💡First, I need to determine the total number of numbers, denoted as ( n ), and the sum of all these numbers, denoted as ( S ). Given that the average of all the numbers is 60, I can express this relationship as:[frac{S}{n} = 60 quad Rightarrow quad S = 60n quad text{(1)}]Next, I know that the average of the first 6 numbers is 57. Therefore, the sum of these first 6 numbers is:[57 times 6 = 342 quad text{(2)}]Similarly, the average of the last 8 numbers is 61, so the sum of these last 8 numbers is:[61 times 8 = 488 quad text{(3)}]The 8th number, which is 50, is included in both the first 6 numbers and the last 8 numbers. This means that when I add the sums of the first 6 numbers and the last 8 numbers, I am double-counting the 8th number. To find the correct total sum of all the numbers, I need to subtract the 8th number once:[S = 342 + 488 - 50 = 780 quad text{(4)}]Now, I can equate equation (1) and equation (4) to solve for ( n ):[60n = 780 quad Rightarrow quad n = frac{780}{60} = 13]Therefore, there are 13 numbers in total.

💡Okay, so I have this problem about a rectangle ABCD. Let me visualize it first. In a rectangle, opposite sides are equal and all angles are right angles. They've given me that AB is 6 centimeters and AD is 4 centimeters. So, if I imagine the rectangle, AB is the length, and AD is the width. Let me label the rectangle with points A, B, C, D in order. So, A is connected to B and D, B is connected to C, and so on.Now, M is the midpoint of BC. Since BC is one of the sides, and in a rectangle, BC should be equal to AD, which is 4 centimeters. Wait, no, actually, in a rectangle, opposite sides are equal. So, AB is 6 cm, so CD is also 6 cm. AD is 4 cm, so BC is also 4 cm. So, BC is 4 cm, and M is the midpoint, so BM and MC are each 2 cm.They want me to find the ratio of BO to BD, where O is the intersection of AM and BD. So, BD is one of the diagonals of the rectangle. In a rectangle, the diagonals are equal in length and they bisect each other. So, the diagonals intersect at their midpoints. But in this case, O is the intersection of AM and BD, not necessarily the midpoint of BD.Hmm, so I need to find where the line AM intersects the diagonal BD, and then find the ratio of the length from B to O compared to the entire diagonal BD.Maybe coordinate geometry would help here. Let me assign coordinates to each of the points. Let's place point A at the origin (0,0). Then, since AB is 6 cm, point B would be at (6,0). AD is 4 cm, so point D would be at (0,4), and point C, which is opposite to A, would be at (6,4).Wait, hold on, actually, in a rectangle, if A is at (0,0), then B would be at (6,0), D at (0,4), and C at (6,4). That makes sense because AB is 6 units long along the x-axis, and AD is 4 units up along the y-axis.Now, M is the midpoint of BC. So, points B and C are at (6,0) and (6,4) respectively. The midpoint M would be the average of their coordinates. So, the x-coordinate is (6 + 6)/2 = 6, and the y-coordinate is (0 + 4)/2 = 2. So, M is at (6,2).Now, I need to find the equations of lines AM and BD to find their intersection point O.First, let's find the equation of line AM. Points A and M are at (0,0) and (6,2) respectively. The slope of AM is (2 - 0)/(6 - 0) = 2/6 = 1/3. So, the slope is 1/3. Using the point-slope form, the equation of AM is y = (1/3)x + b. Since it passes through (0,0), b is 0. So, the equation is y = (1/3)x.Next, let's find the equation of BD. Points B and D are at (6,0) and (0,4) respectively. The slope of BD is (4 - 0)/(0 - 6) = 4/(-6) = -2/3. So, the slope is -2/3. Using point-slope form, let's use point B (6,0). So, y - 0 = (-2/3)(x - 6). Simplifying, y = (-2/3)x + 4.Now, to find the intersection point O of lines AM and BD, we can set their equations equal to each other:(1/3)x = (-2/3)x + 4Let me solve for x:(1/3)x + (2/3)x = 4(3/3)x = 4x = 4Now, plug x = 4 back into the equation of AM to find y:y = (1/3)(4) = 4/3So, point O is at (4, 4/3).Now, I need to find the ratio of BO to BD. First, let's find the length of BD. Since BD is the diagonal of the rectangle, its length can be found using the Pythagorean theorem. The sides are 6 cm and 4 cm, so the diagonal BD is sqrt(6^2 + 4^2) = sqrt(36 + 16) = sqrt(52) = 2*sqrt(13) cm.Alternatively, since we have coordinates for B and D, we can calculate the distance between (6,0) and (0,4):Distance BD = sqrt[(6 - 0)^2 + (0 - 4)^2] = sqrt[36 + 16] = sqrt[52] = 2*sqrt(13) cm.Now, let's find the length of BO. Point B is at (6,0) and point O is at (4, 4/3). So, the distance BO is sqrt[(6 - 4)^2 + (0 - 4/3)^2] = sqrt[(2)^2 + (-4/3)^2] = sqrt[4 + 16/9] = sqrt[(36/9) + (16/9)] = sqrt[52/9] = (sqrt(52))/3 = (2*sqrt(13))/3 cm.So, BO is (2*sqrt(13))/3 cm, and BD is 2*sqrt(13) cm. Therefore, the ratio BO/BD is [(2*sqrt(13))/3] / [2*sqrt(13)] = (2*sqrt(13))/3 * 1/(2*sqrt(13)) = 1/3.Wait, that seems straightforward, but let me double-check my calculations.First, coordinates:A(0,0), B(6,0), C(6,4), D(0,4), M(6,2). Equations:AM: y = (1/3)xBD: y = (-2/3)x + 4Intersection at x=4, y=4/3. So, O(4,4/3).Distance BO: sqrt[(6-4)^2 + (0 - 4/3)^2] = sqrt[4 + 16/9] = sqrt[52/9] = (2*sqrt(13))/3Distance BD: sqrt[6^2 + 4^2] = sqrt[52] = 2*sqrt(13)Ratio BO/BD: (2*sqrt(13)/3) / (2*sqrt(13)) = 1/3Yes, that seems correct. So, the ratio is 1/3.Alternatively, maybe there's a way to do this without coordinates, using similar triangles or properties of rectangles and midpoints.Let me think. In rectangle ABCD, diagonals bisect each other, so the point where the diagonals intersect is the midpoint of both diagonals. But in this case, we're dealing with line AM, not a diagonal.Since M is the midpoint of BC, and we're drawing a line from A to M, which is not a diagonal, but a median. The intersection point O of AM and BD divides BD into segments BO and OD. We need to find the ratio BO/BD.Perhaps using mass point geometry or coordinate geometry as I did before. But since I already got 1/3 using coordinates, and the calculations seem correct, I think 1/3 is the right answer.Wait, but in my initial thought process, I considered the coordinates and got O at (4,4/3). Let me confirm that BD is from (6,0) to (0,4), and O is at (4,4/3). So, from B(6,0) to O(4,4/3), the change in x is -2, and change in y is +4/3. The length is sqrt(4 + 16/9) as before.Alternatively, maybe using vectors. Let me try that approach.Vector approach:Let’s consider point A as the origin (0,0). Then, vector AB is (6,0), vector AD is (0,4). Point M is the midpoint of BC. Since B is (6,0) and C is (6,4), M is (6,2).Line AM goes from A(0,0) to M(6,2). Parametric equations:x = 6ty = 2tfor t from 0 to 1.Line BD goes from B(6,0) to D(0,4). Parametric equations:x = 6 - 6sy = 0 + 4sfor s from 0 to 1.To find intersection O, set the coordinates equal:6t = 6 - 6s2t = 4sFrom the second equation: 2t = 4s => t = 2sSubstitute into the first equation: 6*(2s) = 6 - 6s => 12s = 6 - 6s => 18s = 6 => s = 1/3Then, t = 2*(1/3) = 2/3So, point O is at x = 6*(2/3) = 4, y = 2*(2/3) = 4/3, which matches the previous result.Therefore, BO is the distance from B(6,0) to O(4,4/3). As calculated before, it's (2*sqrt(13))/3.BD is 2*sqrt(13). So, ratio BO/BD = 1/3.So, yes, the ratio is 1/3.Alternatively, using similar triangles.In triangle ABD, line AM intersects BD at O. Let's see if triangles ABO and something else are similar.Wait, maybe not directly. Alternatively, since M is the midpoint of BC, and we're connecting A to M, perhaps we can use the concept of ratios in intersecting lines.In general, when two lines intersect, the ratio can be found by considering the areas or using Menelaus' theorem.Menelaus' theorem states that for a triangle, if a line crosses the three sides (or their extensions), the product of the segment ratios is equal to 1.Let me consider triangle BCD. Wait, no, maybe triangle ABD.Wait, let me think. If I consider triangle ABD, with transversal AMO.Wait, Menelaus' theorem might be a bit complicated here. Alternatively, using coordinate geometry was straightforward and gave me the answer.So, I think the ratio is 1/3.**Final Answer**The ratio of ( BO ) to ( BD ) is boxed{dfrac{1}{3}}.

💡Okay, so I have this problem here: I need to find how many right triangles have integer leg lengths (a) and (b) and a hypotenuse of length (b + 1), where (b < 100). The answer choices are from A to E, with numbers 6 to 10. Hmm, let me try to figure this out step by step.First, I remember that in a right triangle, the Pythagorean theorem applies. That is, (a^2 + b^2 = c^2), where (c) is the hypotenuse. In this case, the hypotenuse is given as (b + 1). So, substituting that into the theorem, I get:[a^2 + b^2 = (b + 1)^2]Let me expand the right side of this equation:[a^2 + b^2 = b^2 + 2b + 1]Hmm, okay, so if I subtract (b^2) from both sides, I can simplify this equation:[a^2 = 2b + 1]So, (a^2) equals (2b + 1). That means (a^2) is an odd number because (2b) is even and adding 1 makes it odd. Therefore, (a) must also be an odd integer because the square of an even number is even, and the square of an odd number is odd.Now, I need to find all integer values of (a) and (b) such that (a^2 = 2b + 1) and (b < 100). Let me solve for (b) in terms of (a):[b = frac{a^2 - 1}{2}]Since (b) must be an integer, (frac{a^2 - 1}{2}) must also be an integer. Given that (a) is odd, let me represent (a) as (2k + 1) where (k) is a non-negative integer. Substituting this into the equation for (b):[b = frac{(2k + 1)^2 - 1}{2} = frac{4k^2 + 4k + 1 - 1}{2} = frac{4k^2 + 4k}{2} = 2k^2 + 2k]So, (b = 2k^2 + 2k). Now, since (b < 100), I can set up the inequality:[2k^2 + 2k < 100]Let me simplify this inequality:[2k^2 + 2k < 100 k^2 + k < 50 k^2 + k - 50 < 0]This is a quadratic inequality. To find the values of (k) that satisfy this, I can solve the quadratic equation (k^2 + k - 50 = 0). Using the quadratic formula:[k = frac{-1 pm sqrt{1 + 200}}{2} = frac{-1 pm sqrt{201}}{2}]Calculating (sqrt{201}), which is approximately 14.177. So,[k = frac{-1 + 14.177}{2} approx frac{13.177}{2} approx 6.588]Since (k) must be a non-negative integer, the maximum value (k) can take is 6 because 6.588 is approximately 6.59, and (k) must be less than this. So, (k) can be 0, 1, 2, 3, 4, 5, or 6.Wait, but if (k = 0), then (b = 0), which doesn't make sense because the length of a side can't be zero. So, (k) should start from 1.Let me list the possible values of (k) from 1 to 6 and compute (b) for each:- For (k = 1): [ b = 2(1)^2 + 2(1) = 2 + 2 = 4 ]- For (k = 2): [ b = 2(2)^2 + 2(2) = 8 + 4 = 12 ]- For (k = 3): [ b = 2(3)^2 + 2(3) = 18 + 6 = 24 ]- For (k = 4): [ b = 2(4)^2 + 2(4) = 32 + 8 = 40 ]- For (k = 5): [ b = 2(5)^2 + 2(5) = 50 + 10 = 60 ]- For (k = 6): [ b = 2(6)^2 + 2(6) = 72 + 12 = 84 ]Let me check if (k = 7) would give (b < 100):- For (k = 7): [ b = 2(7)^2 + 2(7) = 98 + 14 = 112 ] But 112 is greater than 100, so (k = 7) is not allowed. Therefore, (k) can only go up to 6.So, the possible values of (k) are 1 through 6, giving us 6 different triangles. But wait, let me double-check to make sure each of these (b) values is indeed less than 100:- (k = 1): (b = 4) ✔️- (k = 2): (b = 12) ✔️- (k = 3): (b = 24) ✔️- (k = 4): (b = 40) ✔️- (k = 5): (b = 60) ✔️- (k = 6): (b = 84) ✔️All of these are less than 100, so that's good. Each (k) from 1 to 6 gives a valid (b). Therefore, there are 6 such triangles.But wait, hold on. Let me make sure I didn't miss any other possible (a) values. Earlier, I assumed (a = 2k + 1), but is that the only way? Or could there be other odd integers (a) that don't fit into this form?Wait, no, because any odd integer can be expressed as (2k + 1) for some integer (k). So, I think that covers all possible cases. Therefore, each (k) from 1 to 6 gives a unique (a) and (b), so that's 6 triangles.But let me check the answer choices again: A is 6, B is 7, C is 8, D is 9, E is 10. So, 6 is an option, but I want to make sure I didn't miss any.Wait, when (k = 0), (b = 0), which is invalid, so we start at (k = 1). So, 6 values of (k) correspond to 6 triangles. Therefore, the answer should be 6, which is option A.But hold on, let me think again. Maybe there's another way to approach this problem where I can get more triangles. For example, perhaps (a) doesn't have to be expressed as (2k + 1). Maybe I can think of (a) as any integer such that (a^2 = 2b + 1), which would mean (a) is odd, as we saw earlier.So, let's list all odd integers (a) such that (a^2 < 2*100 + 1 = 201). So, (a^2 < 201), so (a < sqrt{201} approx 14.177). Therefore, the possible odd integers (a) are 1, 3, 5, 7, 9, 11, 13.But (a = 1) would give (b = (1^2 - 1)/2 = 0), which is invalid, as before. So, starting from (a = 3):- (a = 3): (b = (9 - 1)/2 = 4)- (a = 5): (b = (25 - 1)/2 = 12)- (a = 7): (b = (49 - 1)/2 = 24)- (a = 9): (b = (81 - 1)/2 = 40)- (a = 11): (b = (121 - 1)/2 = 60)- (a = 13): (b = (169 - 1)/2 = 84)So, that's 6 values of (a) (from 3 to 13, odd numbers), each giving a valid (b < 100). So, again, 6 triangles.Wait, but the answer choices include 7 as an option. Did I miss something? Let me check if (a = 15) would give (b < 100):- (a = 15): (b = (225 - 1)/2 = 112), which is greater than 100, so it's invalid.So, (a = 15) is too big. So, only up to (a = 13). Therefore, 6 triangles.But hold on, maybe I'm missing (k = 0) somehow? No, because (b = 0) is invalid. So, 6 is the correct number.Wait, but let me think again. Maybe there's a different way to parameterize the solutions. Sometimes, Pythagorean triples can be generated using different formulas, but in this case, since the hypotenuse is (b + 1), it's a specific case.Alternatively, maybe I can think of this as a Pell equation or something similar, but I don't think that's necessary here.Wait, another approach: Since (a^2 = 2b + 1), this is equivalent to (a^2 - 2b = 1). So, it's a linear Diophantine equation in two variables. The solutions can be found using the theory of Pell equations, but in this case, since it's a simple equation, we can find all solutions by testing odd integers (a) as I did before.So, I think my initial approach is correct, and the number of solutions is 6. Therefore, the answer should be A) 6.But just to be thorough, let me list all the possible (a) and (b) pairs:1. (a = 3), (b = 4)2. (a = 5), (b = 12)3. (a = 7), (b = 24)4. (a = 9), (b = 40)5. (a = 11), (b = 60)6. (a = 13), (b = 84)That's six pairs. Each of these satisfies (a^2 + b^2 = (b + 1)^2), and (b < 100). So, yes, 6 is the correct number.Wait, but let me check one of these to make sure. Let's take (a = 3), (b = 4):[3^2 + 4^2 = 9 + 16 = 25 = 5^2 = (4 + 1)^2]Yes, that works. Another one: (a = 5), (b = 12):[5^2 + 12^2 = 25 + 144 = 169 = 13^2 = (12 + 1)^2]That works too. And (a = 7), (b = 24):[7^2 + 24^2 = 49 + 576 = 625 = 25^2 = (24 + 1)^2]Yes, correct. Similarly, (a = 9), (b = 40):[9^2 + 40^2 = 81 + 1600 = 1681 = 41^2 = (40 + 1)^2]Good. (a = 11), (b = 60):[11^2 + 60^2 = 121 + 3600 = 3721 = 61^2 = (60 + 1)^2]Perfect. And finally, (a = 13), (b = 84):[13^2 + 84^2 = 169 + 7056 = 7225 = 85^2 = (84 + 1)^2]Yes, that works as well. So, all six pairs satisfy the conditions. Therefore, the number of such triangles is indeed 6.I think I've covered all bases here. I considered the equation, solved for (b), found the constraints on (k), listed all possible values, and even double-checked each pair to ensure they satisfy the Pythagorean theorem. So, I'm confident that the answer is 6, which is option A.

💡Alright, so I have this problem about counterfeit coins. It's divided into two parts, a) and b). Let me start with part a).**Problem a)**: Among 80 coins, there's one counterfeit that's lighter. I need to find it using four weighings on a balance scale without weights. Hmm, okay. I remember that with a balance scale, you can compare groups of coins and determine which side is lighter or if they're equal. Since the counterfeit is lighter, it should tip the scale to its side if it's on one of the pans.I think the standard approach for these kinds of problems is to divide the coins into groups and narrow down the possibilities with each weighing. For example, if I have 80 coins, I can divide them into three groups because each weighing has three possible outcomes: left pan lighter, right pan lighter, or balanced. So, dividing into three groups allows me to use each outcome effectively.Let me try to outline the steps:1. **First Weighing**: Divide the 80 coins into three groups. Since 80 isn't perfectly divisible by 3, I'll have two groups of 27 coins and one group of 26 coins. Weigh the two groups of 27 against each other. - If one side is lighter, the counterfeit is in that group of 27. - If they balance, the counterfeit is in the group of 26.2. **Second Weighing**: Suppose the counterfeit is in a group of 27. I'll divide this 27 into three groups of 9 each. Weigh two groups of 9 against each other. - If one side is lighter, the counterfeit is in that group of 9. - If they balance, the counterfeit is in the remaining group of 9.3. **Third Weighing**: Now, with a group of 9, divide them into three groups of 3 each. Weigh two groups of 3 against each other. - If one side is lighter, the counterfeit is in that group of 3. - If they balance, the counterfeit is in the remaining group of 3.4. **Fourth Weighing**: Finally, with a group of 3, weigh one coin against another. - If one side is lighter, that's the counterfeit. - If they balance, the third coin is the counterfeit.Wait, that seems to work. Each time, I'm dividing the group into three parts and using the outcome of the weighing to narrow down the possibilities. Since 3^4 = 81, which is just enough to cover 80 coins, this should work.But let me double-check. If I have 80 coins, the first weighing reduces it to either 27 or 26. Then, the second weighing reduces it to 9 or 8 (if it was 26 initially). Wait, hold on, 26 divided by 3 is about 8 or 9. Hmm, maybe I need to adjust the group sizes to ensure each weighing optimally narrows it down.Actually, 80 divided by 3 is approximately 26.666, so two groups of 27 and one group of 26 is correct. Then, 27 divided by 3 is 9, and 26 divided by 3 is approximately 8.666, so two groups of 9 and one group of 8. But in my initial plan, I assumed 27, 27, 26, then 9,9,9 or 9,9,8. Maybe I should adjust the group sizes accordingly to make sure each step is as efficient as possible.But regardless, since 3^4 = 81, which is just one more than 80, it should still be possible to find the counterfeit in four weighings. So, I think my initial plan is solid.**Problem b)**: Now, this is a generalization. Given n coins, what's the minimal number k of weighings needed to find the counterfeit, which is lighter.From part a), I saw that with four weighings, I can handle up to 81 coins. So, it seems like the number of weighings relates to powers of 3. Specifically, k weighings can handle up to 3^k coins. So, to find the minimal k such that 3^k is at least n, we can take the logarithm base 3 of n and round up to the nearest integer.Mathematically, k = ceiling(log3(n)). That makes sense because each weighing effectively reduces the problem size by a factor of 3.Let me test this with some examples. If n = 1, then k = 0, which makes sense because there's only one coin, so no weighings needed. If n = 3, k = 1, which is correct because you can weigh one against another, and if they balance, the third is counterfeit. If n = 4, then k = 2, since 3^1 = 3 < 4 ≤ 3^2 = 9. So, two weighings can handle up to 9 coins, which aligns with the standard counterfeit coin problem.Another example: n = 10, k = 2 because 3^2 = 9 < 10 ≤ 27, so k = 3? Wait, no, 3^2 = 9, so 10 would require k = 3? Wait, no, hold on. If n = 10, then 3^2 = 9 < 10 ≤ 27, so k = 3? But wait, in part a), 80 coins required k = 4, and 3^4 = 81. So, for n = 10, since 3^2 = 9 < 10 ≤ 27, k = 3? But actually, 10 coins can be handled in 2 weighings because 3^2 = 9, but 10 is just one more. Hmm, maybe my initial thought isn't entirely accurate.Wait, let me think again. Each weighing can give us three outcomes, so the maximum number of distinguishable outcomes with k weighings is 3^k. Since we have n coins, each of which could be the counterfeit, we need 3^k ≥ n. Therefore, k must satisfy k ≥ log3(n). Since k has to be an integer, we take the ceiling of log3(n).So, for n = 10, log3(10) ≈ 2.095, so ceiling of that is 3. Therefore, k = 3. But wait, can't we do it in 2 weighings? Let's see.If I have 10 coins, first weighing: divide into 3, 3, 4. Weigh the two groups of 3.- If one side is lighter, counterfeit is in that group of 3. Then, second weighing: weigh 1 vs 1. If one is lighter, that's counterfeit; else, the third is counterfeit.- If the two groups of 3 balance, counterfeit is in the group of 4. Then, second weighing: divide the 4 into 1,1,2. Weigh the two single coins. - If one is lighter, that's counterfeit. - If they balance, counterfeit is in the group of 2. Then, third weighing: weigh the two against each other. The lighter one is counterfeit.Wait, so actually, for n = 10, it might require up to 3 weighings in the worst case. So, my initial thought was correct. Therefore, k = ceiling(log3(n)) is the right formula.So, in general, the minimal number of weighings required is the smallest integer k such that 3^k is greater than or equal to n. That is, k = ceiling(log3(n)).Let me verify this with another example. If n = 27, then log3(27) = 3, so k = 3. That makes sense because 3 weighings can handle up to 27 coins. If n = 28, then log3(28) ≈ 3.03, so ceiling is 4, meaning 4 weighings are needed. That aligns with part a), where n = 80 required k = 4 because 3^4 = 81.Therefore, the formula holds. So, for any n, the minimal k is the ceiling of the base-3 logarithm of n.**Final Answer**a) The counterfeit coin can be found in four weighings. (boxed{4})b) The minimum number of weighings required is (boxed{lceil log_3 n rceil}).

💡First, I need to find the expression for ( h(x) ) such that ( f(h(x)) = g(x) ).Given:[ f(x) = 2x + 3 ][ g(x) = 4x - 5 ]I'll start by substituting ( h(x) ) into ( f(x) ):[ f(h(x)) = 2h(x) + 3 ]Next, I'll set this equal to ( g(x) ):[ 2h(x) + 3 = 4x - 5 ]To solve for ( h(x) ), I'll subtract 3 from both sides:[ 2h(x) = 4x - 8 ]Finally, I'll divide both sides by 2 to isolate ( h(x) ):[ h(x) = 2x - 4 ]So, the expression for ( h(x) ) is ( 2x - 4 ).

💡Okay, so I have this problem about an equiangular convex hexagon with a perimeter of 2. The question is asking about the perimeter of a six-pointed star formed by extending the sides of the hexagon. Hmm, let me try to visualize this.First, an equiangular hexagon means all its interior angles are equal. For a regular hexagon, each interior angle is 120 degrees, right? But wait, this hexagon isn't necessarily regular because the sides might not all be equal. However, since it's equiangular, all the angles are still 120 degrees each.Now, the perimeter is 2, so the sum of all its sides is 2. Let's denote the sides as AB, BC, CD, DE, EF, and FA. Since it's convex, all the sides bulge outwards, and when we extend the sides, they intersect each other to form a star. I think this star is similar to the Star of David, which has six points.I need to figure out the perimeter of this star. So, each point of the star is formed by the intersection of two extended sides of the hexagon. Let me try to draw this mentally. Each side of the hexagon, when extended, will meet another extended side at a point, forming a triangle outside the hexagon. Since all the angles are 120 degrees, these triangles might be equilateral or something similar.Wait, if each interior angle is 120 degrees, then the exterior angle would be 60 degrees because the exterior angle is 180 minus the interior angle. So, each exterior angle is 60 degrees. That might be useful.When we extend the sides, the triangles formed outside the hexagon will have angles related to these exterior angles. Maybe each triangle is equilateral because 60 degrees is a key angle for equilateral triangles. If that's the case, then the sides of these triangles would be equal to the sides of the hexagon.But hold on, the hexagon isn't regular, so the sides can be different lengths. However, the angles are all 120 degrees, so when we extend the sides, the triangles formed might still have some proportional relationships.Let me think about one side of the hexagon. Suppose I have side AB. If I extend AB beyond A and beyond B, it will intersect with the extensions of other sides. Specifically, each side will intersect with two non-adjacent sides. For example, side AB might intersect with the extension of side DE and the extension of side EF or something like that.Wait, maybe I should consider the triangles formed by each pair of extended sides. Each triangle would have a base which is a side of the hexagon and two other sides which are extensions. Since the angles at the base are 120 degrees, the triangles formed outside would have base angles of 60 degrees each because 180 - 120 = 60.So, each of these triangles is an isosceles triangle with a base of length equal to a side of the hexagon and two equal sides. The two equal sides can be calculated using trigonometry. If the base is, say, length 'a', and the base angles are 60 degrees, then the two equal sides would each be 'a' divided by (2 times the cosine of 60 degrees). Since cosine of 60 degrees is 0.5, that would make each equal side 'a' divided by 1, so just 'a'. Wait, that means each of those triangles is actually equilateral because all sides are equal.So, each side of the hexagon, when extended, forms an equilateral triangle outside the hexagon. Therefore, the length from the hexagon's vertex to the star's point is equal to the side length of the hexagon.But the star is formed by connecting these points. Each point of the star is the apex of one of these equilateral triangles. So, the sides of the star are actually the sides of these equilateral triangles.But wait, each side of the star is shared by two adjacent triangles. So, if I have six triangles, each contributing two sides to the star, but each side is counted twice, so the total number of sides for the star is six, each corresponding to a side of the hexagon.But the perimeter of the star would then be the sum of all these sides. Since each side of the star is equal to the side of the hexagon, and the hexagon has a perimeter of 2, does that mean the star's perimeter is also 2? But that doesn't seem right because the star is larger than the hexagon.Wait, no. Each side of the star is actually twice the length of the hexagon's side because each side of the star is formed by two sides of the equilateral triangles. Let me clarify.If each triangle has sides equal to the hexagon's side, then each side of the star is actually two sides of these triangles. So, if the hexagon has a side length 'a', each triangle has sides 'a', and the star's side would be 'a + a' = 2a. But wait, that might not be accurate because the triangles are connected at the hexagon's vertices.Alternatively, maybe each side of the star is just equal to the side of the hexagon. But that would make the star's perimeter equal to the hexagon's perimeter, which is 2. But the answer choices don't have 2 as the correct answer because the options are 2, 4, 6, 8, and the answer is supposed to be 4.Wait, maybe I made a mistake. Let me think again.Each side of the hexagon, when extended, forms an equilateral triangle. So, each side of the star is actually the side of this equilateral triangle. If the hexagon's side is 'a', then the triangle's side is 'a', but the star's side is actually the distance between two points where the extensions meet. Since the triangles are equilateral, the distance between two such points would be '2a' because the triangles are placed opposite each other.Wait, no. Maybe it's better to use coordinate geometry to figure this out.Let me place the hexagon in a coordinate system. Let's assume it's centered at the origin, and one of its vertices is at (1, 0). Since it's equiangular, each vertex is separated by 60 degrees in terms of direction. However, since the sides can be different lengths, the distances from the center to each vertex can vary.But this might complicate things. Maybe there's a simpler way.I remember that in an equiangular hexagon, even if the sides are not equal, the sides can be paired such that opposite sides are equal. Wait, is that true? For an equiangular hexagon, if it's convex, the sides can be arranged such that opposite sides are parallel and equal in length. But I'm not sure if that's always the case.Wait, no, that's for regular hexagons. In an equiangular hexagon, opposite sides are parallel, but they might not be equal. So, maybe each pair of opposite sides are parallel but can have different lengths.Given that, when we extend the sides, the intersections forming the star would depend on the lengths of these sides. But since the hexagon is convex and equiangular, the extensions should form a star with consistent proportions.I think the key here is that the perimeter of the star is twice the perimeter of the hexagon. So, if the hexagon has a perimeter of 2, the star would have a perimeter of 4. That seems to make sense because each side of the hexagon contributes twice to the star's perimeter.But why is that? Let me think about it. Each side of the hexagon, when extended, forms two sides of the star. So, for each side of the hexagon, there are two corresponding sides on the star. Therefore, the total perimeter of the star would be twice that of the hexagon.Yes, that seems logical. So, if the hexagon has a perimeter of 2, the star would have a perimeter of 4.Alternatively, considering that each side of the hexagon is extended to form two sides of the star, each of those sides is equal in length to the original side. Therefore, the total perimeter would be 2 times the perimeter of the hexagon, which is 2 times 2, equaling 4.So, I think the answer is 4, which is option B.**Final Answer**boxed{B}