💡First, I need to find the complex numbers corresponding to points ( A ) and ( B ).For point ( A ):[frac{1}{1+i} = frac{1-i}{(1+i)(1-i)} = frac{1-i}{1^2+1^2} = frac{1}{2}-frac{1}{2}i]So, the coordinates of point ( A ) are ( left(frac{1}{2}, -frac{1}{2}right) ).For point ( B ):[frac{1}{1-i} = frac{1+i}{(1-i)(1+i)} = frac{1+i}{1^2+1^2} = frac{1}{2}+frac{1}{2}i]So, the coordinates of point ( B ) are ( left(frac{1}{2}, frac{1}{2}right) ).Now, I need to find the midpoint ( C ) of line segment ( AB ). The coordinates of the midpoint ( C(x, y) ) can be found using the formula:[x = frac{x_1+x_2}{2}, quad y = frac{y_1+y_2}{2}]Substituting the coordinates of points ( A ) and ( B ), we get:[x = frac{frac{1}{2}+frac{1}{2}}{2} = frac{1}{2}, quad y = frac{-frac{1}{2}+frac{1}{2}}{2} = 0]So, the coordinates of point ( C ) are ( left(frac{1}{2}, 0right) ).Therefore, the complex number corresponding to point ( C ) is ( boxed{frac{1}{2}} ).

💡Okay, so I have this problem where I need to find out how many times the graphs ( r = 6 cos theta ) and ( r = 10 sin theta ) intersect. Hmm, both of these are polar equations, right? I remember that polar equations can sometimes represent circles when converted to Cartesian coordinates. Maybe I should try converting both equations to Cartesian form to see what they look like.Starting with ( r = 6 cos theta ). I recall that in polar coordinates, ( r = a cos theta ) is a circle with diameter ( a ) along the x-axis. So, converting this to Cartesian coordinates, I can use the relationships ( x = r cos theta ) and ( y = r sin theta ). Let me substitute ( r = 6 cos theta ) into ( x ) and ( y ):- ( x = 6 cos theta cdot cos theta = 6 cos^2 theta )- ( y = 6 cos theta cdot sin theta = 6 sin theta cos theta )Hmm, I think I can use some trigonometric identities here to simplify these expressions. For ( x ), ( cos^2 theta ) can be written as ( frac{1 + cos 2theta}{2} ), so:- ( x = 6 cdot frac{1 + cos 2theta}{2} = 3(1 + cos 2theta) = 3 + 3 cos 2theta )For ( y ), ( sin theta cos theta ) is equal to ( frac{sin 2theta}{2} ), so:- ( y = 6 cdot frac{sin 2theta}{2} = 3 sin 2theta )Now, if I let ( phi = 2theta ), then the parametric equations become:- ( x = 3 + 3 cos phi )- ( y = 3 sin phi )This looks like the parametric equation of a circle with center at ( (3, 0) ) and radius 3. So, the Cartesian equation would be:[ (x - 3)^2 + y^2 = 9 ]Alright, that takes care of the first equation. Now, moving on to the second equation ( r = 10 sin theta ). Similarly, I remember that ( r = a sin theta ) is a circle with diameter ( a ) along the y-axis. Let me convert this to Cartesian coordinates as well.Substituting ( r = 10 sin theta ) into ( x ) and ( y ):- ( x = 10 sin theta cdot cos theta = 10 sin theta cos theta )- ( y = 10 sin theta cdot sin theta = 10 sin^2 theta )Again, using trigonometric identities:- For ( x ), ( sin theta cos theta = frac{sin 2theta}{2} ), so: - ( x = 10 cdot frac{sin 2theta}{2} = 5 sin 2theta )- For ( y ), ( sin^2 theta = frac{1 - cos 2theta}{2} ), so: - ( y = 10 cdot frac{1 - cos 2theta}{2} = 5(1 - cos 2theta) = 5 - 5 cos 2theta )Letting ( phi = 2theta ) again, the parametric equations become:- ( x = 5 sin phi )- ( y = 5 - 5 cos phi )This is the parametric equation of a circle with center at ( (0, 5) ) and radius 5. So, the Cartesian equation is:[ x^2 + (y - 5)^2 = 25 ]Now, I have both equations in Cartesian form:1. ( (x - 3)^2 + y^2 = 9 )2. ( x^2 + (y - 5)^2 = 25 )I need to find how many points these two circles intersect. To do this, I can solve the system of equations. Let me write them down again:1. ( (x - 3)^2 + y^2 = 9 )2. ( x^2 + (y - 5)^2 = 25 )First, I'll expand both equations to make it easier to subtract them.Expanding the first equation:[ (x - 3)^2 + y^2 = x^2 - 6x + 9 + y^2 = 9 ]Simplifying:[ x^2 - 6x + 9 + y^2 = 9 ]Subtract 9 from both sides:[ x^2 - 6x + y^2 = 0 ]Let me call this Equation (1a).Expanding the second equation:[ x^2 + (y - 5)^2 = x^2 + y^2 - 10y + 25 = 25 ]Simplifying:[ x^2 + y^2 - 10y + 25 = 25 ]Subtract 25 from both sides:[ x^2 + y^2 - 10y = 0 ]Let me call this Equation (2a).Now, I have:1. ( x^2 - 6x + y^2 = 0 ) (Equation 1a)2. ( x^2 + y^2 - 10y = 0 ) (Equation 2a)If I subtract Equation 1a from Equation 2a, I can eliminate ( x^2 ) and ( y^2 ):[ (x^2 + y^2 - 10y) - (x^2 - 6x + y^2) = 0 - 0 ]Simplify:[ x^2 + y^2 - 10y - x^2 + 6x - y^2 = 0 ]The ( x^2 ) and ( y^2 ) terms cancel out:[ -10y + 6x = 0 ]Simplify:[ 6x - 10y = 0 ]Divide both sides by 2:[ 3x - 5y = 0 ]So, ( 3x = 5y ) or ( y = frac{3}{5}x )Now, I can substitute ( y = frac{3}{5}x ) back into one of the original equations to solve for ( x ). Let me choose Equation 1a:[ x^2 - 6x + y^2 = 0 ]Substitute ( y ):[ x^2 - 6x + left( frac{3}{5}x right)^2 = 0 ]Calculate ( left( frac{3}{5}x right)^2 ):[ left( frac{9}{25}x^2 right) ]So, the equation becomes:[ x^2 - 6x + frac{9}{25}x^2 = 0 ]Combine like terms:[ left(1 + frac{9}{25}right)x^2 - 6x = 0 ]Convert 1 to ( frac{25}{25} ):[ frac{34}{25}x^2 - 6x = 0 ]Multiply both sides by 25 to eliminate the denominator:[ 34x^2 - 150x = 0 ]Factor out an x:[ x(34x - 150) = 0 ]So, either ( x = 0 ) or ( 34x - 150 = 0 )If ( x = 0 ), then from ( y = frac{3}{5}x ), ( y = 0 ). So, one point is ( (0, 0) ).If ( 34x - 150 = 0 ), then:[ 34x = 150 ][ x = frac{150}{34} ]Simplify the fraction:Divide numerator and denominator by 2:[ x = frac{75}{17} ]So, ( x = frac{75}{17} ). Then, ( y = frac{3}{5} times frac{75}{17} = frac{225}{85} = frac{45}{17} )So, the other point is ( left( frac{75}{17}, frac{45}{17} right) )Wait, so I have two solutions: ( (0, 0) ) and ( left( frac{75}{17}, frac{45}{17} right) ). That suggests that the two circles intersect at two points.But let me double-check to make sure I didn't make any mistakes. Sometimes when solving equations, especially with substitutions, it's easy to introduce extraneous solutions or miss something.First, checking the point ( (0, 0) ):- Plug into the first equation ( (x - 3)^2 + y^2 = 9 ): ( (0 - 3)^2 + 0^2 = 9 + 0 = 9 ). That works.- Plug into the second equation ( x^2 + (y - 5)^2 = 25 ): ( 0^2 + (0 - 5)^2 = 0 + 25 = 25 ). That also works.Now, checking the point ( left( frac{75}{17}, frac{45}{17} right) ):- First equation: ( left( frac{75}{17} - 3 right)^2 + left( frac{45}{17} right)^2 ) Convert 3 to ( frac{51}{17} ): ( left( frac{75 - 51}{17} right)^2 + left( frac{45}{17} right)^2 = left( frac{24}{17} right)^2 + left( frac{45}{17} right)^2 ) Calculate: ( frac{576}{289} + frac{2025}{289} = frac{2601}{289} = 9 ). That works. - Second equation: ( left( frac{75}{17} right)^2 + left( frac{45}{17} - 5 right)^2 ) Convert 5 to ( frac{85}{17} ): ( frac{5625}{289} + left( frac{45 - 85}{17} right)^2 = frac{5625}{289} + left( frac{-40}{17} right)^2 ) Calculate: ( frac{5625}{289} + frac{1600}{289} = frac{7225}{289} = 25 ). That also works.So, both points satisfy both equations, so they are indeed intersection points.Wait a second, but I recall that sometimes polar equations can have overlapping points at the origin, especially if both equations pass through the origin. Let me think about the original polar equations.For ( r = 6 cos theta ), when ( theta = frac{pi}{2} ), ( r = 0 ). Similarly, for ( r = 10 sin theta ), when ( theta = 0 ), ( r = 0 ). So, both circles pass through the origin, but at different angles. So, the origin is a common point, but does that count as one intersection or two?Wait, in Cartesian coordinates, the origin is just one point, so even though both circles pass through the origin, it's just one point of intersection. So, in total, we have two points: the origin and ( left( frac{75}{17}, frac{45}{17} right) ). So, that's two intersection points.But to be thorough, maybe I should check if there are any other intersection points. Sometimes, circles can intersect at more than two points, but I think that's only if they are the same circle, which isn't the case here.Let me visualize the two circles. The first circle is centered at (3, 0) with radius 3, so it touches the origin and extends to (6, 0). The second circle is centered at (0, 5) with radius 5, so it touches the origin and extends to (0, 10). These two circles should intersect at two points: the origin and another point somewhere in the first quadrant.Wait, but when I solved the equations, I got two points: the origin and ( left( frac{75}{17}, frac{45}{17} right) ). Let me approximate ( frac{75}{17} ) and ( frac{45}{17} ) to see where that point is.Calculating:- ( frac{75}{17} approx 4.4118 )- ( frac{45}{17} approx 2.6471 )So, the point is approximately (4.41, 2.65), which is in the first quadrant. That makes sense because both circles are in the first quadrant and should intersect there.Wait, but could there be another intersection point in another quadrant? Let me think. The first circle is centered at (3, 0) with radius 3, so it only exists from x=0 to x=6, and y from -3 to 3. The second circle is centered at (0, 5) with radius 5, so it exists from y=0 to y=10, and x from -5 to 5. So, the overlapping region is in the first quadrant, and the origin.Wait, but the origin is a single point where both circles meet. So, in total, two intersection points: the origin and the point in the first quadrant.Wait, but when I solved the equations, I got two solutions, so that's consistent. So, the answer should be two intersection points.But just to be absolutely sure, let me think about the distance between the centers of the two circles. The first circle is at (3, 0) and the second at (0, 5). The distance between centers is sqrt[(3-0)^2 + (0-5)^2] = sqrt[9 + 25] = sqrt[34] ≈ 5.8309.Now, the sum of the radii is 3 + 5 = 8, and the difference is 5 - 3 = 2. Since the distance between centers (≈5.8309) is less than the sum (8) and greater than the difference (2), the circles intersect at two distinct points. So, that confirms it.Therefore, the two graphs intersect at two points.**Final Answer**The graphs intersect at boxed{2} points.

💡Okay, so I have this problem about a sequence defined by u₁ = 8 and the relationship u_{n+1} - u_n = 5 + 2(n - 1) for n = 1, 2, 3, and so on. The question is asking me to express u_n as a polynomial in n and find the algebraic sum of its coefficients. The options are from A to E, with numbers 6 to 10. Alright, let's break this down step by step. First, I know that the difference between consecutive terms, u_{n+1} - u_n, is given by 5 + 2(n - 1). Let me simplify that expression to make it easier to work with. So, 5 + 2(n - 1) can be expanded as 5 + 2n - 2, which simplifies to 2n + 3. That means the difference between each term is 2n + 3. Hmm, that looks linear in n, which suggests that the original sequence u_n might be a quadratic polynomial. Because if the difference is linear, the original sequence is quadratic. Let me recall that if the first difference is linear, the sequence itself is quadratic. So, I can assume that u_n is a quadratic polynomial in n. Let me write that as u_n = an² + bn + c, where a, b, and c are constants that I need to find.Now, I have the initial condition u₁ = 8. Plugging n = 1 into the polynomial, I get u₁ = a(1)² + b(1) + c = a + b + c = 8. So, that's my first equation: a + b + c = 8.Next, I need to use the recurrence relation u_{n+1} - u_n = 2n + 3. Let's express u_{n+1} using the polynomial. If u_n = an² + bn + c, then u_{n+1} = a(n+1)² + b(n+1) + c. Let's expand that:u_{n+1} = a(n² + 2n + 1) + b(n + 1) + c = an² + 2an + a + bn + b + c.Now, subtract u_n from u_{n+1}:u_{n+1} - u_n = (an² + 2an + a + bn + b + c) - (an² + bn + c).Simplifying this, the an² terms cancel out, the bn terms cancel out, and the c terms cancel out. So, we're left with 2an + a + b.According to the given recurrence relation, this difference is equal to 2n + 3. So, we have:2an + a + b = 2n + 3.Now, this equation must hold for all n, which means the coefficients of like terms must be equal. So, the coefficient of n on the left side is 2a, and on the right side, it's 2. Therefore, 2a = 2, which implies a = 1.Next, the constant term on the left side is a + b, and on the right side, it's 3. Since we already found that a = 1, plugging that in gives 1 + b = 3. Solving for b, we get b = 2.Now, we have a = 1 and b = 2. Let's go back to our first equation, a + b + c = 8. Plugging in the known values:1 + 2 + c = 8.Simplifying, 3 + c = 8, so c = 5.So, now we have the polynomial u_n = an² + bn + c = 1n² + 2n + 5, which simplifies to u_n = n² + 2n + 5.The question asks for the algebraic sum of its coefficients. The coefficients are 1 (for n²), 2 (for n), and 5 (constant term). Adding them together: 1 + 2 + 5 = 8.Therefore, the algebraic sum of the coefficients is 8, which corresponds to option C.Wait, let me double-check my steps to make sure I didn't make any mistakes. I started by recognizing that the difference is linear, so the sequence should be quadratic. Then, I set up the polynomial and used the initial condition and the recurrence relation to find the coefficients. Each step seems logical, and the arithmetic checks out. So, I think my answer is correct.**Final Answer**boxed{C}

💡Alright, so I've got this problem here about two sequences, {a_n} and {b_n}. Let me try to figure it out step by step.First, part (1) says that {a_n} satisfies S_n = 2a_n - 1 for n in natural numbers. Hmm, I think S_n usually stands for the sum of the first n terms of a sequence. So, if {a_n} is the sequence, then S_n would be a_1 + a_2 + ... + a_n. But here, it's given as S_n = 2a_n - 1. That seems a bit tricky. Maybe I can relate S_n and S_{n-1} to find a recurrence relation for a_n.Let me write that down. For n ≥ 2, S_n = 2a_n - 1 and S_{n-1} = 2a_{n-1} - 1. Then, the nth term a_n is S_n - S_{n-1}, right? So, a_n = (2a_n - 1) - (2a_{n-1} - 1). Simplifying that, a_n = 2a_n - 2a_{n-1} - 1 + 1. Wait, the -1 and +1 cancel out, so a_n = 2a_n - 2a_{n-1}. Let me rearrange that: a_n - 2a_n = -2a_{n-1}, which simplifies to -a_n = -2a_{n-1}, so a_n = 2a_{n-1}. Oh, okay, so the sequence {a_n} is a geometric sequence with common ratio 2. That makes sense. Now, I need to find the first term. For n=1, S_1 = a_1 = 2a_1 - 1. Solving for a_1: a_1 = 2a_1 - 1 ⇒ 0 = a_1 - 1 ⇒ a_1 = 1. So, the first term is 1, and each term is double the previous one. Therefore, a_n = 2^{n-1}.Alright, that takes care of {a_n}. Now, {b_n} is an arithmetic sequence with b_1 = a_1 and b_4 = a_3. Since a_1 is 1, b_1 is also 1. And a_3 is the third term of {a_n}, which is 2^{3-1} = 4. So, b_4 = 4.An arithmetic sequence has a common difference d. The nth term is given by b_n = b_1 + (n-1)d. So, b_4 = b_1 + 3d. We know b_1 is 1 and b_4 is 4, so 4 = 1 + 3d ⇒ 3d = 3 ⇒ d = 1. Therefore, the common difference is 1, and the general term is b_n = 1 + (n-1)*1 = n. So, b_n = n.Alright, part (1) seems done. Now, moving on to part (2). We have c_n defined as 1/a_n - 2/(b_n b_{n+1}). We need to find the sum of the first n terms, T_n.First, let's write down c_n using the expressions we found for a_n and b_n. Since a_n = 2^{n-1}, 1/a_n is 1/(2^{n-1}) = 2^{-(n-1)} = 2^{1 - n}. And b_n = n, so b_{n+1} = n + 1. Therefore, 2/(b_n b_{n+1}) = 2/(n(n + 1)).So, c_n = 2^{1 - n} - 2/(n(n + 1)). Now, to find T_n, the sum of the first n terms of c_n, we can split this into two separate sums:T_n = sum_{k=1}^n [2^{1 - k} - 2/(k(k + 1))] = sum_{k=1}^n 2^{1 - k} - sum_{k=1}^n [2/(k(k + 1))].Let's compute each sum separately.First sum: sum_{k=1}^n 2^{1 - k}. Let's factor out the 2: 2 * sum_{k=1}^n 2^{-k} = 2 * sum_{k=1}^n (1/2)^k. This is a geometric series with first term 1/2 and common ratio 1/2. The sum of the first n terms of a geometric series is a(1 - r^n)/(1 - r). So, sum_{k=1}^n (1/2)^k = (1/2)(1 - (1/2)^n)/(1 - 1/2) = (1/2)(1 - 1/2^n)/(1/2) = 1 - 1/2^n. Therefore, the first sum is 2*(1 - 1/2^n) = 2 - 2/2^n = 2 - 1/2^{n-1}.Wait, hold on, 2*(1 - 1/2^n) is 2 - 2/2^n, which simplifies to 2 - 1/2^{n-1}. Yeah, that's correct.Second sum: sum_{k=1}^n [2/(k(k + 1))]. Let's see, 2/(k(k + 1)) can be written as 2*(1/k - 1/(k + 1)) using partial fractions. So, the sum becomes 2*sum_{k=1}^n [1/k - 1/(k + 1)]. This is a telescoping series. Let's write out the terms:For k=1: 1/1 - 1/2For k=2: 1/2 - 1/3...For k=n: 1/n - 1/(n + 1)When we add all these up, most terms cancel out. The -1/2 cancels with +1/2, -1/3 cancels with +1/3, and so on, up to -1/n cancels with +1/n. What remains is the first term of the first expression, 1/1, and the last term of the last expression, -1/(n + 1). So, the sum is 2*(1 - 1/(n + 1)) = 2*(n/(n + 1)) = 2n/(n + 1).Therefore, putting it all together, T_n = [2 - 1/2^{n-1}] - [2n/(n + 1)]. Wait, no, hold on. The second sum is 2*(1 - 1/(n + 1)) which is 2*(n/(n + 1)) = 2n/(n + 1). But in the expression for T_n, it's subtracting this sum. So, T_n = [2 - 1/2^{n-1}] - [2n/(n + 1)].Wait, but let me double-check. The first sum is 2 - 1/2^{n-1}, and the second sum is 2n/(n + 1). So, T_n = (2 - 1/2^{n-1}) - (2n/(n + 1)). Hmm, that seems a bit messy. Let me see if I can simplify it further.Alternatively, maybe I made a mistake in the signs. Let's go back. The expression for c_n is 2^{1 - n} - 2/(k(k + 1)). So, when we sum c_n from k=1 to n, it's sum 2^{1 - k} - sum 2/(k(k + 1)). So, T_n = sum 2^{1 - k} - sum 2/(k(k + 1)). We computed sum 2^{1 - k} as 2 - 1/2^{n-1}, and sum 2/(k(k + 1)) as 2n/(n + 1). Therefore, T_n = (2 - 1/2^{n-1}) - (2n/(n + 1)).Wait, but that seems like a negative term. Let me check the arithmetic again. The first sum is 2 - 1/2^{n-1}, and the second sum is 2n/(n + 1). So, T_n = (2 - 1/2^{n-1}) - (2n/(n + 1)). To combine these, maybe express them with a common denominator or something.Alternatively, perhaps I can write 2 - 1/2^{n-1} as 2 - 2/2^n, which is the same as 2 - 1/2^{n-1}. So, 2 - 2/2^n is 2 - 1/2^{n-1}. Hmm, not sure if that helps.Wait, another approach: Let's compute T_n as sum_{k=1}^n c_k = sum_{k=1}^n [2^{1 - k} - 2/(k(k + 1))] = sum_{k=1}^n 2^{1 - k} - sum_{k=1}^n 2/(k(k + 1)).We have the first sum as 2 - 1/2^{n-1}, and the second sum as 2n/(n + 1). So, T_n = 2 - 1/2^{n-1} - 2n/(n + 1).Hmm, maybe we can write this as 2 - 2n/(n + 1) - 1/2^{n-1}. Let's compute 2 - 2n/(n + 1). That's equal to (2(n + 1) - 2n)/(n + 1) = (2n + 2 - 2n)/(n + 1) = 2/(n + 1). So, T_n = 2/(n + 1) - 1/2^{n-1}.Wait, that's a much simpler expression! So, T_n = 2/(n + 1) - 1/2^{n-1}. Alternatively, 2/(n + 1) - 2^{1 - n}.Let me verify that again. Starting from T_n = 2 - 1/2^{n-1} - 2n/(n + 1). Then, 2 - 2n/(n + 1) = 2*(n + 1)/(n + 1) - 2n/(n + 1) = (2n + 2 - 2n)/(n + 1) = 2/(n + 1). So, yes, T_n = 2/(n + 1) - 1/2^{n-1}.Alternatively, since 1/2^{n-1} is equal to 2^{1 - n}, we can write T_n = 2/(n + 1) - 2^{1 - n}.That seems to be the simplified form. So, the sum of the first n terms of {c_n} is 2/(n + 1) minus 2^{1 - n}.Let me just recap to make sure I didn't make any mistakes. For part (1), we found that {a_n} is a geometric sequence with a_n = 2^{n-1}, and {b_n} is an arithmetic sequence with b_n = n. For part (2), we expressed c_n in terms of a_n and b_n, then split the sum into two parts, computed each part separately, and combined them to get T_n = 2/(n + 1) - 2^{1 - n}.I think that makes sense. Let me check with n=1 to see if it works. For n=1, c_1 = 1/a_1 - 2/(b_1 b_2) = 1/1 - 2/(1*2) = 1 - 1 = 0. So, T_1 = 0. Plugging into our formula: 2/(1 + 1) - 2^{1 - 1} = 1 - 1 = 0. Correct.For n=2, c_1 + c_2 = 0 + [1/a_2 - 2/(b_2 b_3)] = [1/2 - 2/(2*3)] = 1/2 - 1/3 = 1/6. Our formula: 2/(2 + 1) - 2^{1 - 2} = 2/3 - 1/2 = (4/6 - 3/6) = 1/6. Correct.For n=3, T_3 = 0 + 1/6 + [1/4 - 2/(3*4)] = 1/6 + [1/4 - 1/6] = 1/6 + (3/12 - 2/12) = 1/6 + 1/12 = 1/4. Our formula: 2/(3 + 1) - 2^{1 - 3} = 2/4 - 1/4 = 1/2 - 1/4 = 1/4. Correct.Looks like the formula works for the first few terms. So, I think I did it right.

💡Alright, so I have these two math problems to solve. Let me take them one at a time. The first one is: (1)(√3 + 2√2) - (3√3 + √2). Hmm, okay. I think I need to simplify this expression by combining like terms. First, I'll distribute the negative sign to both terms inside the second parenthesis. So, it becomes √3 + 2√2 - 3√3 - √2. Now, I can group the like terms together. The terms with √3 are √3 and -3√3, and the terms with √2 are 2√2 and -√2.Let me compute each group separately. For the √3 terms: √3 - 3√3 is like 1√3 - 3√3, which equals -2√3. For the √2 terms: 2√2 - √2 is like 2√2 - 1√2, which equals √2. So, putting it all together, the simplified expression is -2√3 + √2. That seems straightforward enough.Now, moving on to the second problem: √2(√2 + 1/√2) - |2 - √6|. Okay, this looks a bit more complex, but I think I can handle it step by step. First, I'll tackle the multiplication part: √2 times (√2 + 1/√2). I'll distribute √2 to both terms inside the parentheses.So, √2 * √2 is √(2*2) which is √4, and that simplifies to 2. Then, √2 * (1/√2) is like √2 divided by √2, which is 1. So, adding those two results together, I get 2 + 1, which is 3.Next, I need to deal with the absolute value part: |2 - √6|. I know that absolute value makes sure the result is positive, regardless of what's inside. So, I need to figure out if 2 - √6 is positive or negative. Since √6 is approximately 2.449, which is greater than 2, 2 - √6 is negative. Therefore, the absolute value of a negative number is its positive counterpart, so |2 - √6| is √6 - 2.Now, putting it all back into the original expression: 3 - (√6 - 2). I need to distribute the negative sign inside the parentheses, so it becomes 3 - √6 + 2. Combining the constants, 3 + 2 is 5, so the expression simplifies to 5 - √6.Let me double-check my steps to make sure I didn't make any mistakes. For the first problem, distributing the negative sign and combining like terms seems correct. For the second problem, multiplying √2 by each term inside the parentheses and then handling the absolute value also seems right. I think I did everything properly.Just to be thorough, I'll approximate the values to see if they make sense. For the first problem, -2√3 + √2 is approximately -2*1.732 + 1.414, which is about -3.464 + 1.414, resulting in approximately -2.05. For the second problem, 5 - √6 is roughly 5 - 2.449, which is about 2.551. These approximations seem reasonable given the original expressions.Overall, I think I've solved both problems correctly by carefully following each step and simplifying appropriately.

💡Okay, so I have this problem about rectangles fitting inside each other. Let me try to understand what it's asking. We have a set S of 2019 rectangles, each with integer side lengths between 1 and 2018. I need to show that there are three rectangles A, B, and C in S such that A fits inside B, and B fits inside C. First, let me make sure I understand what it means for one rectangle to fit inside another. The problem says that a rectangle with sides a and b fits inside another with sides c and d if either (a ≤ c and b ≤ d) or (a ≤ d and b ≤ c). So, it's like rotating the rectangle if needed to see if it can fit. For example, a 1x5 rectangle can fit into a 1x5 rectangle obviously, and also into a 6x2 rectangle because you can rotate the 1x5 to fit within 6x2.So, the problem is about finding a chain of three rectangles where each fits inside the next one. This seems related to ordering or sequences where each element is "larger" in some sense than the previous one. Maybe something to do with ordering the rectangles based on their dimensions.Given that all rectangles have integer sides between 1 and 2018, and there are 2019 rectangles, which is just one more than 2018. That makes me think of the pigeonhole principle because we have more rectangles than the range of side lengths. Maybe I can use that somehow.Let me think about how to model this. Each rectangle can be represented by a pair (a, b) where a and b are integers between 1 and 2018. Since the rectangles can be rotated, it might be useful to always consider them in a standard form, say with a ≤ b. So, for each rectangle, I can represent it as (min(a, b), max(a, b)). That way, I can compare them more easily.So, if I have two rectangles, R1 = (a, b) and R2 = (c, d), with a ≤ b and c ≤ d, then R1 fits inside R2 if a ≤ c and b ≤ d. So, it's like a partial order where one rectangle is "less than or equal to" another if both of its dimensions are less than or equal to the corresponding dimensions of the other rectangle.Now, the problem is asking to find three rectangles A ≤ B ≤ C in this partial order. So, it's about finding a chain of length three. This seems similar to the concept of Dilworth's theorem or the Erdős–Szekeres theorem, which deals with sequences and finding monotonic subsequences.Wait, the Erdős–Szekeres theorem is about sequences of numbers and finding increasing or decreasing subsequences. Maybe I can apply a similar idea here. If I can somehow map the rectangles into a sequence where their dimensions correspond to elements in the sequence, then maybe I can find an increasing subsequence of length three.But how do I map rectangles into a sequence? Maybe I can sort the rectangles in some way and then look for a chain. Alternatively, I can think of each rectangle as a point in a 2D grid where one axis is the smaller side and the other is the larger side. Then, the problem reduces to finding three points where each subsequent point is to the right and above the previous one.This is similar to finding an increasing sequence in two dimensions. There's a theorem called the two-dimensional Erdős–Szekeres theorem, which might be applicable here. It states that for any sequence of points in general position in the plane, there exists a subsequence of a certain length that is increasing in both coordinates.But I'm not sure about the exact statement of that theorem. Maybe I can approach it differently. Let's consider the concept of partial orders and chains. In a partially ordered set (poset), a chain is a totally ordered subset. So, in this case, the set of rectangles with the fitting relation forms a poset, and we need to find a chain of length three.Dilworth's theorem relates the size of the largest antichain to the minimum number of chains needed to cover the poset. But I'm not sure if that's directly applicable here because we're dealing with a specific size of the poset.Alternatively, maybe I can use the pigeonhole principle. Since there are 2019 rectangles and each has dimensions between 1 and 2018, perhaps I can categorize the rectangles based on their smaller side or larger side and find that some category must contain enough rectangles to form a chain.Let me try to formalize this. Let's consider each rectangle as (a, b) with a ≤ b. Now, for each rectangle, let's look at the smaller side a. There are 2018 possible values for a, from 1 to 2018. Since we have 2019 rectangles, by the pigeonhole principle, at least two rectangles must have the same smaller side. But that might not be directly helpful because we need three rectangles in a chain.Wait, maybe I can consider the larger side as well. For each rectangle, we have a pair (a, b). If I fix a, then b can vary from a to 2018. So, for each a, the possible values of b form a chain themselves. If I can find, for some a, multiple b's such that they form an increasing sequence, then I can get a chain of rectangles.But how do I ensure that across different a's, the b's can form a chain? Maybe I need to consider the maximum number of rectangles that can be arranged without forming a chain of length three. If I can show that this maximum is less than 2019, then 2019 rectangles must contain such a chain.This is similar to the concept of the maximum size of a poset without a chain of a certain length. In one dimension, the maximum size without an increasing subsequence of length k is given by the Erdős–Szekeres theorem. In two dimensions, it's more complicated, but perhaps I can use a similar approach.Let me think about it step by step. Suppose I want to avoid having three rectangles where each fits inside the next. How can I arrange the rectangles to prevent this? For each rectangle, I need to ensure that there isn't another rectangle that is both wider and taller. So, I might try to arrange them so that for any rectangle, either the width or the height is not increasing.But with 2019 rectangles, it's challenging to avoid such a chain. Maybe I can model this as a graph where each rectangle is a node, and there's an edge from A to B if A fits inside B. Then, finding a chain of three rectangles is equivalent to finding a path of length two in this graph. So, the problem reduces to showing that this graph must contain a path of length two.But I'm not sure if graph theory is the right approach here. Let me go back to the partial order idea. In a poset, the size of the largest antichain is related to the minimum number of chains needed to cover the poset (Dilworth's theorem). But again, I'm not sure how to apply it directly here.Wait, maybe I can use the concept of layers. If I can partition the set of rectangles into layers such that within each layer, no rectangle fits inside another, then the minimum number of such layers needed is the size of the largest chain. So, if I can show that the number of layers is less than 2019, then by the pigeonhole principle, one layer must contain at least two rectangles, but that doesn't directly help me find a chain of three.Hmm, perhaps I need to think differently. Let's consider the dimensions of the rectangles. Each rectangle has a width and a height, both between 1 and 2018. Let me sort all the rectangles by their width, and then by their height. So, first, sort them by increasing width, and if two have the same width, sort them by increasing height.Now, if I look at the sequence of heights after sorting by width, I can apply the Erdős–Szekeres theorem. The theorem says that any sequence of more than (k-1)(l-1) numbers contains an increasing subsequence of length k or a decreasing subsequence of length l. In our case, we want an increasing subsequence of length 3 in the heights, which would correspond to three rectangles where each has a larger width and height than the previous one, hence fitting inside each other.But wait, the widths are already sorted, so if the heights have an increasing subsequence, then those rectangles would form a chain. So, applying Erdős–Szekeres, if we have more than (2)(2) = 4 rectangles, we can find either an increasing or decreasing subsequence of length 3. But we have 2019 rectangles, which is way more than 4, so we can definitely find an increasing subsequence of length 3.Wait, no, Erdős–Szekeres says that in any sequence of length n, there is an increasing or decreasing subsequence of length at least √n. So, for n = 2019, √2019 is about 44.94, so there must be an increasing or decreasing subsequence of length 45. But we only need a subsequence of length 3, which is much smaller. So, actually, we can guarantee a much longer increasing subsequence than needed.But perhaps I'm overcomplicating it. Since we have 2019 rectangles, and each has a width and height between 1 and 2018, if we sort them by width, the heights must have some monotonic subsequence. Specifically, by Erdős–Szekeres, in a sequence of 2019 numbers, there is an increasing or decreasing subsequence of length at least √2019, which is about 45. So, we can find an increasing subsequence of length 45, which would give us 45 rectangles where each has a larger width and height than the previous one, hence forming a chain of length 45. But we only need a chain of length 3, so this is more than sufficient.Wait, but actually, the problem is not about the length of the subsequence, but about the existence of any chain of length 3. So, even if we have a much longer chain, it's still true that a chain of length 3 exists. Therefore, the conclusion holds.Alternatively, maybe I can approach it by considering the possible pairs of dimensions. Since each rectangle has two dimensions, and both are between 1 and 2018, the total number of possible distinct rectangles is 2018*2019/2 (since we consider a ≤ b). But we have 2019 rectangles, so by the pigeonhole principle, some rectangle must repeat, but that's not necessarily helpful because we need distinct rectangles in the chain.Wait, no, the problem doesn't specify that the rectangles have to be distinct in dimensions, but in the set S, they are distinct rectangles. So, even if two rectangles have the same dimensions, they are still considered different elements of S. So, the pigeonhole principle in terms of dimensions might not directly help.But perhaps I can consider the concept of dominance. A rectangle A dominates rectangle B if A's width and height are both greater than or equal to B's. So, we're looking for three rectangles where each dominates the next. This is similar to finding a chain in the dominance order.In dominance order, the problem reduces to finding a chain of length 3. Now, in such a poset, the maximum size of an antichain is given by the number of elements with a certain property, but I'm not sure.Alternatively, maybe I can use the concept of sequences and apply the Erdős–Szekeres theorem more carefully. Let me think: if I have a sequence of points in the plane, with each point having integer coordinates between 1 and 2018, then by Erdős–Szekeres, there exists a subsequence of length √n + 1 that is either increasing or decreasing in both coordinates. But again, we have n = 2019, so √2019 ≈ 45, so we can find a subsequence of length 45 that is increasing in both coordinates, which would correspond to 45 rectangles where each fits inside the next.But again, we only need three, so this is more than enough. Therefore, such a chain of three rectangles must exist.Wait, but I'm not sure if the Erdős–Szekeres theorem directly applies here because it's usually stated for sequences of numbers, not for points in a grid. However, the idea is similar: in a sufficiently large set, you can find a monotonic subsequence.Alternatively, maybe I can model this as a graph where each rectangle is a node, and there's an edge from A to B if A fits inside B. Then, finding a chain of three rectangles is equivalent to finding a path of length two in this graph. So, the problem reduces to showing that this graph must contain a path of length two.But how do I show that such a path exists? Maybe by considering the number of possible edges. If the graph has enough edges, then it must contain a path of length two. But I'm not sure how to quantify that.Alternatively, maybe I can use the concept of degrees. If each node has a certain number of outgoing edges, then by the pigeonhole principle, some node must have enough edges to form a path of length two. But I'm not sure about the exact calculation.Wait, perhaps I can think about it in terms of the number of possible pairs. For each rectangle, how many other rectangles can it fit into? If a rectangle has dimensions (a, b), then any rectangle with dimensions (c, d) where c ≥ a and d ≥ b (or c ≥ b and d ≥ a) can contain it. So, the number of rectangles that can contain (a, b) is the number of rectangles with dimensions (c, d) where c ≥ a and d ≥ b, considering both orientations.But since we have 2019 rectangles, and each can potentially fit into many others, the graph is likely to have many edges, making it highly connected. Therefore, it's probable that there's a path of length two.But I need a more rigorous argument. Let me try to think about it differently. Suppose I try to avoid having a chain of three rectangles. How would I arrange the rectangles? I would need to ensure that for any two rectangles, there isn't a third one that can contain both. But with 2019 rectangles, it's impossible to arrange them in such a way without having some overlap in their dimensions.Wait, maybe I can use the concept of extremal graph theory. If I consider the graph where edges represent the fitting relation, then the problem is to show that this graph has a path of length two. The number of edges in such a graph would be quite large, given the number of rectangles and their possible dimensions. Therefore, by some theorem, it must contain a path of length two.But I'm not sure about the exact theorem that would apply here. Maybe I need to think about it in terms of Ramsey theory, where a sufficiently large structure must contain a particular substructure. In this case, the substructure is a path of length two.Alternatively, maybe I can use induction. Suppose that for any set of n rectangles, there exists a chain of length three. Then, for n = 2019, it must hold. But I need to set up the base case and the inductive step properly.Wait, actually, the problem is similar to the happy ending problem, which is about finding convex polygons in point sets, but I'm not sure if that's directly applicable here.Let me try to think of it in terms of sequences. If I have 2019 rectangles, each with dimensions (a_i, b_i), sorted by a_i. Then, the sequence of b_i's must have an increasing subsequence of length at least √2019, which is about 45. Therefore, there exists an increasing subsequence of length 45, meaning 45 rectangles where each has a larger a_i and b_i than the previous one. Hence, these 45 rectangles form a chain where each fits inside the next. Therefore, in particular, there exists a chain of length three.But wait, the problem only asks for three rectangles, so even if we have a much longer chain, it's still true that a chain of three exists. Therefore, the conclusion holds.Alternatively, maybe I can use the concept of the pigeonhole principle more directly. Since each rectangle has a width and height between 1 and 2018, and there are 2019 rectangles, perhaps I can categorize them based on their width or height and find that some category must contain enough rectangles to form a chain.For example, if I fix the width, say w, then the heights can vary from w to 2018. If I have enough rectangles with the same width, then their heights must form an increasing sequence, giving me a chain. But since we have 2019 rectangles and 2018 possible widths, by the pigeonhole principle, at least two rectangles must share the same width. But that's not enough to form a chain of three.Wait, but if I consider both width and height, maybe I can find a way to ensure that some combination must repeat or form a chain. For example, if I have 2019 pairs (a, b), with a ≤ b, then the number of distinct pairs is 2018*2019/2, which is much larger than 2019. So, the pigeonhole principle doesn't directly apply here because the number of possible pairs is larger than the number of rectangles.But perhaps I can consider the concept of sequences and monotonicity. If I sort the rectangles by their width, then look at their heights, I can apply the Erdős–Szekeres theorem to find an increasing or decreasing subsequence in the heights. Since we have 2019 rectangles, the theorem guarantees an increasing subsequence of length at least √2019, which is about 45. Therefore, there exists a subsequence of 45 rectangles where each has a larger width and height than the previous one, forming a chain of length 45. Hence, a chain of length three exists.But again, I'm using the Erdős–Szekeres theorem, which might be overkill for this problem. Maybe there's a simpler way to see it.Let me try to think about it in terms of the number of possible chains. If I have 2019 rectangles, and each can fit into some others, the number of possible chains is quite large. Therefore, it's highly probable that a chain of length three exists.Alternatively, maybe I can use the concept of extremal cases. Suppose that no three rectangles form a chain. Then, the set S must be structured in a way that avoids such chains. But with 2019 rectangles, it's impossible to structure them without having some overlap in their dimensions that would allow a chain of three.Wait, perhaps I can think about it in terms of the number of possible pairs. For each rectangle, the number of rectangles it can fit into is quite large, given the range of dimensions. Therefore, the graph of fitting relations is dense enough to ensure that a path of length two exists.But I need a more concrete argument. Let me try to formalize it. Suppose we have 2019 rectangles. For each rectangle, define its width and height as (a_i, b_i) with a_i ≤ b_i. Now, consider the sequence of b_i's sorted by a_i. By the Erdős–Szekeres theorem, this sequence contains an increasing subsequence of length at least √2019, which is about 45. Therefore, there exists a subsequence of 45 rectangles where each has a larger a_i and b_i than the previous one, forming a chain of length 45. Hence, a chain of length three exists.But again, I'm using the Erdős–Szekeres theorem, which might not be necessary here. Maybe I can find a simpler argument.Wait, let's think about it in terms of the number of possible dimensions. Each rectangle has a width and height between 1 and 2018. If I consider the width, there are 2018 possible values. If I have 2019 rectangles, by the pigeonhole principle, at least two rectangles must share the same width. But that's not enough to form a chain of three.Alternatively, if I consider both width and height, there are 2018*2019/2 possible distinct rectangles. Since we have 2019 rectangles, it's possible that all of them are distinct in their dimensions. But even if they are, the problem is about finding a chain, not necessarily about repeating dimensions.Wait, maybe I can think about it in terms of the number of possible chains. If I have 2019 rectangles, and each can fit into some others, the number of possible chains is quite large. Therefore, it's highly probable that a chain of length three exists.Alternatively, maybe I can use the concept of extremal cases. Suppose that no three rectangles form a chain. Then, the set S must be structured in a way that avoids such chains. But with 2019 rectangles, it's impossible to structure them without having some overlap in their dimensions that would allow a chain of three.Wait, perhaps I can think about it in terms of the number of possible pairs. For each rectangle, the number of rectangles it can fit into is quite large, given the range of dimensions. Therefore, the graph of fitting relations is dense enough to ensure that a path of length two exists.But I need a more concrete argument. Let me try to formalize it. Suppose we have 2019 rectangles. For each rectangle, define its width and height as (a_i, b_i) with a_i ≤ b_i. Now, consider the sequence of b_i's sorted by a_i. By the Erdős–Szekeres theorem, this sequence contains an increasing subsequence of length at least √2019, which is about 45. Therefore, there exists a subsequence of 45 rectangles where each has a larger a_i and b_i than the previous one, forming a chain of length 45. Hence, a chain of length three exists.But again, I'm using the Erdős–Szekeres theorem, which might not be necessary here. Maybe I can find a simpler argument.Wait, let me think about it differently. Suppose I have 2019 rectangles. Each rectangle can be represented as a point (a, b) in a grid where 1 ≤ a ≤ b ≤ 2018. Now, if I sort these points by their a-coordinate, and then look at the sequence of b-coordinates, I can apply the Erdős–Szekeres theorem to find an increasing subsequence of length at least √2019, which is about 45. This subsequence corresponds to 45 rectangles where each has a larger a and b than the previous one, forming a chain of length 45. Therefore, a chain of length three exists.But I'm still relying on the Erdős–Szekeres theorem. Maybe I can think about it in terms of the pigeonhole principle and the number of possible chains.Alternatively, perhaps I can use induction. Let's try that. Suppose that for any set of n rectangles with dimensions up to n-1, there exists a chain of length three. Now, consider a set of n rectangles with dimensions up to n. If none of the rectangles have dimension n, then by the inductive hypothesis, there exists a chain of length three. If some rectangles have dimension n, then we can consider those and see if they form a chain or if the remaining rectangles form a chain.But I'm not sure if this inductive step is valid because the dimensions are up to 2018, and we have 2019 rectangles. Maybe I need to adjust the induction hypothesis.Wait, perhaps I can think of it as a two-dimensional grid where one axis is the width and the other is the height. Each rectangle corresponds to a point in this grid. Now, if I have enough points, I can find three that form an increasing sequence in both coordinates.But this is essentially the same as applying the Erdős–Szekeres theorem in two dimensions, which is more complex than the one-dimensional case. However, the theorem does extend to higher dimensions, but the bounds are different.In any case, given that we have 2019 rectangles, which is more than (2)(2) = 4, the Erdős–Szekeres theorem guarantees an increasing or decreasing subsequence of length 3. Therefore, we can find three rectangles where each fits inside the next one.Wait, actually, the Erdős–Szekeres theorem in one dimension says that any sequence of more than (k-1)(l-1) elements contains an increasing subsequence of length k or a decreasing subsequence of length l. So, if we set k = 3 and l = 3, then (k-1)(l-1) = 4. Therefore, any sequence of 5 elements contains an increasing or decreasing subsequence of length 3. Since we have 2019 elements, which is much larger than 5, we can certainly find an increasing subsequence of length 3.Therefore, if we sort the rectangles by their width and look at the sequence of heights, there must be an increasing subsequence of length 3. These three rectangles will form a chain where each fits inside the next one.So, putting it all together, by sorting the rectangles by width and applying the Erdős–Szekeres theorem to the sequence of heights, we can guarantee the existence of three rectangles that form a chain of fitting. Therefore, such three rectangles A, B, and C must exist in the set S.