💡Okay, so I have these two equations: (x = tan^2 alpha) and (y = sin^2 alpha). I need to eliminate (alpha) from them. Hmm, I remember that when you eliminate a parameter, you usually express one variable in terms of the other without the parameter. So, in this case, I need to find a relationship between (x) and (y) that doesn't involve (alpha).First, let me recall some trigonometric identities. I know that (tan alpha = frac{sin alpha}{cos alpha}), so (tan^2 alpha = frac{sin^2 alpha}{cos^2 alpha}). Since (x = tan^2 alpha), I can write (x = frac{sin^2 alpha}{cos^2 alpha}). But I also know that (y = sin^2 alpha), so maybe I can substitute that in.Substituting (y) into the equation for (x), I get (x = frac{y}{cos^2 alpha}). Hmm, so that means (cos^2 alpha = frac{y}{x}). Okay, now I have expressions for both (sin^2 alpha) and (cos^2 alpha) in terms of (x) and (y).I remember another identity: (sin^2 alpha + cos^2 alpha = 1). That's the Pythagorean identity. Since I have both (sin^2 alpha) and (cos^2 alpha) in terms of (x) and (y), I can substitute them into this identity.So substituting, I get (y + frac{y}{x} = 1). Let me write that down: (y + frac{y}{x} = 1). Hmm, that looks like an equation I can solve for one variable in terms of the other.Let me factor out the (y) from the left side: (y left(1 + frac{1}{x}right) = 1). That simplifies to (y left(frac{x + 1}{x}right) = 1). So, multiplying both sides by (frac{x}{x + 1}), I get (y = frac{x}{x + 1}).Wait, is that right? Let me check my steps again. Starting from (y + frac{y}{x} = 1), factoring out (y) gives (y(1 + frac{1}{x}) = 1). Then, (1 + frac{1}{x}) is (frac{x + 1}{x}), so (y = frac{x}{x + 1}). Yeah, that seems correct.But I think I might have made a mistake earlier. Let me go back. I had (x = frac{y}{cos^2 alpha}), so (cos^2 alpha = frac{y}{x}). Then, using (sin^2 alpha + cos^2 alpha = 1), I substituted (y) and (frac{y}{x}) to get (y + frac{y}{x} = 1). That seems right.Alternatively, maybe I can express everything in terms of (sin alpha). Since (y = sin^2 alpha), then (sin alpha = sqrt{y}). Then, (tan alpha = frac{sin alpha}{cos alpha}), so (tan^2 alpha = frac{sin^2 alpha}{cos^2 alpha}). But (cos^2 alpha = 1 - sin^2 alpha = 1 - y). So, (tan^2 alpha = frac{y}{1 - y}). But (x = tan^2 alpha), so (x = frac{y}{1 - y}).Wait, that's a different expression. Let me see. If (x = frac{y}{1 - y}), then multiplying both sides by (1 - y), I get (x(1 - y) = y). Expanding, that's (x - xy = y). Then, bringing all terms to one side, (x - xy - y = 0), which can be written as (x - y(1 + x) = 0). Hmm, that's another form.But earlier, I had (y = frac{x}{x + 1}). Let me see if these are consistent. If (y = frac{x}{x + 1}), then multiplying both sides by (x + 1), I get (y(x + 1) = x), which is (xy + y = x). Rearranging, (x - xy - y = 0), which is the same as before. So both approaches lead to the same equation.Therefore, the relationship between (x) and (y) is (x - y = xy). Let me write that as the final equation: (x - y = xy).I think that's it. I eliminated (alpha) and found a direct relationship between (x) and (y). Let me just double-check by plugging in some values. Suppose (alpha = 45^circ). Then, (tan 45^circ = 1), so (x = 1^2 = 1). (sin 45^circ = frac{sqrt{2}}{2}), so (y = left(frac{sqrt{2}}{2}right)^2 = frac{1}{2}). Plugging into (x - y = xy), we get (1 - frac{1}{2} = 1 times frac{1}{2}), which is (frac{1}{2} = frac{1}{2}). That checks out.Another example: (alpha = 30^circ). Then, (tan 30^circ = frac{1}{sqrt{3}}), so (x = left(frac{1}{sqrt{3}}right)^2 = frac{1}{3}). (sin 30^circ = frac{1}{2}), so (y = left(frac{1}{2}right)^2 = frac{1}{4}). Plugging into (x - y = xy), we get (frac{1}{3} - frac{1}{4} = frac{1}{3} times frac{1}{4}). Calculating the left side: (frac{4}{12} - frac{3}{12} = frac{1}{12}). The right side: (frac{1}{12}). So, it works again.Alright, I think I'm confident that the equation (x - y = xy) correctly eliminates (alpha) from the given equations.

💡Okay, so I need to figure out the least possible cost for buying grass seed when purchasing at least 65 pounds but no more than 80 pounds. The customer must buy at least one of each type of bag: 5-pound, 10-pound, 25-pound, and 40-pound. Also, the total number of bags can't exceed 5. First, I think I should look at the cost per pound for each bag size to see which ones give the best value. That might help me decide how many of each to buy to minimize the cost. Let me calculate the cost per pound:- 5-pound bag: 13.85 / 5 = 2.77 per pound- 10-pound bag: 20.43 / 10 = 2.043 per pound- 25-pound bag: 32.20 / 25 = 1.288 per pound- 40-pound bag: 54.30 / 40 = 1.3575 per poundOkay, so the 25-pound bag is the cheapest per pound, followed by the 40-pound bag, then the 10-pound, and the 5-pound is the most expensive. Since I need to buy at least one of each bag, I'll start by buying one of each:- 5 pounds + 10 pounds + 25 pounds + 40 pounds = 80 poundsThat's exactly 80 pounds, which is within the required range. I've used 4 bags, which is under the 5-bag limit. Let me calculate the total cost:- 1 x 13.85 = 13.85- 1 x 20.43 = 20.43- 1 x 32.20 = 32.20- 1 x 54.30 = 54.30Adding these up: 13.85 + 20.43 + 32.20 + 54.30 = 120.78Hmm, that seems pretty straightforward. But I wonder if I can get a lower total cost by buying more of the cheaper bags and fewer of the expensive ones, while still meeting the requirements. Let me think about other combinations. Maybe buying two 25-pound bags instead of one 40-pound bag? Let's see:- 5 pounds + 10 pounds + 25 pounds + 25 pounds = 65 poundsThat's exactly 65 pounds. But I've only used 4 bags again. The cost would be:- 1 x 13.85 = 13.85- 1 x 20.43 = 20.43- 2 x 32.20 = 64.40Total cost: 13.85 + 20.43 + 64.40 = 98.68Wait, that's cheaper! But does this meet all the requirements? I have at least one of each bag except the 40-pound bag. Oh, I didn't buy the 40-pound bag in this combination. The problem says I must buy at least one of each type of bag. So I can't skip the 40-pound bag. Okay, so I need to include at least one 40-pound bag. Maybe I can adjust the other bags accordingly. Let's try:- 5 pounds + 10 pounds + 25 pounds + 40 pounds = 80 pounds (as before)But that's the same as the first combination, costing 120.78.Is there a way to buy more of the cheaper bags without exceeding the 5-bag limit and still meet the weight requirement? Let's see.If I buy two 25-pound bags and one 40-pound bag, that's 25 + 25 + 40 = 90 pounds, which is over the 80-pound limit. So that's not allowed.What if I buy one 25-pound bag, one 40-pound bag, and then adjust the smaller bags to make up the remaining weight? Let's say I have 25 + 40 = 65 pounds. I need at least 65 pounds, so that's exactly 65. But I also need to buy at least one 5-pound and one 10-pound bag. So:- 5 + 10 + 25 + 40 = 80 poundsAgain, that's the same as the first combination. Alternatively, maybe I can buy one 5-pound, one 10-pound, one 25-pound, and one 40-pound bag, totaling 80 pounds, as before.Is there a way to reduce the number of bags beyond 4? No, because I need at least one of each, so 4 is the minimum. Wait, what if I buy two 5-pound bags instead of one? Let's see:- 5 + 5 + 10 + 25 + 40 = 85 poundsThat's over the 80-pound limit. So that's not allowed.Alternatively, maybe one 5-pound, one 10-pound, one 25-pound, and one 40-pound bag, totaling 80 pounds, as before.I think that's the only combination that meets all the requirements: at least one of each bag, total weight between 65 and 80 pounds, and no more than 5 bags. So, the least possible cost is 120.78.

💡Okay, so I have this problem where a line starts at 1 inch long and grows according to this series: [1 + frac{1}{3}sqrt{3} + frac{1}{3} + frac{1}{9}sqrt{3} + frac{1}{9} + frac{1}{27}sqrt{3} + frac{1}{27} + cdots]I need to find the limit of the length if this process continues forever. The options are given, so maybe I can figure it out step by step.First, looking at the series, it alternates between terms with (sqrt{3}) and terms without. Let me write them out separately to see if they form geometric series.So, separating the terms:Terms without (sqrt{3}): (1 + frac{1}{3} + frac{1}{9} + frac{1}{27} + cdots)Terms with (sqrt{3}): (frac{1}{3}sqrt{3} + frac{1}{9}sqrt{3} + frac{1}{27}sqrt{3} + cdots)Hmm, both of these look like geometric series. A geometric series has the form (a + ar + ar^2 + ar^3 + cdots), where (a) is the first term and (r) is the common ratio.For the first series (without (sqrt{3})):First term (a = 1)Common ratio (r = frac{1}{3}), because each term is (frac{1}{3}) of the previous one.The sum of an infinite geometric series is (S = frac{a}{1 - r}), provided that (|r| < 1). Here, (r = frac{1}{3}), which is less than 1, so it works.Calculating the sum:[S_1 = frac{1}{1 - frac{1}{3}} = frac{1}{frac{2}{3}} = frac{3}{2}]Okay, so the sum of the non-(sqrt{3}) terms is (frac{3}{2}).Now, for the second series (with (sqrt{3})):First term (a = frac{1}{3}sqrt{3})Common ratio (r = frac{1}{3}), since each term is (frac{1}{3}) of the previous one.Again, using the sum formula:[S_2 = frac{frac{1}{3}sqrt{3}}{1 - frac{1}{3}} = frac{frac{1}{3}sqrt{3}}{frac{2}{3}} = frac{sqrt{3}}{2}]So the sum of the (sqrt{3}) terms is (frac{sqrt{3}}{2}).Now, to find the total length, I just add these two sums together:Total length (= S_1 + S_2 = frac{3}{2} + frac{sqrt{3}}{2})Combining the terms over a common denominator:[frac{3 + sqrt{3}}{2}]Looking at the options, that's option B.Wait, let me double-check to make sure I didn't make any mistakes. I separated the series correctly, identified the geometric series, calculated the sums, and added them up. The calculations seem right. Yeah, I think that's correct.**Final Answer**boxed{B}

💡First, I need to evaluate the function ( g(3) ). Given that ( g(x) = x + 1 ), substituting ( x = 3 ) gives ( g(3) = 3 + 1 = 4 ).Next, I calculate ( 1 + g(3) ), which is ( 1 + 4 = 5 ).Finally, I substitute this result into the function ( f(x) ). Since ( f(x) = 3x - 5 ), substituting ( x = 5 ) gives ( f(5) = 3(5) - 5 = 15 - 5 = 10 ).

💡Okay, so I have this problem here: A line L with an inclination angle of 45 degrees passes through the focus F of the parabola y² = 8x and intersects the parabola at points A and B. I need to determine the value of |FA|·|FB|. Hmm, okay, let me try to figure this out step by step.First, I remember that the standard form of a parabola that opens to the right is y² = 4ax, where 'a' is the distance from the vertex to the focus. In this case, the equation is y² = 8x, so comparing it to y² = 4ax, I can see that 4a = 8, which means a = 2. So, the focus F of the parabola is at (a, 0), which is (2, 0). Got that down.Next, the line L has an inclination angle of 45 degrees. I know that the inclination angle is the angle the line makes with the positive direction of the x-axis, measured counterclockwise. So, a 45-degree angle would mean the slope of the line is tan(45°), which is 1. Therefore, the slope of line L is 1.Since the line passes through the focus F(2, 0), I can write its equation using the point-slope form. The point-slope form is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a point the line passes through. Plugging in the values, we get y - 0 = 1*(x - 2), which simplifies to y = x - 2. So, the equation of line L is y = x - 2.Now, this line intersects the parabola y² = 8x at points A and B. To find these points of intersection, I need to solve the system of equations:1. y = x - 22. y² = 8xI can substitute the expression for y from the first equation into the second equation. So, substituting y = x - 2 into y² = 8x gives:(x - 2)² = 8xLet me expand the left side:(x - 2)² = x² - 4x + 4So, the equation becomes:x² - 4x + 4 = 8xNow, let's bring all terms to one side to form a quadratic equation:x² - 4x + 4 - 8x = 0Combine like terms:x² - 12x + 4 = 0Alright, so we have a quadratic equation in terms of x: x² - 12x + 4 = 0. Let me denote the roots of this equation as x₁ and x₂, which correspond to the x-coordinates of points A and B.From quadratic equation theory, I know that for an equation ax² + bx + c = 0, the sum of the roots is -b/a and the product is c/a. So, in this case:Sum of roots, x₁ + x₂ = 12 (since -(-12)/1 = 12)Product of roots, x₁x₂ = 4 (since 4/1 = 4)Okay, so x₁ + x₂ = 12 and x₁x₂ = 4.Now, I need to find |FA|·|FB|. Let me recall that |FA| and |FB| are the distances from the focus F(2, 0) to points A and B, respectively.Since points A and B lie on the parabola y² = 8x, their coordinates satisfy this equation. Also, since they lie on the line y = x - 2, their coordinates also satisfy this equation. So, each point A and B has coordinates (x₁, y₁) and (x₂, y₂), where y₁ = x₁ - 2 and y₂ = x₂ - 2.To find |FA| and |FB|, I can use the distance formula. The distance between two points (x, y) and (a, b) is √[(x - a)² + (y - b)²]. So, applying this to points A and F:|FA| = √[(x₁ - 2)² + (y₁ - 0)²] = √[(x₁ - 2)² + y₁²]But since y₁ = x₁ - 2, we can substitute that in:|FA| = √[(x₁ - 2)² + (x₁ - 2)²] = √[2*(x₁ - 2)²] = √2 * |x₁ - 2|Similarly, |FB| = √2 * |x₂ - 2|Therefore, |FA|·|FB| = (√2 * |x₁ - 2|) * (√2 * |x₂ - 2|) = 2 * |x₁ - 2| * |x₂ - 2|So, now I need to compute |x₁ - 2| * |x₂ - 2|. Let me denote this as |(x₁ - 2)(x₂ - 2)|.Expanding (x₁ - 2)(x₂ - 2):(x₁ - 2)(x₂ - 2) = x₁x₂ - 2x₁ - 2x₂ + 4We already know x₁ + x₂ = 12 and x₁x₂ = 4, so let's substitute:= 4 - 2*(x₁ + x₂) + 4= 4 - 2*12 + 4= 4 - 24 + 4= -16But since we're taking the absolute value, |(x₁ - 2)(x₂ - 2)| = | -16 | = 16Therefore, |FA|·|FB| = 2 * 16 = 32Wait, hold on. Let me double-check that expansion:(x₁ - 2)(x₂ - 2) = x₁x₂ - 2x₁ - 2x₂ + 4Yes, that's correct. Then substituting:x₁x₂ = 4, x₁ + x₂ = 12So, 4 - 2*(12) + 4 = 4 - 24 + 4 = -16Absolute value is 16, so 2*16=32. That seems right.Alternatively, maybe there's another way to think about this without expanding. Let me see.I remember that for a parabola, the product of the distances from the focus to two points on the parabola can sometimes be related to the properties of the parabola. But I'm not sure if that's a standard formula. Maybe it's better to stick with the algebra.Wait, another thought: Since both points A and B lie on the parabola, and the focus is F(2,0), perhaps we can use the definition of a parabola, which is the set of points equidistant from the focus and the directrix.But in this case, since we're dealing with distances from the focus, I don't know if that directly helps. Maybe not necessary here.Alternatively, maybe parametric equations could help. Let me recall that for the parabola y² = 4ax, parametric equations are x = at², y = 2at. In our case, a=2, so x=2t², y=4t.So, points on the parabola can be written as (2t², 4t). Then, the line y = x - 2 intersects the parabola at points where 4t = 2t² - 2.So, 4t = 2t² - 2Bring all terms to one side:2t² - 4t - 2 = 0Divide both sides by 2:t² - 2t - 1 = 0Solving this quadratic equation for t:t = [2 ± √(4 + 4)] / 2 = [2 ± √8]/2 = [2 ± 2√2]/2 = 1 ± √2So, the parameters are t₁ = 1 + √2 and t₂ = 1 - √2.Therefore, the coordinates of points A and B are:For t = 1 + √2:x = 2*(1 + √2)² = 2*(1 + 2√2 + 2) = 2*(3 + 2√2) = 6 + 4√2y = 4*(1 + √2) = 4 + 4√2For t = 1 - √2:x = 2*(1 - √2)² = 2*(1 - 2√2 + 2) = 2*(3 - 2√2) = 6 - 4√2y = 4*(1 - √2) = 4 - 4√2So, points A and B are (6 + 4√2, 4 + 4√2) and (6 - 4√2, 4 - 4√2).Now, let's compute |FA| and |FB|.First, |FA| is the distance from F(2,0) to A(6 + 4√2, 4 + 4√2):|FA| = √[(6 + 4√2 - 2)² + (4 + 4√2 - 0)²]= √[(4 + 4√2)² + (4 + 4√2)²]= √[2*(4 + 4√2)²]Let me compute (4 + 4√2)²:= 16 + 32√2 + 32= 48 + 32√2So, |FA| = √[2*(48 + 32√2)] = √[96 + 64√2]Similarly, |FB| is the distance from F(2,0) to B(6 - 4√2, 4 - 4√2):|FB| = √[(6 - 4√2 - 2)² + (4 - 4√2 - 0)²]= √[(4 - 4√2)² + (4 - 4√2)²]= √[2*(4 - 4√2)²]Compute (4 - 4√2)²:= 16 - 32√2 + 32= 48 - 32√2So, |FB| = √[2*(48 - 32√2)] = √[96 - 64√2]Now, to compute |FA|·|FB|, we have:|FA|·|FB| = √[96 + 64√2] * √[96 - 64√2]This is of the form √(a + b) * √(a - b) = √(a² - b²)So, let's compute a² - b² where a = 96 and b = 64√2:a² = 96² = 9216b² = (64√2)² = 64² * 2 = 4096 * 2 = 8192Thus, a² - b² = 9216 - 8192 = 1024Therefore, |FA|·|FB| = √1024 = 32Okay, so that confirms the earlier result. So, whether I approach it by solving the quadratic equation directly and using the relationships between roots or by parametrizing the parabola and computing the distances, I end up with |FA|·|FB| = 32.Wait, just to make sure I didn't make any calculation errors, let me go through the parametric approach again.We had t² - 2t - 1 = 0, so t = [2 ± √(4 + 4)]/2 = [2 ± √8]/2 = 1 ± √2. Correct.Then, x = 2t², so for t = 1 + √2:t² = (1 + √2)² = 1 + 2√2 + 2 = 3 + 2√2So, x = 2*(3 + 2√2) = 6 + 4√2. Correct.Similarly, for t = 1 - √2:t² = (1 - √2)² = 1 - 2√2 + 2 = 3 - 2√2x = 2*(3 - 2√2) = 6 - 4√2. Correct.Then, y = 4t, so for t = 1 + √2, y = 4 + 4√2; for t = 1 - √2, y = 4 - 4√2. Correct.Then, computing |FA|:√[(6 + 4√2 - 2)² + (4 + 4√2 - 0)²] = √[(4 + 4√2)² + (4 + 4√2)²] = √[2*(4 + 4√2)²]. Correct.Similarly for |FB|. Then, when multiplying, using the difference of squares, we get √(96² - (64√2)²) = √(9216 - 8192) = √1024 = 32. Perfect.Alternatively, going back to the first method, where I had |FA|·|FB| = 2 * |(x₁ - 2)(x₂ - 2)|, and expanding that gave me 32 as well. So both methods agree.I think that solidifies the answer as 32.**Final Answer**boxed{32}

💡Okay, so I need to figure out the ratio of the shortest diagonal to the longest diagonal in a regular octagon. Hmm, let me start by recalling what a regular octagon is. It's an eight-sided polygon where all sides are equal in length and all internal angles are equal too. Each internal angle in a regular octagon is 135 degrees, right?Now, diagonals in a polygon are lines connecting non-adjacent vertices. In a regular octagon, there are diagonals of different lengths. I remember that in a regular polygon, the number of diagonals can be calculated, but I don't think that's necessary here. What I need to focus on is identifying the shortest and longest diagonals.So, in a regular octagon, the shortest diagonal would be the one that connects two vertices that are next to each other but not adjacent. Wait, actually, no. If they're not adjacent, then they are connected by a side, which isn't a diagonal. So, the shortest diagonal would connect two vertices with one vertex in between them. Let me visualize this: imagine an octagon, and if I pick a vertex, the next vertex is connected by a side, the one after that is connected by a diagonal. So, that's the shortest diagonal.On the other hand, the longest diagonal would connect two vertices that are directly opposite each other in the octagon. Since it's regular, these opposite vertices are as far apart as possible, making that diagonal the longest.Alright, so now I need to find the lengths of these diagonals. I think the best way to approach this is by using some trigonometry. Let me consider the regular octagon inscribed in a circle. If I can find the radius of the circumscribed circle, I can use that to find the lengths of the diagonals.Let me denote the side length of the octagon as 's'. The radius 'R' of the circumscribed circle can be related to the side length. I remember that in a regular polygon with 'n' sides, the side length 's' is given by s = 2R * sin(π/n). For an octagon, n=8, so s = 2R * sin(π/8). Therefore, R = s / (2 * sin(π/8)).But maybe there's a simpler way without involving the radius. Let me think about the coordinates of the vertices. If I place the octagon on a coordinate system with its center at the origin, the vertices can be represented using cosine and sine functions. Each vertex would be at an angle of 45 degrees apart since 360/8=45.So, the coordinates of the vertices would be (R*cos(k*45°), R*sin(k*45°)) for k=0,1,2,...,7.Now, to find the length of a diagonal, I can use the distance formula between two such points. For the shortest diagonal, let's say between vertex 0 and vertex 2. The angle between them is 90 degrees (since each step is 45 degrees). So, the distance between these two points would be sqrt[(R*cos(90°)-R*cos(0°))² + (R*sin(90°)-R*sin(0°))²].Calculating that, cos(90°)=0, cos(0°)=1, sin(90°)=1, sin(0°)=0. So, the distance becomes sqrt[(0 - R)² + (R - 0)²] = sqrt[R² + R²] = sqrt[2R²] = R*sqrt(2).Wait, that seems too straightforward. Is that correct? Let me double-check. The angle between vertex 0 and vertex 2 is indeed 90 degrees, so the chord length formula is 2R*sin(θ/2), where θ is the central angle. So, chord length = 2R*sin(45°) = 2R*(√2/2) = R*√2. Yeah, that matches. So, the shortest diagonal is R*√2.Now, for the longest diagonal, which connects opposite vertices, say vertex 0 and vertex 4. The angle between them is 180 degrees. Using the chord length formula again, chord length = 2R*sin(180°/2) = 2R*sin(90°) = 2R*1 = 2R.So, the longest diagonal is 2R.Wait a minute, so the ratio of the shortest diagonal to the longest diagonal would be (R*√2)/(2R) = √2/2. Simplifying that, √2/2 is equal to 1/√2, but usually, we rationalize the denominator, so it becomes √2/2.But hold on, is that correct? Because I feel like I might have made a mistake here. Let me think again. The chord length for the shortest diagonal was R*√2, and the longest was 2R. So, their ratio is √2/2, which is approximately 0.707. That seems reasonable because the shortest diagonal should be shorter than the longest.But wait, in a regular octagon, aren't there diagonals of different lengths? For example, connecting every other vertex gives a shorter diagonal, and connecting opposite vertices gives the longest. So, maybe my initial approach is correct.Alternatively, I remember that in a regular octagon, the ratio of the diagonal to the side length can be expressed in terms of the golden ratio or something similar, but I'm not sure. Let me try another approach using the side length.Let me denote the side length as 's'. I can relate the side length to the radius R as s = 2R*sin(π/8). So, R = s/(2*sin(π/8)).Now, the shortest diagonal, which spans two sides, would correspond to a central angle of 2*(45°) = 90°, so its length is 2R*sin(45°) = 2*(s/(2*sin(π/8)))*sin(45°) = (s/sin(π/8))*sin(45°).Similarly, the longest diagonal spans four sides, corresponding to a central angle of 180°, so its length is 2R*sin(90°) = 2*(s/(2*sin(π/8)))*1 = s/sin(π/8).Therefore, the ratio of the shortest diagonal to the longest diagonal is [(s/sin(π/8))*sin(45°)] / [s/sin(π/8)] = sin(45°) = √2/2.Wait, that's the same result as before. So, it seems consistent. Therefore, the ratio is √2/2.But let me verify this with another method to be sure. Maybe using the properties of the octagon and some geometric constructions.I recall that a regular octagon can be thought of as a square with its corners cut off, each corner being a right-angled isosceles triangle. If I consider the side length of the octagon as 's', then the length of the legs of those triangles would be 'a', and the hypotenuse would be 'a√2'. The side of the square would then be s + 2a, but I'm not sure if that helps directly.Alternatively, perhaps using the law of cosines in the triangles formed by the diagonals. Let me consider the triangle formed by two adjacent vertices and the vertex connected by the shortest diagonal. The sides of this triangle would be 's', 's', and the diagonal 'd1'. The angle between the two sides of length 's' would be 135°, since each internal angle of the octagon is 135°.Wait, no. Actually, when considering the triangle formed by two adjacent vertices and the vertex connected by the shortest diagonal, the angle at the center would be 90°, not 135°. Because the central angle between two vertices separated by one vertex is 2*45°=90°. So, using the law of cosines, the diagonal d1 can be found as:d1² = s² + s² - 2*s*s*cos(90°) = 2s² - 0 = 2s², so d1 = s√2.Wait, that's the same as before. But that would mean the shortest diagonal is s√2, which seems longer than the side length 's', which makes sense because diagonals are longer than sides.But then, the longest diagonal, which spans across the octagon, would be the distance between two opposite vertices. The central angle here is 180°, so using the law of cosines again:d2² = s² + s² - 2*s*s*cos(180°) = 2s² - 2s²*(-1) = 2s² + 2s² = 4s², so d2 = 2s.Therefore, the ratio of d1 to d2 is (s√2)/(2s) = √2/2.Hmm, so that's consistent with my previous calculations. So, it seems that the ratio is indeed √2/2.But wait, I just realized something. In a regular octagon, the diagonals can be of different lengths depending on how many vertices they skip. So, the shortest diagonal skips one vertex, the next one skips two, and so on. But in an octagon, skipping four vertices brings you back to the starting point, so the longest diagonal is the one that skips three vertices, which is the same as skipping four in the other direction, hence the longest diagonal.Wait, actually, in an octagon, the number of vertices you skip determines the length of the diagonal. So, skipping one vertex gives the shortest diagonal, skipping two gives a longer one, and skipping three gives the longest diagonal, which is the same as skipping four in the opposite direction.But in my earlier calculations, I considered the central angles as 90° and 180°, which correspond to skipping one and four vertices, respectively. But skipping two vertices would correspond to a central angle of 135°, and skipping three would be 180°, which is the same as skipping four in the opposite direction.Wait, no. Let me clarify. In an octagon, each vertex is 45° apart. So, skipping one vertex would be 2*45°=90°, skipping two vertices would be 3*45°=135°, skipping three vertices would be 4*45°=180°, and so on. So, the central angles for the diagonals are 90°, 135°, and 180°, corresponding to skipping 1, 2, and 3 vertices, respectively.Therefore, the diagonals have lengths corresponding to these central angles. So, the shortest diagonal corresponds to 90°, the next to 135°, and the longest to 180°.So, using the chord length formula, which is 2R*sin(θ/2), where θ is the central angle.For the shortest diagonal (θ=90°): 2R*sin(45°) = 2R*(√2/2) = R√2.For the diagonal skipping two vertices (θ=135°): 2R*sin(67.5°). Hmm, sin(67.5°) is sin(45°+22.5°). I can use the sine addition formula: sin(A+B) = sinA cosB + cosA sinB.So, sin(67.5°) = sin(45°+22.5°) = sin45*cos22.5 + cos45*sin22.5.I know that sin45 = cos45 = √2/2, and sin22.5 and cos22.5 can be expressed using half-angle formulas.sin22.5° = sin(45°/2) = √[(1 - cos45°)/2] = √[(1 - √2/2)/2] = √[(2 - √2)/4] = √(2 - √2)/2.Similarly, cos22.5° = √[(1 + cos45°)/2] = √[(1 + √2/2)/2] = √[(2 + √2)/4] = √(2 + √2)/2.So, sin(67.5°) = (√2/2)*(√(2 + √2)/2) + (√2/2)*(√(2 - √2)/2).Let me compute that:First term: (√2/2)*(√(2 + √2)/2) = (√2 * √(2 + √2))/4.Second term: (√2/2)*(√(2 - √2)/2) = (√2 * √(2 - √2))/4.So, sin(67.5°) = [√2*(√(2 + √2) + √(2 - √2))]/4.This seems complicated, but maybe it simplifies. Let me square sin(67.5°) to see if that helps.Wait, maybe there's a better way. Alternatively, I can use the exact value of sin(67.5°), which is sin(45° + 22.5°). But I think it's known that sin(67.5°) = √(2 + √2)/2. Let me verify that.Yes, because sin(67.5°) = sin(45° + 22.5°) = sin45*cos22.5 + cos45*sin22.5.We already have expressions for sin22.5 and cos22.5.So, plugging in:sin(67.5°) = (√2/2)*(√(2 + √2)/2) + (√2/2)*(√(2 - √2)/2).Let me compute each term:First term: (√2/2)*(√(2 + √2)/2) = (√2 * √(2 + √2))/4.Similarly, second term: (√2/2)*(√(2 - √2)/2) = (√2 * √(2 - √2))/4.So, sin(67.5°) = [√2*(√(2 + √2) + √(2 - √2))]/4.Let me compute the numerator:√2*(√(2 + √2) + √(2 - √2)).Let me square this expression to see if it simplifies:[√2*(√(2 + √2) + √(2 - √2))]^2 = 2*( (√(2 + √2) + √(2 - √2))^2 ).Expanding the square inside:(√(2 + √2) + √(2 - √2))^2 = (2 + √2) + (2 - √2) + 2*√{(2 + √2)(2 - √2)}.Simplify:= 2 + √2 + 2 - √2 + 2*√{(4 - (√2)^2)}.= 4 + 2*√(4 - 2).= 4 + 2*√2.So, the squared term is 2*(4 + 2√2) = 8 + 4√2.Therefore, the original expression is sqrt(8 + 4√2). Let me see if that can be simplified.sqrt(8 + 4√2) = sqrt(4*(2 + √2)) = 2*sqrt(2 + √2).So, going back, sin(67.5°) = [√2*(√(2 + √2) + √(2 - √2))]/4 = [2*sqrt(2 + √2)]/4 = sqrt(2 + √2)/2.Yes, that matches the known value. So, sin(67.5°) = sqrt(2 + √2)/2.Therefore, the diagonal skipping two vertices (θ=135°) has length 2R*sin(67.5°) = 2R*(sqrt(2 + √2)/2) = R*sqrt(2 + √2).Similarly, the longest diagonal (θ=180°) is 2R*sin(90°) = 2R*1 = 2R.So, now I have the lengths of the diagonals:- Shortest diagonal (skipping one vertex, θ=90°): R*sqrt(2).- Next diagonal (skipping two vertices, θ=135°): R*sqrt(2 + √2).- Longest diagonal (skipping three vertices, θ=180°): 2R.But wait, the problem asks for the ratio of the shortest diagonal to the longest diagonal. So, that would be (R*sqrt(2)) / (2R) = sqrt(2)/2.But hold on, earlier I considered the diagonal skipping one vertex as the shortest, but in reality, in a regular octagon, there are two different diagonals: the ones that skip one vertex and the ones that skip two vertices. The one that skips one vertex is indeed shorter than the one that skips two, which is shorter than the longest diagonal.So, the shortest diagonal is R*sqrt(2), the next is R*sqrt(2 + √2), and the longest is 2R.Therefore, the ratio of the shortest to the longest is sqrt(2)/2.But let me confirm this with another approach. Maybe using the side length 's' instead of the radius 'R'.We know that in a regular octagon, the side length 's' is related to the radius 'R' by s = 2R*sin(π/8). So, R = s/(2*sin(π/8)).Now, the shortest diagonal is R*sqrt(2), so substituting R:Shortest diagonal = (s/(2*sin(π/8)))*sqrt(2) = s*sqrt(2)/(2*sin(π/8)).Similarly, the longest diagonal is 2R = 2*(s/(2*sin(π/8))) = s/sin(π/8).So, the ratio is [s*sqrt(2)/(2*sin(π/8))] / [s/sin(π/8)] = sqrt(2)/2.Again, the same result. So, it seems consistent.But wait, I just realized that sin(π/8) is sin(22.5°), which is equal to sqrt(2 - √2)/2. Let me verify that.Yes, sin(22.5°) = sin(45°/2) = sqrt[(1 - cos45°)/2] = sqrt[(1 - √2/2)/2] = sqrt[(2 - √2)/4] = sqrt(2 - √2)/2.So, sin(π/8) = sqrt(2 - √2)/2.Therefore, the ratio sqrt(2)/2 can also be expressed in terms of sin(π/8), but I don't think that's necessary here.So, putting it all together, the ratio of the shortest diagonal to the longest diagonal in a regular octagon is sqrt(2)/2.But wait, the problem asks to express the answer as a common fraction in simplest form. So, sqrt(2)/2 is already a common fraction, but perhaps it can be rationalized or expressed differently.Alternatively, sqrt(2)/2 is equal to 1/sqrt(2), but usually, we rationalize the denominator, so it's better to write it as sqrt(2)/2.Therefore, the ratio is sqrt(2)/2.But let me just make sure I didn't confuse the diagonals. In a regular octagon, the diagonals that skip one vertex are shorter than those that skip two, which are shorter than the longest. So, the ratio of the shortest to the longest is indeed sqrt(2)/2.Alternatively, I can think of the regular octagon as being composed of isosceles triangles from the center. Each triangle has a central angle of 45°, and the sides are radii of length R. The base of each triangle is the side length 's' of the octagon.Using the law of cosines on these triangles, s² = R² + R² - 2R²*cos(45°) = 2R²(1 - cos45°).So, s² = 2R²(1 - √2/2) = 2R² - R²√2.Therefore, s = R*sqrt(2 - √2).Wait, that's the side length in terms of R. So, s = R*sqrt(2 - √2).Now, the shortest diagonal, which skips one vertex, corresponds to a central angle of 90°, so its length is 2R*sin(45°) = 2R*(√2/2) = R√2.Similarly, the longest diagonal is 2R.So, the ratio is R√2 / 2R = √2/2.Consistent again.Alternatively, expressing everything in terms of 's':Since s = R*sqrt(2 - √2), then R = s / sqrt(2 - √2).So, the shortest diagonal is R√2 = (s / sqrt(2 - √2)) * sqrt(2) = s*sqrt(2)/sqrt(2 - √2).Similarly, the longest diagonal is 2R = 2*(s / sqrt(2 - √2)).So, the ratio is [s*sqrt(2)/sqrt(2 - √2)] / [2s / sqrt(2 - √2)] = sqrt(2)/2.Same result.Therefore, after multiple approaches and verifications, I'm confident that the ratio of the shortest diagonal to the longest diagonal in a regular octagon is sqrt(2)/2.But just to make sure, let me check an external source or formula. Wait, I can't access external resources, but I recall that in a regular octagon, the ratio of the diagonal to the side length is 1 + sqrt(2), but that might be for a different diagonal.Wait, actually, the diagonal that skips two vertices (the longer diagonal) is s*(1 + sqrt(2)), where 's' is the side length. So, if the side length is 's', then the longer diagonal is s*(1 + sqrt(2)), and the longest diagonal is 2s.Wait, that contradicts my earlier findings. Let me see.If the side length is 's', then the diagonal that skips two vertices is s*(1 + sqrt(2)), which is approximately 2.414s, and the longest diagonal is 2s.But according to my previous calculations, the diagonal skipping two vertices was R*sqrt(2 + √2), which, when expressed in terms of 's', is s*sqrt(2 + √2)/sqrt(2 - √2).Wait, let me compute sqrt(2 + √2)/sqrt(2 - √2).Multiply numerator and denominator by sqrt(2 + √2):[sqrt(2 + √2)*sqrt(2 + √2)] / [sqrt(2 - √2)*sqrt(2 + √2)] = (2 + √2)/sqrt((2)^2 - (√2)^2) = (2 + √2)/sqrt(4 - 2) = (2 + √2)/sqrt(2) = (2 + √2)/√2 = (2/√2) + (√2/√2) = sqrt(2) + 1.Ah, so sqrt(2 + √2)/sqrt(2 - √2) = 1 + sqrt(2).Therefore, the diagonal skipping two vertices is s*(1 + sqrt(2)), which is approximately 2.414s, and the longest diagonal is 2s.Wait, so that means the diagonal skipping two vertices is longer than the side length but shorter than the longest diagonal.So, in that case, the shortest diagonal is the one skipping one vertex, which is s*sqrt(2 - √2)/sqrt(2 - √2) = s*sqrt(2)/sqrt(2 - √2).Wait, no, earlier we had s = R*sqrt(2 - √2), so R = s / sqrt(2 - √2).Then, the shortest diagonal is R*sqrt(2) = (s / sqrt(2 - √2)) * sqrt(2) = s*sqrt(2)/sqrt(2 - √2).As we saw earlier, sqrt(2)/sqrt(2 - √2) = sqrt(2)*(sqrt(2 + √2))/sqrt((2 - √2)(2 + √2)) = sqrt(2)*sqrt(2 + √2)/sqrt(4 - 2) = sqrt(2)*sqrt(2 + √2)/sqrt(2) = sqrt(2 + √2).Wait, that can't be. Let me compute sqrt(2)/sqrt(2 - √2).Multiply numerator and denominator by sqrt(2 + √2):sqrt(2)*sqrt(2 + √2) / sqrt((2 - √2)(2 + √2)) = sqrt(2)*sqrt(2 + √2)/sqrt(4 - 2) = sqrt(2)*sqrt(2 + √2)/sqrt(2) = sqrt(2 + √2).So, sqrt(2)/sqrt(2 - √2) = sqrt(2 + √2).Therefore, the shortest diagonal is s*sqrt(2 + √2).Wait, but that contradicts my earlier conclusion that the shortest diagonal is s*sqrt(2). Hmm, I must have made a mistake somewhere.Let me recap:- The side length 's' is related to the radius 'R' by s = 2R*sin(π/8) = 2R*sin(22.5°).- The shortest diagonal skips one vertex, central angle 90°, so its length is 2R*sin(45°) = 2R*(√2/2) = R√2.- The diagonal skipping two vertices has central angle 135°, length 2R*sin(67.5°) = 2R*(sqrt(2 + √2)/2) = R*sqrt(2 + √2).- The longest diagonal skips three vertices, central angle 180°, length 2R.Expressed in terms of 's':Since s = 2R*sin(22.5°), R = s/(2*sin(22.5°)).So, shortest diagonal = R√2 = (s/(2*sin(22.5°)))*√2.Similarly, longest diagonal = 2R = 2*(s/(2*sin(22.5°))) = s/sin(22.5°).So, the ratio is [ (s/(2*sin(22.5°)))*√2 ] / [ s/sin(22.5°) ] = (√2/2).Therefore, the ratio is √2/2.But earlier, when expressing the diagonal skipping two vertices in terms of 's', I got s*(1 + sqrt(2)), which is longer than the side length but shorter than the longest diagonal (2s). So, that seems consistent.Wait, but if the shortest diagonal is R√2, and R = s/(2*sin(22.5°)), then the shortest diagonal is (s/(2*sin(22.5°)))*√2.But sin(22.5°) = sqrt(2 - √2)/2, so:Shortest diagonal = (s/(2*(sqrt(2 - √2)/2)))*√2 = (s / sqrt(2 - √2)) * √2 = s*√2 / sqrt(2 - √2).As we saw earlier, sqrt(2)/sqrt(2 - √2) = sqrt(2 + √2), so the shortest diagonal is s*sqrt(2 + √2).Wait, that can't be, because sqrt(2 + √2) is approximately 1.847, which is less than 2, so s*1.847 is less than 2s, which is the longest diagonal. So, that makes sense.But earlier, I thought the shortest diagonal was s*sqrt(2), which is approximately 1.414s, which is less than s*sqrt(2 + √2) ≈1.847s, which is less than 2s.Wait, so actually, the shortest diagonal is s*sqrt(2), which is approximately 1.414s, the next diagonal is s*sqrt(2 + √2) ≈1.847s, and the longest is 2s.But according to the previous calculation, the shortest diagonal is s*sqrt(2 + √2), which contradicts this.I think I made a mistake in the substitution.Let me clarify:We have s = 2R*sin(22.5°).Therefore, R = s/(2*sin(22.5°)).The shortest diagonal is 2R*sin(45°) = 2*(s/(2*sin(22.5°)))*sin(45°) = (s/sin(22.5°))*sin(45°).Now, sin(45°) = sqrt(2)/2, and sin(22.5°) = sqrt(2 - √2)/2.Therefore, the shortest diagonal is (s / (sqrt(2 - √2)/2)) * (sqrt(2)/2) = (2s / sqrt(2 - √2)) * (sqrt(2)/2) = (s*sqrt(2)) / sqrt(2 - √2).As before, sqrt(2)/sqrt(2 - √2) = sqrt(2 + √2).Therefore, the shortest diagonal is s*sqrt(2 + √2).Wait, but that's longer than s*sqrt(2). So, that contradicts my earlier assumption that the shortest diagonal is s*sqrt(2).I think the confusion arises from the definition of the diagonals. In a regular octagon, the diagonals that skip one vertex are indeed shorter than those that skip two vertices, but when expressed in terms of 's', they might seem longer because of the relationship between 's' and 'R'.Wait, no. Let me think again. If s = 2R*sin(22.5°), then R = s/(2*sin(22.5°)).The shortest diagonal is 2R*sin(45°) = 2*(s/(2*sin(22.5°)))*sin(45°) = (s/sin(22.5°))*sin(45°).Now, sin(45°)/sin(22.5°) = [sqrt(2)/2] / [sqrt(2 - √2)/2] = sqrt(2)/sqrt(2 - √2) = sqrt(2 + √2).Therefore, the shortest diagonal is s*sqrt(2 + √2).Wait, but that's longer than the side length 's', which makes sense because diagonals are longer than sides. But earlier, I thought the shortest diagonal was s*sqrt(2), which is approximately 1.414s, but according to this, it's s*sqrt(2 + √2) ≈1.847s.So, which one is correct?I think the confusion comes from the fact that in a regular octagon, the term "shortest diagonal" can be ambiguous because there are diagonals of different lengths. However, in standard terminology, the shortest diagonal is the one that connects two vertices with one vertex in between, which corresponds to a central angle of 90°, and its length is 2R*sin(45°) = R*sqrt(2).But when expressed in terms of the side length 's', it's s*sqrt(2 + √2), which is longer than 's*sqrt(2)'.Wait, that doesn't make sense because sqrt(2 + √2) ≈1.847, which is greater than sqrt(2)≈1.414.So, perhaps I made a mistake in the substitution.Let me re-express everything step by step.Given a regular octagon with side length 's', the radius 'R' of the circumscribed circle is R = s/(2*sin(π/8)) = s/(2*sin(22.5°)).The shortest diagonal (skipping one vertex) has a central angle of 90°, so its length is 2R*sin(45°) = 2*(s/(2*sin(22.5°)))*sin(45°) = (s/sin(22.5°))*sin(45°).Now, sin(45°) = sqrt(2)/2, and sin(22.5°) = sqrt(2 - √2)/2.Therefore, the shortest diagonal is (s / (sqrt(2 - √2)/2)) * (sqrt(2)/2) = (2s / sqrt(2 - √2)) * (sqrt(2)/2) = (s*sqrt(2)) / sqrt(2 - √2).As before, sqrt(2)/sqrt(2 - √2) = sqrt(2 + √2).Therefore, the shortest diagonal is s*sqrt(2 + √2).Wait, but that's longer than the side length 's', which is correct, but it's also longer than s*sqrt(2), which is approximately 1.414s.But according to this, the shortest diagonal is s*sqrt(2 + √2) ≈1.847s, which is longer than s*sqrt(2).So, that suggests that the shortest diagonal is actually longer than s*sqrt(2), which contradicts my earlier assumption.I think the confusion arises from the fact that in a regular octagon, the term "shortest diagonal" refers to the diagonal that skips one vertex, which is indeed longer than the side length but shorter than the diagonal that skips two vertices.Wait, no, in a regular octagon, the diagonals that skip one vertex are shorter than those that skip two vertices. So, the shortest diagonal is the one skipping one vertex, and the next is skipping two, and the longest is skipping three (opposite vertices).But according to the calculations, the length of the diagonal skipping one vertex is s*sqrt(2 + √2), which is approximately 1.847s, and the diagonal skipping two vertices is s*(1 + sqrt(2)) ≈2.414s, and the longest is 2s.Wait, that can't be because 2s is longer than 2.414s. Wait, no, 2s is approximately 2s, which is less than 2.414s. Wait, no, 2.414s is greater than 2s.Wait, no, 2.414 is approximately 1 + sqrt(2) ≈2.414, which is greater than 2. So, 2s is less than s*(1 + sqrt(2)).Wait, that can't be. Let me compute 1 + sqrt(2) ≈1 + 1.414≈2.414, which is greater than 2. So, s*(1 + sqrt(2))≈2.414s, which is longer than 2s.But that contradicts the fact that the longest diagonal should be the one connecting opposite vertices, which is 2R.Wait, let's compute 2R in terms of 's'.Since R = s/(2*sin(22.5°)) ≈s/(2*0.38268)≈s/0.76536≈1.30656s.Therefore, 2R≈2.61312s.But s*(1 + sqrt(2))≈2.414s, which is less than 2R≈2.613s.So, the longest diagonal is indeed 2R≈2.613s, which is longer than s*(1 + sqrt(2))≈2.414s.Therefore, the diagonals in order from shortest to longest are:1. Skipping one vertex: s*sqrt(2 + √2)≈1.847s.2. Skipping two vertices: s*(1 + sqrt(2))≈2.414s.3. Skipping three vertices (longest): 2R≈2.613s.Therefore, the shortest diagonal is s*sqrt(2 + √2), and the longest is 2R.But wait, earlier I thought the shortest diagonal was s*sqrt(2), but that seems incorrect because when expressed in terms of 's', it's actually s*sqrt(2 + √2).So, perhaps my initial assumption that the shortest diagonal was s*sqrt(2) was wrong. Instead, it's s*sqrt(2 + √2).But then, how does that reconcile with the chord length formula?Wait, chord length for central angle θ is 2R*sin(θ/2).For θ=90°, chord length=2R*sin(45°)=2R*(√2/2)=R√2.But R = s/(2*sin(22.5°)), so chord length= (s/(2*sin(22.5°)))*√2= s*√2/(2*sin(22.5°)).As before, sin(22.5°)=sqrt(2 - √2)/2, so chord length= s*√2/(2*(sqrt(2 - √2)/2))= s*√2/sqrt(2 - √2)=s*sqrt(2 + √2).Therefore, the chord length for θ=90° is s*sqrt(2 + √2).Similarly, for θ=135°, chord length=2R*sin(67.5°)=2*(s/(2*sin(22.5°)))*sin(67.5°)= (s/sin(22.5°))*sin(67.5°).Sin(67.5°)=sqrt(2 + √2)/2, so chord length= (s / (sqrt(2 - √2)/2))*(sqrt(2 + √2)/2)= (2s / sqrt(2 - √2))*(sqrt(2 + √2)/2)= s*sqrt(2 + √2)/sqrt(2 - √2)=s*(sqrt(2 + √2)/sqrt(2 - √2)).As before, sqrt(2 + √2)/sqrt(2 - √2)=sqrt((2 + √2)/(2 - √2))=sqrt((2 + √2)^2/(4 - 2))=sqrt((4 + 4√2 + 2)/2)=sqrt((6 + 4√2)/2)=sqrt(3 + 2√2)=1 + sqrt(2).Therefore, chord length= s*(1 + sqrt(2)).And for θ=180°, chord length=2R=2*(s/(2*sin(22.5°)))=s/sin(22.5°)=s/(sqrt(2 - √2)/2)=2s/sqrt(2 - √2)=2s*sqrt(2 + √2)/sqrt((2 - √2)(2 + √2))=2s*sqrt(2 + √2)/sqrt(4 - 2)=2s*sqrt(2 + √2)/sqrt(2)=2s*sqrt(2 + √2)/sqrt(2)=2s*sqrt((2 + √2)/2)=2s*sqrt(1 + (√2)/2).Wait, that seems more complicated, but we know that 2R≈2.613s, which is greater than s*(1 + sqrt(2))≈2.414s.Therefore, the order of diagonals from shortest to longest is:1. Skipping one vertex: s*sqrt(2 + √2)≈1.847s.2. Skipping two vertices: s*(1 + sqrt(2))≈2.414s.3. Skipping three vertices (longest): 2R≈2.613s.Therefore, the ratio of the shortest diagonal to the longest diagonal is [s*sqrt(2 + √2)] / [2R].But 2R = s/sin(22.5°)=s/(sqrt(2 - √2)/2)=2s/sqrt(2 - √2).Therefore, the ratio is [s*sqrt(2 + √2)] / [2s/sqrt(2 - √2)] = [sqrt(2 + √2) * sqrt(2 - √2)] / 2.Multiplying the numerators:sqrt(2 + √2)*sqrt(2 - √2)=sqrt{(2 + √2)(2 - √2)}=sqrt{4 - (√2)^2}=sqrt{4 - 2}=sqrt{2}.Therefore, the ratio is sqrt(2)/2.So, despite the confusion earlier, the ratio simplifies to sqrt(2)/2.Therefore, the ratio of the shortest diagonal to the longest diagonal in a regular octagon is sqrt(2)/2.But to express this as a common fraction in simplest form, sqrt(2)/2 is already a common fraction, but it's often rationalized as 1/sqrt(2), but since the problem asks for a common fraction, sqrt(2)/2 is acceptable.Alternatively, if we rationalize, it's 1/sqrt(2), but usually, we rationalize denominators, so sqrt(2)/2 is preferred.Therefore, the final answer is sqrt(2)/2.