💡Okay, so I have this problem here: If ( g(x) = g(3 - x) ) for all ( x ), then what line is necessarily an axis of symmetry of the graph of ( y = g(x) )? Hmm, interesting. I need to figure out which line is the axis of symmetry for this function. Alright, let me start by recalling what an axis of symmetry means. If a graph has an axis of symmetry, it means that if you reflect the graph over that line, it will look the same. For example, a parabola has a vertical line of symmetry, and if you reflect it over that line, the two sides mirror each other perfectly.In this case, the function is given by ( g(x) = g(3 - x) ). So, for any input ( x ), the output is the same as the output when the input is ( 3 - x ). That seems like some kind of symmetry. Let me think about what kind of transformation this is.If I take a point ( (x, y) ) on the graph of ( y = g(x) ), then according to the equation ( g(x) = g(3 - x) ), the point ( (3 - x, y) ) should also be on the graph. So, for every point ( (x, y) ), there's a corresponding point ( (3 - x, y) ). What does this transformation look like? If I have a point ( x ), and I map it to ( 3 - x ), that's like reflecting it over some line. Let me see, reflection over a vertical line. Because if I have a vertical line at some point ( a ), then reflecting ( x ) over ( a ) would give me ( 2a - x ). Wait, that's similar to what we have here.In our case, the reflection is ( 3 - x ). So, if I set ( 2a - x = 3 - x ), then ( 2a = 3 ), so ( a = 1.5 ). Therefore, the vertical line ( x = 1.5 ) is the axis of symmetry.Let me double-check that. If I take a point ( x ), its reflection over ( x = 1.5 ) should be ( 3 - x ). Let's test with a specific value. Suppose ( x = 0 ). Then ( 3 - 0 = 3 ). So, the point ( (0, g(0)) ) reflects to ( (3, g(3)) ). Since ( g(0) = g(3) ), that makes sense. Similarly, if ( x = 1 ), then ( 3 - 1 = 2 ). So, ( (1, g(1)) ) reflects to ( (2, g(2)) ), and since ( g(1) = g(2) ), that also holds.What if I pick a point exactly at ( x = 1.5 )? Then ( 3 - 1.5 = 1.5 ), so it reflects to itself. That makes sense because the axis of symmetry should pass through the vertex or the midpoint of the graph.Let me think of another example. Suppose ( x = 2 ). Then ( 3 - 2 = 1 ). So, ( (2, g(2)) ) reflects to ( (1, g(1)) ). Again, since ( g(2) = g(1) ), it works out.So, it seems consistent that reflecting over ( x = 1.5 ) maintains the equality ( g(x) = g(3 - x) ). Therefore, the axis of symmetry must be the vertical line ( x = 1.5 ).I wonder if there's another way to see this. Maybe by rewriting the equation ( g(x) = g(3 - x) ) in terms of transformations. If I let ( h(x) = g(x) ), then the equation becomes ( h(x) = h(3 - x) ). This suggests that the function is symmetric about the line ( x = 1.5 ), as we've already deduced.Alternatively, if I consider the midpoint between ( x ) and ( 3 - x ), it should be the axis of symmetry. The midpoint between ( x ) and ( 3 - x ) is ( frac{x + (3 - x)}{2} = frac{3}{2} = 1.5 ). So, that's another way to confirm that the axis of symmetry is ( x = 1.5 ).Is there a graphical way to visualize this? If I plot ( y = g(x) ), and for every point ( (x, y) ), there's a corresponding point ( (3 - x, y) ), then the graph should be a mirror image across the line ( x = 1.5 ). So, if I were to fold the graph along ( x = 1.5 ), both sides would overlap perfectly.Let me think about a specific function that satisfies ( g(x) = g(3 - x) ). For example, consider ( g(x) = (x - 1.5)^2 ). Then, ( g(3 - x) = (3 - x - 1.5)^2 = (1.5 - x)^2 = (x - 1.5)^2 = g(x) ). So, this function is symmetric about ( x = 1.5 ), as expected.Another example: ( g(x) = |x - 1.5| ). Then, ( g(3 - x) = |3 - x - 1.5| = |1.5 - x| = |x - 1.5| = g(x) ). Again, it's symmetric about ( x = 1.5 ).What if I take a more complicated function, like ( g(x) = sin(pi(x - 1.5)) )? Then, ( g(3 - x) = sin(pi(3 - x - 1.5)) = sin(pi(1.5 - x)) = sin(-pi(x - 1.5)) = -sin(pi(x - 1.5)) ). Wait, that's not equal to ( g(x) ). Hmm, so this function doesn't satisfy ( g(x) = g(3 - x) ). So, not all functions symmetric about ( x = 1.5 ) will satisfy this condition, but those that do will have this property.Wait, actually, in this case, ( sin(pi(x - 1.5)) ) is an odd function about ( x = 1.5 ), meaning it's symmetric about that point but in a rotational sense, not reflectional. So, that's different. So, reflectional symmetry would require ( g(x) = g(3 - x) ), which is a mirror image, not a rotational symmetry.So, to sum up, the condition ( g(x) = g(3 - x) ) implies that for every point ( (x, y) ) on the graph, the point ( (3 - x, y) ) is also on the graph. This reflection across the vertical line ( x = 1.5 ) ensures that the graph is symmetric about that line. Therefore, the axis of symmetry is ( x = 1.5 ).I think I've covered different angles to approach this problem: algebraic manipulation, midpoint calculation, specific examples, and even a graphical interpretation. All of them point to the same conclusion that the axis of symmetry is ( x = 1.5 ). I don't see any flaws in this reasoning, so I'm pretty confident about the answer.**Final Answer**The line necessarily an axis of symmetry of the graph of ( y = g(x) ) is boxed{x = 1.5}.

💡First, I notice that the denominator is a difference of squares, which can be factored using the formula ( a^2 - b^2 = (a - b)(a + b) ).Applying this formula to ( 262^2 - 258^2 ), we get:[262^2 - 258^2 = (262 - 258)(262 + 258) = 4 times 520 = 2080]Next, I calculate the numerator ( 900^2 ):[900^2 = 810000]Now, I divide the numerator by the denominator:[frac{810000}{2080} = 389.42]Rounding to one decimal place, the final value is:[389.4]

💡Okay, so I have this geometry problem here about triangle ABC. The midpoints of sides AB and AC are D and E, respectively. I need to prove that the bisectors of angles BDE and CED intersect at side BC if the length of BC is the arithmetic mean of AB and AC. Hmm, okay, let me try to visualize this.First, let me draw triangle ABC. Let me label the vertices A, B, and C. Now, D is the midpoint of AB, so I'll mark that point halfway between A and B. Similarly, E is the midpoint of AC, so I'll mark that point halfway between A and C. So, now I have points D and E as midpoints.Now, I need to consider the angles at points D and E. Specifically, angle BDE and angle CED. So, angle BDE is the angle at point D between points B and E, and angle CED is the angle at point E between points C and D. I need to find the bisectors of these two angles and show that they intersect on side BC.Given that BC is the arithmetic mean of AB and AC, that means BC = (AB + AC)/2. That's an important condition. I wonder how that plays into the problem.Maybe I should recall some properties of midpoints and angle bisectors. Since D and E are midpoints, DE is the midline of triangle ABC, which means DE is parallel to BC and half its length. So, DE || BC and DE = (1/2)BC. That's a useful piece of information.Since DE is parallel to BC, maybe there are some similar triangles involved here. Let me think. If DE is parallel to BC, then triangle ADE is similar to triangle ABC by the Basic Proportionality Theorem, also known as Thales' theorem. So, triangle ADE ~ triangle ABC with a ratio of 1:2.Now, considering the angles at D and E. Since DE is parallel to BC, the angles formed by transversals with these lines should be equal. So, angle BDE should be equal to angle ABC, and angle CED should be equal to angle ACB. Wait, is that correct?Let me double-check. If DE is parallel to BC, then the corresponding angles should be equal. So, angle BDE corresponds to angle ABC, and angle CED corresponds to angle ACB. So, yes, angle BDE = angle ABC and angle CED = angle ACB.Therefore, the bisectors of angles BDE and CED are essentially bisecting angles that are equal to angles ABC and ACB, respectively. So, maybe the bisectors of BDE and CED are related to the bisectors of ABC and ACB?But I need to show that these bisectors intersect on BC. Hmm. Maybe I can use the Angle Bisector Theorem here. The Angle Bisector Theorem states that the bisector of an angle in a triangle divides the opposite side into segments proportional to the adjacent sides.But in this case, the bisectors are not of the angles at B and C, but rather at D and E. So, perhaps I need to apply the Angle Bisector Theorem in triangles BDE and CED.Let me consider triangle BDE first. The bisector of angle BDE will divide side BE into segments proportional to BD and DE. Similarly, in triangle CED, the bisector of angle CED will divide side CD into segments proportional to CE and DE.Wait, but BD and CE are both midpoints, so BD = AB/2 and CE = AC/2. Also, DE = BC/2, as DE is the midline.Given that BC = (AB + AC)/2, so BC is the average of AB and AC. Let me denote AB = c, AC = b, and BC = a. Then, according to the given condition, a = (b + c)/2.So, BC is the arithmetic mean of AB and AC. That might be useful in setting up proportions.Let me try to apply the Angle Bisector Theorem to triangle BDE. The bisector of angle BDE will meet BE at some point, say F. Then, BF/FE = BD/DE.Similarly, in triangle CED, the bisector of angle CED will meet CD at some point, say G. Then, CG/GD = CE/DE.Given that BD = c/2, CE = b/2, and DE = a/2.So, BF/FE = (c/2)/(a/2) = c/a.Similarly, CG/GD = (b/2)/(a/2) = b/a.But since a = (b + c)/2, we can substitute a in terms of b and c.So, BF/FE = c/( (b + c)/2 ) = 2c/(b + c).Similarly, CG/GD = b/( (b + c)/2 ) = 2b/(b + c).Hmm, interesting. So, the ratios BF/FE and CG/GD are 2c/(b + c) and 2b/(b + c), respectively.Now, I need to see if these bisectors intersect on BC. Let me denote the intersection point as P. So, P lies on BC, and it's the intersection of the bisectors of angles BDE and CED.If I can show that P divides BC in a certain ratio, maybe I can confirm that it lies on BC.Alternatively, perhaps I can use coordinate geometry to solve this problem. Let me assign coordinates to the triangle.Let me place point A at (0, 0), point B at (2c, 0), and point C at (2b, 2d), just to make the midpoints easier. Wait, but since D and E are midpoints, maybe it's better to assign coordinates such that D and E have integer coordinates.Alternatively, let me place A at (0, 0), B at (2, 0), and C at (0, 2), so that D is at (1, 0) and E is at (0, 1). Then, DE would be the line from (1, 0) to (0, 1), which has a slope of -1.But in this case, BC would be from (2, 0) to (0, 2), which has a length of sqrt((2)^2 + (2)^2) = sqrt(8). But AB is 2, AC is 2, so BC should be (2 + 2)/2 = 2. But in this case, BC is sqrt(8), which is not equal to 2. So, this coordinate system doesn't satisfy the given condition that BC is the arithmetic mean of AB and AC.Hmm, maybe I need to choose coordinates where BC is indeed the arithmetic mean of AB and AC.Let me try again. Let me let AB = c, AC = b, and BC = (b + c)/2. Let me place point A at (0, 0), point B at (c, 0), and point C somewhere in the plane. Then, D is the midpoint of AB, so D is at (c/2, 0), and E is the midpoint of AC, so E is at (b/2, d/2), where C is at (b, d).Wait, maybe it's getting too complicated. Perhaps I should use vectors or coordinate geometry more carefully.Alternatively, maybe I can use mass point geometry. Since D and E are midpoints, the masses can be assigned accordingly.But perhaps another approach is to use Ceva's Theorem. Ceva's Theorem states that for concurrent cevians in a triangle, the product of certain ratios equals 1.But in this case, the cevians are the angle bisectors of BDE and CED. Hmm, not sure if Ceva's Theorem applies directly here.Wait, maybe I can consider triangle BDE and triangle CED separately and see where their angle bisectors meet.Alternatively, since DE is parallel to BC, maybe there's a homothety or similarity transformation that can relate these triangles.Wait, if DE is parallel to BC, then triangles BDE and BAC are similar, as well as triangles CED and CAB. Hmm, not sure.Wait, actually, since DE is the midline, triangle ADE is similar to triangle ABC with a ratio of 1:2. So, angles at D and E are similar to angles at B and C.Therefore, angle BDE is equal to angle ABC, and angle CED is equal to angle ACB.So, the bisectors of angles BDE and CED are essentially bisecting angles equal to angles ABC and ACB.Therefore, maybe the intersection point P of these bisectors lies on BC because of some proportionalities.Alternatively, perhaps I can use trigonometric properties. Let me denote the angles.Let me denote angle ABC as β and angle ACB as γ. Then, angle BDE = β and angle CED = γ.So, the bisector of angle BDE will split β into two angles of β/2 each, and the bisector of angle CED will split γ into two angles of γ/2 each.Now, if I can find the point P on BC such that the bisectors from D and E meet at P, then perhaps I can use the Angle Bisector Theorem in triangles BDE and CED.Wait, in triangle BDE, the bisector of angle BDE (which is β) will divide side BE into segments proportional to BD and DE.Similarly, in triangle CED, the bisector of angle CED (which is γ) will divide side CD into segments proportional to CE and DE.Given that BD = AB/2, CE = AC/2, and DE = BC/2, and BC = (AB + AC)/2.Let me denote AB = c, AC = b, so BC = (b + c)/2.Then, BD = c/2, CE = b/2, DE = (b + c)/4.Wait, no, DE is the midline, so DE = BC/2. Since BC = (b + c)/2, then DE = (b + c)/4.Wait, that doesn't seem right. If DE is the midline, it should be parallel to BC and half its length. So, if BC = (b + c)/2, then DE = (b + c)/4.Wait, but actually, DE is the midline, so DE = BC/2 regardless of the length of BC. So, if BC = (b + c)/2, then DE = (b + c)/4.But in triangle ABC, DE is the midline, so DE = BC/2. Therefore, DE = (b + c)/4.Wait, but in triangle BDE, BD = c/2, DE = (b + c)/4, and BE is the distance from B to E.Wait, E is the midpoint of AC, so coordinates might help here.Let me assign coordinates to the triangle. Let me place point A at (0, 0), point B at (c, 0), and point C at (0, b). Then, D is the midpoint of AB, so D is at (c/2, 0), and E is the midpoint of AC, so E is at (0, b/2).Then, DE is the line connecting (c/2, 0) to (0, b/2). The equation of DE can be found.Similarly, BC is the line from (c, 0) to (0, b). The equation of BC is y = (-b/c)x + b.Now, the bisector of angle BDE. Point D is at (c/2, 0). Angle BDE is the angle at D between points B and E.So, the bisector of angle BDE will be a line starting at D and going towards some point P on BC.Similarly, the bisector of angle CED is the angle at E between points C and D, so it's a line starting at E and going towards point P on BC.So, if I can find the equations of these bisectors and show that they intersect on BC, that would solve the problem.First, let me find the coordinates of points:A: (0, 0)B: (c, 0)C: (0, b)D: (c/2, 0)E: (0, b/2)Now, let me find the vectors for angle BDE.At point D: vectors DB and DE.Vector DB is from D to B: (c - c/2, 0 - 0) = (c/2, 0)Vector DE is from D to E: (0 - c/2, b/2 - 0) = (-c/2, b/2)Similarly, at point E: vectors EC and ED.Vector EC is from E to C: (0 - 0, b - b/2) = (0, b/2)Vector ED is from E to D: (c/2 - 0, 0 - b/2) = (c/2, -b/2)Now, to find the angle bisectors at D and E, I can use the direction vectors of the bisectors.The angle bisector at D will be a vector that is a unit vector in the direction of the sum of the unit vectors of DB and DE.Similarly, the angle bisector at E will be a vector that is a unit vector in the direction of the sum of the unit vectors of EC and ED.First, let's compute the unit vectors.At point D:Vector DB: (c/2, 0), length = c/2Unit vector: (1, 0)Vector DE: (-c/2, b/2), length = sqrt( (c/2)^2 + (b/2)^2 ) = (1/2)sqrt(c² + b²)Unit vector: (-c/sqrt(c² + b²), b/sqrt(c² + b²))So, the sum of the unit vectors is:(1 - c/sqrt(c² + b²), 0 + b/sqrt(c² + b²)) = (1 - c/sqrt(c² + b²), b/sqrt(c² + b²))This is the direction vector of the bisector at D.Similarly, at point E:Vector EC: (0, b/2), length = b/2Unit vector: (0, 1)Vector ED: (c/2, -b/2), length = sqrt( (c/2)^2 + (b/2)^2 ) = (1/2)sqrt(c² + b²)Unit vector: (c/sqrt(c² + b²), -b/sqrt(c² + b²))Sum of unit vectors:(0 + c/sqrt(c² + b²), 1 - b/sqrt(c² + b²)) = (c/sqrt(c² + b²), 1 - b/sqrt(c² + b²))This is the direction vector of the bisector at E.Now, the bisector at D passes through D (c/2, 0) and has direction vector (1 - c/sqrt(c² + b²), b/sqrt(c² + b²)).Similarly, the bisector at E passes through E (0, b/2) and has direction vector (c/sqrt(c² + b²), 1 - b/sqrt(c² + b²)).Now, I need to find the equations of these two lines and see if they intersect on BC.Let me denote sqrt(c² + b²) as k for simplicity.So, direction vector at D: (1 - c/k, b/k)Direction vector at E: (c/k, 1 - b/k)Now, the parametric equation of the bisector from D is:x = c/2 + t(1 - c/k)y = 0 + t(b/k)Similarly, the parametric equation of the bisector from E is:x = 0 + s(c/k)y = b/2 + s(1 - b/k)Now, to find the intersection point P, we need to solve for t and s such that:c/2 + t(1 - c/k) = s(c/k)andt(b/k) = b/2 + s(1 - b/k)Let me write these equations:1) c/2 + t(1 - c/k) = s(c/k)2) t(b/k) = b/2 + s(1 - b/k)Let me solve equation 2 for t:t(b/k) - s(1 - b/k) = b/2Let me express t from equation 1:From equation 1:t(1 - c/k) = s(c/k) - c/2So,t = [s(c/k) - c/2] / (1 - c/k)Now, substitute t into equation 2:[ [s(c/k) - c/2] / (1 - c/k) ] * (b/k) - s(1 - b/k) = b/2This seems complicated, but let me try to simplify.Let me denote m = c/k and n = b/k, so m² + n² = 1 because k = sqrt(c² + b²).So, m = c/k, n = b/k, m² + n² = 1.Then, equation 1 becomes:c/2 + t(1 - m) = s mEquation 2 becomes:t n = b/2 + s(1 - n)But since b = n k, so b/2 = (n k)/2.Similarly, c = m k.So, equation 1:(m k)/2 + t(1 - m) = s mEquation 2:t n = (n k)/2 + s(1 - n)Let me express equation 1:t(1 - m) = s m - (m k)/2So,t = [s m - (m k)/2] / (1 - m)Similarly, equation 2:t n = (n k)/2 + s(1 - n)Substitute t from equation 1:[ (s m - (m k)/2 ) / (1 - m) ] * n = (n k)/2 + s(1 - n)Multiply both sides by (1 - m):(s m - (m k)/2 ) * n = [ (n k)/2 + s(1 - n) ] * (1 - m)Let me expand both sides:Left side: s m n - (m n k)/2Right side: (n k)/2 (1 - m) + s(1 - n)(1 - m)Now, let me collect terms with s on the left and constants on the right:s m n - s(1 - n)(1 - m) = (n k)/2 (1 - m) + (m n k)/2Factor s:s [ m n - (1 - n)(1 - m) ] = (n k)/2 [ (1 - m) + m ]Simplify the right side:(1 - m) + m = 1, so right side becomes (n k)/2Now, the left side:m n - (1 - n)(1 - m) = m n - [1 - m - n + m n] = m n - 1 + m + n - m n = m + n - 1So, equation becomes:s (m + n - 1) = (n k)/2Therefore,s = (n k)/[2(m + n - 1)]Now, recall that m² + n² = 1, so m + n = ?Wait, but m + n is not necessarily equal to anything specific. Hmm.But we have the condition that BC = (AB + AC)/2.Given that AB = c, AC = b, and BC = (b + c)/2.But in our coordinate system, AB is from (0,0) to (c,0), so AB = c.AC is from (0,0) to (0,b), so AC = b.BC is from (c,0) to (0,b), so BC = sqrt(c² + b²) = k.Given that BC = (AB + AC)/2, so k = (c + b)/2.Therefore, k = (b + c)/2.So, we have k = (b + c)/2, and m = c/k = 2c/(b + c), n = b/k = 2b/(b + c).So, m = 2c/(b + c), n = 2b/(b + c).So, m + n = 2c/(b + c) + 2b/(b + c) = 2(b + c)/(b + c) = 2.Therefore, m + n = 2.So, going back to the expression for s:s = (n k)/[2(m + n - 1)] = (n k)/[2(2 - 1)] = (n k)/2But n = 2b/(b + c), and k = (b + c)/2.So, n k = (2b/(b + c)) * ( (b + c)/2 ) = b.Therefore, s = b / 2.So, s = b/2.Now, let's find t from equation 1:t = [s m - (m k)/2 ] / (1 - m )We have s = b/2, m = 2c/(b + c), k = (b + c)/2.So,s m = (b/2) * (2c/(b + c)) = (b c)/(b + c)m k /2 = (2c/(b + c)) * ( (b + c)/2 ) / 2 = (2c/(b + c)) * ( (b + c)/2 ) /2 = (c)/2So,t = [ (b c)/(b + c) - c/2 ] / (1 - 2c/(b + c))Simplify numerator:(b c)/(b + c) - c/2 = c [ b/(b + c) - 1/2 ] = c [ (2b - (b + c)) / (2(b + c)) ] = c [ (b - c) / (2(b + c)) ] = c(b - c)/(2(b + c))Denominator:1 - 2c/(b + c) = (b + c - 2c)/(b + c) = (b - c)/(b + c)So,t = [ c(b - c)/(2(b + c)) ] / [ (b - c)/(b + c) ] = [ c(b - c)/(2(b + c)) ] * [ (b + c)/(b - c) ] = c/2So, t = c/2.Now, let's find the coordinates of point P.From the bisector at D:x = c/2 + t(1 - m) = c/2 + (c/2)(1 - 2c/(b + c)) = c/2 + (c/2)( (b + c - 2c)/(b + c) ) = c/2 + (c/2)( (b - c)/(b + c) )Similarly,y = t(b/k) = (c/2)( b / ( (b + c)/2 ) ) = (c/2)( 2b/(b + c) ) = (c b)/(b + c)Let me compute x:x = c/2 + (c/2)( (b - c)/(b + c) ) = (c/2)[ 1 + (b - c)/(b + c) ] = (c/2)[ (b + c + b - c)/(b + c) ] = (c/2)( 2b/(b + c) ) = (c b)/(b + c)So, point P has coordinates ( (c b)/(b + c), (c b)/(b + c) )Wait, that's interesting. Both x and y coordinates are equal, so P lies on the line y = x.But in our coordinate system, BC is the line from (c, 0) to (0, b). The equation of BC is y = (-b/c)x + b.So, does point P lie on BC?Let me check if y = (-b/c)x + b holds for P.Given that x = (c b)/(b + c), y = (c b)/(b + c).So,y = (-b/c)x + bSubstitute x:y = (-b/c)(c b/(b + c)) + b = (-b²/(b + c)) + b = (-b² + b(b + c))/(b + c) = (-b² + b² + b c)/(b + c) = (b c)/(b + c)Which is equal to y. So, yes, point P lies on BC.Therefore, the bisectors of angles BDE and CED intersect at point P on BC.Hence, we've proved that under the condition that BC is the arithmetic mean of AB and AC, the bisectors of angles BDE and CED intersect at side BC.**Final Answer**boxed{BC}

💡First, I need to determine the length of the train that crosses a pole in 18 seconds while traveling at a speed of 60 km/hr.To do this, I'll convert the speed from kilometers per hour to meters per second. This conversion is necessary because the time is given in seconds, and I want the length in meters.The conversion formula is:[ text{Speed in m/s} = text{Speed in km/hr} times frac{1000 text{ m}}{1 text{ km}} times frac{1 text{ hr}}{3600 text{ s}} ]Plugging in the values:[ text{Speed in m/s} = 60 times frac{1000}{3600} = 60 times frac{5}{18} approx 16.67 text{ m/s} ]Now, using the formula for distance:[ text{Distance} = text{Speed} times text{Time} ][ text{Distance} = 16.67 text{ m/s} times 18 text{ s} approx 300.06 text{ meters} ]Therefore, the length of the train is approximately 300.06 meters.

💡Okay, so I have this problem where I need to cut a 5x5 square into 1x3 and 1x4 rectangles. Hmm, let me think about how to approach this. First, I should probably figure out the total area of the square and see if it's possible to divide it into these smaller rectangles.The area of the 5x5 square is 25. Each 1x3 rectangle has an area of 3, and each 1x4 rectangle has an area of 4. So, I need to find how many of each rectangle I can fit into the square such that their total area adds up to 25.Let me denote the number of 1x3 rectangles as x and the number of 1x4 rectangles as y. Then, the equation would be:3x + 4y = 25I need to find integer solutions for x and y. Let's try different values for y and see if x comes out as an integer.If y = 1:3x + 4(1) = 253x + 4 = 253x = 21x = 7Okay, so x = 7 and y = 1 is a solution. That means I can have 7 rectangles of 1x3 and 1 rectangle of 1x4. Let me check if this works in terms of fitting into the 5x5 grid.Each 1x3 rectangle occupies 3 units in one direction and 1 unit in the other. Similarly, the 1x4 rectangle occupies 4 units in one direction and 1 unit in the other. Since the square is 5x5, I need to arrange these rectangles without overlapping and covering the entire area.Let me try to visualize the grid. Maybe I can place the 1x4 rectangle horizontally in the first row, covering columns 1 to 4. Then, in the remaining column 5, I can place 1x3 rectangles vertically. But wait, column 5 is only 5 units tall, and each 1x3 rectangle is 3 units tall. If I place one 1x3 rectangle vertically in column 5, it will cover rows 1 to 3, leaving rows 4 and 5. That's only 2 units, which isn't enough for another 1x3 rectangle. Hmm, that doesn't seem to work.Maybe I should try placing the 1x4 rectangle vertically instead. If I place it in column 1, covering rows 1 to 4, then I have row 5 in column 1 left. But that's only 1 unit, which doesn't fit either rectangle. So that might not work either.Perhaps I need to mix horizontal and vertical placements. Let me try placing the 1x4 rectangle horizontally in the first row, covering columns 1 to 4. Then, in column 5, I can place a 1x3 rectangle vertically from row 1 to row 3. That leaves rows 4 and 5 in column 5, which is 2 units. I still can't fit another 1x3 or 1x4 rectangle there.Wait, maybe I can rearrange the rectangles. What if I place the 1x4 rectangle in the first row, columns 2 to 5? Then, column 1 would have 5 units, which could be covered by one 1x3 rectangle from row 1 to row 3 and another 1x3 rectangle from row 4 to row 5. But that would require two 1x3 rectangles in column 1, which is fine. Then, in the remaining columns 2 to 5, I have a 1x4 rectangle in the first row, and the rest of the area needs to be covered by 1x3 rectangles.Let me see. After placing the 1x4 rectangle in row 1, columns 2-5, and two 1x3 rectangles in column 1, rows 1-3 and 4-5, I have the area from rows 2-5 in columns 2-5 left. That's a 4x4 area. But 4x4 is 16 units, and each 1x3 rectangle covers 3 units. 16 isn't divisible by 3, so that doesn't work. Hmm.Maybe I need to use a different combination. Let me try placing the 1x4 rectangle vertically in column 1, covering rows 1-4. Then, in row 5, column 1, I have 1 unit left, which doesn't fit any rectangle. So that's not helpful.What if I place the 1x4 rectangle vertically in column 5, covering rows 1-4? Then, row 5, column 5 is left, which is 1 unit, again not useful.Perhaps I need to use multiple 1x4 rectangles? Wait, the equation only allows for one 1x4 rectangle because y=1. So I can't use more than one.Maybe I need to think differently. Instead of trying to place the 1x4 rectangle first, I can try arranging the 1x3 rectangles and see where the 1x4 can fit.If I place four 1x3 rectangles horizontally in rows 1-4, each covering columns 1-3, that would leave columns 4-5 in each row. But each row would have 2 units left, which isn't enough for a 1x3 or 1x4 rectangle.Alternatively, if I place 1x3 rectangles vertically in columns 1-3, covering rows 1-3, that leaves rows 4-5 in columns 1-3. That's 2 units in height, which isn't enough for another 1x3 rectangle. Hmm.Wait, maybe I can combine horizontal and vertical placements. Let me try placing some 1x3 rectangles horizontally and some vertically.Suppose I place a 1x3 rectangle horizontally in row 1, columns 1-3. Then, in row 1, columns 4-5, I have 2 units left. That's not enough for a 1x3 or 1x4 rectangle.Alternatively, place a 1x3 rectangle vertically in column 1, rows 1-3. Then, in column 1, rows 4-5, I have 2 units left. Again, not enough.This is tricky. Maybe I need to use the 1x4 rectangle in a way that helps fill the gaps.Let me try placing the 1x4 rectangle horizontally in row 1, columns 1-4. Then, in column 5, rows 1-3, I can place a 1x3 rectangle vertically. That leaves column 5, rows 4-5, which is 2 units. I can't fit a 1x3 or 1x4 there.Wait, what if I place the 1x4 rectangle vertically in column 5, rows 1-4. Then, in column 5, row 5, I have 1 unit left. Not helpful.Alternatively, place the 1x4 rectangle horizontally in row 5, columns 1-4. Then, in column 5, rows 1-4, I can place a 1x3 rectangle vertically, covering rows 1-3. That leaves row 4, column 5, which is 1 unit. Still not helpful.Hmm, maybe I need to use the 1x4 rectangle in a different position. Let me try placing it in the middle somewhere.Suppose I place the 1x4 rectangle horizontally in row 3, columns 1-4. Then, in row 3, column 5, I have 1 unit left. Not useful.Alternatively, place it vertically in column 3, rows 1-4. Then, in column 3, row 5, I have 1 unit left.This isn't working. Maybe I need to adjust the number of 1x3 and 1x4 rectangles. Wait, the equation only gives x=7 and y=1 as a solution. Are there other solutions?Let me check. If y=2:3x + 4(2) = 253x + 8 = 253x = 17x ≈ 5.666Not an integer. y=3:3x + 12 = 253x = 13x ≈ 4.333Nope. y=4:3x + 16 = 253x = 9x=3So, x=3 and y=4 is another solution. Let me see if that works.So, 3 rectangles of 1x3 and 4 rectangles of 1x4. Total area: 3*3 + 4*4 = 9 + 16 = 25. Perfect.Now, can I arrange 3 1x3 and 4 1x4 rectangles in a 5x5 grid?Let me try. Maybe place the 1x4 rectangles horizontally in rows 1-4, each covering columns 1-4. Then, in column 5, rows 1-4, I can place 1x3 rectangles vertically, but that would require 3 units, leaving row 5, column 5 empty. That's 1 unit, which isn't enough.Alternatively, place the 1x4 rectangles vertically in columns 1-4, each covering rows 1-4. Then, in column 5, rows 1-4, I can place 1x3 rectangles vertically, but again, row 5, column 5 is left.Wait, maybe I need to mix horizontal and vertical placements.Let me try placing two 1x4 rectangles horizontally in rows 1 and 2, covering columns 1-4. Then, in rows 1 and 2, column 5, I have 2 units left. That's not enough for a 1x3 or 1x4.Alternatively, place two 1x4 rectangles vertically in columns 1 and 2, covering rows 1-4. Then, in columns 1 and 2, row 5, I have 2 units left. Not helpful.This is getting complicated. Maybe I need to look for a different approach.I recall that sometimes, tiling problems can be approached by coloring the grid in a way that makes it impossible to tile with certain rectangles. Maybe I can use a checkerboard pattern or something similar.Let me try coloring the 5x5 grid in three colors in a repeating pattern. For example, color the grid with colors A, B, C, A, B in each row, shifting each row by one color. This way, each 1x3 rectangle will cover one of each color, and each 1x4 rectangle will cover two of one color and one of another.But I'm not sure if this will help. Maybe I need to think about the dimensions.The 5x5 grid has both dimensions as 5, which is not a multiple of 3 or 4. So, arranging the rectangles might be challenging.Wait, maybe I can divide the grid into regions that can be covered by the rectangles. For example, if I divide the grid into a 3x5 area and a 2x5 area, but 2x5 can't be covered by 1x3 or 1x4 rectangles.Alternatively, divide it into a 4x5 area and a 1x5 area. The 1x5 area can't be covered by 1x3 or 1x4.Hmm.Maybe I need to use a combination of horizontal and vertical rectangles to fill the space.Let me try placing a 1x4 rectangle horizontally in row 1, columns 1-4. Then, in row 1, column 5, I have 1 unit left. I can't place anything there.Alternatively, place a 1x3 rectangle vertically in column 1, rows 1-3. Then, in column 1, rows 4-5, I have 2 units left. Not helpful.Wait, what if I place a 1x4 rectangle vertically in column 1, covering rows 1-4. Then, in column 1, row 5, I have 1 unit left. Not useful.This is frustrating. Maybe I need to try a different solution. Let me go back to the equation.We have two solutions: x=7, y=1 and x=3, y=4.I tried x=7, y=1 and couldn't find a way to fit them. Maybe x=3, y=4 is the way to go.Let me try arranging 3 1x3 and 4 1x4 rectangles.Suppose I place two 1x4 rectangles horizontally in rows 1 and 2, covering columns 1-4. Then, in rows 1 and 2, column 5, I have 2 units left. I can't place a 1x3 or 1x4 there.Alternatively, place two 1x4 rectangles vertically in columns 1 and 2, covering rows 1-4. Then, in columns 1 and 2, row 5, I have 2 units left. Still not helpful.Wait, maybe I can place the 1x4 rectangles in a way that they overlap in different columns.Let me try placing a 1x4 rectangle horizontally in row 1, columns 1-4. Then, place another 1x4 rectangle horizontally in row 2, columns 2-5. This way, they overlap in column 2-4 of row 2. But this creates a 2x4 area, which might not be helpful.Alternatively, place a 1x4 rectangle horizontally in row 1, columns 1-4, and another 1x4 rectangle vertically in column 5, rows 1-4. Then, in row 5, columns 1-4, I have 4 units left, which can be covered by a 1x4 rectangle. But that would require a third 1x4 rectangle, which we have (y=4). Then, in column 5, row 5, I have 1 unit left, which is still a problem.Wait, but we have y=4, so we can place another 1x4 rectangle in row 5, columns 1-4. Then, column 5, row 5 is still 1 unit. Hmm.Alternatively, maybe I can place the 1x4 rectangles in a way that they cover the problematic areas.Let me try placing a 1x4 rectangle vertically in column 5, rows 1-4. Then, in column 5, row 5, I have 1 unit left. I can't do anything about that.Alternatively, place a 1x4 rectangle horizontally in row 5, columns 1-4. Then, column 5, rows 1-4 can be covered by a 1x3 rectangle vertically, covering rows 1-3, leaving row 4, column 5, which is 1 unit.This seems to be a recurring issue. Maybe I need to use the 1x3 rectangles to cover the gaps.Wait, if I place a 1x3 rectangle vertically in column 5, covering rows 1-3, then I have rows 4-5 left in column 5. That's 2 units. I can't place a 1x3 or 1x4 there.Alternatively, place a 1x3 rectangle horizontally in row 5, columns 1-3. Then, in row 5, columns 4-5, I have 2 units left. Not helpful.This is really challenging. Maybe I need to think outside the box. Perhaps the rectangles can be placed in a way that they overlap in different rows and columns, but without overlapping each other.Wait, maybe I can place the 1x4 rectangles in a staggered manner.Let me try placing a 1x4 rectangle horizontally in row 1, columns 1-4. Then, place another 1x4 rectangle horizontally in row 2, columns 2-5. This covers columns 1-5 in rows 1-2, except for column 1 in row 2 and column 5 in row 1. But those are single units, which I can't cover.Alternatively, place a 1x4 rectangle horizontally in row 1, columns 1-4, and another 1x4 rectangle vertically in column 5, rows 1-4. Then, in row 5, columns 1-4, I can place another 1x4 rectangle. That leaves column 5, row 5, which is 1 unit.But we have y=4, so we can place another 1x4 rectangle in row 5, columns 1-4. Then, column 5, row 5 is still 1 unit. I'm stuck again.Wait, maybe I need to use the 1x3 rectangles to cover some of these gaps.If I place a 1x3 rectangle vertically in column 5, covering rows 1-3, then I have rows 4-5 left in column 5. That's 2 units. I can't place a 1x3 or 1x4 there.Alternatively, place a 1x3 rectangle horizontally in row 5, columns 1-3. Then, in row 5, columns 4-5, I have 2 units left. Not helpful.This is really frustrating. Maybe I need to try a different approach altogether.Let me try dividing the 5x5 grid into smaller sections that can be covered by the rectangles.For example, if I divide the grid into a 3x5 area and a 2x5 area. The 3x5 area can be covered by 5 1x3 rectangles placed horizontally. Then, the 2x5 area can be covered by... Wait, 2x5 is 10 units. Each 1x4 rectangle covers 4 units, so 10 isn't divisible by 4. That doesn't work.Alternatively, divide the grid into a 4x5 area and a 1x5 area. The 4x5 area can be covered by 5 1x4 rectangles placed horizontally. Then, the 1x5 area can't be covered by 1x3 or 1x4.Hmm.Wait, maybe I can divide the grid into a 3x3 area and a 2x3 area, but that's not helpful either.Alternatively, think of the grid as a combination of 1x3 and 1x4 rectangles.Wait, maybe I can place the 1x4 rectangle in a corner and then fill the rest with 1x3 rectangles.Let me try placing the 1x4 rectangle vertically in column 1, rows 1-4. Then, in column 1, row 5, I have 1 unit left. I can't do anything about that.Alternatively, place the 1x4 rectangle horizontally in row 1, columns 1-4. Then, in row 1, column 5, I have 1 unit left. Not helpful.Wait, maybe I can place the 1x4 rectangle in such a way that it covers parts of multiple rows and columns.For example, place a 1x4 rectangle starting at row 1, column 1, going down to row 4, column 1. Then, in row 5, column 1, I have 1 unit left. Not helpful.Alternatively, place a 1x4 rectangle starting at row 1, column 1, going right to row 1, column 4. Then, in row 1, column 5, I have 1 unit left.This seems to be a recurring issue. Maybe I need to use multiple 1x4 rectangles to cover the problematic areas.Wait, if I have y=4, I can place four 1x4 rectangles. Maybe I can arrange them in a way that they cover the entire grid except for a 3x3 area, which can then be covered by 1x3 rectangles.Let me try placing four 1x4 rectangles around the edges of the grid.Place one 1x4 rectangle horizontally in row 1, columns 1-4.Place another 1x4 rectangle horizontally in row 5, columns 1-4.Place another 1x4 rectangle vertically in column 5, rows 1-4.Place the last 1x4 rectangle vertically in column 1, rows 1-4.Wait, but that would overlap in column 1, row 1-4 and column 5, row 1-4, and rows 1 and 5, columns 1-4. This creates overlapping areas, which isn't allowed.Alternatively, place two 1x4 rectangles horizontally in rows 1 and 5, columns 1-4. Then, place two 1x4 rectangles vertically in columns 1 and 5, rows 2-4. This way, they don't overlap.So, in row 1, columns 1-4: 1x4 rectangle.In row 5, columns 1-4: 1x4 rectangle.In column 1, rows 2-4: 1x4 rectangle.In column 5, rows 2-4: 1x4 rectangle.Now, what's left is the central 3x3 area in rows 2-4, columns 2-4. That's 9 units. Each 1x3 rectangle covers 3 units, so we need 3 of them.Yes! That works.So, the arrangement would be:- 1x4 rectangle in row 1, columns 1-4.- 1x4 rectangle in row 5, columns 1-4.- 1x4 rectangle in column 1, rows 2-4.- 1x4 rectangle in column 5, rows 2-4.- Then, in the central 3x3 area (rows 2-4, columns 2-4), place three 1x3 rectangles horizontally in each row.This covers the entire 5x5 grid without overlapping.So, the final arrangement is:- Four 1x4 rectangles placed around the edges.- Three 1x3 rectangles placed in the central 3x3 area.This should solve the problem.

💡First, I notice that each term in the expression (2a^{2} - 4ab + 2b^{2}) has a common factor of 2. I'll start by factoring out this 2.This leaves me with (2(a^{2} - 2ab + b^{2})). Now, I recognize that the expression inside the parentheses is a perfect square trinomial. Specifically, (a^{2} - 2ab + b^{2}) can be written as ((a - b)^{2}).Substituting this back into the expression gives me (2(a - b)^{2}). Therefore, the factorized form of the original expression is (2(a - b)^{2}).