💡Okay, so I have this problem where Andrea inscribed a circle inside a regular pentagon and circumscribed another circle around it, then calculated the area between the two circles. Charles did the same with a regular octagon. The pentagon has a side length of 3, and the octagon has a side length of 2. I need to figure out which area between the circles is greater.First, I need to recall some properties of regular polygons. A regular polygon has all sides equal and all interior angles equal. When a circle is inscribed in a regular polygon, it touches all the sides, and the radius of this inscribed circle is called the apothem. When a circle is circumscribed around a regular polygon, it passes through all the vertices, and the radius of this circumscribed circle is called the circumradius.The area between the two circles would be the area of the circumscribed circle minus the area of the inscribed circle. So, for each polygon, I need to find the circumradius and the apothem, then calculate the areas of the respective circles and find the difference.Let me start with the pentagon. A regular pentagon has 5 sides. The formula for the apothem (a) of a regular polygon is given by:[ a = frac{s}{2 tan(pi/n)} ]where ( s ) is the side length and ( n ) is the number of sides. For the pentagon, ( n = 5 ) and ( s = 3 ). So,[ a_{text{pentagon}} = frac{3}{2 tan(pi/5)} ]Similarly, the formula for the circumradius (R) is:[ R = frac{s}{2 sin(pi/n)} ]So,[ R_{text{pentagon}} = frac{3}{2 sin(pi/5)} ]Now, I need to compute these values. Let me calculate ( tan(pi/5) ) and ( sin(pi/5) ). Since ( pi ) radians is 180 degrees, ( pi/5 ) is 36 degrees.Using a calculator:[ tan(36^circ) approx 0.7265 ][ sin(36^circ) approx 0.5878 ]So,[ a_{text{pentagon}} = frac{3}{2 times 0.7265} approx frac{3}{1.453} approx 2.063 ][ R_{text{pentagon}} = frac{3}{2 times 0.5878} approx frac{3}{1.1756} approx 2.552 ]Now, the area between the circles for the pentagon is:[ pi R_{text{pentagon}}^2 - pi a_{text{pentagon}}^2 = pi (R_{text{pentagon}}^2 - a_{text{pentagon}}^2) ]Calculating ( R_{text{pentagon}}^2 ) and ( a_{text{pentagon}}^2 ):[ R_{text{pentagon}}^2 approx (2.552)^2 approx 6.513 ][ a_{text{pentagon}}^2 approx (2.063)^2 approx 4.255 ]So,[ pi (6.513 - 4.255) = pi (2.258) approx 7.09 ]Now, moving on to the octagon. A regular octagon has 8 sides. The side length is 2. Using the same formulas:[ a_{text{octagon}} = frac{2}{2 tan(pi/8)} = frac{1}{tan(pi/8)} ][ R_{text{octagon}} = frac{2}{2 sin(pi/8)} = frac{1}{sin(pi/8)} ]Calculating ( tan(pi/8) ) and ( sin(pi/8) ). ( pi/8 ) is 22.5 degrees.Using a calculator:[ tan(22.5^circ) approx 0.4142 ][ sin(22.5^circ) approx 0.3827 ]So,[ a_{text{octagon}} = frac{1}{0.4142} approx 2.414 ][ R_{text{octagon}} = frac{1}{0.3827} approx 2.613 ]Now, the area between the circles for the octagon is:[ pi R_{text{octagon}}^2 - pi a_{text{octagon}}^2 = pi (R_{text{octagon}}^2 - a_{text{octagon}}^2) ]Calculating ( R_{text{octagon}}^2 ) and ( a_{text{octagon}}^2 ):[ R_{text{octagon}}^2 approx (2.613)^2 approx 6.827 ][ a_{text{octagon}}^2 approx (2.414)^2 approx 5.828 ]So,[ pi (6.827 - 5.828) = pi (0.999) approx 3.14 ]Comparing the two areas:- Pentagon: approximately 7.09- Octagon: approximately 3.14Therefore, the area between the circles for the pentagon is greater than that for the octagon.

💡Okay, so I have this geometry problem here. Let me try to visualize it first. There's a square ABCD. I know that in a square, all sides are equal and all angles are 90 degrees. So, let me label the square: A is the top-left corner, B is the top-right, C is the bottom-right, and D is the bottom-left. So, it's a standard square.Now, the problem says to draw an equilateral triangle BCD on the outside of the square. Hmm, wait, BCD? So, points B, C, D. But in the square, B, C, D are three consecutive corners. So, if I draw an equilateral triangle on BC, outside the square, that would mean the triangle is attached to side BC. But wait, the problem says BCD, so is it triangle BCD? So, points B, C, D. But in the square, B, C, D are already connected. So, maybe it's an equilateral triangle on side BC, but with the third vertex outside the square. Let me clarify.Wait, the square is ABCD, so sides are AB, BC, CD, DA. So, side BC is from B to C. If I draw an equilateral triangle on BC outside the square, the third vertex would be somewhere outside. Let me call that point S. So, triangle BCS is equilateral, with S outside the square.Wait, the problem says "over the side BC draw an equilateral triangle BCD on the outside." Hmm, maybe it's triangle BCD, meaning points B, C, D, but D is already a corner of the square. So, maybe the equilateral triangle is BCD, but D is not the same as the square's D? Wait, that can't be. The square is ABCD, so D is fixed.Wait, maybe it's a typo? Or maybe it's triangle BCD, but D is the same as the square's D. But in the square, angle at D is 90 degrees, so triangle BCD can't be equilateral because BC is equal to CD in the square, but BD would be longer. So, that can't be.Wait, maybe the triangle is on BC, so the base is BC, and the third vertex is S, outside the square. So, triangle BCS is equilateral. That makes more sense. So, let me assume that. So, S is a new point outside the square, forming an equilateral triangle with BC.So, points: square ABCD, equilateral triangle BCS outside the square. Then, the midpoint of segment AS is N, and the midpoint of side CD is H. We need to prove that angle NHC is 60 degrees.Alright, let's try to sketch this mentally. Square ABCD: A at top-left, B at top-right, C at bottom-right, D at bottom-left. Equilateral triangle BCS: so from B to C to S, forming a triangle outside the square. So, S is somewhere below the square, forming a 60-degree angle at C.Midpoint of AS is N. So, A is top-left, S is below the square, so segment AS connects A to S, and N is the midpoint. Midpoint of CD is H. Since CD is the bottom side from C to D, H is the midpoint, so it's halfway between C and D.We need to show that angle NHC is 60 degrees. So, points N, H, C. So, connecting N to H to C, forming triangle NHC, and we need to show that the angle at H is 60 degrees.Hmm, okay. Let's think about coordinates. Maybe assigning coordinates to each point would help. Let me place the square on a coordinate system. Let me assume the square has side length 2 for simplicity, so that midpoints will have integer coordinates.Let me assign coordinates:- Let’s put point A at (0, 2)- Point B at (2, 2)- Point C at (2, 0)- Point D at (0, 0)So, square ABCD with side length 2.Now, equilateral triangle BCS. Since BC is from (2,2) to (2,0). So, BC is a vertical line segment of length 2. To form an equilateral triangle on BC outside the square, we need to find point S such that BCS is equilateral.Since BC is vertical, the equilateral triangle can be either to the left or right of BC. But since it's outside the square, and the square is on the left side of BC (since D is at (0,0)), so outside would be to the right of BC. Wait, but BC is the right side of the square, so outside would be to the right.Wait, but BC is vertical from (2,2) to (2,0). So, to form an equilateral triangle outside the square, point S would be to the right of BC.To find the coordinates of S, we can calculate it. Since BC is vertical, the equilateral triangle will have point S either to the right or left. Since it's outside the square, which is on the left, so S should be to the right.The length of BC is 2. So, the distance from B to S and from C to S should also be 2.Let me calculate the coordinates of S. Since BC is vertical from (2,2) to (2,0). The midpoint of BC is at (2,1). To find point S such that triangle BCS is equilateral, we can move from the midpoint in the horizontal direction.The height of an equilateral triangle with side length 2 is √3. So, from midpoint (2,1), moving √3 units to the right, we get point S at (2 + √3, 1).Wait, let me verify that. The height h of an equilateral triangle with side length a is h = (√3/2)a. So, for a = 2, h = √3. So, yes, moving √3 units from midpoint (2,1) to the right gives us S at (2 + √3, 1).So, coordinates:- S: (2 + √3, 1)Now, midpoint of AS is N. Point A is at (0,2), point S is at (2 + √3, 1). So, midpoint N is the average of the coordinates.Midpoint formula: ((x1 + x2)/2, (y1 + y2)/2)So, N_x = (0 + 2 + √3)/2 = (2 + √3)/2N_y = (2 + 1)/2 = 3/2So, N is at ((2 + √3)/2, 3/2)Midpoint of CD is H. CD is from (2,0) to (0,0). Midpoint H is at ((2 + 0)/2, (0 + 0)/2) = (1, 0)So, H is at (1, 0)Now, we need to find angle NHC, which is the angle at H between points N, H, and C.So, points:- N: ((2 + √3)/2, 3/2)- H: (1, 0)- C: (2, 0)We need to find angle at H: ∠NHC.To find this angle, we can use vectors or coordinate geometry.Let me compute vectors HN and HC.First, vector HN: from H to N.HN_x = N_x - H_x = ((2 + √3)/2) - 1 = (2 + √3 - 2)/2 = √3/2HN_y = N_y - H_y = (3/2) - 0 = 3/2So, vector HN is (√3/2, 3/2)Vector HC: from H to C.HC_x = C_x - H_x = 2 - 1 = 1HC_y = C_y - H_y = 0 - 0 = 0So, vector HC is (1, 0)Now, to find angle between vectors HN and HC, we can use the dot product formula:cosθ = (HN • HC) / (|HN| |HC|)Compute the dot product:HN • HC = (√3/2)(1) + (3/2)(0) = √3/2 + 0 = √3/2Compute |HN|:|HN| = sqrt( (√3/2)^2 + (3/2)^2 ) = sqrt( (3/4) + (9/4) ) = sqrt(12/4) = sqrt(3) ≈ 1.732Compute |HC|:|HC| = sqrt(1^2 + 0^2) = 1So,cosθ = (√3/2) / (sqrt(3) * 1) = (√3/2) / sqrt(3) = (1/2)So, θ = arccos(1/2) = 60 degrees.Therefore, angle NHC is 60 degrees.Wait, that seems straightforward. Let me double-check my steps.1. Assigned coordinates correctly? Yes, square ABCD with side length 2, points at (0,2), (2,2), (2,0), (0,0).2. Equilateral triangle BCS: calculated S correctly? Yes, midpoint of BC is (2,1), moved √3 to the right, so S is at (2 + √3, 1).3. Midpoint N of AS: A is (0,2), S is (2 + √3,1). Midpoint is ((2 + √3)/2, 3/2). Correct.4. Midpoint H of CD: CD is from (2,0) to (0,0). Midpoint is (1,0). Correct.5. Vectors HN and HC: Calculated correctly. HN is (√3/2, 3/2), HC is (1,0). Correct.6. Dot product: √3/2. Correct.7. Magnitudes: |HN| is sqrt( (√3/2)^2 + (3/2)^2 ) = sqrt(3/4 + 9/4) = sqrt(12/4) = sqrt(3). Correct. |HC| is 1. Correct.8. cosθ = (√3/2)/sqrt(3) = 1/2. So θ = 60 degrees. Correct.So, all steps seem correct. Therefore, angle NHC is indeed 60 degrees.I think that's a solid proof using coordinate geometry. Alternatively, one could use complex numbers or synthetic geometry, but coordinate geometry seems straightforward here.Another way could be using transformations or properties of midpoints and equilateral triangles, but since I already have a coordinate proof, I think that suffices.**Final Answer**boxed{60^circ}

💡Okay, so I'm trying to figure out how much each pair of sunglasses costs the vendor to buy. Let's see, the vendor sells each pair for 30 and sells 10 pairs in a day. That means his total revenue is 10 times 30, which is 300. Now, he uses half of his profits to buy a new sign that costs 20. So, if half of his profits is 20, then his total profits must be 40 because 20 times 2 is 40. To find out how much he spent on buying the sunglasses, I need to subtract his profits from his total revenue. So, 300 minus 40 equals 260. That means he spent 260 to buy all 10 pairs of sunglasses. Finally, to find out how much each pair costs him, I divide the total cost by the number of pairs. So, 260 divided by 10 is 26. Therefore, each pair of sunglasses costs him 26 to buy.

💡Alright, so I have this problem about an odd function f(x) defined on all real numbers. When x is greater than 0, f(x) is given as x squared minus 2x. There are two parts to the problem: first, to find the analytical expression of f(x) for all real numbers, and second, to determine the range of a real number a such that f(x) is monotonically decreasing on the interval [-1, a-2].Starting with part (1), since f(x) is an odd function, I remember that odd functions satisfy the property f(-x) = -f(x). That means if I know f(x) for positive x, I can find f(x) for negative x by reflecting it over both axes. So, for x > 0, f(x) is given as x² - 2x. For x < 0, I should compute f(-x) first and then negate it to get f(x).Let me write that down:For x > 0, f(x) = x² - 2x.For x < 0, let’s let x be negative, so -x is positive. Then f(-x) = (-x)² - 2*(-x) = x² + 2x. Since f is odd, f(x) = -f(-x) = -(x² + 2x) = -x² - 2x.At x = 0, since f is odd, f(0) must be 0 because f(-0) = -f(0) implies f(0) = -f(0), which only holds if f(0) is 0.So putting it all together, the function f(x) is:f(x) = x² - 2x for x > 0,f(x) = 0 for x = 0,f(x) = -x² - 2x for x < 0.That should be the analytical expression for f(x) on all real numbers.Moving on to part (2), we need to find the range of a such that f(x) is monotonically decreasing on the interval [-1, a-2]. First, I need to understand where f(x) is decreasing.Since f(x) is defined piecewise, I should analyze its behavior on different intervals. Let me consider the derivative of f(x) to determine where it's increasing or decreasing.For x > 0, f(x) = x² - 2x. The derivative f’(x) is 2x - 2. Setting this equal to zero gives critical points at x = 1. So, for x > 1, f’(x) is positive, meaning f(x) is increasing. For 0 < x < 1, f’(x) is negative, so f(x) is decreasing.For x < 0, f(x) = -x² - 2x. The derivative f’(x) is -2x - 2. Let's find where this derivative is zero: -2x - 2 = 0 ⇒ x = -1. So, for x < -1, f’(x) = -2x - 2. If x is less than -1, say x = -2, f’(-2) = -2*(-2) - 2 = 4 - 2 = 2, which is positive. So f(x) is increasing for x < -1. For -1 < x < 0, f’(x) = -2x - 2. If x is between -1 and 0, say x = -0.5, f’(-0.5) = -2*(-0.5) - 2 = 1 - 2 = -1, which is negative. So f(x) is decreasing for -1 < x < 0.At x = -1, the derivative is zero, so it's a critical point.Putting this together, f(x) is decreasing on the intervals (-1, 0) and (0, 1). Wait, but at x=0, the function is continuous, right? Let me check the value at x approaching 0 from the left and right.As x approaches 0 from the right, f(x) approaches 0² - 2*0 = 0.As x approaches 0 from the left, f(x) approaches -0² - 2*0 = 0. So yes, f(x) is continuous at x=0.Now, considering the entire real line, f(x) is decreasing on [-1, 1]. Because from -1 to 0, it's decreasing, and from 0 to 1, it's also decreasing. Beyond x=1, it starts increasing, and for x less than -1, it's increasing as well.So, the interval where f(x) is monotonically decreasing is [-1, 1].Now, the problem states that f(x) is monotonically decreasing on [-1, a-2]. So, this interval must be a subset of the interval where f(x) is decreasing, which is [-1, 1].Therefore, the interval [-1, a-2] must satisfy two conditions:1. The left endpoint is -1, which is already fixed.2. The right endpoint a-2 must be less than or equal to 1 because beyond 1, f(x) starts increasing.Also, since the interval is [-1, a-2], a-2 must be greater than -1 because the interval must have a lower bound less than the upper bound.So, we have:-1 < a - 2 ≤ 1Let me solve these inequalities for a.First inequality: a - 2 > -1Adding 2 to both sides: a > 1Second inequality: a - 2 ≤ 1Adding 2 to both sides: a ≤ 3So, combining these, a must satisfy 1 < a ≤ 3.Therefore, the range of real number a is (1, 3].Wait, let me double-check if a-2 can be equal to 1. If a-2 = 1, then a = 3. At x=1, f(x) is at a critical point where the derivative is zero. So, is the function decreasing at x=1? Since the derivative is zero, it's neither increasing nor decreasing at that exact point. But when considering the interval up to a-2, if a-2 is 1, does that mean the function is decreasing on [-1, 1]? Well, at x=1, it's a critical point, so technically, the function stops decreasing there. So, including 1 in the interval is acceptable because the function is decreasing up to that point, even though it's not decreasing beyond it.Similarly, if a-2 is greater than 1, then the interval would extend beyond 1, where the function starts increasing, which would violate the condition of being monotonically decreasing on the entire interval. Therefore, a-2 must be less than or equal to 1.Also, a-2 must be greater than -1 because otherwise, the interval would start at -1 and end at a point less than or equal to -1, which doesn't make sense since the upper bound must be greater than the lower bound. So, a-2 > -1 is necessary.So, yes, the range of a is (1, 3].I think that's it. Let me recap:1. For part (1), since f(x) is odd, we used the definition to find f(x) for x < 0 and x = 0.2. For part (2), we found the intervals where f(x) is decreasing by analyzing the derivative, determined that f(x) is decreasing on [-1, 1], and then set up inequalities to ensure that the given interval [-1, a-2] lies within [-1, 1], leading to the range of a being (1, 3].I don't see any mistakes in this reasoning, so I think this is the correct solution.**Final Answer**(1) The analytical expression of ( f(x) ) on ( mathbb{R} ) is (boxed{f(x) = begin{cases} -x^2 - 2x & text{if } x < 0, 0 & text{if } x = 0, x^2 - 2x & text{if } x > 0. end{cases}})(2) The range of real number ( a ) is (boxed{(1, 3]}).

💡Okay, so I have this problem where I'm given a sequence defined by (0 < a < 1), and the sequence is (x_1 = a), (x_2 = a^{x_1}), (x_3 = a^{x_2}), and so on. I need to figure out whether the sequence is increasing, decreasing, or if the odd or even terms are increasing or decreasing. The options are A, B, C, or D.First, I remember that when (0 < a < 1), the function (y = a^x) is a decreasing function. That means as (x) increases, (a^x) decreases. So, if I have two numbers where one is bigger than the other, the smaller number will give a larger result when plugged into (a^x).Let me start by calculating the first few terms to see what's happening.Starting with (x_1 = a). Since (0 < a < 1), (x_1) is between 0 and 1.Next, (x_2 = a^{x_1}). Since (x_1 = a), this becomes (x_2 = a^a). Now, (a^a) is interesting because when (0 < a < 1), (a^a) is actually greater than (a). Wait, is that right? Let me think. For example, if (a = 0.5), then (a^a = 0.5^{0.5} = sqrt{0.5} approx 0.707), which is greater than 0.5. So, yes, (x_2 > x_1).Now, (x_3 = a^{x_2}). Since (x_2 > x_1), and because (a^x) is decreasing, (a^{x_2} < a^{x_1}), which means (x_3 < x_2). So, (x_3 < x_2).Continuing, (x_4 = a^{x_3}). Since (x_3 < x_2), and (a^x) is decreasing, (a^{x_3} > a^{x_2}), so (x_4 > x_3).Hmm, so it seems like the sequence is alternating: (x_1 < x_2 > x_3 < x_4 > x_5), and so on. So, the odd terms are increasing, and the even terms are decreasing? Or is it the other way around?Wait, let's list them out:- (x_1 = a)- (x_2 = a^a) (which is greater than (x_1))- (x_3 = a^{x_2}) (which is less than (x_2))- (x_4 = a^{x_3}) (which is greater than (x_3))- (x_5 = a^{x_4}) (which is less than (x_4))So, the odd terms: (x_1 < x_3 < x_5 < dots)The even terms: (x_2 > x_4 > x_6 > dots)Wait, so the odd-indexed terms are increasing, and the even-indexed terms are decreasing. That would make option C correct.But let me double-check. Maybe I made a mistake in the direction.Since (0 < a < 1), (a^x) is decreasing. So, if (x_n) increases, (x_{n+1} = a^{x_n}) decreases, and vice versa.Starting with (x_1 = a), then (x_2 = a^{x_1}). Since (x_1 = a), and (a < 1), (x_2 = a^a > a = x_1). So, (x_2 > x_1).Then, (x_3 = a^{x_2}). Since (x_2 > x_1), and (a^x) is decreasing, (x_3 = a^{x_2} < a^{x_1} = x_2). So, (x_3 < x_2).Next, (x_4 = a^{x_3}). Since (x_3 < x_2), and (a^x) is decreasing, (x_4 = a^{x_3} > a^{x_2} = x_3). So, (x_4 > x_3).Similarly, (x_5 = a^{x_4}). Since (x_4 > x_3), (x_5 = a^{x_4} < a^{x_3} = x_4). So, (x_5 < x_4).This pattern continues, so the odd terms are increasing: (x_1 < x_3 < x_5 < dots), and the even terms are decreasing: (x_2 > x_4 > x_6 > dots).Therefore, the correct answer should be option C: Odd-indexed terms are increasing, and even-indexed terms are decreasing.

💡Okay, so I have this problem here: I need to find the base ten equivalent of the number 54321 in base six. Hmm, base conversions can be a bit tricky, but I think I can handle this. Let me start by recalling what base six means. In base six, each digit represents a power of six, just like in base ten each digit represents a power of ten. So, the rightmost digit is six to the power of zero, which is 1, then the next one to the left is six to the power of one, which is six, and so on. Given the number 54321 in base six, I need to break it down digit by digit, starting from the right. Let me write down the positions and their corresponding powers of six:- The rightmost digit is 1, which is in the 6^0 position.- Moving left, the next digit is 2, which is in the 6^1 position.- Then we have 3 in the 6^2 position.- Next is 4 in the 6^3 position.- Finally, the leftmost digit is 5 in the 6^4 position.Now, I need to calculate the value of each digit multiplied by its corresponding power of six and then add them all up to get the base ten equivalent.Starting with the rightmost digit:1 * 6^0 = 1 * 1 = 1Next:2 * 6^1 = 2 * 6 = 12Then:3 * 6^2 = 3 * 36 = 108After that:4 * 6^3 = 4 * 216 = 864Finally:5 * 6^4 = 5 * 1296 = 6480Now, I'll add all these values together:1 + 12 = 1313 + 108 = 121121 + 864 = 985985 + 6480 = 7465So, adding all these up, I get 7465 in base ten. Let me just double-check my calculations to make sure I didn't make any mistakes.First, 6^0 is 1, so 1 * 1 is definitely 1.6^1 is 6, so 2 * 6 is 12.6^2 is 36, so 3 * 36 is 108.6^3 is 216, so 4 * 216 is 864.6^4 is 1296, so 5 * 1296 is 6480.Adding them up step by step:1 + 12 = 1313 + 108 = 121121 + 864 = 985985 + 6480 = 7465Everything seems to check out. I don't see any errors in my calculations. So, I'm confident that the base ten equivalent of 54321 in base six is 7465.