💡First, I need to determine the actual distance the person traveled at the original speed of 12 km/hr.I'll denote the time spent walking as ( t ) hours.At 12 km/hr, the distance traveled is ( 12t ) km.If the person walked at 20 km/hr instead, the distance would be ( 20t ) km.According to the problem, walking at 20 km/hr results in 30 km more than walking at 12 km/hr. This gives the equation:[ 20t = 12t + 30 ]Solving for ( t ):[ 20t - 12t = 30 ][ 8t = 30 ][ t = frac{30}{8} ][ t = 3.75 ] hoursNow, calculate the actual distance traveled at 12 km/hr:[ text{Distance} = 12t ][ text{Distance} = 12 times 3.75 ][ text{Distance} = 45 ] kmTherefore, the actual distance traveled by the person is 45 km.

💡Okay, so I have this problem where every point in the plane is colored either red, green, or blue. I need to show that there exists a rectangle whose vertices are all the same color. Hmm, interesting. I remember something about the pigeonhole principle being useful in these kinds of problems. Let me think about how to approach this.First, maybe I should consider a grid of points. If I can find a grid where the coloring leads to a monochromatic rectangle, that would solve the problem. Let me try to visualize a grid with several rows and columns. Suppose I have a grid with 4 rows and some number of columns. Each column has 4 points, each colored either red, green, or blue.Now, for each column, there are 3 colors, and each of the 4 points can be any of these colors. So, the number of possible colorings for a single column is 3^4, which is 81. That means there are 81 different ways to color a column with 4 points using 3 colors.If I have more than 81 columns, by the pigeonhole principle, at least two columns must have the same coloring. So, if I have 82 columns, there must be at least two columns that are identical in color. Let's say these two columns are column A and column B.Now, looking at these two identical columns, each has 4 points colored the same way. So, for each row, the color in column A is the same as in column B. Now, if I look at the rows, there are 4 rows, each with two points (one in column A and one in column B). Since there are 4 rows and only 3 colors, by the pigeonhole principle again, at least two of these rows must have the same color in both columns.Let's say rows 1 and 2 are both red in both columns A and B. Then, the points (A,1), (B,1), (A,2), and (B,2) form a rectangle, and all four points are red. So, that's a monochromatic rectangle.Wait, but what if the two identical columns don't have two rows with the same color? Let me check. If each column has 4 points, and each point can be one of three colors, then in each column, there must be at least two points of the same color, right? Because with 4 points and 3 colors, by the pigeonhole principle, at least one color must be used at least twice.So, in each column, there are at least two points of the same color. Now, if two columns are identical, then those two points in the same rows will also be the same color. So, if I have two columns where the same two rows are colored the same, then those four points form a rectangle.But wait, what if the two identical columns have different pairs of rows with the same color? For example, column A has two reds in rows 1 and 2, and column B has two reds in rows 3 and 4. Then, do we still get a rectangle? Hmm, in that case, the points (A,1), (B,1), (A,2), (B,2) would be red in column A but not necessarily in column B. So, maybe that doesn't form a rectangle.Wait, no. If the columns are identical, then the coloring in column B is exactly the same as in column A. So, if column A has two reds in rows 1 and 2, column B also has two reds in rows 1 and 2. Therefore, the points (A,1), (B,1), (A,2), (B,2) are all red, forming a rectangle.But what if the two identical columns have different color distributions? For example, column A has two reds in rows 1 and 2, and column B has two reds in rows 3 and 4. But since the columns are identical, column B must have the same coloring as column A. So, if column A has two reds in rows 1 and 2, column B must also have two reds in rows 1 and 2. Therefore, the rectangle is formed.Wait, I think I confused myself earlier. If the columns are identical, then their colorings are exactly the same. So, if column A has two reds in rows 1 and 2, column B must also have two reds in rows 1 and 2. Therefore, the four points (A,1), (B,1), (A,2), (B,2) are all red, forming a rectangle.So, in general, if we have a grid with 4 rows and 82 columns, each column colored with 3 colors, then by the pigeonhole principle, two columns must be identical. In those two columns, since they have the same coloring, there must be at least two rows with the same color, forming a rectangle.But wait, the problem says "each point in the plane," not just a grid. So, does this argument extend to the entire plane? I think so, because the plane contains infinitely many points, but we can always find a grid within it. So, by considering a sufficiently large grid, we can guarantee the existence of a monochromatic rectangle.Alternatively, maybe I can think of it as coloring the entire plane, which includes all possible grids. So, by the same argument, there must be a grid somewhere in the plane with the required properties.Hmm, I think this makes sense. By using the pigeonhole principle twice—first to find two identical columns, and then to find two rows within those columns with the same color—we can guarantee the existence of a monochromatic rectangle.Let me try to summarize the steps:1. Consider a grid with 4 rows and 82 columns.2. Each column has 4 points, each colored red, green, or blue.3. The number of possible colorings for a column is 3^4 = 81.4. With 82 columns, by the pigeonhole principle, at least two columns must have the same coloring.5. In these two identical columns, look at the colors in each row.6. Since there are 4 rows and 3 colors, by the pigeonhole principle, at least two rows must have the same color in both columns.7. These four points form a rectangle with all vertices of the same color.Yes, that seems to cover it. I think I've got the gist of the proof. It relies heavily on the pigeonhole principle applied twice—once for the columns and once for the rows within those columns.I wonder if there's a way to generalize this for more colors or different grid sizes. For example, if we had more colors, say k colors, and we wanted to find a monochromatic rectangle, how would the grid dimensions change? Probably, the number of columns needed would increase exponentially with the number of colors, similar to how 3^4 = 81 columns were needed for 3 colors.Also, I'm curious if this problem is related to Ramsey theory, which deals with conditions under which order must appear. It seems similar because we're looking for a structured object (a rectangle) within a colored plane, which is a kind of Ramsey-type problem.But for now, focusing on the problem at hand, I think the argument using the pigeonhole principle is solid. It ensures that with a sufficiently large grid, we can find the required monochromatic rectangle.Just to double-check, let's plug in the numbers:- 4 rows, 82 columns.- Each column: 4 points, 3 colors.- Number of colorings per column: 3^4 = 81.- With 82 columns, two must be identical.- In those two columns, 4 rows, 3 colors: at least two rows share the same color.- Thus, four points forming a rectangle are monochromatic.Yes, that all adds up. I don't see any gaps in the logic. The key steps are applying the pigeonhole principle twice and ensuring that the grid is large enough to force the existence of the rectangle.So, in conclusion, by constructing a grid with enough columns and applying the pigeonhole principle, we can guarantee the existence of a monochromatic rectangle in the plane.

💡First, I need to determine the new duration John can run after increasing his original time by 75%. The original duration is 8 hours. To find 75% of 8 hours:[0.75 times 8 = 6 text{ hours}]Adding this increase to the original duration:[8 text{ hours} + 6 text{ hours} = 14 text{ hours}]Next, I'll calculate John's new speed after increasing his original speed of 8 mph by 4 mph:[8 text{ mph} + 4 text{ mph} = 12 text{ mph}]Finally, to find out how far John can run with his new duration and speed, I'll multiply the two:[text{Distance} = text{Duration} times text{Speed} = 14 text{ hours} times 12 text{ mph} = 168 text{ miles}]

💡First, I need to simplify the complex number ( z = frac{4 + 2i}{(1 + i)^2} ). To do this, I'll start by expanding the denominator.Calculating ( (1 + i)^2 ):[(1 + i)^2 = 1^2 + 2 cdot 1 cdot i + i^2 = 1 + 2i + (-1) = 2i]Now, substitute the simplified denominator back into the expression for ( z ):[z = frac{4 + 2i}{2i}]To simplify further, I'll multiply both the numerator and the denominator by the complex conjugate of the denominator, which is (-i):[z = frac{(4 + 2i)(-i)}{2i cdot (-i)} = frac{-4i - 2i^2}{-2i^2}]Since ( i^2 = -1 ), this becomes:[z = frac{-4i - 2(-1)}{-2(-1)} = frac{-4i + 2}{2}]Simplifying the numerator:[z = frac{2 - 4i}{2} = 1 - 2i]The complex number ( z = 1 - 2i ) corresponds to the point ( (1, -2) ) in the complex plane.Next, I'll substitute ( x = 1 ) and ( y = -2 ) into the line equation ( x - 2y + m = 0 ):[1 - 2(-2) + m = 0][1 + 4 + m = 0][5 + m = 0][m = -5]Therefore, the value of ( m ) is (-5).

💡First, I need to determine the number of men in the first group, denoted as ( M ).The total amount of work can be expressed in terms of man-days. If ( M ) men can complete the work in 25 days, the total work is ( M times 25 ) man-days.Similarly, if 20 men can complete the same work in 20 days, the total work is ( 20 times 20 ) man-days.Since both expressions represent the same amount of work, I can set them equal to each other:[M times 25 = 20 times 20]Solving for ( M ):[M = frac{20 times 20}{25} = frac{400}{25} = 16]Therefore, there were 16 men in the first group.

💡Okay, so I have this triangle ABC where angle A is 30 degrees, angle B is 45 degrees, and side a is the square root of 2. I need to find the length of side b and the area of the triangle. Hmm, let's see. First, I remember that in any triangle, the sum of the angles is 180 degrees. So, if angle A is 30 and angle B is 45, then angle C must be 180 minus 30 minus 45. Let me calculate that: 180 - 30 is 150, and 150 - 45 is 105. So angle C is 105 degrees. Okay, got that down.Now, I need to find the length of side b. I recall the Law of Sines, which says that in any triangle, the ratio of a side to the sine of its opposite angle is constant. So, that means a over sin A equals b over sin B. Let me write that down:a / sin A = b / sin BGiven that a is sqrt(2), angle A is 30 degrees, and angle B is 45 degrees, I can plug those values in:sqrt(2) / sin(30) = b / sin(45)I know that sin(30) is 0.5 and sin(45) is sqrt(2)/2. Let me substitute those values in:sqrt(2) / 0.5 = b / (sqrt(2)/2)Simplifying the left side: sqrt(2) divided by 0.5 is the same as sqrt(2) multiplied by 2, which is 2*sqrt(2). So now I have:2*sqrt(2) = b / (sqrt(2)/2)To solve for b, I can multiply both sides by sqrt(2)/2:b = 2*sqrt(2) * (sqrt(2)/2)Let me compute that. The 2 in the numerator and the 2 in the denominator cancel out, leaving sqrt(2) * sqrt(2), which is 2. So, b equals 2. Okay, that seems straightforward.Now, onto the area of the triangle. I remember there are a few formulas for the area of a triangle. One common one is (1/2)*base*height, but I don't know the height here. Another formula is using two sides and the sine of the included angle: (1/2)*a*b*sin(C). Since I know sides a and b, and I know angle C, this should work.So, plugging in the values I have:Area = (1/2) * a * b * sin(C)We know a is sqrt(2), b is 2, and angle C is 105 degrees. So:Area = (1/2) * sqrt(2) * 2 * sin(105)Simplify that: the 1/2 and the 2 cancel each other out, so we're left with sqrt(2) * sin(105). Now, I need to figure out what sin(105) is.I remember that 105 degrees can be expressed as 60 degrees plus 45 degrees. So, sin(105) is sin(60 + 45). There's a sine addition formula that says sin(A + B) = sin A cos B + cos A sin B. Let me apply that:sin(105) = sin(60 + 45) = sin(60)*cos(45) + cos(60)*sin(45)I know the values of these sines and cosines:sin(60) = sqrt(3)/2cos(45) = sqrt(2)/2cos(60) = 1/2sin(45) = sqrt(2)/2Plugging these in:sin(105) = (sqrt(3)/2)*(sqrt(2)/2) + (1/2)*(sqrt(2)/2)Let me compute each term:First term: (sqrt(3)/2)*(sqrt(2)/2) = (sqrt(6))/4Second term: (1/2)*(sqrt(2)/2) = (sqrt(2))/4So, sin(105) = (sqrt(6) + sqrt(2))/4Now, going back to the area:Area = sqrt(2) * (sqrt(6) + sqrt(2))/4Let me multiply sqrt(2) with each term in the numerator:sqrt(2)*sqrt(6) = sqrt(12) = 2*sqrt(3)sqrt(2)*sqrt(2) = sqrt(4) = 2So, the numerator becomes 2*sqrt(3) + 2, and the denominator is 4. So:Area = (2*sqrt(3) + 2)/4I can factor out a 2 in the numerator:Area = 2*(sqrt(3) + 1)/4Simplify by dividing numerator and denominator by 2:Area = (sqrt(3) + 1)/2So, that's the area. Let me recap:- I used the Law of Sines to find side b, which came out to 2.- Then, I used the area formula with sides a and b and the included angle C to find the area, which simplified to (sqrt(3) + 1)/2.I think that's it. I don't see any mistakes in my calculations, but let me double-check the sine of 105 degrees. Using the calculator, sin(105) is approximately 0.9659. Let me compute (sqrt(6) + sqrt(2))/4:sqrt(6) is about 2.449, sqrt(2) is about 1.414. Adding them gives about 3.863. Dividing by 4 gives approximately 0.9659, which matches. So that part checks out.Also, when I multiplied sqrt(2) by sin(105), I got sqrt(2)*(sqrt(6) + sqrt(2))/4, which simplifies correctly to (sqrt(3) + 1)/2. So, that seems right too.Therefore, I'm confident that side b is 2 and the area is (sqrt(3) + 1)/2.**Final Answer**The length of side ( b ) is (boxed{2}) and the area of triangle ( ABC ) is (boxed{dfrac{sqrt{3} + 1}{2}}).