💡Okay, so I have this problem here about a ship and cargo. Let me read it again to make sure I understand what's being asked. "A ship is filled with 5973 tons of cargo. It stops in the Bahamas, where sailors load some tons of cargo onboard. The ship now holds 14696 tons of cargo. How many tons of cargo were loaded in the Bahamas?"Alright, so the ship starts with 5973 tons of cargo. Then, it stops in the Bahamas, and some additional cargo is loaded onto the ship. After loading, the total cargo on the ship is 14696 tons. The question is asking how much cargo was added in the Bahamas.Hmm, so I think this is a subtraction problem. If the ship had 5973 tons initially and ended up with 14696 tons after loading more cargo, then the difference between these two numbers should be the amount of cargo loaded in the Bahamas. Let me write that down to visualize it better:Final cargo amount = 14696 tonsInitial cargo amount = 5973 tonsCargo loaded in Bahamas = Final cargo amount - Initial cargo amountSo, Cargo loaded in Bahamas = 14696 tons - 5973 tonsNow, let's do the subtraction. I'll write it out step by step to make sure I don't make any mistakes.First, align the numbers by their place values:``` 14696- 5973```Starting from the rightmost digit (the units place):6 - 3 = 3Next, the tens place:9 - 7 = 2Then, the hundreds place:6 - 9. Wait, 6 is less than 9, so I need to borrow from the thousands place.The thousands place is 4, so I'll borrow 1 from there, making it 3, and add 10 to the hundreds place, making it 16.Now, 16 - 9 = 7Moving to the thousands place:3 - 5. Again, 3 is less than 5, so I need to borrow from the ten-thousands place.The ten-thousands place is 1, so I'll borrow 1 from there, making it 0, and add 10 to the thousands place, making it 13.Now, 13 - 5 = 8Finally, the ten-thousands place:0 - 0 = 0Putting it all together, the result is 8723 tons.So, the amount of cargo loaded in the Bahamas is 8723 tons.Wait, let me double-check my subtraction to make sure I didn't make any errors. Starting with 14696 - 5973:14696 minus 5000 is 9696.9696 minus 900 is 8796.8796 minus 70 is 8726.8726 minus 3 is 8723.Yes, that matches my earlier result. So, I'm confident that 8723 tons is the correct amount of cargo loaded in the Bahamas.

💡Alright, so I have this problem here where I need to find constants ( a ), ( b ), and ( c ) such that the equation[1 cdot 2^2 + 2 cdot 3^2 + cdots + n(n+1)^2 = frac{n(n+1)}{12} left( a n^2 + b n + c right)]holds true for all natural numbers ( n ). Hmm, okay. Let me try to break this down step by step.First, I need to understand what the left side of the equation represents. It's a sum of terms where each term is of the form ( k(k+1)^2 ) for ( k ) ranging from 1 to ( n ). So, the sum is:[sum_{k=1}^{n} k(k+1)^2]I think I can expand this term ( k(k+1)^2 ) to make it easier to sum. Let me try expanding it:[k(k+1)^2 = k(k^2 + 2k + 1) = k^3 + 2k^2 + k]Okay, so the sum becomes:[sum_{k=1}^{n} (k^3 + 2k^2 + k) = sum_{k=1}^{n} k^3 + 2sum_{k=1}^{n} k^2 + sum_{k=1}^{n} k]Now, I remember there are formulas for the sums of ( k ), ( k^2 ), and ( k^3 ). Let me recall them:1. Sum of the first ( n ) natural numbers: [ sum_{k=1}^{n} k = frac{n(n+1)}{2} ]2. Sum of the squares of the first ( n ) natural numbers: [ sum_{k=1}^{n} k^2 = frac{n(n+1)(2n+1)}{6} ]3. Sum of the cubes of the first ( n ) natural numbers: [ sum_{k=1}^{n} k^3 = left( frac{n(n+1)}{2} right)^2 ]Great, so I can substitute these into the expression:[sum_{k=1}^{n} k^3 + 2sum_{k=1}^{n} k^2 + sum_{k=1}^{n} k = left( frac{n(n+1)}{2} right)^2 + 2 cdot frac{n(n+1)(2n+1)}{6} + frac{n(n+1)}{2}]Let me simplify each term one by one.First term:[left( frac{n(n+1)}{2} right)^2 = frac{n^2(n+1)^2}{4}]Second term:[2 cdot frac{n(n+1)(2n+1)}{6} = frac{2n(n+1)(2n+1)}{6} = frac{n(n+1)(2n+1)}{3}]Third term:[frac{n(n+1)}{2}]So, putting it all together:[frac{n^2(n+1)^2}{4} + frac{n(n+1)(2n+1)}{3} + frac{n(n+1)}{2}]Now, I need to combine these terms into a single fraction. To do that, I should find a common denominator. The denominators are 4, 3, and 2. The least common multiple of 4, 3, and 2 is 12. So, I'll convert each term to have a denominator of 12.First term:[frac{n^2(n+1)^2}{4} = frac{3n^2(n+1)^2}{12}]Second term:[frac{n(n+1)(2n+1)}{3} = frac{4n(n+1)(2n+1)}{12}]Third term:[frac{n(n+1)}{2} = frac{6n(n+1)}{12}]Now, combining all three terms:[frac{3n^2(n+1)^2 + 4n(n+1)(2n+1) + 6n(n+1)}{12}]I can factor out ( n(n+1) ) from each term in the numerator:[frac{n(n+1) [3n(n+1) + 4(2n+1) + 6]}{12}]Let me simplify inside the brackets:First, expand ( 3n(n+1) ):[3n(n+1) = 3n^2 + 3n]Next, expand ( 4(2n+1) ):[4(2n+1) = 8n + 4]And the last term is just 6.So, combining these:[3n^2 + 3n + 8n + 4 + 6 = 3n^2 + 11n + 10]Therefore, the entire expression becomes:[frac{n(n+1)(3n^2 + 11n + 10)}{12}]Which is exactly the form given in the problem:[frac{n(n+1)}{12} left( a n^2 + b n + c right)]By comparing the two expressions, it's clear that:[a = 3, quad b = 11, quad c = 10]So, the constants ( a ), ( b ), and ( c ) that satisfy the equation for all natural numbers ( n ) are 3, 11, and 10 respectively.**Final Answer**The constants are ( a = 3 ), ( b = 11 ), and ( c = 10 ). Thus, the solution is boxed{a = 3, b = 11, c = 10}.

💡First, I need to solve the inequality ((x-2)(6+2x) > 0). To do this, I'll identify the critical points where each factor equals zero.Setting (x - 2 = 0) gives (x = 2), and setting (6 + 2x = 0) gives (x = -3). These points divide the number line into three intervals: (x < -3), (-3 < x < 2), and (x > 2).Next, I'll test each interval to determine the sign of the product ((x-2)(6+2x)) within that interval.1. **Interval (x < -3):** - Choose (x = -4). - (( -4 - 2)(6 + 2(-4)) = (-6)(-2) = 12), which is positive. - So, the inequality holds in this interval.2. **Interval (-3 < x < 2):** - Choose (x = 0). - ((0 - 2)(6 + 2(0)) = (-2)(6) = -12), which is negative. - So, the inequality does not hold in this interval.3. **Interval (x > 2):** - Choose (x = 3). - ((3 - 2)(6 + 2(3)) = (1)(12) = 12), which is positive. - So, the inequality holds in this interval.Combining the intervals where the inequality holds, the solution is (x < -3) or (x > 2).

💡Okay, so I have this problem here about two lines intersecting and forming a quadrilateral, and I need to find its area. Let me try to understand what's going on step by step.First, there's a line with a slope of -3 that intersects the positive x-axis at point A and the positive y-axis at point B. Then, there's another line that intersects the x-axis at point C, which is given as (10, 0), and the y-axis at some point D. These two lines intersect each other at point E, which is (5, 5). The quadrilateral we're interested in is OBEC, and we need to find its area.Alright, so I think the first thing I need to do is figure out the equations of both lines. Once I have the equations, I can find the coordinates of points A, B, and D. Then, with all these points, I can determine the vertices of the quadrilateral OBEC and calculate its area.Starting with the first line that has a slope of -3. Since it intersects the x-axis at A and the y-axis at B, it must be a straight line crossing both axes. The general equation of a line is y = mx + b, where m is the slope and b is the y-intercept. Here, the slope m is -3, so the equation is y = -3x + b. But we also know that this line passes through point E, which is (5, 5). So, I can plug these coordinates into the equation to find b. Let's do that:5 = -3(5) + b 5 = -15 + b Adding 15 to both sides: 5 + 15 = b 20 = bSo, the equation of the first line is y = -3x + 20. Now, to find the x-intercept A, we set y = 0 and solve for x:0 = -3x + 20 3x = 20 x = 20/3 ≈ 6.666...So, point A is at (20/3, 0). Point B is the y-intercept, which we already found as (0, 20).Alright, that takes care of the first line. Now, let's move on to the second line that passes through point C (10, 0) and point D on the y-axis, and it also passes through point E (5, 5). Since this line passes through E (5, 5) and C (10, 0), we can find its slope first. The slope m is given by (y2 - y1)/(x2 - x1). Let's compute that:m = (0 - 5)/(10 - 5) m = (-5)/5 m = -1So, the slope of the second line is -1. Now, using the point-slope form of the equation, which is y - y1 = m(x - x1), and using point E (5, 5):y - 5 = -1(x - 5) y - 5 = -x + 5 Adding 5 to both sides: y = -x + 10So, the equation of the second line is y = -x + 10. To find the y-intercept D, we can set x = 0:y = -0 + 10 y = 10Therefore, point D is at (0, 10).Now, let me summarize the coordinates of all the points involved:- Point O is the origin, (0, 0).- Point B is the y-intercept of the first line, (0, 20).- Point E is the intersection point, (5, 5).- Point C is given as (10, 0).So, the quadrilateral OBEC has vertices at O(0, 0), B(0, 20), E(5, 5), and C(10, 0). To find the area of quadrilateral OBEC, I can use the shoelace formula, which is a method for determining the area of a polygon when the coordinates of its vertices are known. The formula is:Area = |(x1y2 + x2y3 + x3y4 + x4y1) - (y1x2 + y2x3 + y3x4 + y4x1)| / 2But since this is a quadrilateral, I need to make sure the points are listed in order, either clockwise or counterclockwise, without crossing. Let me list the points in order: O(0, 0), B(0, 20), E(5, 5), C(10, 0), and back to O(0, 0).Plugging these into the shoelace formula:First, list the coordinates in order:O: (0, 0) B: (0, 20) E: (5, 5) C: (10, 0) O: (0, 0)Now, compute the sum of the products of the coordinates going one way:S1 = (0*20) + (0*5) + (5*0) + (10*0) S1 = 0 + 0 + 0 + 0 S1 = 0Wait, that doesn't seem right. Maybe I misapplied the formula. Let me double-check.Actually, the shoelace formula is:Area = |(x1y2 + x2y3 + x3y4 + x4y1) - (y1x2 + y2x3 + y3x4 + y4x1)| / 2So, plugging in the values:x1 = 0, y1 = 0 x2 = 0, y2 = 20 x3 = 5, y3 = 5 x4 = 10, y4 = 0Compute S1 = x1y2 + x2y3 + x3y4 + x4y1 S1 = (0*20) + (0*5) + (5*0) + (10*0) S1 = 0 + 0 + 0 + 0 S1 = 0Compute S2 = y1x2 + y2x3 + y3x4 + y4x1 S2 = (0*0) + (20*5) + (5*10) + (0*0) S2 = 0 + 100 + 50 + 0 S2 = 150Then, Area = |S1 - S2| / 2 = |0 - 150| / 2 = 150 / 2 = 75Wait, that gives me an area of 75, but I have a feeling that might not be correct because when I visualize the quadrilateral, it seems larger. Maybe I made a mistake in applying the shoelace formula.Alternatively, perhaps I should divide the quadrilateral into two triangles and calculate their areas separately. Let me try that approach.Quadrilateral OBEC can be divided into triangle OBE and triangle OEC.First, triangle OBE has vertices at O(0, 0), B(0, 20), and E(5, 5). The area of triangle OBE can be found using the formula for the area of a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3):Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2|Plugging in the coordinates:Area = |(0*(20 - 5) + 0*(5 - 0) + 5*(0 - 20)) / 2| = |(0 + 0 + 5*(-20)) / 2| = |(-100) / 2| = | -50 | = 50So, the area of triangle OBE is 50.Next, triangle OEC has vertices at O(0, 0), E(5, 5), and C(10, 0).Using the same formula:Area = |(0*(5 - 0) + 5*(0 - 0) + 10*(0 - 5)) / 2| = |(0 + 0 + 10*(-5)) / 2| = |(-50) / 2| = | -25 | = 25So, the area of triangle OEC is 25.Adding the areas of the two triangles: 50 + 25 = 75.Hmm, so both methods give me an area of 75. But earlier, I thought it might be larger. Let me double-check my calculations.Wait, perhaps I made a mistake in identifying the points or in the shoelace formula. Let me try the shoelace formula again, but this time, I'll list the points in a different order to ensure they are in the correct sequence.Points in order: O(0,0), B(0,20), E(5,5), C(10,0), O(0,0)Compute S1 = (0*20) + (0*5) + (5*0) + (10*0) = 0 + 0 + 0 + 0 = 0 Compute S2 = (0*0) + (20*5) + (5*10) + (0*0) = 0 + 100 + 50 + 0 = 150 Area = |0 - 150| / 2 = 75Same result. So, it seems consistent. Maybe my initial intuition was wrong, and the area is indeed 75.But wait, looking back at the problem, it mentions the quadrilateral OBEC. Let me make sure I have the correct points. O is (0,0), B is (0,20), E is (5,5), and C is (10,0). Yes, that seems correct.Alternatively, maybe I should use vectors or another method to confirm. Let's try using the area formula for a quadrilateral by breaking it into two triangles as I did before.Another way is to use the formula for the area of a trapezoid, but since OBEC is not necessarily a trapezoid, that might not apply here.Wait, actually, looking at the coordinates, O(0,0), B(0,20), E(5,5), and C(10,0), it's a quadrilateral that is not a trapezoid because only OB and EC are not necessarily parallel. So, the shoelace formula is the most reliable here.Given that both methods give me 75, I think that must be the correct area.But just to be thorough, let me plot these points mentally:- O is at the origin.- B is straight up the y-axis to (0,20).- E is at (5,5), which is somewhere in the middle.- C is at (10,0), which is on the x-axis.So, connecting these points, the quadrilateral is a four-sided figure with vertices at these points. The area seems to be 75.Wait, but in the initial problem, the first line has a slope of -3, which we used to find point A at (20/3, 0) and point B at (0,20). The second line has a slope of -1, passing through (10,0) and (5,5), giving us point D at (0,10).But in the quadrilateral OBEC, we don't use point D or A, so maybe that's why the area is 75.Alternatively, perhaps I should calculate the area using integration or another method, but that might be overcomplicating things.Wait, another approach: the area of quadrilateral OBEC can be found by subtracting the area of triangle EDC from the area of triangle OBC.Let me see:First, find the area of triangle OBC. Points O(0,0), B(0,20), C(10,0).Area of OBC can be calculated as (base * height)/2. The base is OC, which is 10 units, and the height is OB, which is 20 units.Area = (10 * 20)/2 = 100.Now, find the area of triangle EDC. Points E(5,5), D(0,10), C(10,0).Using the shoelace formula for triangle EDC:Coordinates: E(5,5), D(0,10), C(10,0), back to E(5,5).Compute S1 = (5*10) + (0*0) + (10*5) = 50 + 0 + 50 = 100 Compute S2 = (5*0) + (10*10) + (0*5) = 0 + 100 + 0 = 100 Area = |100 - 100| / 2 = 0 / 2 = 0Wait, that can't be right. There must be a mistake here. Let me recalculate.Actually, for triangle EDC, the coordinates are E(5,5), D(0,10), C(10,0).Using the formula:Area = |(5*(10 - 0) + 0*(0 - 5) + 10*(5 - 10)) / 2| = |(5*10 + 0*(-5) + 10*(-5)) / 2| = |(50 + 0 - 50) / 2| = |0 / 2| = 0Hmm, that's odd. It seems like the area is zero, which would imply that the points are colinear, but they aren't. I must have made a mistake in the formula.Wait, the correct formula for the area of a triangle with vertices (x1,y1), (x2,y2), (x3,y3) is:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2|So, plugging in E(5,5), D(0,10), C(10,0):Area = |5*(10 - 0) + 0*(0 - 5) + 10*(5 - 10)| / 2 = |5*10 + 0*(-5) + 10*(-5)| / 2 = |50 + 0 - 50| / 2 = |0| / 2 = 0This still gives zero, which is incorrect. I think the issue is that I'm using the wrong order of points. Let me try a different order.Let's list the points as D(0,10), E(5,5), C(10,0).Area = |0*(5 - 0) + 5*(0 - 10) + 10*(10 - 5)| / 2 = |0 + 5*(-10) + 10*5| / 2 = |0 - 50 + 50| / 2 = |0| / 2 = 0Still zero. This is confusing. Maybe I should use the shoelace formula correctly by ensuring the points are listed in order without crossing.Alternatively, perhaps triangle EDC is degenerate, but that can't be because the points are not colinear.Wait, let me check if points E, D, and C are colinear. The line from D(0,10) to C(10,0) has a slope of (0 - 10)/(10 - 0) = -10/10 = -1. The line from E(5,5) to C(10,0) also has a slope of (0 - 5)/(10 - 5) = -5/5 = -1. So, E lies on the line DC. Therefore, triangle EDC is actually a straight line, making its area zero. That's why the area is zero.So, going back, if I try to subtract the area of triangle EDC from triangle OBC, I get 100 - 0 = 100, which is not the area of quadrilateral OBEC. Therefore, this approach is flawed.Perhaps I should instead consider the quadrilateral OBEC as a polygon and use the shoelace formula correctly. Let me try that again.List the coordinates in order: O(0,0), B(0,20), E(5,5), C(10,0), back to O(0,0).Compute S1 = (0*20) + (0*5) + (5*0) + (10*0) = 0 + 0 + 0 + 0 = 0 Compute S2 = (0*0) + (20*5) + (5*10) + (0*0) = 0 + 100 + 50 + 0 = 150 Area = |S1 - S2| / 2 = |0 - 150| / 2 = 75So, the area is indeed 75. Wait, but earlier when I thought of the area as 75, I felt it was too small, but considering the coordinates, it might actually be correct. Let me visualize it again.From O(0,0) to B(0,20): that's a vertical line up the y-axis. Then to E(5,5): that's a line going down to the right. Then to C(10,0): that's a line going further down to the right, and back to O(0,0). So, the shape is a quadrilateral that is somewhat diamond-shaped but not symmetric.Given that the shoelace formula gives 75, and the triangle areas also add up to 75, I think that must be the correct answer.But just to be absolutely sure, let me try another method. Maybe using vectors or coordinates to find the area.Another approach is to use the formula for the area of a polygon by summing up the areas of trapezoids formed with the x-axis. But since the quadrilateral is not necessarily convex or aligned in a particular way, that might complicate things.Alternatively, I can use the determinant method for each triangle and sum them up, which I did earlier, giving me 50 + 25 = 75.Given that multiple methods are giving me the same result, I think it's safe to conclude that the area of quadrilateral OBEC is 75.Wait, but in the initial problem, the first line intersects the x-axis at A(20/3, 0) and y-axis at B(0,20). The second line intersects the x-axis at C(10,0) and y-axis at D(0,10). They intersect at E(5,5). So, the quadrilateral OBEC is formed by points O, B, E, and C.Yes, that's correct. So, the area is 75.But just to be thorough, let me calculate the area using vectors.Define vectors OB, BE, EC, and CO.But that might not be straightforward. Alternatively, I can use the coordinates to calculate the area.Wait, another method: the area can be found by calculating the area under the first line from O to B to E and subtracting the area under the second line from O to E to C.But that might not be the right approach either.Alternatively, since we have the equations of both lines, we can find the area between them from x=0 to x=5 and from x=5 to x=10.Wait, that might work. Let me try.From x=0 to x=5, the upper boundary is the first line y = -3x + 20, and the lower boundary is the second line y = -x + 10.From x=5 to x=10, the upper boundary is the second line y = -x + 10, and the lower boundary is the x-axis y=0.So, the area can be calculated as the integral from 0 to 5 of (upper - lower) dx plus the integral from 5 to 10 of (upper - lower) dx.Let's compute that.First integral from 0 to 5:Upper function: y = -3x + 20 Lower function: y = -x + 10 Difference: (-3x + 20) - (-x + 10) = (-3x + 20) + x - 10 = (-2x + 10)Integral from 0 to 5 of (-2x + 10) dx = [ -x^2 + 10x ] from 0 to 5 = [ -(25) + 50 ] - [ 0 + 0 ] = 25Second integral from 5 to 10:Upper function: y = -x + 10 Lower function: y = 0 Difference: (-x + 10) - 0 = -x + 10Integral from 5 to 10 of (-x + 10) dx = [ (-0.5x^2) + 10x ] from 5 to 10 At x=10: (-0.5*100) + 100 = -50 + 100 = 50 At x=5: (-0.5*25) + 50 = -12.5 + 50 = 37.5 Difference: 50 - 37.5 = 12.5Total area = 25 + 12.5 = 37.5Wait, that's only 37.5, which is half of 75. That can't be right. I must have made a mistake in setting up the integrals.Wait, no, actually, the area between the two lines from 0 to 5 is 25, and the area under the second line from 5 to 10 is 12.5, but the quadrilateral OBEC includes the area from O to B to E to C, which is more than just the area between the two lines.Wait, perhaps I need to consider the area above the x-axis as well. Let me think.Alternatively, maybe I should calculate the area of the quadrilateral by subtracting the area below the x-axis, but since all points are in the positive quadrant, that's not necessary.Wait, perhaps I'm overcomplicating it. Given that both the shoelace formula and the triangle area method give me 75, I think that is correct. The integral approach might not be the right method here because it's calculating the area between the curves, not the quadrilateral itself.So, to conclude, after multiple methods and double-checking, the area of quadrilateral OBEC is 75.

💡Alright, so I'm trying to tackle this geometry problem where I have to reflect the lines PA, PB, and PC about the angle bisectors of triangle ABC and show that these reflected lines meet at a single point P'. Then, I need to figure out what P' is when P is the circumcenter, centroid, and orthocenter of triangle ABC. Hmm, okay, let's break this down step by step.First, let me visualize triangle ABC with point P somewhere inside it. I know that reflecting a line over an angle bisector involves some kind of symmetry. The angle bisector of angle BAC, for example, splits angle A into two equal parts. So, reflecting PA over this bisector should give me a new line that's symmetric with respect to the bisector.I remember that reflecting a line over another line can be done using some properties of angles. Maybe I can use the concept that the angle between the original line and the bisector is equal to the angle between the reflected line and the bisector. So, if I reflect PA over the angle bisector of angle A, the new line should make the same angle with the bisector as PA does, but on the other side.Similarly, I'll have to do this for PB and PC over their respective angle bisectors at B and C. After reflecting all three lines, I need to show that these three new lines intersect at a single point P'. That sounds like I might need to use some concurrency theorem or maybe properties of triangle centers.Now, considering specific cases where P is the circumcenter, centroid, or orthocenter. Let me recall what these points are:1. **Circumcenter (O)**: The point where the perpendicular bisectors of the sides meet; it's equidistant from all three vertices.2. **Centroid (G)**: The intersection of the medians; it divides each median in a 2:1 ratio.3. **Orthocenter (H)**: The intersection of the altitudes of the triangle.I need to figure out what P' is in each case. Maybe reflecting these specific points over the angle bisectors will lead to known centers or other significant points.Let me start with the circumcenter. If P is the circumcenter O, then OA = OB = OC. Reflecting OA over the angle bisector of A... Hmm, since OA is equal to OB and OC, reflecting it over the bisector might relate to some symmetry. Maybe the reflection of O over the angle bisector is another known point?Wait, reflecting the circumcenter over the angle bisector might lead to the ex-circumcenter or something related. But I'm not sure. Maybe I should think about the properties of reflections in angle bisectors.Another approach: when you reflect a point over a line, the line is the perpendicular bisector of the segment joining the original point and its reflection. So, if I reflect PA over the angle bisector, the angle bisector is the perpendicular bisector of the segment joining P and its reflection P'.But wait, in this case, we're reflecting the line PA, not the point P. So, maybe the reflection of the line PA over the angle bisector is another line that is symmetric to PA with respect to the bisector.I think I need to use some coordinate geometry here. Maybe placing triangle ABC in a coordinate system would help. Let me assign coordinates to A, B, and C, and then find the equations of the angle bisectors and the reflections.But before diving into coordinates, perhaps there's a synthetic geometry approach. I remember that reflecting lines over angle bisectors can relate to isogonal conjugates. Isogonal conjugate points are such that the reflections of their cevians over the angle bisectors concur at another point.So, if P is a point inside the triangle, reflecting its cevians PA, PB, PC over the angle bisectors should give me the cevians of its isogonal conjugate. Therefore, the point P' is the isogonal conjugate of P.Ah! So, if that's the case, then P' is the isogonal conjugate of P. That would explain why the reflected lines concur at P'. So, the first part of the problem is essentially showing that the reflections of PA, PB, PC over the angle bisectors concur at the isogonal conjugate of P.Now, for the specific cases:1. **When P is the circumcenter (O)**: The isogonal conjugate of the circumcenter is the orthocenter (H). So, P' should be the orthocenter.2. **When P is the centroid (G)**: The isogonal conjugate of the centroid is the symmedian point (also known as the Lemoine point). So, P' should be the symmedian point.3. **When P is the orthocenter (H)**: The isogonal conjugate of the orthocenter is the circumcenter (O). So, P' should be the circumcenter.Wait, let me verify this. I think the isogonal conjugate of the centroid is indeed the symmedian point. The centroid has trilinear coordinates proportional to 1:1:1, and the symmedian point has trilinear coordinates proportional to a:b:c, which are the isogonal conjugates.Similarly, the circumcenter and orthocenter are isogonal conjugates in an acute triangle. In an obtuse triangle, the orthocenter lies outside, but the relationship still holds in terms of their isogonal conjugates.So, putting it all together:- If P is the circumcenter, P' is the orthocenter.- If P is the centroid, P' is the symmedian point.- If P is the orthocenter, P' is the circumcenter.I think that makes sense. The isogonal conjugate relationship swaps certain triangle centers in this way.But just to make sure, let me think about reflecting the circumcenter over the angle bisectors. The circumcenter is equidistant from all vertices, so reflecting it over the angle bisectors might lead to a point that has equal distances from the sides, which is the incenter. Wait, no, the incenter is the intersection of the angle bisectors, not necessarily the reflection of the circumcenter.Wait, maybe I confused something. The isogonal conjugate of the incenter is the ex-incenter? Or is it something else?No, actually, the incenter is its own isogonal conjugate because reflecting its cevians (which are the angle bisectors) over themselves doesn't change them. So, the incenter is self-isogonal.But in our case, reflecting the circumcenter's cevians over the angle bisectors leads to the orthocenter's cevians. So, yes, P' is the orthocenter when P is the circumcenter.Similarly, reflecting the centroid's cevians (which are the medians) over the angle bisectors leads to the symmedian point's cevians (which are the symmedians).And reflecting the orthocenter's cevians (which are the altitudes) over the angle bisectors leads back to the circumcenter's cevians (which are the perpendicular bisectors).Okay, that seems consistent. So, I think I've got it.**Final Answer**When ( P ) is the circumcenter, centroid, and orthocenter, the point ( P' ) is the orthocenter, symmedian point, and circumcenter respectively. Thus:- If ( P ) is the circumcenter, then ( P' ) is the boxed{text{orthocenter}}.- If ( P ) is the centroid, then ( P' ) is the boxed{text{symmedian point}}.- If ( P ) is the orthocenter, then ( P' ) is the boxed{text{circumcenter}}.

💡Okay, so I have this problem where I need to find the number of ordered triples (x, y, z) of non-negative integers such that x ≤ y ≤ z and x + y + z ≤ 100. Hmm, that seems a bit tricky, but let me try to break it down step by step.First, I know that x, y, and z are non-negative integers, which means they can be 0 or any positive integer. The conditions are that x is less than or equal to y, which is less than or equal to z, and their sum is less than or equal to 100. So, I need to count all such triples that satisfy these inequalities.I remember that when dealing with inequalities like x ≤ y ≤ z, it's often helpful to use a substitution to simplify the problem. Maybe I can express y and z in terms of x and some other variables. Let me think... If x ≤ y ≤ z, then y can be written as x + a, where a is a non-negative integer. Similarly, z can be written as y + b, which would be x + a + b, where b is also a non-negative integer. So, substituting these into the original variables, I get:x = x y = x + a z = x + a + bWhere a and b are non-negative integers. That makes sense because this substitution ensures that x ≤ y ≤ z by construction.Now, let's substitute these expressions into the inequality x + y + z ≤ 100. Plugging in the values, we get:x + (x + a) + (x + a + b) ≤ 100Simplifying this, we have:x + x + a + x + a + b ≤ 100 3x + 2a + b ≤ 100So, now the problem reduces to finding the number of non-negative integer solutions (x, a, b) to the inequality 3x + 2a + b ≤ 100.Hmm, okay. So, I need to count the number of triples (x, a, b) such that 3x + 2a + b ≤ 100. This seems more manageable because now I can think of this as a problem of counting the number of non-negative integer solutions to an inequality with three variables.I recall that for inequalities like this, generating functions can be a useful tool. A generating function is a way to encode the number of solutions to an equation or inequality by treating each variable as a power of a formal variable, usually x. The coefficient of x^n in the generating function gives the number of solutions for that particular n.In this case, since we have three variables x, a, and b with coefficients 3, 2, and 1 respectively, the generating function would be the product of the generating functions for each variable. Specifically, the generating function for x would be 1 + x^3 + x^6 + x^9 + ..., since x can take values 0, 1, 2, ... multiplied by 3. Similarly, the generating function for a would be 1 + x^2 + x^4 + x^6 + ..., and for b it would be 1 + x + x^2 + x^3 + ..., since b can be any non-negative integer.So, the generating function G(x) would be:G(x) = (1 + x^3 + x^6 + ...) * (1 + x^2 + x^4 + ...) * (1 + x + x^2 + x^3 + ...)This can be written in a more compact form using the formula for the sum of a geometric series:G(x) = 1 / [(1 - x^3)(1 - x^2)(1 - x)]Now, I need to find the coefficient of x^k in this generating function for k from 0 to 100 and sum them up to get the total number of solutions. However, calculating this directly seems complicated, especially for such a high degree like 100.I remember that partial fraction decomposition can be used to simplify the generating function into a form that is easier to expand as a power series. Let me try that.First, let's factor the denominator:(1 - x^3)(1 - x^2)(1 - x) = (1 - x)(1 + x + x^2)(1 - x)(1 + x)(1 - x)Wait, that seems a bit messy. Let me see if I can write it as:(1 - x)^3 * (1 + x) * (1 + x + x^2)Yes, because (1 - x^3) factors into (1 - x)(1 + x + x^2), and (1 - x^2) factors into (1 - x)(1 + x). So, altogether, we have (1 - x)^3 * (1 + x) * (1 + x + x^2).So, the generating function is:G(x) = 1 / [(1 - x)^3 * (1 + x) * (1 + x + x^2)]Now, to perform partial fraction decomposition, I need to express G(x) as a sum of simpler fractions. The general form for partial fractions when dealing with repeated linear factors and irreducible quadratic factors is:G(x) = A / (1 - x) + B / (1 - x)^2 + C / (1 - x)^3 + D / (1 + x) + (E x + F) / (1 + x + x^2)Where A, B, C, D, E, and F are constants to be determined.Multiplying both sides by the denominator, we get:1 = A (1 - x)^2 (1 + x)(1 + x + x^2) + B (1 - x)(1 + x)(1 + x + x^2) + C (1 + x)(1 + x + x^2) + D (1 - x)^3 (1 + x + x^2) + (E x + F)(1 - x)^3 (1 + x)This looks quite complicated, but maybe we can find the constants by plugging in suitable values of x.Let me try plugging in x = 1. However, x = 1 would cause some denominators to become zero, but let's see:If x = 1, the left side is 1. On the right side, terms with (1 - x) will become zero, except for the terms multiplied by A, B, C, D, E, F. Wait, actually, when x = 1, (1 - x) becomes zero, so all terms except those with (1 - x) in the denominator will vanish. But since we multiplied by the denominator, actually, plugging x = 1 would cause the entire equation to be 1 = 0 + 0 + 0 + 0 + (E*1 + F)(0), which is 1 = 0, which is not helpful. Maybe I need another approach.Alternatively, I can equate coefficients of powers of x on both sides. But that might be tedious. Alternatively, maybe I can use the method of residues or other techniques, but I'm not sure.Wait, perhaps I can use the fact that the generating function can be expressed in terms of known series expansions. For example, 1 / (1 - x)^3 is the generating function for combinations with repetition, which is (n + 2 choose 2). Similarly, 1 / (1 + x) is the generating function for (-1)^n, and 1 / (1 + x + x^2) can be expressed in terms of roots of unity.Alternatively, maybe I can use generating functions in a different way. Let me think about the original inequality 3x + 2a + b ≤ 100. Maybe I can fix x and a, and then b can range from 0 to 100 - 3x - 2a. So, for each x and a, the number of possible b is 100 - 3x - 2a + 1, provided that 3x + 2a ≤ 100.Therefore, the total number of solutions is the sum over x and a of (100 - 3x - 2a + 1), where 3x + 2a ≤ 100.So, mathematically, this is:Total = Σ (from x=0 to floor(100/3)) Σ (from a=0 to floor((100 - 3x)/2)) (101 - 3x - 2a)Hmm, that seems manageable, but it's still a double summation. Maybe I can compute it by changing the order of summation or finding a closed-form expression.Let me try to compute the inner sum first for a fixed x. For a fixed x, the inner sum is over a from 0 to floor((100 - 3x)/2) of (101 - 3x - 2a). Let me denote k = 101 - 3x, so the inner sum becomes Σ (from a=0 to m) (k - 2a), where m = floor((100 - 3x)/2).This is an arithmetic series where the first term is k and the last term is k - 2m. The number of terms is m + 1. The sum of an arithmetic series is (number of terms) * (first term + last term) / 2.So, the inner sum S(x) is:S(x) = (m + 1) * (k + (k - 2m)) / 2 = (m + 1) * (2k - 2m) / 2 = (m + 1)(k - m)Substituting back k = 101 - 3x and m = floor((100 - 3x)/2):S(x) = (floor((100 - 3x)/2) + 1) * (101 - 3x - floor((100 - 3x)/2))Hmm, this is getting a bit complicated, but maybe I can express it in terms of (100 - 3x). Let me denote n = 100 - 3x, so m = floor(n / 2). Then,S(x) = (floor(n / 2) + 1) * (101 - 3x - floor(n / 2))But n = 100 - 3x, so 101 - 3x = n + 1. Therefore,S(x) = (floor(n / 2) + 1) * (n + 1 - floor(n / 2))Let me compute this expression for both even and odd n.Case 1: n is even. Let n = 2k. Then floor(n / 2) = k.So,S(x) = (k + 1) * (2k + 1 - k) = (k + 1)(k + 1) = (k + 1)^2Case 2: n is odd. Let n = 2k + 1. Then floor(n / 2) = k.So,S(x) = (k + 1) * (2k + 2 - k) = (k + 1)(k + 2)Therefore, depending on whether n is even or odd, S(x) is either (k + 1)^2 or (k + 1)(k + 2), where k = floor(n / 2).But n = 100 - 3x, so let's see when n is even or odd.n = 100 - 3x. Since 3x mod 2 is equal to x mod 2 (because 3 is odd), so n mod 2 = (100 - x) mod 2. Since 100 is even, n mod 2 = (0 - x) mod 2 = (-x) mod 2 = x mod 2.Therefore, n is even if x is even, and n is odd if x is odd.So, depending on whether x is even or odd, S(x) will be either (k + 1)^2 or (k + 1)(k + 2), where k = floor(n / 2) = floor((100 - 3x)/2).But let's express this in terms of x.If x is even: Let x = 2m. Then n = 100 - 6m.So, k = floor((100 - 6m)/2) = 50 - 3m.Thus, S(x) = (50 - 3m + 1)^2 = (51 - 3m)^2.If x is odd: Let x = 2m + 1. Then n = 100 - 3(2m + 1) = 100 - 6m - 3 = 97 - 6m.So, k = floor((97 - 6m)/2) = 48 - 3m (since 97 - 6m is odd, floor division gives 48 - 3m).Thus, S(x) = (48 - 3m + 1)(48 - 3m + 2) = (49 - 3m)(50 - 3m).Therefore, the total number of solutions is the sum over x of S(x), where S(x) is as above depending on whether x is even or odd.So, let's separate the sum into even and odd x.Let me denote x = 2m for even x, where m ranges from 0 to floor(100 / 6) = 16 (since 3x ≤ 100 => x ≤ 33.333, so m ≤ 16.666, so m up to 16).Similarly, for odd x, x = 2m + 1, where m ranges from 0 to floor((100 - 3)/6) = floor(97/6) = 16 (since 3x ≤ 100 => x ≤ 33.333, so m ≤ 16.166, so m up to 16).Wait, actually, for x even: x = 2m, m from 0 to floor(100 / 3) / 2, but 100 / 3 ≈ 33.333, so m up to 16 (since 2*16=32 ≤ 33.333). Similarly, for x odd: x = 2m + 1, m from 0 to 16 (since 2*16 +1=33 ≤ 33.333).So, total sum is:Total = Σ (m=0 to 16) S(2m) + Σ (m=0 to 16) S(2m +1)Where S(2m) = (51 - 3m)^2 and S(2m +1) = (49 - 3m)(50 - 3m)So, let's compute these two sums separately.First, compute Σ (m=0 to 16) (51 - 3m)^2.Let me denote t = 51 - 3m. When m=0, t=51; when m=16, t=51 - 48=3. So, t decreases by 3 each time. So, the sum is Σ (t=51, 48, ..., 3) t^2.Similarly, for the second sum, Σ (m=0 to16) (49 - 3m)(50 - 3m).Let me denote u = 49 - 3m and v = 50 - 3m. So, u = v -1. So, (u)(v) = (v -1)v = v^2 - v.But v = 50 - 3m. When m=0, v=50; when m=16, v=50 - 48=2. So, v decreases by 3 each time. So, the sum is Σ (v=50,47,...,2) (v^2 - v).Therefore, the total sum becomes:Total = Σ (t=51,48,...,3) t^2 + Σ (v=50,47,...,2) (v^2 - v)Let me compute these two sums.First, compute Σ (t=51,48,...,3) t^2.This is an arithmetic sequence of t where t starts at 51, decreases by 3 each time, until t=3. The number of terms is ((51 - 3)/3) +1 = (48/3)+1=16+1=17 terms.Similarly, for the second sum, Σ (v=50,47,...,2) (v^2 - v). Here, v starts at 50, decreases by 3 each time, until v=2. The number of terms is ((50 - 2)/3)+1=(48/3)+1=16+1=17 terms.So, both sums have 17 terms.Let me compute the first sum: Σ t^2 where t=51,48,...,3.This is equivalent to Σ (3k + 3)^2 where k goes from 0 to 16, but wait, let's see:Wait, t starts at 51 and decreases by 3 each time. So, t = 51 - 3m, where m=0 to16.So, t = 51 - 3m, so t^2 = (51 - 3m)^2.So, the sum is Σ (m=0 to16) (51 - 3m)^2.Similarly, the second sum is Σ (v=50,47,...,2) (v^2 - v) = Σ (m=0 to16) (50 - 3m)^2 - (50 - 3m).So, let's compute both sums.First sum: S1 = Σ (m=0 to16) (51 - 3m)^2.Let me expand (51 - 3m)^2 = 51^2 - 2*51*3m + (3m)^2 = 2601 - 306m + 9m^2.So, S1 = Σ (2601 - 306m + 9m^2) from m=0 to16.This can be split into:S1 = 17*2601 - 306*Σ m + 9*Σ m^2.Compute each term:17*2601 = 17*2600 + 17*1 = 44200 + 17 = 44217.Σ m from m=0 to16 is (16*17)/2 = 136.So, 306*136 = Let's compute 300*136 + 6*136 = 40800 + 816 = 41616.Σ m^2 from m=0 to16 is (16)(16+1)(2*16+1)/6 = 16*17*33/6.Compute 16*17=272, 272*33=8976, 8976/6=1496.So, 9*1496=13464.Therefore, S1 = 44217 - 41616 + 13464.Compute 44217 - 41616 = 2601.Then, 2601 + 13464 = 16065.So, S1 = 16065.Now, compute the second sum: S2 = Σ (v=50,47,...,2) (v^2 - v) = Σ (m=0 to16) (50 - 3m)^2 - (50 - 3m).Let me expand (50 - 3m)^2 - (50 - 3m) = (2500 - 300m + 9m^2) - 50 + 3m = 2500 - 300m + 9m^2 -50 +3m = 2450 - 297m + 9m^2.So, S2 = Σ (2450 - 297m + 9m^2) from m=0 to16.Again, split into:S2 = 17*2450 - 297*Σ m + 9*Σ m^2.Compute each term:17*2450 = 17*2000 + 17*450 = 34000 + 7650 = 41650.Σ m from m=0 to16 is 136, as before.297*136 = Let's compute 300*136 - 3*136 = 40800 - 408 = 40392.Σ m^2 from m=0 to16 is 1496, as before.So, 9*1496=13464.Therefore, S2 = 41650 - 40392 + 13464.Compute 41650 - 40392 = 1258.Then, 1258 + 13464 = 14722.So, S2 = 14722.Therefore, the total number of solutions is S1 + S2 = 16065 + 14722 = 30787.Wait, that seems like a lot, but let me double-check my calculations.First, S1 was computed as 16065, which came from 44217 - 41616 + 13464. Let me verify:44217 - 41616 = 2601.2601 + 13464 = 16065. Correct.Then, S2 was 41650 - 40392 + 13464.41650 - 40392 = 1258.1258 + 13464 = 14722. Correct.So, 16065 + 14722 = 30787. Yes, that seems right.Therefore, the number of ordered triples (x, y, z) of non-negative integers satisfying x ≤ y ≤ z and x + y + z ≤ 100 is 30787.I think that's the answer. It was a bit involved, but breaking it down into even and odd cases and then computing the sums separately helped manage the complexity.