💡Alright, so I've got this problem here: I need to find the smallest positive integer ( n ) such that there exist positive real numbers ( a ) and ( b ) where ( (a + bi)^n = (a - bi)^n ). Then, I have to compute the ratio ( frac{b}{a} ).Hmm, okay. Let me start by understanding what this equation means. ( a + bi ) and ( a - bi ) are complex conjugates of each other. Raising them to the same power ( n ) and setting them equal... That seems interesting.I remember that when you raise complex numbers to powers, you can use De Moivre's theorem, which relates complex numbers in polar form to their powers. Maybe I should try expressing ( a + bi ) and ( a - bi ) in polar form.First, let's recall that any complex number ( z = a + bi ) can be written in polar form as ( r(cos theta + i sin theta) ), where ( r ) is the modulus (or magnitude) of ( z ), given by ( r = sqrt{a^2 + b^2} ), and ( theta ) is the argument (or angle) of ( z ), given by ( theta = arctanleft(frac{b}{a}right) ).Similarly, the complex conjugate ( overline{z} = a - bi ) would have the same modulus ( r ) but an argument of ( -theta ).So, using De Moivre's theorem, ( z^n = r^n (cos ntheta + i sin ntheta) ) and ( overline{z}^n = r^n (cos (-ntheta) + i sin (-ntheta)) ).Simplifying ( overline{z}^n ), since cosine is even and sine is odd, this becomes ( r^n (cos ntheta - i sin ntheta) ).So, setting ( z^n = overline{z}^n ), we have:[r^n (cos ntheta + i sin ntheta) = r^n (cos ntheta - i sin ntheta)]Since ( r ) is positive, we can divide both sides by ( r^n ), giving:[cos ntheta + i sin ntheta = cos ntheta - i sin ntheta]Subtracting ( cos ntheta ) from both sides:[i sin ntheta = -i sin ntheta]Bringing all terms to one side:[2i sin ntheta = 0]Since ( i ) is not zero, this implies that ( sin ntheta = 0 ).So, ( sin ntheta = 0 ) means that ( ntheta ) must be an integer multiple of ( pi ). That is:[ntheta = kpi quad text{for some integer } k]But ( theta ) is the argument of ( z ), which is ( arctanleft(frac{b}{a}right) ). So,[n arctanleft(frac{b}{a}right) = kpi]Hmm, okay. So, ( arctanleft(frac{b}{a}right) = frac{kpi}{n} ).Since ( a ) and ( b ) are positive, ( theta ) is in the first quadrant, so ( 0 < theta < frac{pi}{2} ). Therefore, ( 0 < frac{kpi}{n} < frac{pi}{2} ), which implies that ( 0 < k < frac{n}{2} ).But ( k ) must be an integer, so the smallest positive integer ( k ) is 1. Therefore, ( arctanleft(frac{b}{a}right) = frac{pi}{n} ).Thus, ( frac{b}{a} = tanleft(frac{pi}{n}right) ).Now, we need to find the smallest positive integer ( n ) such that ( tanleft(frac{pi}{n}right) ) is a positive real number, which it always is for ( n geq 2 ), but we need to ensure that ( a ) and ( b ) are positive real numbers. So, as long as ( n ) is an integer greater than or equal to 2, ( frac{b}{a} ) will be positive.But wait, the problem says "the smallest positive integer ( n )". So, we need to find the smallest ( n ) for which such ( a ) and ( b ) exist. Let's test ( n = 1 ), ( n = 2 ), ( n = 3 ), etc., to see when this equation holds.Starting with ( n = 1 ):If ( n = 1 ), then ( (a + bi)^1 = (a - bi)^1 ) implies ( a + bi = a - bi ). Subtracting ( a ) from both sides, we get ( bi = -bi ), which implies ( 2bi = 0 ). Since ( b ) is positive, this is impossible. So, ( n = 1 ) doesn't work.Next, ( n = 2 ):( (a + bi)^2 = (a - bi)^2 ).Expanding both sides:Left side: ( a^2 + 2abi - b^2 ).Right side: ( a^2 - 2abi - b^2 ).Setting them equal:( a^2 + 2abi - b^2 = a^2 - 2abi - b^2 ).Subtracting ( a^2 - b^2 ) from both sides:( 4abi = 0 ).Again, since ( a ) and ( b ) are positive, this implies ( ab = 0 ), which is impossible. So, ( n = 2 ) doesn't work either.Moving on to ( n = 3 ):( (a + bi)^3 = (a - bi)^3 ).Let me compute both sides.First, ( (a + bi)^3 ):Using the binomial theorem:( (a + bi)^3 = a^3 + 3a^2(bi) + 3a(bi)^2 + (bi)^3 ).Simplify each term:- ( a^3 ) remains ( a^3 ).- ( 3a^2(bi) = 3a^2b i ).- ( 3a(bi)^2 = 3a b^2 i^2 = 3a b^2 (-1) = -3a b^2 ).- ( (bi)^3 = b^3 i^3 = b^3 (-i) = -b^3 i ).So, combining all terms:( a^3 + 3a^2b i - 3a b^2 - b^3 i ).Similarly, ( (a - bi)^3 ):( (a - bi)^3 = a^3 - 3a^2(bi) + 3a(bi)^2 - (bi)^3 ).Simplify each term:- ( a^3 ) remains ( a^3 ).- ( -3a^2(bi) = -3a^2b i ).- ( 3a(bi)^2 = 3a b^2 (-1) = -3a b^2 ).- ( -(bi)^3 = -b^3 (-i) = b^3 i ).So, combining all terms:( a^3 - 3a^2b i - 3a b^2 + b^3 i ).Now, setting ( (a + bi)^3 = (a - bi)^3 ):Left side: ( a^3 + 3a^2b i - 3a b^2 - b^3 i ).Right side: ( a^3 - 3a^2b i - 3a b^2 + b^3 i ).Subtracting the right side from the left side:( (a^3 + 3a^2b i - 3a b^2 - b^3 i) - (a^3 - 3a^2b i - 3a b^2 + b^3 i) = 0 ).Simplify term by term:- ( a^3 - a^3 = 0 ).- ( 3a^2b i - (-3a^2b i) = 6a^2b i ).- ( -3a b^2 - (-3a b^2) = 0 ).- ( -b^3 i - b^3 i = -2b^3 i ).So, combining:( 6a^2b i - 2b^3 i = 0 ).Factor out ( 2b i ):( 2b i (3a^2 - b^2) = 0 ).Since ( b ) is positive, ( 2b i ) is not zero. Therefore, we must have:( 3a^2 - b^2 = 0 ).So,( 3a^2 = b^2 ).Taking square roots (since ( a ) and ( b ) are positive):( b = a sqrt{3} ).Therefore, ( frac{b}{a} = sqrt{3} ).So, for ( n = 3 ), such positive real numbers ( a ) and ( b ) exist, and the ratio ( frac{b}{a} ) is ( sqrt{3} ).But wait, before I conclude, I should check if ( n = 3 ) is indeed the smallest such integer. We saw that ( n = 1 ) and ( n = 2 ) don't work, so ( n = 3 ) is the next candidate. Since ( n = 3 ) works, it must be the smallest positive integer satisfying the condition.Therefore, the answer is ( sqrt{3} ).**Final Answer**boxed{sqrt{3}}

💡Alright, so I've got this problem here about two cars: a new model and an old model. The new car is supposed to be more fuel-efficient, right? It says it travels 4.2 kilometers more per liter of gasoline than the old model. Plus, the fuel consumption for the new model is 2 liters less per 100 km. I need to find out how many liters of gasoline the new car consumes per 100 km. Hmm, okay, let's break this down.First, I think I need to define some variables to represent the unknowns here. Let me call the fuel consumption of the new car "x" liters per 100 km. That means the old car would consume "x + 2" liters per 100 km because it's 2 liters more than the new one. Okay, that makes sense.Now, fuel consumption is liters per 100 km, but the problem also mentions how many kilometers the car can travel per liter. So, I guess I need to relate these two things. If the new car consumes "x" liters per 100 km, then its fuel efficiency in kilometers per liter would be 100 divided by x, right? Because if you use "x" liters to go 100 km, then per liter, you go 100/x km.Similarly, for the old car, which consumes "x + 2" liters per 100 km, its fuel efficiency would be 100 divided by (x + 2) km per liter.The problem states that the new car travels 4.2 kilometers more per liter than the old car. So, the fuel efficiency of the new car minus the fuel efficiency of the old car equals 4.2 km/l. Putting that into an equation:100/x - 100/(x + 2) = 4.2Okay, so now I have an equation that I need to solve for x. Let's see how to do that. First, I can find a common denominator for the two fractions on the left side. The common denominator would be x(x + 2). So, rewriting the fractions:(100(x + 2) - 100x) / [x(x + 2)] = 4.2Simplifying the numerator:100x + 200 - 100x = 200So, the equation becomes:200 / [x(x + 2)] = 4.2Now, I can cross-multiply to get rid of the fraction:200 = 4.2 * x(x + 2)Let's expand the right side:200 = 4.2x^2 + 8.4xHmm, now I have a quadratic equation. To make it easier, I can divide both sides by 4.2 to simplify:200 / 4.2 = x^2 + 2xCalculating 200 divided by 4.2:200 / 4.2 ≈ 47.619So, the equation is approximately:47.619 = x^2 + 2xLet's rearrange it to standard quadratic form:x^2 + 2x - 47.619 = 0Now, I can use the quadratic formula to solve for x. The quadratic formula is:x = [-b ± sqrt(b^2 - 4ac)] / (2a)In this equation, a = 1, b = 2, and c = -47.619. Plugging these values in:x = [-2 ± sqrt((2)^2 - 4*1*(-47.619))] / (2*1)Calculating inside the square root:sqrt(4 + 190.476) = sqrt(194.476) ≈ 13.945So, the solutions are:x = [-2 + 13.945] / 2 ≈ 11.945 / 2 ≈ 5.9725andx = [-2 - 13.945] / 2 ≈ -15.945 / 2 ≈ -7.9725Since fuel consumption can't be negative, we discard the negative solution. Therefore, x ≈ 5.9725 liters per 100 km.Rounding to the nearest hundredth, that's approximately 5.97 liters per 100 km.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting from the equation:100/x - 100/(x + 2) = 4.2I found a common denominator and simplified to:200 / [x(x + 2)] = 4.2Then cross-multiplied to get:200 = 4.2x(x + 2)Expanded to:200 = 4.2x^2 + 8.4xDivided by 4.2:47.619 ≈ x^2 + 2xThen rearranged to:x^2 + 2x - 47.619 = 0Applied quadratic formula:x = [-2 ± sqrt(4 + 190.476)] / 2sqrt(194.476) ≈ 13.945So, x ≈ (11.945)/2 ≈ 5.9725Yes, that seems correct. So, the new car consumes approximately 5.97 liters per 100 km.

💡Alright, so I have this problem about a convex pentagon, which is a five-sided polygon where all the interior angles are less than 180 degrees. The pentagon is labeled ABCDE, and I need to find the measure of angle D. Let me read the problem again and jot down the important details.First, it says that angle D is 30 degrees more than angle A. So, if I let the measure of angle A be some value, then angle D would be that value plus 30 degrees. Similarly, angle E is 50 degrees more than angle A. So, angle E is angle A plus 50 degrees.Next, it mentions that angles B and C are equal. That means whatever measure angle B has, angle C has the same measure. Additionally, it says that angle A is 45 degrees less than angle B. So, if I know angle B, I can find angle A by subtracting 45 degrees from it. Alternatively, if I know angle A, angle B would be angle A plus 45 degrees.Since it's a convex pentagon, the sum of all its interior angles should be 540 degrees. I remember that the formula for the sum of interior angles of an n-sided polygon is (n-2)*180 degrees. So, for a pentagon, it's (5-2)*180 = 540 degrees. That's a key point.Let me try to structure this information step by step.1. Let the measure of angle A be x degrees.2. Then, angle D is x + 30 degrees.3. Angle E is x + 50 degrees.4. Angles B and C are equal. Let me denote their measure as y degrees each.5. It's given that angle A is 45 degrees less than angle B. So, angle A = angle B - 45 degrees. Which means x = y - 45. Therefore, y = x + 45 degrees.Now, I can express all angles in terms of x.- Angle A: x- Angle B: x + 45- Angle C: x + 45- Angle D: x + 30- Angle E: x + 50Now, the sum of all these angles should be 540 degrees.So, let's write the equation:x + (x + 45) + (x + 45) + (x + 30) + (x + 50) = 540Let me simplify this equation step by step.First, combine like terms. Let's count how many x terms there are. There's one x in angle A, and then each of the other angles adds another x. So, that's 1 + 1 + 1 + 1 + 1 = 5x.Now, the constants: 45 (from angle B) + 45 (from angle C) + 30 (from angle D) + 50 (from angle E). Let's add those up.45 + 45 = 9090 + 30 = 120120 + 50 = 170So, the equation simplifies to:5x + 170 = 540Now, I need to solve for x. Let's subtract 170 from both sides.5x = 540 - 1705x = 370Now, divide both sides by 5.x = 370 / 5x = 74So, angle A is 74 degrees. Now, since angle D is 30 degrees more than angle A, angle D is 74 + 30 = 104 degrees.Wait, let me double-check my calculations to make sure I didn't make any mistakes.Starting with the equation:x + (x + 45) + (x + 45) + (x + 30) + (x + 50) = 540Combine like terms:5x + (45 + 45 + 30 + 50) = 5405x + 170 = 540Subtract 170:5x = 370Divide by 5:x = 74Yes, that seems correct. So, angle A is 74 degrees, and angle D is 74 + 30 = 104 degrees.Just to be thorough, let me calculate all the angles and make sure they add up to 540 degrees.- Angle A: 74 degrees- Angle B: 74 + 45 = 119 degrees- Angle C: 119 degrees (same as angle B)- Angle D: 74 + 30 = 104 degrees- Angle E: 74 + 50 = 124 degreesNow, adding them up:74 + 119 + 119 + 104 + 124Let me add them step by step.74 + 119 = 193193 + 119 = 312312 + 104 = 416416 + 124 = 540Perfect, that adds up to 540 degrees, which is the total sum for a pentagon. So, my calculations seem correct.Therefore, the measure of angle D is 104 degrees.**Final Answer**The measure of angle ( D ) is boxed{104} degrees.

💡Okay, so I have this problem about a quadrilateral with sides a, b, c, d, and I need to prove that its area S is less than or equal to (ac + bd)/2. And equality holds only when it's an inscribed quadrilateral with perpendicular diagonals. Hmm, okay, let me try to figure this out step by step.First, I remember that for a quadrilateral, the area can be calculated in different ways depending on the type of quadrilateral. For example, for a rectangle, it's just length times width. For a parallelogram, it's base times height. But this is a general quadrilateral, so it might not have any specific properties like right angles or parallel sides.Wait, the problem mentions an inscribed quadrilateral, which I think means a cyclic quadrilateral, where all four vertices lie on a circle. And it also mentions that the diagonals are perpendicular. So, maybe I need to relate the area to the product of the diagonals?I recall that for a quadrilateral with perpendicular diagonals, the area is half the product of the diagonals. So, if the diagonals are p and q, then the area S is (p*q)/2. But in this problem, the inequality is in terms of the sides, not the diagonals. So, maybe I need to express the diagonals in terms of the sides?Alternatively, maybe I can use some inequality like the Cauchy-Schwarz inequality or AM-GM inequality to relate the sides and the area. Let me think about that.Another approach is to use coordinate geometry. Maybe I can place the quadrilateral on a coordinate plane and express the area in terms of coordinates, then try to maximize it under the given side lengths. But that might get complicated.Wait, the problem says "equality holds only for an inscribed quadrilateral whose diagonals are perpendicular." So, maybe if I can show that for such a quadrilateral, the area is exactly (ac + bd)/2, and for any other quadrilateral, the area is less than that.Let me try to visualize an inscribed quadrilateral with perpendicular diagonals. So, all four vertices lie on a circle, and the diagonals intersect at right angles. In such a case, the area is indeed (p*q)/2, where p and q are the lengths of the diagonals.But how does that relate to (ac + bd)/2? Maybe I need to express p and q in terms of a, b, c, d.I remember that in a cyclic quadrilateral, there is a relation called Ptolemy's theorem, which states that the product of the diagonals is equal to the sum of the products of the opposite sides. So, p*q = ac + bd. Wait, that's exactly the expression we have in the inequality!So, if the quadrilateral is cyclic with perpendicular diagonals, then its area S is (p*q)/2, and by Ptolemy's theorem, p*q = ac + bd. Therefore, S = (ac + bd)/2. So, equality holds for such a quadrilateral.But what about other quadrilaterals? How can I show that their area is less than or equal to (ac + bd)/2?Maybe I can use some kind of inequality that relates the area to the sides. I know that for triangles, Heron's formula relates the area to the sides, but for quadrilaterals, it's more complicated. Maybe I can divide the quadrilateral into two triangles and sum their areas.Let me try that. Suppose I divide the quadrilateral into two triangles by one of its diagonals, say AC. Then the area S is the sum of the areas of triangles ABC and ADC.The area of triangle ABC can be expressed as (1/2)*ab*sin(theta1), where theta1 is the angle between sides a and b. Similarly, the area of triangle ADC is (1/2)*cd*sin(theta2), where theta2 is the angle between sides c and d.So, the total area S = (1/2)*ab*sin(theta1) + (1/2)*cd*sin(theta2).Hmm, but I need to relate this to (ac + bd)/2. Maybe I can use some trigonometric identities or inequalities.I know that sin(theta) <= 1 for any angle theta. So, the maximum area occurs when both theta1 and theta2 are 90 degrees, making the triangles right-angled. But wait, in that case, the quadrilateral would have both triangles as right-angled, which might not necessarily make it cyclic or have perpendicular diagonals.Wait, maybe I need to consider the maximum possible area given the sides. So, if I fix the sides a, b, c, d, what configuration gives the maximum area?I recall that for a quadrilateral with given sides, the maximum area is achieved when it is cyclic. So, maybe the maximum area is when it's cyclic, and in that case, the area is (ac + bd)/2.But wait, earlier I thought that for a cyclic quadrilateral with perpendicular diagonals, the area is (ac + bd)/2. So, maybe the maximum area is indeed (ac + bd)/2, achieved when the quadrilateral is cyclic with perpendicular diagonals.But how can I formalize this? Maybe I can use the fact that in any quadrilateral, the area is less than or equal to the area of the cyclic quadrilateral with the same sides.Alternatively, I can use the formula for the area of a quadrilateral in terms of its sides and angles. But that might be too involved.Wait, another idea: use the Cauchy-Schwarz inequality. Let me think about how to apply it here.If I consider the sides a, c and b, d, maybe I can pair them up somehow. Let me write the inequality as S <= (ac + bd)/2.So, I need to show that the area is bounded by the average of the products of opposite sides.Hmm, maybe I can express the area in terms of the products of sides and some trigonometric functions, then apply an inequality.Let me go back to the expression S = (1/2)*ab*sin(theta1) + (1/2)*cd*sin(theta2).I need to relate this to (ac + bd)/2. Maybe I can find a way to bound this sum.Alternatively, perhaps I can use the fact that in any quadrilateral, the product of the diagonals is less than or equal to ac + bd, with equality when the quadrilateral is cyclic with perpendicular diagonals.Wait, that might be the key. If I can show that p*q <= ac + bd, then since the area S = (p*q)/2 sin(theta), where theta is the angle between the diagonals, and sin(theta) <=1, then S <= (p*q)/2 <= (ac + bd)/2.But wait, actually, for a general quadrilateral, the area is (1/2)*p*q*sin(theta), where theta is the angle between the diagonals. So, if I can show that p*q <= ac + bd, then since sin(theta) <=1, it would follow that S <= (ac + bd)/2.But how can I show that p*q <= ac + bd?Wait, in a cyclic quadrilateral, by Ptolemy's theorem, p*q = ac + bd. So, for a cyclic quadrilateral, p*q = ac + bd. For non-cyclic quadrilaterals, is p*q <= ac + bd?I think so, because Ptolemy's inequality states that for any quadrilateral, p*q <= ac + bd, with equality if and only if the quadrilateral is cyclic.Yes, that's right! Ptolemy's inequality says that in any quadrilateral, the product of the diagonals is less than or equal to the sum of the products of the opposite sides. So, p*q <= ac + bd.Therefore, since the area S = (1/2)*p*q*sin(theta), and sin(theta) <=1, we have S <= (1/2)*p*q <= (1/2)*(ac + bd).So, that proves the inequality S <= (ac + bd)/2.Now, for equality, when does S = (ac + bd)/2? That would happen when both p*q = ac + bd and sin(theta) =1. So, p*q = ac + bd implies the quadrilateral is cyclic (by Ptolemy's theorem), and sin(theta) =1 implies that the angle between the diagonals is 90 degrees, i.e., the diagonals are perpendicular.Therefore, equality holds if and only if the quadrilateral is cyclic and has perpendicular diagonals.So, putting it all together, we've shown that the area of a quadrilateral is less than or equal to (ac + bd)/2, with equality if and only if the quadrilateral is cyclic with perpendicular diagonals.I think that makes sense. I used Ptolemy's inequality to relate the product of the diagonals to the sides, and then used the formula for the area in terms of the diagonals and the angle between them. The maximum area occurs when both conditions for equality in Ptolemy's inequality and the sine function are met, which corresponds to a cyclic quadrilateral with perpendicular diagonals.

💡Okay, so I have this problem with four statements, labeled A to D, and I need to figure out which ones are correct. Let me take them one by one and think through each carefully.Starting with **Statement A**: If ( a > b ) and ( c < 0 ), then ( a^{2}c < b^{2}c ).Hmm, okay. So, ( a ) is greater than ( b ), and ( c ) is negative. I know that when you multiply both sides of an inequality by a negative number, the inequality sign flips. But here, it's not just ( a ) and ( b ) being multiplied by ( c ); it's their squares. So, ( a^2 ) and ( b^2 ) are both positive, regardless of whether ( a ) and ( b ) are positive or negative. Let me try plugging in some numbers to test this. Suppose ( a = 3 ), ( b = 2 ), and ( c = -1 ). Then:- ( a^2c = 9 times (-1) = -9 )- ( b^2c = 4 times (-1) = -4 )So, ( -9 < -4 ), which means ( a^2c < b^2c ) is true in this case.Wait, what if ( a ) and ( b ) are negative? Let's say ( a = -2 ), ( b = -3 ), and ( c = -1 ). Then:- ( a^2c = 4 times (-1) = -4 )- ( b^2c = 9 times (-1) = -9 )Here, ( -4 > -9 ), so ( a^2c > b^2c ). That contradicts the statement. So, in this case, Statement A isn't necessarily true. It depends on whether ( a ) and ( b ) are positive or negative. Since the statement doesn't specify, I think Statement A isn't always correct.Moving on to **Statement B**: If ( a > b ) and ( c < 0 ), then ( a^{3}c < b^{3}c ).Alright, similar setup, but now it's cubes instead of squares. Cubes preserve the sign of the original number, unlike squares which are always positive. So, if ( a > b ), then ( a^3 > b^3 ) if both ( a ) and ( b ) are positive or both are negative. Let's test with positive numbers first. Let ( a = 3 ), ( b = 2 ), ( c = -1 ):- ( a^3c = 27 times (-1) = -27 )- ( b^3c = 8 times (-1) = -8 )So, ( -27 < -8 ), which fits the statement.Now, let's try with negative numbers. Let ( a = -2 ), ( b = -3 ), ( c = -1 ):- ( a^3c = (-8) times (-1) = 8 )- ( b^3c = (-27) times (-1) = 27 )Here, ( 8 < 27 ) is false because ( 8 < 27 ) is actually true, but wait, ( a^3c = 8 ) and ( b^3c = 27 ), so ( 8 < 27 ), which means ( a^3c < b^3c ) is true. Hmm, interesting. Even with negatives, it still holds because the cubes maintain the inequality direction, and multiplying by a negative flips it again.Wait, let me double-check. If ( a = -2 ) and ( b = -3 ), then ( a > b ) because ( -2 > -3 ). Their cubes are ( (-2)^3 = -8 ) and ( (-3)^3 = -27 ). So, ( -8 > -27 ). Then, multiplying both by ( c = -1 ):- ( a^3c = (-8) times (-1) = 8 )- ( b^3c = (-27) times (-1) = 27 )So, ( 8 < 27 ), which is true. So, in both cases, whether ( a ) and ( b ) are positive or negative, as long as ( a > b ) and ( c < 0 ), ( a^3c < b^3c ) holds. Therefore, Statement B seems correct.Next, **Statement C**: If ( c > a > b > 0 ), then ( frac{a}{c - a} > frac{b}{c - b} ).Okay, so all variables are positive, and ( c ) is the largest. Let me try plugging in some numbers. Let ( c = 5 ), ( a = 3 ), ( b = 2 ):- ( frac{a}{c - a} = frac{3}{5 - 3} = frac{3}{2} = 1.5 )- ( frac{b}{c - b} = frac{2}{5 - 2} = frac{2}{3} approx 0.666 )So, ( 1.5 > 0.666 ), which is true.Let me try another set. ( c = 10 ), ( a = 6 ), ( b = 4 ):- ( frac{a}{c - a} = frac{6}{10 - 6} = frac{6}{4} = 1.5 )- ( frac{b}{c - b} = frac{4}{10 - 4} = frac{4}{6} approx 0.666 )Again, ( 1.5 > 0.666 ).Wait, is this always true? Let me see. Since ( c > a > b > 0 ), ( c - a ) and ( c - b ) are both positive. Also, ( a > b ) and ( c - a < c - b ) because ( a > b ). So, the denominators ( c - a ) and ( c - b ) satisfy ( c - a < c - b ).Thus, ( frac{a}{c - a} ) and ( frac{b}{c - b} ). Since ( a > b ) and ( c - a < c - b ), both numerators and denominators are increasing, but the denominators are decreasing. Hmm, it's a bit tricky.Alternatively, cross-multiplying to compare:( frac{a}{c - a} > frac{b}{c - b} )Multiply both sides by ( (c - a)(c - b) ), which is positive since ( c > a ) and ( c > b ):( a(c - b) > b(c - a) )Expanding both sides:( ac - ab > bc - ab )Simplify:( ac > bc )Since ( c > 0 ), we can divide both sides by ( c ):( a > b )Which is given. Therefore, the inequality ( frac{a}{c - a} > frac{b}{c - b} ) is indeed true. So, Statement C is correct.Finally, **Statement D**: If ( a > b ) and ( frac{1}{a} > frac{1}{b} ), then ( a > 0 ), ( b < 0 ).Alright, so we have ( a > b ) and ( frac{1}{a} > frac{1}{b} ). Let's see what this implies.First, ( frac{1}{a} > frac{1}{b} ). Let's rearrange this inequality:( frac{1}{a} - frac{1}{b} > 0 )Combine the fractions:( frac{b - a}{ab} > 0 )So, ( frac{b - a}{ab} > 0 ). Since ( a > b ), ( b - a ) is negative. Therefore, the numerator is negative. For the entire fraction to be positive, the denominator ( ab ) must also be negative. A negative denominator implies that ( a ) and ( b ) have opposite signs. Given that ( a > b ), if ( a ) is positive, ( b ) must be negative. If ( a ) were negative, ( b ) would have to be more negative, but then ( frac{1}{a} ) would be negative and ( frac{1}{b} ) would also be negative, but ( frac{1}{a} > frac{1}{b} ) would require ( a ) to be closer to zero than ( b ), which might not necessarily hold. Wait, let's test with numbers. Suppose ( a = 2 ), ( b = -1 ):- ( a > b ): ( 2 > -1 ) is true.- ( frac{1}{a} = 0.5 ), ( frac{1}{b} = -1 )- ( 0.5 > -1 ) is true.So, ( a > 0 ), ( b < 0 ).Another example: ( a = -1 ), ( b = -2 ):- ( a > b ): ( -1 > -2 ) is true.- ( frac{1}{a} = -1 ), ( frac{1}{b} = -0.5 )- ( -1 > -0.5 ) is false.So, in this case, ( frac{1}{a} ) is not greater than ( frac{1}{b} ). Therefore, if ( a > b ) and ( frac{1}{a} > frac{1}{b} ), ( a ) must be positive and ( b ) must be negative. Hence, Statement D is correct.In summary, after analyzing each statement:- **A**: Not always correct.- **B**: Correct.- **C**: Correct.- **D**: Correct.So, the correct statements are B, C, and D.

💡To determine the number of integer values of ( x ) that form a non-degenerate triangle with sides 15 and 40, we apply the triangle inequality theorem. This theorem states that the sum of any two sides must be greater than the third side.1. **First Inequality:** ( x + 15 > 40 ) - Solving for ( x ), we subtract 15 from both sides: ( x > 25 ).2. **Second Inequality:** ( x + 40 > 15 ) - Simplifying, this becomes ( x > -25 ). Since ( x ) must be positive, this condition is automatically satisfied.3. **Third Inequality:** ( 15 + 40 > x ) - This simplifies to ( 55 > x ) or ( x < 55 ).Combining the relevant inequalities, we find that ( 25 < x < 55 ). The integer values of ( x ) within this range are 26, 27, 28, ..., 54. To count the number of integers, subtract the lower bound from the upper bound and add one: ( 54 - 26 + 1 = 29 ).Therefore, there are 29 possible integer values for ( x ).