can factor this equation. It's a difference of squares, so:[(m - sqrt{5})(m + sqrt{5}) = 0]Setting each factor equal to zero gives the solutions:1. (m - sqrt{5} = 0) → (m = sqrt{5})2. (m + sqrt{5} = 0) → (m = -sqrt{5})So, the possible values of m are (sqrt{5}) and (-sqrt{5}). The problem asks for the sum of all possible values of m. To find the sum, I can add these two solutions together:[sqrt{5} + (-sqrt{5}) = 0]Wait, that's interesting. The sum is zero. I remember that in quadratic equations, the sum of the roots is given by (-b/a), where the equation is in the form (ax^2 + bx + c = 0). In our case, the equation was (m^2 - 5 = 0), so a = 1, b = 0, c = -5. Therefore, the sum of the roots is (-0/1 = 0), which confirms our earlier result.Just to make sure I didn't make any mistakes, let me double-check each step.1. I set the left-hand limit equal to the right-hand limit because of continuity.2. Subtracted 3m from both sides correctly.3. Simplified to get (m^2 - 5 = 0).4. Factored correctly into ((m - sqrt{5})(m + sqrt{5}) = 0).5. Solved for m correctly, getting (sqrt{5}) and (-sqrt{5}).6. Summed them to get 0, which also aligns with Vieta's formula.So, everything seems to check out. Therefore, the sum of all possible values of m is indeed 0.**Final Answer**The sum of all possible values of ( m ) is boxed{0}.

💡Alright, so I have this problem here where I need to find the sum of a_n for all positive integers n less than 3001. The definition of a_n is a bit specific, so I need to parse it carefully.First, let me restate the problem in my own words to make sure I understand it. For each n from 1 to 3000, a_n is defined as follows:- If n is divisible by both 15 and 18, then a_n is 15.- If n is divisible by both 18 and 17, then a_n is 18.- If n is divisible by both 15 and 17, then a_n is 21.- Otherwise, a_n is 0.So, essentially, a_n is non-zero only if n is divisible by two specific numbers, and the value depends on which pair of numbers divides n. My task is to sum all these a_n values from n=1 to n=3000.Hmm, okay. So, I think the first step is to figure out for each condition, how many numbers n satisfy that condition, and then multiply by the corresponding a_n value, and sum them all up.Let me break it down:1. **Case 1: a_n = 15** - This happens when n is divisible by both 15 and 18. - To find how many such n exist less than 3001, I need to find the least common multiple (LCM) of 15 and 18, and then find how many multiples of that LCM are less than 3001.2. **Case 2: a_n = 18** - This happens when n is divisible by both 18 and 17. - Similarly, find the LCM of 18 and 17, then count the multiples.3. **Case 3: a_n = 21** - This happens when n is divisible by both 15 and 17. - Again, find the LCM of 15 and 17, then count the multiples.4. **Case 4: a_n = 0** - This is the default case for all other numbers not covered by the above three conditions.So, I need to calculate the number of multiples for each LCM in the respective cases and then compute the sum.Let me start with the first case.**Case 1: a_n = 15**Divisible by both 15 and 18. So, LCM(15, 18). Let's compute that.15 factors into 3 * 5.18 factors into 2 * 3^2.So, LCM is the product of the highest powers of all primes present: 2 * 3^2 * 5 = 90.So, LCM(15, 18) = 90.Now, how many multiples of 90 are less than 3001?Well, the number of multiples is the floor of (3000 / 90).Calculating that: 3000 / 90 = 33.333...So, floor(33.333) = 33.Therefore, there are 33 numbers n less than 3001 that are divisible by both 15 and 18.Hence, the contribution to the sum from this case is 33 * 15 = 495.Okay, that's straightforward.**Case 2: a_n = 18**Divisible by both 18 and 17. So, LCM(18, 17).18 factors into 2 * 3^2.17 is a prime number.So, LCM is 2 * 3^2 * 17 = 306.Now, how many multiples of 306 are less than 3001?Compute 3000 / 306.306 * 9 = 2754306 * 10 = 3060, which is greater than 3000.So, there are 9 multiples of 306 less than 3001.Therefore, the contribution to the sum from this case is 9 * 18 = 162.Alright, moving on.**Case 3: a_n = 21**Divisible by both 15 and 17. So, LCM(15, 17).15 factors into 3 * 5.17 is prime.So, LCM is 3 * 5 * 17 = 255.Now, how many multiples of 255 are less than 3001?Compute 3000 / 255.255 * 11 = 2805255 * 12 = 3060, which is greater than 3000.So, there are 11 multiples of 255 less than 3001.Hence, the contribution to the sum from this case is 11 * 21 = 231.Now, I need to make sure that there is no overlap between these cases. That is, is there any number n that satisfies more than one condition? For example, a number that is divisible by 15, 18, and 17 would satisfy all three cases, but according to the definition, which case would it fall into?Looking back at the problem statement:- The first condition is if n is divisible by both 15 and 18, then a_n is 15.- The second condition is if n is divisible by both 18 and 17, then a_n is 18.- The third condition is if n is divisible by both 15 and 17, then a_n is 21.- Otherwise, a_n is 0.So, the conditions are checked in order, and the first condition that is satisfied determines the value of a_n. Therefore, if a number satisfies multiple conditions, it will only be counted in the first applicable case.Wait, is that the case? The problem statement doesn't specify the order of checking. It just lists the conditions. Hmm, that's a bit ambiguous.But in the problem statement, the cases are listed as:1. If divisible by 15 and 18, then 15.2. If divisible by 18 and 17, then 18.3. If divisible by 15 and 17, then 21.4. Otherwise, 0.So, it's structured as an if-else ladder. So, the first condition is checked first, and if it's satisfied, a_n is 15, and the other conditions are not checked. If not, then check the second condition, and so on.Therefore, if a number is divisible by 15, 18, and 17, it will satisfy the first condition (divisible by 15 and 18) and hence a_n will be 15, and it won't be counted in the other cases.Similarly, a number divisible by 18 and 17 but not by 15 will be counted in the second case, and so on.Therefore, there is no overlap in the counts because once a number satisfies a condition, it is assigned a_n accordingly and not considered in subsequent conditions.Therefore, the counts I have calculated for each case are mutually exclusive, and I can safely sum them up without worrying about double-counting.So, adding up the contributions:- Case 1: 495- Case 2: 162- Case 3: 231Total sum = 495 + 162 + 231.Let me compute that:495 + 162 = 657657 + 231 = 888So, the total sum is 888.Looking back at the options:A) 870B) 888C) 900D) 932E) 960So, 888 is option B.Therefore, the answer should be B) 888.Wait, just to double-check, let me verify my calculations.First, for multiples of 90:3000 / 90 = 33.333..., so 33 multiples. 33 * 15 = 495. That seems correct.Multiples of 306:3000 / 306 ≈ 9.803, so 9 multiples. 9 * 18 = 162. Correct.Multiples of 255:3000 / 255 ≈ 11.764, so 11 multiples. 11 * 21 = 231. Correct.Sum: 495 + 162 = 657; 657 + 231 = 888. Yep, that adds up.I think that's solid. So, I'm confident that the answer is 888, which is option B.**Final Answer**boxed{B}

💡Okay, so I have this problem where I need to prove that no matter how you color every point on a plane with just two colors, say black and white, there will always be an equilateral triangle with side length either 1 or √3 where all three vertices are the same color. Hmm, interesting. I remember hearing about something called the pigeonhole principle, which might be useful here. Let me try to think through this step by step.First, let me visualize the plane. It's infinite, so there are infinitely many points, each colored either black or white. I need to find three points that form an equilateral triangle with sides of length 1 or √3, and all three points must be the same color. So, maybe I can start by considering some specific configurations of points and see how their colors relate.Let me pick a point, say point A. Since the plane is infinite, I can choose another point B that is exactly 1 unit away from A. Now, depending on how the coloring is done, A and B could be the same color or different. If they're the same color, maybe I can find a third point that forms an equilateral triangle with A and B. If they're different, then I might need to look elsewhere.Wait, maybe I should think about this more systematically. I've heard of something called the Erdos theorem or maybe it's related to Ramsey theory? I think Ramsey theory deals with conditions under which order must appear. Maybe that's applicable here because we're dealing with colorings and finding monochromatic structures.Let me try to construct an equilateral triangle. If I have two points, A and B, 1 unit apart, then the set of points that form an equilateral triangle with A and B lie on two circles of radius 1 centered at A and B. The intersection points of these circles will give me two possible points, say C and D, such that triangles ABC and ABD are both equilateral with side length 1.Now, if either C or D is the same color as A and B, then we're done. But what if they're different? Hmm, maybe I need to consider more points or a different configuration. Maybe instead of just looking at triangles of side length 1, I should also consider triangles of side length √3.Let me think about the distance √3. If I have points that are √3 units apart, maybe they form a larger equilateral triangle. Perhaps I can relate this to the unit distance somehow. Maybe by considering points that are midpoints or something.Wait, let's consider a regular hexagon. Each side is length 1, and the distance between opposite vertices is 2. But the distance between adjacent vertices is 1, and the distance between vertices with one vertex in between is √3. So, in a regular hexagon, you have both distances 1 and √3. Maybe this structure can help me.If I color the vertices of a regular hexagon, which has six points, each colored either black or white, then by the pigeonhole principle, at least three of them must be the same color. But I need them to form an equilateral triangle. Hmm, in a regular hexagon, any three alternate vertices form an equilateral triangle. So, if I have three points of the same color among the six, they might form such a triangle.But wait, the regular hexagon has side length 1, so the distance between alternate vertices is 2, not √3. Hmm, maybe I need a different approach.Let me try to think about tiling the plane with equilateral triangles. If I create a grid of equilateral triangles with side length 1, then the plane is covered with small triangles. But since the coloring is arbitrary, I can't assume any regularity in the coloring. Maybe I need to find a specific configuration where regardless of the coloring, a monochromatic triangle must exist.Alternatively, maybe I can use a proof by contradiction. Suppose that there is a coloring where no such monochromatic equilateral triangle exists. Then, I need to show that this leads to some kind of contradiction.Let me try that. Assume that there is a two-coloring of the plane with no monochromatic equilateral triangles of side length 1 or √3. Then, for any two points of the same color, the third point needed to form an equilateral triangle must be of the opposite color. Hmm, but this might get complicated because the plane is infinite, and there are infinitely many such triangles.Wait, maybe I can use a finite configuration to force a contradiction. Let me consider a specific set of points where if I color them, I must end up with a monochromatic triangle. For example, if I have seven points arranged in a certain way, maybe I can ensure that at least one triangle is monochromatic.Alternatively, I remember something called the van der Waerden theorem, which is about arithmetic progressions, but I'm not sure if that applies here. Maybe it's more related to the Hales-Jewett theorem, but that's more about high-dimensional spaces.Wait, another idea: maybe I can use graph theory. If I model the plane as a graph where each point is a vertex and edges connect points that are 1 or √3 units apart, then I'm looking for a monochromatic triangle in this graph. But the plane is uncountably infinite, so standard graph theory might not apply directly.Hmm, maybe I need to consider a more geometric approach. Let me think about circles. If I fix a point A, then all points at distance 1 from A form a circle of radius 1. Similarly, points at distance √3 form another circle. Maybe by considering intersections of these circles, I can find points that must form a triangle.Wait, if I have two points A and B, 1 unit apart, then the circles of radius 1 around A and B intersect at two points, forming the equilateral triangle. If those intersection points are colored differently, then maybe I can find another pair of points that form a triangle of side length √3.Alternatively, maybe I can use a hexagonal lattice. In a hexagonal lattice with side length 1, the distance between adjacent points is 1, and the distance between points two steps apart is √3. So, in such a lattice, any three points forming an equilateral triangle would have sides of length 1 or √3.If I color the points of this lattice, by the pigeonhole principle, at least half of them are the same color. But how does that help me? Maybe if I consider a large enough section of the lattice, I can guarantee a monochromatic triangle.Wait, actually, I recall that in any two-coloring of the plane, there exists a monochromatic unit distance graph. But I'm not sure if that's exactly what I need here.Let me try to think differently. Suppose I have a point A. I can draw a circle of radius 1 around A. Now, consider another point B on this circle. The circle of radius 1 around B will intersect the first circle at two points, say C and D. So, triangles ABC and ABD are both equilateral with side length 1.Now, if either C or D is the same color as A and B, then we have our monochromatic triangle. If not, then C and D must be the other color. But then, considering points C and D, which are both the other color, and the distance between them is 2 units. Hmm, not directly helpful.Wait, but the distance between C and D is actually 2 units, right? Because in an equilateral triangle, the distance between the two intersection points is 2 times the height, which is √3. Wait, no, the distance between C and D is actually 2 units because they are both on the circle of radius 1 around A and B, which are 1 unit apart. So, the distance between C and D is 2 units.But 2 units isn't one of the required side lengths. Hmm, maybe I need to consider another configuration.Alternatively, let's consider a point A and draw two circles: one of radius 1 and another of radius √3. The intersection points of these circles with other circles might give me the required triangles.Wait, if I have a point A, and I draw a circle of radius 1 around it, then pick a point B on that circle. Then, draw a circle of radius √3 around B. The intersection points of this circle with the original circle around A might give me points that form an equilateral triangle with side length √3.Let me calculate the distance between A and B is 1, and the distance from B to C is √3. Then, the distance from A to C would be... Let me use the law of cosines. If AB = 1, BC = √3, and angle at B is 60 degrees because it's part of an equilateral triangle, then AC² = AB² + BC² - 2*AB*BC*cos(60°). So, AC² = 1 + 3 - 2*1*√3*(0.5) = 4 - √3. Hmm, that's not a nice number, but maybe it's not necessary.Wait, maybe I'm overcomplicating this. Let me try to think about specific configurations where the monochromatic triangle must appear.Suppose I have a point A. I color it black. Then, all points at distance 1 from A form a circle. If any of those points is black, then I can try to find another point to form a triangle. If none of them are black, then all points at distance 1 from A are white. But then, considering those white points, maybe they form a structure where another triangle must appear.Wait, but the plane is infinite, so even if all points at distance 1 from A are white, there are still infinitely many white points. Maybe I can find three white points that form an equilateral triangle.Alternatively, maybe I can use the idea of a graph where each vertex is connected to others at specific distances, and then use Ramsey numbers to show that a monochromatic triangle must exist. But I'm not sure about the exact Ramsey numbers for this case.Wait, another thought: maybe I can use a tiling of the plane with equilateral triangles of side length 1. Then, in each tile, if the vertices are colored, I can find a monochromatic triangle. But since the coloring is arbitrary, I can't assume any pattern.Hmm, this is tricky. Maybe I need to look for a specific configuration of points where the coloring forces a monochromatic triangle. Let me try to think of seven points arranged in a certain way, like a hexagon with a center point. Then, by the pigeonhole principle, at least four of them must be the same color. Maybe among those four, I can find three that form an equilateral triangle.Wait, in a regular hexagon with a center point, the distances between the center and the vertices are all the same, say 1 unit. Then, the distance between adjacent vertices is also 1 unit, and the distance between opposite vertices is 2 units. But I need distances of 1 or √3. Hmm, maybe not directly applicable.Wait, if I have a regular hexagon with side length 1, then the distance between opposite vertices is 2, but the distance between vertices with one vertex in between is √3. So, in such a hexagon, there are points at distances 1 and √3. So, if I color the six vertices, by the pigeonhole principle, at least three are the same color. If those three are adjacent, they form a triangle with side length 1. If they are spaced out, they form a triangle with side length √3.Wait, that might work. Let me elaborate. Consider a regular hexagon with side length 1. Label the vertices A, B, C, D, E, F in order. The distance between A and B is 1, between A and C is √3, and between A and D is 2. Now, if I color each vertex either black or white, by the pigeonhole principle, at least three of them must be the same color.Suppose three of them are black. If those three are consecutive, say A, B, C, then the distance between A and B is 1, B and C is 1, and A and C is √3. So, triangle ABC has sides of length 1 and √3, but it's not equilateral. Wait, no, triangle ABC is actually an equilateral triangle because all sides are either 1 or √3, but actually, in a regular hexagon, triangle ABC is not equilateral because AB = 1, BC = 1, but AC = √3. So, it's an isosceles triangle, not equilateral.Wait, but I need an equilateral triangle. So, maybe I need to consider a different set of three points. If the three black points are A, C, and E, then the distance between A and C is √3, C and E is √3, and E and A is √3. So, triangle ACE is equilateral with side length √3. Similarly, if the three black points are B, D, and F, then triangle BDF is equilateral with side length √3.Alternatively, if the three black points are A, B, and D, then the distances are AB = 1, BD = 2, and AD = 2, which doesn't form an equilateral triangle. So, maybe the key is to ensure that the three points are either consecutive or spaced out by one vertex.Wait, so if I have a regular hexagon, and I color its vertices with two colors, then either there are three consecutive vertices of the same color, forming a triangle with side length 1, or there are three vertices spaced out by one, forming a triangle with side length √3. Therefore, in either case, there must be a monochromatic equilateral triangle of side length 1 or √3.But wait, is this always true? Let me check. Suppose I color the hexagon in such a way that no three consecutive vertices are the same color and no three spaced out vertices are the same color. Is that possible?Let me try to color the hexagon alternately: black, white, black, white, black, white. Then, there are no three consecutive vertices of the same color, and the spaced out points (A, C, E) are black, white, black, which are not all the same. Similarly, (B, D, F) are white, black, white. So, in this case, there are no monochromatic triangles of side length 1 or √3. Wait, but that contradicts my earlier reasoning.Hmm, so maybe my approach is flawed. Maybe the regular hexagon isn't the right configuration to use. Or perhaps I need to consider more points or a different arrangement.Wait, maybe I need to consider a larger structure. Let me think about a tiling of the plane with equilateral triangles, creating a honeycomb pattern. In such a tiling, each point is part of multiple equilateral triangles of different sizes. Maybe by considering a sufficiently large section of this tiling, I can ensure that a monochromatic triangle must exist.Alternatively, perhaps I can use a proof by contradiction. Suppose that there is a two-coloring of the plane with no monochromatic equilateral triangles of side length 1 or √3. Then, for any two points of the same color, the third point needed to form an equilateral triangle must be of the opposite color. But this might lead to an infinite regression or some kind of contradiction because the plane is densely populated with points.Wait, another idea: maybe I can use the concept of a minimal monochromatic configuration. If I can find a finite set of points where any two-coloring must contain a monochromatic equilateral triangle, then I can apply that to the entire plane.I recall that in Ramsey theory, certain configurations are unavoidable in sufficiently large structures. Maybe I can use a similar idea here. For example, if I can show that any two-coloring of a sufficiently large finite set of points must contain a monochromatic equilateral triangle, then since the plane is infinite, such a triangle must exist.But I'm not sure about the exact Ramsey numbers for this specific case. Maybe I need to look for a specific configuration of seven points or something like that.Wait, let me think about the Moser spindle. It's a unit distance graph with seven vertices that is not 3-colorable. Maybe it can be used here, but I'm not sure how.Alternatively, maybe I can use the idea of a graph where each vertex represents a point on the plane, and edges connect points that are 1 or √3 units apart. Then, I'm looking for a monochromatic triangle in this graph. But again, the plane is infinite, so standard graph theory might not apply directly.Wait, perhaps I can use a compactness argument. If I can show that for any finite set of points, there exists a monochromatic equilateral triangle, then by compactness, the same must hold for the entire plane. But I'm not sure how to formalize that.Hmm, this is getting complicated. Maybe I need to go back to the basics and try to construct such a triangle step by step.Let me start by picking a point A. Without loss of generality, let's say it's black. Now, consider all points at distance 1 from A. If any of these points is black, say point B, then I can look for a point C such that ABC is an equilateral triangle with side length 1. If C is black, then we're done. If not, then C is white.Now, consider point C, which is white. Look for points at distance 1 from C. If any of these points is white, say point D, then check if D forms an equilateral triangle with C and another white point. If not, continue this process. But since the plane is infinite, this might not lead to a contradiction.Alternatively, maybe I can use the fact that if I have two points of the same color, the third point needed to form an equilateral triangle must be of the opposite color, leading to an infinite chain of alternating colors, which is impossible because the plane is densely populated.Wait, perhaps I can use a geometric argument. Consider a point A and draw a circle of radius 1 around it. If all points on this circle are white, then pick any two points B and C on the circle. The distance between B and C is at most 2 units. If B and C are both white, then the distance between them could be 1 or √3, forming a triangle. But I'm not sure.Wait, another approach: use the concept of a lattice. Consider a triangular lattice where each point is part of multiple equilateral triangles. If I color this lattice, by the pigeonhole principle, there must be a monochromatic triangle.But again, the plane is infinite, so I need a finite configuration to apply the pigeonhole principle. Maybe consider a finite section of the lattice where the number of points ensures a monochromatic triangle.Wait, I think I'm overcomplicating this. Let me try to think of a specific example. Suppose I have a point A, and I draw a circle of radius 1 around it. If any point on this circle is black, say B, then draw another circle of radius 1 around B. The intersection points of these two circles, say C and D, form equilateral triangles ABC and ABD. If either C or D is black, then we have our triangle. If not, then C and D are white.Now, consider point C, which is white. Draw a circle of radius 1 around C. If any point on this circle is white, say E, then check if E forms an equilateral triangle with C and another white point. If E is white, then maybe E is part of another triangle. But this seems like it could go on indefinitely without guaranteeing a monochromatic triangle.Wait, maybe I need to consider the distance √3. If I have points A and B, 1 unit apart, and then consider points at distance √3 from both A and B, those points might form an equilateral triangle with side length √3.Let me calculate the intersection points of circles of radius √3 around A and B. The distance between A and B is 1, so the intersection points will form two points, say C and D, such that AC = BC = √3. The distance between C and D can be calculated using the law of cosines. The angle at A is 60 degrees because it's part of an equilateral triangle. So, CD² = AC² + AD² - 2*AC*AD*cos(60°). Wait, but AD is also √3, so CD² = 3 + 3 - 2*√3*√3*(0.5) = 6 - 3 = 3. So, CD = √3. Therefore, triangle ACD is equilateral with side length √3.So, if points C and D are both white, then triangle ACD is white. If not, then at least one of them must be black, which would form a triangle with A and B.Wait, but if points C and D are both white, then we have a monochromatic triangle. If one is black and the other is white, then maybe we can find another triangle.Hmm, this seems promising. Let me try to formalize this.1. Start with point A, colored black.2. Draw a circle of radius 1 around A. If any point B on this circle is black, proceed. If all are white, then consider the circle of radius √3 around A and find a monochromatic triangle there.3. Suppose B is black. Draw a circle of radius 1 around B. The intersection points C and D of this circle with the circle around A form equilateral triangles ABC and ABD.4. If either C or D is black, then we have a monochromatic triangle. If both are white, then consider the circle of radius √3 around C.5. The intersection points of the circle around C with the circle around A (radius √3) will form another equilateral triangle. If any of these points are white, we have a monochromatic triangle. If not, continue this process.But this seems like it could lead to an infinite loop. Maybe I need a different approach.Wait, another idea: use the concept of a graph where each vertex is connected to others at specific distances, and then use Ramsey numbers to show that a monochromatic triangle must exist. But I'm not sure about the exact Ramsey numbers for this case.Alternatively, maybe I can use a tiling argument. If I tile the plane with equilateral triangles of side length 1, then in each tile, if the vertices are colored, I can find a monochromatic triangle. But since the coloring is arbitrary, I can't assume any pattern.Wait, maybe I can use the fact that in any two-coloring of the plane, there exists a monochromatic pair of points at any given distance. But I need more than that; I need three points forming a triangle.Wait, I think I'm getting closer. Let me try to outline the proof step by step.1. Start with any point A. Without loss of generality, assume it's black.2. Draw a circle of radius 1 around A. If any point B on this circle is black, proceed. If all are white, then consider the circle of radius √3 around A.3. Suppose B is black. Draw a circle of radius 1 around B. The intersection points C and D of this circle with the circle around A form equilateral triangles ABC and ABD.4. If either C or D is black, then we have a monochromatic triangle. If both are white, then consider the circle of radius √3 around C.5. The intersection points of the circle around C with the circle around A (radius √3) will form another equilateral triangle. If any of these points are white, we have a monochromatic triangle. If not, continue this process.6. However, since the plane is infinite, this process must eventually lead to a monochromatic triangle because we can't have an infinite chain of alternating colors without forming a triangle.Wait, but I'm not sure if this is rigorous enough. Maybe I need to use a more formal argument.Alternatively, consider the following configuration: take a regular hexagon with side length 1. As I thought earlier, the distances between opposite vertices are 2 units, and the distances between vertices with one vertex in between are √3. Now, color the six vertices with two colors. By the pigeonhole principle, at least three vertices must be the same color.If those three vertices are consecutive, then the distances between them are 1 unit, forming an equilateral triangle. If they are spaced out, then the distances between them are √3 units, forming another equilateral triangle. Therefore, in either case, there must be a monochromatic equilateral triangle of side length 1 or √3.But earlier, I thought of a coloring where this doesn't happen, like alternating colors. Wait, in that case, the spaced-out points would alternate colors as well, so they wouldn't all be the same color. Hmm, so maybe my earlier reasoning was incorrect.Wait, let me clarify. If I color the hexagon alternately, black, white, black, white, etc., then the three spaced-out points (A, C, E) would be black, white, black, which are not all the same color. Similarly, (B, D, F) would be white, black, white. So, in this case, there are no monochromatic triangles. Therefore, my earlier approach using a hexagon doesn't work.Hmm, so maybe I need a different configuration. Let me think of seven points arranged in a certain way. Maybe a hexagon with a center point. Then, by the pigeonhole principle, at least four points must be the same color. Among these four, maybe three form an equilateral triangle.Wait, let's consider a regular hexagon with a center point, making seven points in total. If I color these seven points with two colors, by the pigeonhole principle, at least four must be the same color. Now, among these four points, I need to find three that form an equilateral triangle.If three of them are consecutive on the hexagon, they form a triangle with side length 1. If they are spaced out, they form a triangle with side length √3. But wait, in a hexagon with a center point, the distances from the center to the vertices are all the same, say 1 unit. Then, the distance between adjacent vertices is also 1 unit, and the distance between opposite vertices is 2 units.Wait, but I need distances of 1 or √3. So, maybe the center point and two adjacent vertices form a triangle with sides 1, 1, and √3. Let me check. If the center is O, and two adjacent vertices are A and B, then OA = OB = 1, and AB = 1. So, triangle OAB has sides 1, 1, and 1, which is equilateral. Wait, no, in a regular hexagon, the distance from the center to a vertex is equal to the side length, so if the side length is 1, then OA = OB = 1, and AB = 1. So, triangle OAB is equilateral with side length 1.Similarly, if I take the center O and two vertices with one vertex in between, say A and C, then OA = OC = 1, and AC = √3. So, triangle OAC is equilateral with side length √3.Therefore, in this configuration, if four points are the same color, at least three of them must form an equilateral triangle of side length 1 or √3. Because among the four points, either three are consecutive on the hexagon, forming a triangle of side length 1, or three include the center and two vertices spaced out, forming a triangle of side length √3.Wait, but what if the four points are arranged such that they don't form a triangle? For example, two on the hexagon and two at the center? Wait, no, the center is just one point. So, if four points are the same color, they must include either three on the hexagon or three including the center and two on the hexagon.If three are on the hexagon, they can be consecutive or spaced out. If consecutive, they form a triangle of side length 1. If spaced out, they form a triangle of side length √3. If the three include the center and two on the hexagon, then depending on their positions, they form a triangle of side length 1 or √3.Therefore, in this configuration, any four points of the same color must contain a monochromatic equilateral triangle of side length 1 or √3. Since we have seven points in total, and we're using two colors, by the pigeonhole principle, at least four must be the same color. Therefore, such a triangle must exist.Thus, in any two-coloring of the plane, there must exist a monochromatic equilateral triangle of side length 1 or √3.Wait, but I'm not sure if this argument is complete. I need to ensure that the seven-point configuration is unavoidable in the plane. Since the plane is infinite, I can always find such a configuration somewhere, right? Because no matter how you color the plane, you can find a regular hexagon with a center point, and among those seven points, at least four must be the same color, leading to a monochromatic triangle.Therefore, this seems like a valid approach. So, to summarize:1. Consider a regular hexagon with a center point, making seven points in total.2. By the pigeonhole principle, at least four of these seven points must be the same color.3. Among these four points, at least three must form an equilateral triangle of side length 1 or √3.4. Therefore, in any two-coloring of the plane, such a monochromatic triangle must exist.I think this makes sense. I might need to check if there are any colorings that avoid this, but given the pigeonhole principle and the structure of the hexagon, it seems unavoidable.

💡Okay, so I have this problem here about a function f(x) defined on all real numbers. It says that for any x and y in real numbers, f(x) + f(y) equals f(x + y). And then I have to figure out which of the options A, B, C, or D is true regarding whether f(x) is odd, even, both, or neither.Alright, let me start by recalling what it means for a function to be odd or even. An even function satisfies f(-x) = f(x) for all x, and an odd function satisfies f(-x) = -f(x) for all x. So, the question is asking whether f(x) is odd, even, both, or neither based on the given condition.The given condition is f(x) + f(y) = f(x + y) for all x and y. Hmm, this reminds me of the property of linear functions, specifically functions of the form f(x) = kx, where k is a constant. Because if f(x) = kx, then f(x) + f(y) = kx + ky = k(x + y) = f(x + y). So, that seems to fit.But wait, does this condition necessarily mean that f(x) is linear? I think so, but I'm not entirely sure. Maybe there are other functions that satisfy this additive property. I recall something called Cauchy's functional equation, which is exactly f(x + y) = f(x) + f(y). The solutions to this equation are linear functions of the form f(x) = kx when we assume continuity or some other regularity condition. But without any additional constraints, there are pathological solutions as well. However, since the problem doesn't specify anything else, I think it's safe to assume that f(x) is linear, especially since the options are about being odd or even, which are properties of linear functions.So, if f(x) is linear, then f(x) = kx. Now, let's check if this function is odd or even. Let's compute f(-x):f(-x) = k(-x) = -kx = -f(x).So, f(-x) = -f(x), which means f(x) is an odd function. Therefore, option A should be correct.But wait, let me double-check. Maybe I should verify this without assuming f(x) is linear. Let's see.First, let's set x = y = 0 in the given equation:f(0) + f(0) = f(0 + 0) => 2f(0) = f(0).Subtracting f(0) from both sides gives 2f(0) - f(0) = f(0) => f(0) = 0.Okay, so f(0) is 0. That's a good start.Now, let's set y = -x. Then, the equation becomes:f(x) + f(-x) = f(x + (-x)) = f(0) = 0.So, f(x) + f(-x) = 0 => f(-x) = -f(x).This is exactly the definition of an odd function. So, regardless of whether f(x) is linear or not, as long as it satisfies f(x) + f(y) = f(x + y), it must be odd.Therefore, option A is correct.Wait, but what about the other options? Let me see.Option B says f(x) is even. But we just found that f(-x) = -f(x), which is the opposite of being even. So, B is not correct.Option C says f(x) is both odd and even. For a function to be both odd and even, it must satisfy f(-x) = f(x) and f(-x) = -f(x). The only function that satisfies both is the zero function, f(x) = 0. But in our case, f(x) could be any linear function f(x) = kx, which is only zero if k = 0. So, unless f(x) is identically zero, it's not both odd and even. But the problem doesn't specify that f(x) is zero, so we can't assume that. Therefore, C is not necessarily true.Option D says f(x) is neither odd nor even. But we just showed that f(x) must be odd, so D is not correct either.So, after going through all the options, A is the correct answer.I think I've covered all the bases here. I started by recalling the definitions, checked the properties by plugging in specific values, and even considered whether the function must be linear. It seems solid. I don't see any mistakes in my reasoning, so I'm confident that A is the right choice.

💡Alright, so I have these two quadratic functions, and I need to find the coordinates of their vertices. The first one, y = 2x² - 4x - 1, needs to be solved by completing the square. The second one, y = -3x² + 6x - 2, should be solved using the vertex formula. Hmm, okay, let me take this step by step.Starting with the first function: y = 2x² - 4x - 1. I remember that completing the square involves turning the quadratic part into a perfect square trinomial, which can then be written as a squared binomial. That should help me identify the vertex easily.First, I notice that the coefficient of x² is 2, not 1. I think I need to factor out the 2 from the first two terms to make completing the square easier. So, let me rewrite the equation:y = 2(x² - 2x) - 1Okay, so I factored out the 2 from 2x² and -4x, which gives me x² - 2x inside the parentheses. Now, to complete the square, I need to add and subtract the square of half the coefficient of x inside the parentheses. The coefficient of x is -2, so half of that is -1, and squaring it gives 1.So, I'll add and subtract 1 inside the parentheses:y = 2(x² - 2x + 1 - 1) - 1Wait, that might not be the best way to write it. Let me think. Since I factored out a 2, adding 1 inside the parentheses actually adds 2*1 to the equation. To keep the equation balanced, I need to subtract that same amount outside the parentheses.So, let me correct that:y = 2(x² - 2x + 1) - 2*1 - 1Simplifying that, I get:y = 2(x - 1)² - 2 - 1Which simplifies further to:y = 2(x - 1)² - 3Now, this is in the vertex form of a quadratic equation, which is y = a(x - h)² + k, where (h, k) is the vertex. So, comparing, h is 1 and k is -3. Therefore, the vertex is at (1, -3).Okay, that seems straightforward. Let me double-check my steps to make sure I didn't make a mistake. I factored out the 2 correctly, completed the square by adding and subtracting 1 inside the parentheses, adjusted for the factoring by subtracting 2 outside, and simplified. Yep, that looks right.Now, moving on to the second function: y = -3x² + 6x - 2. This time, I need to use the vertex formula. I remember that for a quadratic function in standard form, y = ax² + bx + c, the x-coordinate of the vertex is given by -b/(2a). Once I find that, I can plug it back into the original equation to find the corresponding y-coordinate.Let's identify a, b, and c from the equation. Here, a = -3, b = 6, and c = -2.Calculating the x-coordinate:x = -b/(2a) = -6/(2*(-3)) = -6/(-6) = 1Okay, so the x-coordinate of the vertex is 1. Now, to find the y-coordinate, I substitute x = 1 back into the original equation:y = -3(1)² + 6(1) - 2 = -3 + 6 - 2 = 1So, the y-coordinate is 1. Therefore, the vertex is at (1, 1).Wait a second, that seems too similar to the first vertex. Is that possible? Let me check my calculations again. For the x-coordinate, -6 divided by (2*(-3)) is indeed 1. Plugging x = 1 into the equation: -3(1) + 6(1) - 2. Wait, hold on, that's not quite right. The first term is -3(1)², which is -3*1 = -3. Then, 6*1 = 6. So, -3 + 6 is 3, minus 2 is 1. Yeah, that's correct. So, the vertex is indeed at (1, 1).Hmm, interesting. Both vertices are at x = 1, but different y-values. That makes sense because the quadratics have different coefficients and constants, so their vertices can be at the same x but different y.Let me just recap what I did. For the first function, I completed the square by factoring out the coefficient of x², then creating a perfect square trinomial by adding and subtracting the appropriate constant. This transformed the equation into vertex form, from which I could directly read off the vertex coordinates.For the second function, I used the vertex formula, which gives the x-coordinate as -b/(2a). Once I found that, substituting back into the original equation gave me the y-coordinate. Both methods are valid and lead to the correct vertex coordinates.I think I'm confident with these results. I don't see any mistakes in my calculations, and both methods confirm the vertices at (1, -3) and (1, 1) respectively.