💡Alright, so I've got this problem about a royal family—king, queen, their three sons, and some daughters. The problem gives me some information about their ages and asks me to figure out how many children they have, with the condition that there are no more than 20 children. Let me try to break this down step by step.First, let's list out what we know:1. The king and queen are both 35 years old on September 1st of a certain year.2. They have three sons and at least one daughter.3. On that same September 1st, the total age of all their children is 35 years.4. Several years later, on another September 1st, the combined age of the king and queen equals the total age of all their children.5. No new children were born, and no one passed away during this time.Our goal is to find out how many children the royal couple has, given that there are no more than 20.Alright, let's start by trying to model this situation mathematically.Let's denote:- ( n ) as the number of years that pass between the two September 1sts.- ( d ) as the number of daughters.Since they have three sons and at least one daughter, the total number of children is ( 3 + d ).Initially, the king and queen are both 35 years old, so their combined age is ( 35 + 35 = 70 ) years.The total age of their children is 35 years.After ( n ) years, the king and queen will each be ( 35 + n ) years old, so their combined age will be ( 2 times (35 + n) = 70 + 2n ) years.Similarly, each child will be ( n ) years older, so the total age of all the children will be ( 35 + n times (3 + d) ) years.According to the problem, after ( n ) years, the combined age of the king and queen equals the total age of all their children. So, we can set up the equation:[70 + 2n = 35 + n(3 + d)]Let me write that out again for clarity:[70 + 2n = 35 + n(3 + d)]Now, let's simplify this equation to solve for ( d ).First, subtract 35 from both sides:[70 + 2n - 35 = n(3 + d)]Which simplifies to:[35 + 2n = n(3 + d)]Now, let's isolate ( n(3 + d) ):[n(3 + d) = 35 + 2n]To solve for ( d ), let's divide both sides by ( n ):[3 + d = frac{35}{n} + 2]Subtract 3 from both sides:[d = frac{35}{n} + 2 - 3]Simplify:[d = frac{35}{n} - 1]So, ( d = frac{35}{n} - 1 ).Since ( d ) must be a positive integer (they have at least one daughter), ( frac{35}{n} - 1 ) must also be a positive integer. Therefore, ( frac{35}{n} ) must be an integer greater than 1.This means that ( n ) must be a divisor of 35. Let's list the positive divisors of 35:1, 5, 7, 35Now, let's consider each possible value of ( n ):1. **Case 1: ( n = 1 )** Plugging into the equation: [ d = frac{35}{1} - 1 = 35 - 1 = 34 ] So, ( d = 34 ). Therefore, the total number of children is ( 3 + 34 = 37 ). However, the problem states that there are no more than 20 children. So, this case is invalid.2. **Case 2: ( n = 5 )** Plugging into the equation: [ d = frac{35}{5} - 1 = 7 - 1 = 6 ] So, ( d = 6 ). Therefore, the total number of children is ( 3 + 6 = 9 ). This is within the limit of 20 children, so this is a valid solution.3. **Case 3: ( n = 7 )** Plugging into the equation: [ d = frac{35}{7} - 1 = 5 - 1 = 4 ] So, ( d = 4 ). Therefore, the total number of children is ( 3 + 4 = 7 ). This is also within the limit of 20 children, so this is another valid solution.4. **Case 4: ( n = 35 )** Plugging into the equation: [ d = frac{35}{35} - 1 = 1 - 1 = 0 ] So, ( d = 0 ). However, the problem states that they have at least one daughter. Therefore, this case is invalid.So, from the above cases, the only valid solutions are when ( n = 5 ) and ( n = 7 ), resulting in 9 and 7 children respectively.Wait a minute, let me double-check these calculations to make sure I haven't made any mistakes.For ( n = 5 ):- King and queen's combined age after 5 years: ( 70 + 2 times 5 = 80 ) years.- Children's total age after 5 years: ( 35 + 5 times 9 = 35 + 45 = 80 ) years.- This checks out.For ( n = 7 ):- King and queen's combined age after 7 years: ( 70 + 2 times 7 = 84 ) years.- Children's total age after 7 years: ( 35 + 7 times 7 = 35 + 49 = 84 ) years.- This also checks out.Okay, so both solutions are valid.But wait, the problem says "no more than 20" children, so technically, 7 and 9 are both within this limit. However, I need to make sure there are no other possible values of ( n ) that could result in a valid number of children.Looking back at the divisors of 35, we've considered all possible positive integer divisors: 1, 5, 7, 35. The only ones that result in a valid number of daughters (at least 1) and a total number of children no more than 20 are ( n = 5 ) and ( n = 7 ).Therefore, the possible number of children is either 7 or 9.Just to be thorough, let's consider if there could be any other values of ( n ) that are not divisors of 35 but could still satisfy the equation ( d = frac{35}{n} - 1 ) with ( d ) being a positive integer.For ( d ) to be an integer, ( frac{35}{n} ) must be an integer. Therefore, ( n ) must be a divisor of 35. Hence, there are no other possible values of ( n ) that would result in ( d ) being an integer.Therefore, the only possible numbers of children are 7 and 9.**Final Answer**The royal couple could have either boxed{7} or boxed{9} children.

💡Okay, so I have this problem where I need to prove that the projection of the left arc on the OX-axis is shorter than the projection of the right arc. The setup is that the angle formed by the rays y = x and y = 2x for x ≥ 0 cuts out two arcs on the parabola y = x² + px + q. These arcs are then projected onto the OX-axis, and I need to show that the left arc's projection is shorter.First, I need to visualize this. There's a parabola, and two lines y = x and y = 2x. These lines form an angle, and where they intersect the parabola, they create two arcs. When we project these arcs onto the x-axis, we get two intervals. The claim is that the interval corresponding to the left arc is shorter than the one corresponding to the right arc.So, to approach this, I think I need to find the points where the lines y = x and y = 2x intersect the parabola y = x² + px + q. Once I have these intersection points, I can determine the intervals on the x-axis where these arcs project to. Then, by comparing the lengths of these intervals, I can show which one is shorter.Let me start by finding the intersection points.For the line y = x:Set x = x² + px + q.This simplifies to x² + (p - 1)x + q = 0.Similarly, for the line y = 2x:Set 2x = x² + px + q.This simplifies to x² + (p - 2)x + q = 0.So, I have two quadratic equations:1. x² + (p - 1)x + q = 0 (from y = x)2. x² + (p - 2)x + q = 0 (from y = 2x)Let me denote the roots of the first equation as x₁ and x₂, and the roots of the second equation as x₃ and x₄.Using Vieta's formulas, for the first equation, the sum of the roots is x₁ + x₂ = 1 - p, and the product is x₁x₂ = q.For the second equation, the sum of the roots is x₃ + x₄ = 2 - p, and the product is x₃x₄ = q.Hmm, interesting. So both quadratics have the same constant term q, but different coefficients for x.Now, since both lines intersect the parabola, each quadratic should have two real roots, meaning the discriminant should be positive.For the first quadratic: discriminant D₁ = (p - 1)² - 4q > 0.For the second quadratic: discriminant D₂ = (p - 2)² - 4q > 0.So, both discriminants must be positive for real intersections.Now, moving on. The projections of the arcs onto the x-axis would be the intervals between the roots of each quadratic.But wait, actually, the arcs are between the two lines y = x and y = 2x. So, the left arc would be between the leftmost intersection with y = x and the leftmost intersection with y = 2x, and the right arc would be between the rightmost intersection with y = 2x and the rightmost intersection with y = x.Wait, is that correct? Let me think.If I imagine the parabola opening upwards (since the coefficient of x² is positive), and the two lines y = x and y = 2x, which are straight lines passing through the origin with slopes 1 and 2, respectively.The angle between them is formed for x ≥ 0, so we're looking at the region in the first quadrant where these two lines intersect the parabola.So, the parabola will intersect each line twice, giving four points of intersection: two with y = x and two with y = 2x.These four points divide the parabola into two arcs: one between the two intersections with y = x, and another between the two intersections with y = 2x. But actually, since the lines form an angle, the arcs are the parts of the parabola that lie within the angle formed by the two lines.Wait, perhaps it's better to think of the arcs as the portions of the parabola that are inside the angle between y = x and y = 2x. So, the left arc would be the part of the parabola between the left intersection points, and the right arc would be between the right intersection points.But I need to be precise. Let me try to sketch this mentally.The parabola y = x² + px + q. Depending on p and q, it can be shifted left or right and up or down. The lines y = x and y = 2x are straight lines through the origin with different slopes.Assuming that the parabola intersects both lines at two points each, we have four intersection points. Let's denote the intersection points with y = x as (x₁, x₁) and (x₂, x₂), and with y = 2x as (x₃, 2x₃) and (x₄, 2x₄).Now, the arcs on the parabola between these intersection points are the left arc and the right arc. The left arc would be between the leftmost intersection points, say (x₁, x₁) and (x₃, 2x₃), and the right arc would be between the rightmost intersection points, say (x₂, x₂) and (x₄, 2x₄).But actually, the arcs are the portions of the parabola that lie within the angle formed by y = x and y = 2x. So, the left arc is the part of the parabola between the two left intersection points, and the right arc is between the two right intersection points.Wait, perhaps it's more accurate to say that the left arc is the part of the parabola between the two intersection points with the smaller x-values, and the right arc is between the two intersection points with the larger x-values.But to be precise, I need to consider the order of the roots.Let me denote the roots of the first quadratic (from y = x) as x₁ and x₂, and the roots of the second quadratic (from y = 2x) as x₃ and x₄.Assuming that x₁ < x₂ and x₃ < x₄, which is a reasonable assumption since the parabola is symmetric about its vertex, and the lines y = x and y = 2x are increasing.Now, depending on the values of p and q, the roots x₁, x₂, x₃, x₄ can be ordered in different ways on the x-axis.But for the purpose of projection, I need to find the intervals on the x-axis that correspond to the arcs.The projection of the left arc would be the interval between the leftmost intersection point and the next one, and the projection of the right arc would be between the next intersection point and the rightmost one.But actually, the projection of an arc onto the x-axis is simply the interval between the x-coordinates of the endpoints of the arc.So, if the left arc is between (x₁, x₁) and (x₃, 2x₃), then its projection is from x₁ to x₃. Similarly, the right arc is between (x₂, x₂) and (x₄, 2x₄), so its projection is from x₂ to x₄.Therefore, the length of the left projection is x₃ - x₁, and the length of the right projection is x₄ - x₂.We need to show that x₃ - x₁ < x₄ - x₂.So, the goal is to prove that x₃ - x₁ < x₄ - x₂.Alternatively, we can show that (x₄ - x₂) - (x₃ - x₁) > 0.Let me compute this difference:(x₄ - x₂) - (x₃ - x₁) = x₄ - x₂ - x₃ + x₁ = (x₁ + x₄) - (x₂ + x₃).From Vieta's formulas, we know that:For the first quadratic (y = x): x₁ + x₂ = 1 - p.For the second quadratic (y = 2x): x₃ + x₄ = 2 - p.So, substituting these into the expression:(x₁ + x₄) - (x₂ + x₃) = (x₁ + x₄) - (x₂ + x₃) = (x₁ + x₄) - (x₂ + x₃).But we can express x₁ + x₄ and x₂ + x₃ in terms of the sums from Vieta's formulas.Wait, let's see:We have x₁ + x₂ = 1 - p and x₃ + x₄ = 2 - p.So, x₁ + x₄ = (x₁ + x₂) + (x₃ + x₄) - (x₂ + x₃) = (1 - p) + (2 - p) - (x₂ + x₃).But that might not be helpful.Alternatively, perhaps we can express x₁ + x₄ and x₂ + x₃ in terms of the known sums.Wait, let me think differently.We have:x₁ + x₂ = 1 - px₃ + x₄ = 2 - pSo, if I subtract the first equation from the second, I get:(x₃ + x₄) - (x₁ + x₂) = (2 - p) - (1 - p) = 1.So, x₃ + x₄ - x₁ - x₂ = 1.But in our expression, we have (x₁ + x₄) - (x₂ + x₃).Let me rearrange the terms:(x₁ + x₄) - (x₂ + x₃) = (x₁ - x₂) + (x₄ - x₃).Hmm, not sure if that helps.Alternatively, let's consider that:(x₁ + x₄) - (x₂ + x₃) = (x₁ - x₂) + (x₄ - x₃).But from the earlier result, we know that x₃ + x₄ - x₁ - x₂ = 1, which can be rewritten as (x₃ - x₁) + (x₄ - x₂) = 1.Wait, that's interesting.So, (x₃ - x₁) + (x₄ - x₂) = 1.But we need to find (x₄ - x₂) - (x₃ - x₁).Let me denote A = x₃ - x₁ and B = x₄ - x₂.From the above, we have A + B = 1.We need to find B - A.So, B - A = ?But we don't have enough information yet. Maybe we can find another relation.Alternatively, perhaps we can express x₁, x₂, x₃, x₄ in terms of the roots.Wait, let me consider the roots of the quadratics.For the first quadratic: x² + (p - 1)x + q = 0.The roots are x₁ and x₂, so x₁ + x₂ = 1 - p and x₁x₂ = q.For the second quadratic: x² + (p - 2)x + q = 0.The roots are x₃ and x₄, so x₃ + x₄ = 2 - p and x₃x₄ = q.So, both quadratics have the same product of roots, q.That's interesting. So, x₁x₂ = x₃x₄ = q.Now, let's consider the difference between the sums:(x₃ + x₄) - (x₁ + x₂) = (2 - p) - (1 - p) = 1.So, x₃ + x₄ = x₁ + x₂ + 1.Therefore, x₃ + x₄ = x₁ + x₂ + 1.Now, let's go back to our expression:(x₁ + x₄) - (x₂ + x₃) = ?Let me rearrange the terms:(x₁ - x₂) + (x₄ - x₃).But from x₃ + x₄ = x₁ + x₂ + 1, we can write x₄ = x₁ + x₂ + 1 - x₃.Substituting into the expression:(x₁ - x₂) + (x₁ + x₂ + 1 - x₃ - x₃) = (x₁ - x₂) + (x₁ + x₂ + 1 - 2x₃).Simplify:x₁ - x₂ + x₁ + x₂ + 1 - 2x₃ = 2x₁ + 1 - 2x₃.So, (x₁ + x₄) - (x₂ + x₃) = 2x₁ + 1 - 2x₃.Hmm, not sure if that helps.Alternatively, perhaps I can consider the difference B - A = (x₄ - x₂) - (x₃ - x₁) = x₄ - x₂ - x₃ + x₁ = (x₁ + x₄) - (x₂ + x₃).But from earlier, we have x₃ + x₄ = x₁ + x₂ + 1.So, x₄ = x₁ + x₂ + 1 - x₃.Substituting into (x₁ + x₄):x₁ + x₄ = x₁ + (x₁ + x₂ + 1 - x₃) = 2x₁ + x₂ + 1 - x₃.Similarly, x₂ + x₃ is just x₂ + x₃.So, (x₁ + x₄) - (x₂ + x₃) = (2x₁ + x₂ + 1 - x₃) - (x₂ + x₃) = 2x₁ + 1 - 2x₃.So, we have:B - A = 2x₁ + 1 - 2x₃.But I need to find whether this is positive or negative.Wait, but I don't know the relationship between x₁ and x₃.Alternatively, perhaps I can consider the fact that the roots x₁ and x₂ are from the first quadratic, and x₃ and x₄ from the second.Given that the second quadratic has a larger coefficient for x (since p - 2 < p - 1 if p is fixed), the roots might be shifted.Wait, actually, the quadratic with the smaller coefficient for x (i.e., p - 1 vs p - 2) will have roots that are further apart or closer?Wait, the quadratic equation x² + bx + c = 0 has roots at (-b ± sqrt(b² - 4c))/2.So, for the first quadratic, the roots are [-(p - 1) ± sqrt((p - 1)² - 4q)]/2.For the second quadratic, the roots are [-(p - 2) ± sqrt((p - 2)² - 4q)]/2.So, comparing the roots, the second quadratic has a smaller coefficient for x (since p - 2 < p - 1), so the roots are shifted.Wait, actually, the roots are [ (1 - p) ± sqrt((p - 1)² - 4q) ] / 2 for the first quadratic, and [ (2 - p) ± sqrt((p - 2)² - 4q) ] / 2 for the second quadratic.So, the second quadratic has a larger constant term in the numerator (2 - p vs 1 - p), so the roots are shifted to the right compared to the first quadratic.Therefore, x₃ and x₄ are greater than x₁ and x₂.Wait, is that necessarily true? Let me think.Actually, the roots depend on both the linear coefficient and the discriminant.But given that both quadratics have the same constant term q, and the second quadratic has a larger linear coefficient (p - 2 vs p - 1, but wait, p - 2 is actually smaller than p - 1 since 2 > 1. So, the linear coefficient is more negative in the second quadratic.Wait, no, the linear coefficient in the first quadratic is (p - 1), and in the second quadratic, it's (p - 2). So, if p is fixed, then (p - 2) < (p - 1), meaning the second quadratic has a more negative linear coefficient.Therefore, the roots of the second quadratic are shifted to the right compared to the first quadratic.Wait, actually, the roots are given by [-b ± sqrt(b² - 4c)]/2.So, for the first quadratic, b = p - 1, so the roots are [-(p - 1) ± sqrt((p - 1)² - 4q)]/2.For the second quadratic, b = p - 2, so the roots are [-(p - 2) ± sqrt((p - 2)² - 4q)]/2.So, comparing the roots, the second quadratic has a smaller b (since p - 2 < p - 1), so the roots are shifted to the right.Therefore, x₃ and x₄ are greater than x₁ and x₂.So, x₁ < x₃ and x₂ < x₄.Wait, but that might not necessarily be the case. The roots could interleave.Wait, perhaps x₁ < x₃ < x₂ < x₄, or x₃ < x₁ < x₄ < x₂, depending on the values of p and q.But given that the second quadratic has a more negative linear coefficient, its roots are shifted to the right.So, likely, x₁ < x₃ < x₂ < x₄.Wait, let me test with specific values.Suppose p = 0 and q = 0.Then the first quadratic becomes x² - x = 0, roots at x = 0 and x = 1.The second quadratic becomes x² - 2x = 0, roots at x = 0 and x = 2.So, in this case, x₁ = 0, x₂ = 1, x₃ = 0, x₄ = 2.So, the roots are x₁ = 0, x₃ = 0, x₂ = 1, x₄ = 2.So, in this case, x₁ = x₃ = 0, x₂ = 1, x₄ = 2.So, the projection of the left arc is from x = 0 to x = 0, which is zero length, and the projection of the right arc is from x = 1 to x = 2, which is length 1.But this is a special case where the parabola passes through the origin, so the intersection points coincide.But in general, if p and q are such that the parabola doesn't pass through the origin, the roots will be distinct.Let me choose p = 1 and q = 0.Then the first quadratic becomes x² = 0, so x₁ = x₂ = 0.The second quadratic becomes x² - x = 0, roots at x = 0 and x = 1.So, again, x₁ = x₂ = 0, x₃ = 0, x₄ = 1.So, the projection of the left arc is from 0 to 0, zero length, and the right arc is from 0 to 1, length 1.But this is another special case.Let me choose p = 3 and q = 2.Then the first quadratic is x² + 2x + 2 = 0. Wait, discriminant is 4 - 8 = -4, which is negative. So no real roots. Not good.Let me choose p = 1 and q = -1.Then the first quadratic is x² = 0, so x = 0 (double root). The second quadratic is x² - x -1 = 0, roots at [1 ± sqrt(5)]/2 ≈ 1.618 and -0.618. Since x ≥ 0, we take x ≈ 1.618.So, x₁ = x₂ = 0, x₃ ≈ 1.618.So, projection of left arc is from 0 to 1.618, but since x₁ = x₂ = 0, the arc is just a single point, so projection is zero. The right arc is from 0 to 1.618, but since x₂ = 0, it's the same.Hmm, maybe I need a different approach.Alternatively, perhaps I can consider the difference between the projections.We have:Projection of left arc: x₃ - x₁Projection of right arc: x₄ - x₂We need to show that x₃ - x₁ < x₄ - x₂.Which is equivalent to x₄ - x₂ - (x₃ - x₁) > 0.So, x₄ - x₂ - x₃ + x₁ > 0.Which is (x₁ + x₄) - (x₂ + x₃) > 0.From Vieta's formulas, we have:x₁ + x₂ = 1 - px₃ + x₄ = 2 - pSo, x₁ + x₂ = 1 - px₃ + x₄ = 2 - pTherefore, (x₁ + x₄) - (x₂ + x₃) = (x₁ + x₄) - (x₂ + x₃) = (x₁ + x₄) - (x₂ + x₃).But from x₃ + x₄ = 2 - p and x₁ + x₂ = 1 - p, we can write:x₄ = 2 - p - x₃x₂ = 1 - p - x₁Substituting into (x₁ + x₄) - (x₂ + x₃):x₁ + (2 - p - x₃) - (1 - p - x₁ + x₃) = x₁ + 2 - p - x₃ -1 + p + x₁ - x₃ = 2x₁ +1 - 2x₃.So, (x₁ + x₄) - (x₂ + x₃) = 2x₁ +1 - 2x₃.We need to show that this is positive, i.e., 2x₁ +1 - 2x₃ > 0.Which simplifies to x₁ + 0.5 > x₃.So, if we can show that x₁ > x₃ - 0.5, then the projection of the left arc is shorter.But I'm not sure how to proceed from here.Alternatively, perhaps I can consider the difference between the roots.Let me denote the roots as follows:For the first quadratic: x₁ and x₂, with x₁ < x₂.For the second quadratic: x₃ and x₄, with x₃ < x₄.We need to show that x₃ - x₁ < x₄ - x₂.Which is equivalent to x₄ - x₂ > x₃ - x₁.Let me consider the difference x₄ - x₂ - (x₃ - x₁) = x₄ - x₂ - x₃ + x₁.From Vieta's formulas, we have:x₁ + x₂ = 1 - px₃ + x₄ = 2 - pSo, x₄ = 2 - p - x₃x₂ = 1 - p - x₁Substituting into the expression:x₄ - x₂ - x₃ + x₁ = (2 - p - x₃) - (1 - p - x₁) - x₃ + x₁Simplify:2 - p - x₃ -1 + p + x₁ - x₃ + x₁ = (2 -1) + (-p + p) + (-x₃ - x₃) + (x₁ + x₁) = 1 - 2x₃ + 2x₁.So, the expression simplifies to 1 + 2(x₁ - x₃).We need to show that this is greater than 0, i.e., 1 + 2(x₁ - x₃) > 0.Which simplifies to x₁ - x₃ > -0.5.But I don't know if x₁ - x₃ is greater than -0.5.Alternatively, perhaps I can consider the fact that the roots of the second quadratic are shifted to the right compared to the first quadratic.Given that the second quadratic has a more negative linear coefficient, its roots are shifted to the right.So, x₃ > x₁ and x₄ > x₂.Therefore, x₃ - x₁ > 0 and x₄ - x₂ > 0.But we need to compare x₃ - x₁ and x₄ - x₂.Wait, perhaps I can consider the difference between the roots.For the first quadratic, the roots are x₁ and x₂, so the difference is x₂ - x₁.For the second quadratic, the roots are x₃ and x₄, so the difference is x₄ - x₃.We can compute these differences using the quadratic formula.For the first quadratic, the difference is sqrt[(p - 1)² - 4q].For the second quadratic, the difference is sqrt[(p - 2)² - 4q].So, the difference between the roots for the second quadratic is sqrt[(p - 2)² - 4q], which is greater than sqrt[(p - 1)² - 4q] if (p - 2)² > (p - 1)².But (p - 2)² = p² -4p +4(p - 1)² = p² -2p +1So, (p - 2)² - (p - 1)² = (p² -4p +4) - (p² -2p +1) = -2p +3.So, if -2p +3 > 0, i.e., p < 1.5, then (p - 2)² > (p - 1)², so the difference between roots for the second quadratic is greater.If p > 1.5, then (p - 2)² < (p - 1)², so the difference is smaller.But I don't know if this helps.Alternatively, perhaps I can consider the midpoints of the roots.For the first quadratic, the midpoint is (x₁ + x₂)/2 = (1 - p)/2.For the second quadratic, the midpoint is (x₃ + x₄)/2 = (2 - p)/2.So, the midpoint of the second quadratic is (2 - p)/2, which is greater than (1 - p)/2, the midpoint of the first quadratic.Therefore, the roots of the second quadratic are shifted to the right compared to the first quadratic.So, x₃ > x₁ and x₄ > x₂.But how does this affect the projections?The projection of the left arc is x₃ - x₁, and the projection of the right arc is x₄ - x₂.We need to show that x₃ - x₁ < x₄ - x₂.Which is equivalent to x₄ - x₂ - (x₃ - x₁) > 0.From earlier, we have:x₄ - x₂ - x₃ + x₁ = 1 + 2(x₁ - x₃).So, we need to show that 1 + 2(x₁ - x₃) > 0.Which is equivalent to x₁ - x₃ > -0.5.But since x₃ > x₁, x₁ - x₃ is negative, so we need to show that this negative number is greater than -0.5, i.e., x₁ - x₃ > -0.5.Which is equivalent to x₃ - x₁ < 0.5.So, if we can show that the difference between x₃ and x₁ is less than 0.5, then the projection of the left arc is shorter.But I don't know if this is always true.Alternatively, perhaps I can consider the function f(x) = x² + px + q and the lines y = x and y = 2x.The angle between the lines y = x and y = 2x is such that the slope of y = 2x is steeper than y = x.So, the parabola intersects y = x at two points and y = 2x at two points.The left arc is between the left intersection points, and the right arc is between the right intersection points.The projection of the left arc is from x₁ to x₃, and the projection of the right arc is from x₂ to x₄.We need to show that x₃ - x₁ < x₄ - x₂.From Vieta's formulas, we have:x₁ + x₂ = 1 - px₃ + x₄ = 2 - pSo, x₃ + x₄ = x₁ + x₂ + 1.Therefore, x₄ = x₁ + x₂ + 1 - x₃.Substituting into x₄ - x₂:x₄ - x₂ = x₁ + x₂ + 1 - x₃ - x₂ = x₁ + 1 - x₃.So, x₄ - x₂ = x₁ + 1 - x₃.Therefore, the projection of the right arc is x₁ + 1 - x₃.The projection of the left arc is x₃ - x₁.So, we need to show that x₃ - x₁ < x₁ + 1 - x₃.Which simplifies to:x₃ - x₁ < x₁ + 1 - x₃Bring all terms to one side:x₃ - x₁ - x₁ - 1 + x₃ < 02x₃ - 2x₁ - 1 < 0Divide both sides by 2:x₃ - x₁ - 0.5 < 0So, x₃ - x₁ < 0.5Which is the same as x₃ - x₁ < 0.5.So, we need to show that the difference between x₃ and x₁ is less than 0.5.But is this always true?Wait, from the earlier expression, we have:x₄ - x₂ = x₁ + 1 - x₃So, x₄ - x₂ = (x₁ - x₃) + 1Therefore, x₄ - x₂ = 1 - (x₃ - x₁)So, the projection of the right arc is 1 minus the projection of the left arc.Therefore, if we denote L = x₃ - x₁ (projection of left arc), then the projection of the right arc is 1 - L.Therefore, the projection of the right arc is 1 - L, which is greater than L if L < 0.5.So, if L < 0.5, then 1 - L > L, which implies that the projection of the right arc is longer.But we need to show that L < 0.5.Wait, but from the above, we have:x₄ - x₂ = 1 - (x₃ - x₁) = 1 - LSo, the projection of the right arc is 1 - L.Therefore, the projection of the right arc is 1 - L, and the projection of the left arc is L.We need to show that L < 1 - L, which is equivalent to L < 0.5.So, if we can show that L < 0.5, then the projection of the left arc is shorter.But how can we show that L = x₃ - x₁ < 0.5?From the quadratic equations, we have:x₁ and x₂ are roots of x² + (p - 1)x + q = 0x₃ and x₄ are roots of x² + (p - 2)x + q = 0So, the roots x₁ and x₂ satisfy:x₁ + x₂ = 1 - px₁x₂ = qSimilarly, x₃ + x₄ = 2 - px₃x₄ = qSo, both quadratics have the same product of roots, q.Let me consider the difference between the roots.For the first quadratic, the difference is sqrt[(p - 1)² - 4q]For the second quadratic, the difference is sqrt[(p - 2)² - 4q]We need to compare these differences.But I'm not sure if this helps.Alternatively, perhaps I can consider the function f(x) = x² + px + q and the lines y = x and y = 2x.The points of intersection are where f(x) = x and f(x) = 2x.So, f(x) - x = x² + (p - 1)x + q = 0f(x) - 2x = x² + (p - 2)x + q = 0Let me denote these as f₁(x) = x² + (p - 1)x + qf₂(x) = x² + (p - 2)x + qSo, f₂(x) = f₁(x) - xTherefore, f₂(x) = f₁(x) - xSo, the roots of f₂(x) are the roots of f₁(x) - x = 0.But I'm not sure if this helps.Alternatively, perhaps I can consider the derivative of f(x).The derivative f’(x) = 2x + p.At the points of intersection, the slope of the parabola is equal to the slope of the line.So, at the points where f(x) = x, the slope is 1.So, 2x + p = 1 => x = (1 - p)/2.Similarly, at the points where f(x) = 2x, the slope is 2.So, 2x + p = 2 => x = (2 - p)/2.Therefore, the points of intersection where the parabola is tangent to the lines y = x and y = 2x would be at x = (1 - p)/2 and x = (2 - p)/2, respectively.But in our case, the parabola intersects the lines at two points each, so these are not points of tangency.But this gives us the x-coordinates where the parabola has the same slope as the lines.So, the points of intersection are symmetric around these x-values.Wait, perhaps I can use this to find the roots.For the first quadratic, f₁(x) = x² + (p - 1)x + q = 0.The roots are symmetric around x = -(p - 1)/2 = (1 - p)/2.Similarly, for the second quadratic, f₂(x) = x² + (p - 2)x + q = 0.The roots are symmetric around x = -(p - 2)/2 = (2 - p)/2.So, the roots of the first quadratic are equidistant from (1 - p)/2, and the roots of the second quadratic are equidistant from (2 - p)/2.Let me denote the roots as follows:For the first quadratic:x₁ = (1 - p)/2 - dx₂ = (1 - p)/2 + dWhere d is the distance from the midpoint.Similarly, for the second quadratic:x₃ = (2 - p)/2 - ex₄ = (2 - p)/2 + eWhere e is the distance from the midpoint.So, x₁ = (1 - p)/2 - dx₂ = (1 - p)/2 + dx₃ = (2 - p)/2 - ex₄ = (2 - p)/2 + eNow, the projection of the left arc is x₃ - x₁, and the projection of the right arc is x₄ - x₂.So, let's compute these:x₃ - x₁ = [(2 - p)/2 - e] - [(1 - p)/2 - d] = [(2 - p)/2 - (1 - p)/2] - e + d = [ (2 - p -1 + p)/2 ] - e + d = (1)/2 - e + dSimilarly, x₄ - x₂ = [(2 - p)/2 + e] - [(1 - p)/2 + d] = [(2 - p)/2 - (1 - p)/2] + e - d = (1)/2 + e - dSo, we have:Projection of left arc: 0.5 - e + dProjection of right arc: 0.5 + e - dWe need to show that 0.5 - e + d < 0.5 + e - dSubtract 0.5 from both sides:-e + d < e - dBring all terms to one side:-e + d - e + d < 0-2e + 2d < 0Divide by 2:-e + d < 0So, d < eTherefore, if d < e, then the projection of the left arc is shorter.But is d < e?From the definitions:d is the distance from the midpoint for the first quadratic, and e is the distance for the second quadratic.But the midpoints are different.The midpoint for the first quadratic is (1 - p)/2, and for the second quadratic, it's (2 - p)/2.So, the midpoints are separated by 0.5 units.But how does this affect d and e?Alternatively, perhaps I can consider the discriminants.For the first quadratic, d = sqrt[(p - 1)² - 4q]/2For the second quadratic, e = sqrt[(p - 2)² - 4q]/2So, d = sqrt[(p - 1)² - 4q]/2e = sqrt[(p - 2)² - 4q]/2We need to compare d and e.So, d < e if sqrt[(p - 1)² - 4q] < sqrt[(p - 2)² - 4q]Which is equivalent to (p - 1)² - 4q < (p - 2)² - 4qSimplify:(p - 1)² < (p - 2)²Which is true if (p - 1)² < (p - 2)²Expanding both sides:p² - 2p +1 < p² -4p +4Subtract p² from both sides:-2p +1 < -4p +4Add 4p to both sides:2p +1 < 4Subtract 1:2p < 3So, p < 1.5Therefore, if p < 1.5, then d < e, and hence the projection of the left arc is shorter.But what if p ≥ 1.5?If p ≥ 1.5, then (p - 1)² ≥ (p - 2)², so d ≥ e.But in that case, the projection of the left arc would be x₃ - x₁ = 0.5 - e + dAnd the projection of the right arc would be 0.5 + e - dIf d ≥ e, then 0.5 - e + d ≥ 0.5 + e - dWhich would imply that the projection of the left arc is ≥ the projection of the right arc.But the problem statement says to prove that the projection of the left arc is shorter than the projection of the right arc.So, this suggests that the statement is only true when p < 1.5.But the problem doesn't specify any constraints on p and q, other than the parabola intersecting both lines.Wait, but the problem says "for x ≥ 0", so we are only considering the right half of the parabola.Therefore, the roots x₁, x₂, x₃, x₄ must be non-negative.So, x₁, x₂, x₃, x₄ ≥ 0.From Vieta's formulas:x₁ + x₂ = 1 - p ≥ 0 => p ≤ 1Similarly, x₃ + x₄ = 2 - p ≥ 0 => p ≤ 2But since p ≤ 1 from the first condition, we have p ≤ 1.Therefore, p ≤ 1.So, p < 1.5 is automatically satisfied because p ≤ 1.Therefore, in this problem, p ≤ 1, so p < 1.5.Therefore, d < e, so the projection of the left arc is shorter.Therefore, the projection of the left arc is shorter than the projection of the right arc.So, in conclusion, the projection of the left arc is shorter because the difference between the roots of the second quadratic is larger than the difference between the roots of the first quadratic, given that p ≤ 1.

💡First, I need to understand the problem. We have a set of odd integers ({1, 3, 5, 7, 9}), and we're choosing two distinct integers (a) and (b) from this set. The goal is to find the probability that the expression (ab - a - b) is even.I recall that the parity of a number (whether it's even or odd) is determined by its remainder when divided by 2. So, I should analyze the expression (ab - a - b) modulo 2.Let's break down the expression:[ab - a - b]I can factor this expression:[ab - a - b = (a - 1)(b - 1) - 1]This factoring helps because it simplifies the analysis. Now, I need to determine when ((a - 1)(b - 1) - 1) is even.Since we're dealing with parity, let's consider the values of (a) and (b) modulo 2. All numbers in the set ({1, 3, 5, 7, 9}) are odd, so:[a equiv 1 pmod{2} quad text{and} quad b equiv 1 pmod{2}]Subtracting 1 from each:[a - 1 equiv 0 pmod{2} quad text{and} quad b - 1 equiv 0 pmod{2}]This means both (a - 1) and (b - 1) are even numbers. The product of two even numbers is also even:[(a - 1)(b - 1) equiv 0 pmod{2}]Now, subtracting 1 from this product:[(a - 1)(b - 1) - 1 equiv 0 - 1 equiv 1 pmod{2}]This shows that (ab - a - b) is always odd, regardless of which two distinct integers (a) and (b) we choose from the set.Since all possible pairs result in an odd value for (ab - a - b), there are no pairs where the expression is even. Therefore, the probability of (ab - a - b) being even is zero.To summarize:- Total number of ways to choose two distinct integers from the set: (binom{5}{2} = 10)- Number of favorable outcomes (where (ab - a - b) is even): (0)- Probability: (frac{0}{10} = 0)

💡First, I need to simplify the given ratio of the areas of the two squares. The ratio is 135/45, which simplifies to 3/1.Next, to find the ratio of their side lengths, I take the square root of the simplified area ratio. The square root of 3/1 is simply the square root of 3.Since the problem asks for the ratio in the form a√b/c with a rationalized denominator, I express √3 as √3/1. Here, a is 1, b is 3, and c is 1.Finally, I add the values of a, b, and c together: 1 + 3 + 1 equals 5.

💡Okay, so I have this problem about triangle ABC where AB is 13, BC is 14, and CA is 15. The question is about dividing the segment BC into n+1 congruent parts using n points. Among these points, there should be the feet of the altitude, median, and angle bisector from point A. I need to find the smallest possible value of n.Hmm, let me break this down. First, I need to figure out where exactly the feet of the altitude, median, and angle bisector from A are located on BC. Then, I need to determine how to divide BC into equal segments such that these three points are among the division points. The smallest n would correspond to the least number of segments that can include all three feet.Starting with the median. The median from A to BC should split BC into two equal parts since a median connects a vertex to the midpoint of the opposite side. Since BC is 14, the midpoint would be at 7 units from B and 7 units from C. So, the foot of the median is at 7.Next, the angle bisector. The Angle Bisector Theorem tells us that the angle bisector divides the opposite side into segments proportional to the adjacent sides. So, the ratio of the lengths of the two segments will be equal to the ratio of the other two sides of the triangle. In this case, AB is 13 and AC is 15, so the ratio is 13:15.Let me denote the foot of the angle bisector as D. Then, BD/DC = AB/AC = 13/15. Since BD + DC = BC = 14, I can set up the equations:BD = (13/28)*14 = 13/2 = 6.5DC = (15/28)*14 = 15/2 = 7.5Wait, that doesn't seem right. Let me check. If BD/DC = 13/15, then BD = (13/(13+15)) * BC = (13/28)*14 = 13/2 = 6.5, and DC = 14 - 6.5 = 7.5. Yeah, that's correct. So, the foot of the angle bisector is at 6.5 units from B.Now, the altitude. To find the foot of the altitude from A to BC, I might need to use some area calculations. First, I can find the area of triangle ABC using Heron's formula because I know all three sides.The semi-perimeter, s, is (13 + 14 + 15)/2 = 21. Then, the area, K, is sqrt[s(s - AB)(s - BC)(s - AC)] = sqrt[21*(21 - 13)*(21 - 14)*(21 - 15)] = sqrt[21*8*7*6]. Let me compute that: 21*8 is 168, 168*7 is 1176, 1176*6 is 7056. So sqrt[7056] is 84. So the area is 84.The area can also be expressed as (1/2)*base*height. Taking BC as the base, which is 14, then the height h from A is given by (1/2)*14*h = 84, so 7h = 84, which means h = 12. So the altitude is 12 units long.Now, to find where the foot of the altitude is on BC. Let me denote the foot as E. Then, BE and EC can be found using the Pythagorean theorem in triangles ABE and AEC.In triangle ABE: AB^2 = AE^2 + BE^2 => 13^2 = 12^2 + BE^2 => 169 = 144 + BE^2 => BE^2 = 25 => BE = 5.Similarly, in triangle AEC: AC^2 = AE^2 + EC^2 => 15^2 = 12^2 + EC^2 => 225 = 144 + EC^2 => EC^2 = 81 => EC = 9.So, the foot of the altitude is at 5 units from B.So, summarizing, the three feet are at 5, 6.5, and 7 units from B on BC.Now, the problem is to divide BC into n+1 equal segments such that these three points (at 5, 6.5, and 7) are among the division points. So, the length of each segment must be a common measure that divides 14 into equal parts, and 5, 6.5, and 7 must be integer multiples of this segment length.In other words, the segment length must be a common divisor of 5, 6.5, 7, and 14. But since 6.5 is a fraction, let me convert everything into halves to make it easier. So, 5 is 10/2, 6.5 is 13/2, 7 is 14/2, and 14 is 28/2.So, now, the problem reduces to finding the greatest common divisor (GCD) of 10, 13, 14, and 28. The GCD of these numbers will give me the largest possible segment length in halves, which when converted back to the original units will be the segment length.Looking at 10, 13, 14, and 28:- The prime factors of 10 are 2 and 5.- 13 is a prime number.- The prime factors of 14 are 2 and 7.- The prime factors of 28 are 2, 2, and 7.The only common factor among all four numbers is 1. So, the GCD is 1. Therefore, the smallest segment length in halves is 1, which means the segment length in original units is 0.5.So, each segment is 0.5 units long. Therefore, the number of segments is 14 / 0.5 = 28. Hence, n+1 = 28, so n = 27.Wait, let me make sure I didn't make a mistake here. If the segment length is 0.5, then the points are at 0.5, 1.0, 1.5, ..., 13.5, 14.0. So, 5 is at the 10th point (5 / 0.5 = 10), 6.5 is at the 13th point (6.5 / 0.5 = 13), and 7 is at the 14th point (7 / 0.5 = 14). So, yes, all three feet are included in the division points.Is there a smaller n? Let's see. If I try a larger segment length, say 1.0, then the points are at 1, 2, ..., 14. Then, 5 is included, 7 is included, but 6.5 is not. So, that doesn't work. Similarly, if I try 0.25, which would make n+1 = 56, n=55, but that's larger than 27, so 27 is smaller.Alternatively, maybe I can find a common measure that is larger than 0.5 but still divides 5, 6.5, and 7. Let's see. 5 is 10/2, 6.5 is 13/2, 7 is 14/2. So, in terms of halves, we have 10, 13, 14. The GCD of 10,13,14 is 1, so the only common measure is 1/2. Therefore, 0.5 is indeed the smallest possible segment length.Therefore, the smallest n is 27.**Final Answer**boxed{27}

💡First, I need to find pairs of odd integers ( a ) and ( b ) such that ( a + b = 40 ).Since ( a ) and ( b ) are odd, they can be expressed as ( a = 2k + 1 ) and ( b = 2j + 1 ), where ( k ) and ( j ) are non-negative integers.Substituting these into the equation:[(2k + 1) + (2j + 1) = 40]Simplifying:[2k + 2j + 2 = 40][2k + 2j = 38][k + j = 19]Now, ( k ) and ( j ) can take integer values where the minimum value for each is ( 0 ). Therefore, ( k ) and ( j ) can independently range from ( 0 ) to ( 19 ), as long as their sum is ( 19 ). This provides ( 20 ) possible pairs ( (k, j) ), corresponding to ( k = 0, 1, 2, ldots, 19 ) and ( j = 19, 18, 17, ldots, 0 ).Thus, there are ( boxed{20} ) distinct ordered-pair solutions ( (a, b) ) where both ( a ) and ( b ) are odd positive integers summing to 40.