💡Okay, so I have this problem about a right triangle DEF with sides DE=12, EF=16, and DF=20. A square PQRS is inscribed in the triangle such that points P and Q are on DF, R is on DE, and S is on EF. I need to find the side length of the square. Hmm, let me try to visualize this.First, since DEF is a right triangle, I should figure out which sides are the legs and which is the hypotenuse. Given the lengths 12, 16, and 20, I remember that 12-16-20 is a Pythagorean triple because 12² + 16² = 144 + 256 = 400, which is 20². So, DEF is a right triangle with the right angle at E, right? That means DE and EF are the legs, and DF is the hypotenuse.Now, the square PQRS is inscribed in triangle DEF. Points P and Q are on the hypotenuse DF, R is on DE, and S is on EF. So, the square is sitting inside the triangle with two vertices on the hypotenuse, one on each leg. I think drawing a diagram would help, but since I can't draw, I'll try to imagine it.Let me denote the side length of the square as t. So, each side of the square PQRS is t. Since P and Q are on DF, the hypotenuse, the segment PQ is part of DF and has length t. Then, from P, the square goes up to R on DE, and from Q, it goes to S on EF. So, the square is kind of "standing" on the hypotenuse.I think I can use similar triangles here. When the square is inscribed in the triangle, it creates smaller similar triangles within the original triangle. Specifically, the triangle above the square (the one with vertex E) should be similar to triangle DEF.Let me denote the altitude from E to DF as h. Since DEF is a right triangle, the area can be calculated in two ways: (1/2)*DE*EF and (1/2)*DF*h. So, setting them equal:(1/2)*12*16 = (1/2)*20*hSimplifying:96 = 10*hSo, h = 9.6. Wait, 96 divided by 10 is 9.6, which is 48/5 or 9 and 3/5. Hmm, okay, so the altitude from E to DF is 48/5.Now, when the square is inscribed, the altitude from E to DF is split into two parts: one part is the side of the square, t, and the other part is the remaining altitude above the square. So, the remaining altitude is h - t = 48/5 - t.Since the triangles are similar, the ratio of their corresponding sides should be equal. The original triangle DEF has base DF=20 and height h=48/5. The smaller triangle above the square has base PQ, which is equal to t, and height h - t.Wait, no, actually, the base of the smaller triangle isn't t. Let me think again. The square has side t, so the base of the smaller triangle should be proportional. Maybe I need to consider the similar triangles more carefully.Alternatively, I can think about the similar triangles formed by the square and the sides of the original triangle. The square creates a smaller right triangle on top, which is similar to DEF.Let me denote the side of the square as t. Then, the height from E to DF is h = 48/5. The square reduces this height by t, so the height of the smaller triangle is h - t = 48/5 - t.Since the triangles are similar, the ratio of their heights should be equal to the ratio of their bases. The base of the original triangle is DF=20, and the base of the smaller triangle is the segment from P to Q on DF, which is t.Wait, no, the base of the smaller triangle isn't t. The base of the smaller triangle is actually the projection of the square onto DF, but since the square is standing on DF, the base of the smaller triangle should be less than 20. Maybe I need to express the base of the smaller triangle in terms of t.Alternatively, perhaps I should consider the similar triangles formed by the legs DE and EF. Let me try that.Looking at side DE=12. The square touches DE at point R, so the distance from D to R is some length, say x. Similarly, on EF=16, the square touches at point S, so the distance from E to S is some length, say y.Since PQRS is a square, the sides PR and QS are both equal to t. So, from point R on DE, moving down t units along DE gives us point P on DF. Similarly, from point S on EF, moving down t units along EF gives us point Q on DF.Wait, maybe I need to express x and y in terms of t.Let me consider the coordinates. Maybe assigning coordinates to the triangle would help. Let me place point D at (0,0), E at (0,12), and F at (16,0). Wait, no, because DE=12, EF=16, and DF=20. Wait, actually, if it's a right triangle at E, then DE and EF are the legs.Wait, hold on, DE=12, EF=16, and DF=20. So, if the right angle is at E, then DE and EF are the legs, so DE is one leg, EF is the other leg, and DF is the hypotenuse.So, placing point E at (0,0), point D at (0,12), and point F at (16,0). Then, DF connects (0,12) to (16,0), which should have length sqrt((16)^2 + (12)^2) = sqrt(256 + 144) = sqrt(400) = 20, which matches DF=20. Okay, so that's a good coordinate system.So, points:- E: (0,0)- D: (0,12)- F: (16,0)Now, the square PQRS is inscribed in triangle DEF with P and Q on DF, R on DE, and S on EF.Let me denote the coordinates:- P: somewhere on DF- Q: somewhere on DF- R: somewhere on DE- S: somewhere on EFSince PQRS is a square, the sides PQ, QR, RS, and SP are all equal and at right angles.Let me denote the coordinates of P as (x1, y1) and Q as (x2, y2). Since PQ is a side of the square, the vector from P to Q should be (t, 0) if the square is aligned along the x-axis, but since the square is inside the triangle, it's likely rotated.Wait, maybe I need a different approach. Let me think about the lines.The line DF goes from (0,12) to (16,0). Its equation can be found. The slope is (0 - 12)/(16 - 0) = -12/16 = -3/4. So, the equation is y = (-3/4)x + 12.Similarly, the line DE is the vertical line x=0 from (0,0) to (0,12).The line EF is the horizontal line y=0 from (0,0) to (16,0).Wait, no, EF is from E(0,0) to F(16,0), so it's along the x-axis.Wait, but in the problem statement, S is on EF, which is the base from E(0,0) to F(16,0). So, point S is somewhere on the x-axis between E and F.Similarly, point R is on DE, which is the vertical line from E(0,0) to D(0,12).So, the square PQRS has:- P and Q on DF: the hypotenuse.- R on DE: the vertical leg.- S on EF: the horizontal leg.So, the square is sitting such that two of its vertices are on the hypotenuse, one on the vertical leg, and one on the horizontal leg.Let me denote the coordinates:- Let’s say point R is on DE at (0, a), so its coordinates are (0, a).- Point S is on EF at (b, 0), so its coordinates are (b, 0).Since PQRS is a square, the vector from R to S should be equal to the vector from S to Q, but rotated 90 degrees. Wait, maybe it's better to think in terms of slopes and distances.Alternatively, since PQRS is a square, the sides RS and SP are equal and perpendicular. So, the vector from R to S is (b - 0, 0 - a) = (b, -a). The vector from S to P should be perpendicular to RS and of the same length. So, the vector from S to P is either (a, b) or (-a, -b), depending on the direction.But since P is on DF, which is the line y = (-3/4)x + 12, the coordinates of P must satisfy this equation.Wait, let me think again. If R is at (0, a) and S is at (b, 0), then the vector RS is (b, -a). Since PQRS is a square, the next vector SP should be a 90-degree rotation of RS. Rotating RS = (b, -a) by 90 degrees counterclockwise gives (a, b). So, the vector SP is (a, b). Therefore, point P is S + vector SP = (b + a, 0 + b) = (b + a, b).But point P lies on DF, whose equation is y = (-3/4)x + 12. So, substituting x = b + a and y = b into the equation:b = (-3/4)(b + a) + 12Simplify:b = (-3/4)b - (3/4)a + 12Bring all terms to one side:b + (3/4)b + (3/4)a - 12 = 0Combine like terms:(7/4)b + (3/4)a - 12 = 0Multiply both sides by 4 to eliminate denominators:7b + 3a - 48 = 0So, 7b + 3a = 48. Let's keep this as equation (1).Now, since PQRS is a square, the length of RS should be equal to the length of SP. The length of RS is sqrt(b² + a²). The length of SP is sqrt(a² + b²). Wait, that's the same, so that doesn't give us new information.Alternatively, since PQRS is a square, the side length t is equal to both |RS| and |SP|. So, t = sqrt(b² + a²). But we also know that the side PQ is on DF, so the distance between P and Q should also be t.Wait, but P and Q are both on DF. So, the distance between P and Q along DF should be t. But DF is a straight line, so the Euclidean distance between P and Q is t. However, since DF is the hypotenuse, which is sloped, the distance between P and Q is t, but their coordinates are separated by some amount along DF.Alternatively, maybe I can express the coordinates of Q in terms of P. Since PQ is a side of the square, and PQ is along DF, which has a slope of -3/4, the vector from P to Q should be in the direction of DF but scaled appropriately.Wait, this might be getting complicated. Maybe I need another approach.Let me consider the similar triangles again. The square creates a smaller similar triangle on top. The original triangle DEF has base DF=20 and height h=48/5=9.6. The smaller triangle above the square has base PQ and height h - t.Since the triangles are similar, the ratio of their bases is equal to the ratio of their heights:PQ / DF = (h - t) / hBut PQ is the side of the square, which is t. Wait, no, PQ is a segment on DF, but it's not necessarily equal to t because DF is sloped. Hmm, maybe this is not the right way.Wait, actually, the length of PQ along DF is t, but the Euclidean distance between P and Q is t*sqrt(1 + (slope)^2). Wait, no, the Euclidean distance between P and Q is t, but along DF, which has a slope, so the actual length along DF would be longer.Wait, maybe I need to express the coordinates of P and Q in terms of t.Let me denote the coordinates of P as (x, y) and Q as (x', y'). Since PQ is a side of the square, the distance between P and Q is t, and the vector PQ is perpendicular to the vector QR.But this might get too involved. Maybe I need to use the similar triangles approach more carefully.Let me think about the triangle above the square. It's similar to DEF, so the ratio of their sides is the same. Let me denote the side length of the square as t. Then, the height from E to DF is h=48/5. The height from E to the top of the square is h - t.Since the triangles are similar, the ratio of the heights is equal to the ratio of the bases:(h - t)/h = (base of small triangle)/DFBut what is the base of the small triangle? It's the segment from P to Q on DF, which is t. Wait, no, the base of the small triangle is not t because t is the side of the square, which is perpendicular to DF. Hmm, maybe I'm confusing the base.Wait, in the original triangle DEF, the base is DF=20, and the height is h=48/5. In the smaller triangle above the square, the base is the segment from P to Q on DF, which is t, and the height is h - t.But since the triangles are similar, the ratio of their heights should equal the ratio of their bases:(h - t)/h = t/20Plugging in h=48/5:(48/5 - t)/(48/5) = t/20Simplify the left side:(48/5 - t)/(48/5) = 1 - (5t)/48So,1 - (5t)/48 = t/20Multiply both sides by 48*20 to eliminate denominators:48*20*(1) - 48*20*(5t)/48 = 48*20*(t)/20Simplify:960 - 100t = 48tCombine like terms:960 = 148tSo,t = 960 / 148Simplify:Divide numerator and denominator by 4:240 / 37 ≈ 6.486Wait, but 240 divided by 37 is approximately 6.486, but I think the exact value is 240/37, which can be simplified further? Wait, 240 and 37 have no common factors, so it's 240/37.But wait, earlier when I calculated h=48/5=9.6, and then set up the ratio (h - t)/h = t/20, leading to t=240/37≈6.486. But I'm not sure if this is correct because I might have made a mistake in identifying the base of the smaller triangle.Wait, actually, the base of the smaller triangle is not t. The base of the smaller triangle is the projection of the square onto DF, but since the square is standing on DF, the base of the smaller triangle should be less than 20. Maybe I need to express the base in terms of t and the slope of DF.Let me consider the slope of DF, which is -3/4. So, for every unit increase in x, y decreases by 3/4. The square has side t, so the horizontal component from P to Q is t / cos(theta), where theta is the angle between DF and the x-axis. Since the slope is -3/4, tan(theta) = 3/4, so cos(theta) = 4/5.Therefore, the length along DF between P and Q is t / (4/5) = (5/4)t. So, the base of the smaller triangle is (5/4)t.Now, using similar triangles:(h - t)/h = (5/4)t / DFPlugging in h=48/5 and DF=20:(48/5 - t)/(48/5) = (5/4)t / 20Simplify left side:1 - (5t)/48 = (5t)/80Multiply both sides by 48*80 to eliminate denominators:48*80 - 5t*80 = 5t*48Simplify:3840 - 400t = 240tCombine like terms:3840 = 640tSo,t = 3840 / 640 = 6Wait, t=6? That seems too clean. Let me check my steps.1. I found h=48/5=9.6.2. The base of the smaller triangle is (5/4)t because the length along DF is (5/4)t.3. Using similar triangles: (h - t)/h = (5/4)t / DFPlugging in:(48/5 - t)/(48/5) = (5/4)t / 20Simplify left side:1 - (5t)/48Right side:(5t)/80So,1 - (5t)/48 = (5t)/80Multiply both sides by 240 (LCM of 48 and 80):240 - 25t = 15t240 = 40tt=6Yes, that seems correct. So, the side length of the square is 6.Wait, but earlier when I tried a different approach, I got t=240/37≈6.486, which is different. So, I must have made a mistake in that approach.Let me see where I went wrong. In the first approach, I considered the base of the smaller triangle as t, but actually, it's not t because t is the side of the square, which is perpendicular to DF. The actual base of the smaller triangle is the projection of the square onto DF, which is longer than t. Since the slope of DF is -3/4, the horizontal component is 4/5 of the hypotenuse, so the base is (5/4)t.Therefore, the correct approach is the second one, leading to t=6.Wait, but let me verify this with coordinates.If t=6, then the base of the smaller triangle is (5/4)*6=7.5. So, the remaining base on DF is 20 - 7.5=12.5. The height of the smaller triangle is h - t=48/5 -6=48/5 -30/5=18/5=3.6.Now, check if the ratio of heights equals the ratio of bases:(18/5)/(48/5)=18/48=3/8And (7.5)/20=0.375=3/8Yes, so the ratios are equal, confirming that t=6 is correct.Therefore, the side length of the square is 6.But wait, earlier when I tried using coordinates and got t=240/37≈6.486, which is approximately 6.486, but the correct answer is 6. So, I must have made a mistake in that coordinate approach.Let me go back to the coordinate system.Points:- E(0,0)- D(0,12)- F(16,0)Line DF: y = (-3/4)x +12Square PQRS with P and Q on DF, R on DE, S on EF.Let me denote R as (0, a) and S as (b, 0). Then, vector RS is (b, -a). The vector SP should be a 90-degree rotation of RS, which would be (a, b). So, point P is S + vector SP = (b + a, 0 + b) = (a + b, b).Since P lies on DF, it must satisfy y = (-3/4)x +12.So,b = (-3/4)(a + b) +12Multiply both sides by 4:4b = -3(a + b) +484b = -3a -3b +48Bring all terms to left:4b +3a +3b -48=07b +3a -48=0Which is the same as equation (1): 7b +3a=48.Now, since PQRS is a square, the length of RS is equal to the length of SP, which is t.Length of RS: sqrt(b² +a²)=tLength of SP: sqrt(a² +b²)=tSo, same as before, no new info.But also, the distance between P and Q is t. Since P and Q are on DF, which is a straight line, the distance between them is t. But the coordinates of P are (a + b, b) and Q is another point on DF.Wait, actually, Q is the next point after P on DF, but since PQ is a side of the square, the vector PQ should be equal to the vector RS rotated 90 degrees.Wait, this is getting too convoluted. Maybe I can express Q in terms of P.Since PQ is a side of the square, and PQ is along DF, which has a slope of -3/4, the vector PQ should be in the direction of DF but scaled to length t.The direction vector of DF is (16, -12), which simplifies to (4, -3). So, the unit vector in the direction of DF is (4/5, -3/5).Therefore, the vector PQ is t*(4/5, -3/5). So, if P is (x, y), then Q is (x + (4t)/5, y - (3t)/5).But both P and Q lie on DF, so their coordinates must satisfy y = (-3/4)x +12.So, for point P: y = (-3/4)x +12For point Q: y - (3t)/5 = (-3/4)(x + (4t)/5) +12Simplify the equation for Q:y - (3t)/5 = (-3/4)x - (3/4)*(4t/5) +12Simplify:y - (3t)/5 = (-3/4)x - (3t)/5 +12Add (3t)/5 to both sides:y = (-3/4)x +12Which is the same as the equation for P. So, this doesn't give us new information.Alternatively, since PQ is a side of the square, the vector PQ is perpendicular to the vector QR.Wait, maybe I need to use the fact that the vector PQ is perpendicular to the vector QR.Vector PQ is (4t/5, -3t/5)Vector QR is the vector from Q to R, which is (0 - (x + 4t/5), a - (y - 3t/5)) = (-x -4t/5, a - y +3t/5)Since PQ and QR are perpendicular, their dot product is zero:(4t/5)(-x -4t/5) + (-3t/5)(a - y +3t/5) =0But this seems too complicated. Maybe I need to find another relation.Wait, from earlier, we have 7b +3a=48, and t= sqrt(a² +b²). So, we have two equations:1. 7b +3a=482. t= sqrt(a² +b²)We need a third equation to relate a and b. Let me think.From the coordinates, point P is (a + b, b). Since P lies on DF, which is y = (-3/4)x +12, we have:b = (-3/4)(a + b) +12Which simplifies to:b = (-3/4)a - (3/4)b +12Multiply both sides by 4:4b = -3a -3b +48Bring all terms to left:4b +3a +3b -48=07b +3a -48=0Which is the same as equation (1). So, we only have two equations and three variables (a, b, t). But t is related to a and b by t= sqrt(a² +b²).So, we can write:t= sqrt(a² +b²)And from equation (1):7b +3a=48Let me solve for a in terms of b:3a=48 -7ba=(48 -7b)/3Now, substitute into t:t= sqrt( ((48 -7b)/3 )² + b² )Simplify:t= sqrt( ( (48 -7b)² )/9 + b² )= sqrt( (2304 - 672b +49b²)/9 + b² )= sqrt( (2304 -672b +49b² +9b²)/9 )= sqrt( (2304 -672b +58b²)/9 )= (1/3)sqrt(58b² -672b +2304)Now, we need another equation to relate b and t. Wait, perhaps from the coordinates of Q.Point Q is the next point after P on DF, which is t units away from P along DF. Since DF has a direction vector (4, -3), the unit vector is (4/5, -3/5). So, moving from P by t units along DF gives Q.So, coordinates of Q:x_Q = x_P + (4/5)ty_Q = y_P - (3/5)tBut Q is also a vertex of the square, so the vector from Q to R should be equal to the vector from S to P, which is (a, b). Wait, no, the vector from Q to R should be equal to the vector from S to P rotated 90 degrees.Wait, this is getting too tangled. Maybe I need to use the fact that the vector QR is perpendicular to PQ.Alternatively, since PQRS is a square, the vector QR is equal to the vector SP rotated 90 degrees.Wait, I think I'm overcomplicating this. Let me try to express everything in terms of b.From equation (1):a=(48 -7b)/3From t= sqrt(a² +b²):t= sqrt( ((48 -7b)/3 )² +b² )Let me compute this:t²= ((48 -7b)/3 )² +b²= (2304 - 672b +49b²)/9 +b²= (2304 -672b +49b² +9b²)/9= (2304 -672b +58b²)/9So,t= (1/3)sqrt(58b² -672b +2304)Now, we need another relation involving t and b. Let's think about the coordinates of Q.Point Q is t units away from P along DF. So, the coordinates of Q are:x_Q = x_P + (4/5)ty_Q = y_P - (3/5)tBut x_P = a + b = (48 -7b)/3 + b = (48 -7b +3b)/3 = (48 -4b)/3y_P = bSo,x_Q = (48 -4b)/3 + (4/5)ty_Q = b - (3/5)tBut Q is also a vertex of the square, so the vector from Q to R should be equal to the vector from S to P rotated 90 degrees.Wait, vector QR is from Q to R: (0 - x_Q, a - y_Q) = (-x_Q, a - y_Q)Vector SP is from S to P: (a + b - b, b -0) = (a, b)Since QR is a rotation of SP by 90 degrees, their dot product should be zero, and their magnitudes should be equal.So,QR • SP =0(-x_Q)(a) + (a - y_Q)(b)=0Also,|QR|=|SP|sqrt(x_Q² + (a - y_Q)² )=sqrt(a² +b² )But this is getting too involved. Maybe instead, since QR is perpendicular to SP, their slopes are negative reciprocals.Slope of SP: (b -0)/(a + b - b)=b/aSlope of QR: (a - y_Q - a)/(0 - x_Q -0)= (-y_Q)/(-x_Q)= y_Q/x_QSince they are perpendicular:(b/a) * (y_Q/x_Q) = -1So,(b/a)*(y_Q/x_Q) = -1From earlier, y_Q = b - (3/5)t and x_Q = (48 -4b)/3 + (4/5)tSo,(b/a)*( (b - (3/5)t ) / ( (48 -4b)/3 + (4/5)t ) ) = -1This is a complicated equation, but let's substitute a=(48 -7b)/3 and t from earlier.First, compute a=(48 -7b)/3Then, compute y_Q = b - (3/5)tx_Q = (48 -4b)/3 + (4/5)tNow, plug into the slope equation:(b / ((48 -7b)/3 )) * ( (b - (3/5)t ) / ( (48 -4b)/3 + (4/5)t ) ) = -1Simplify:(3b/(48 -7b)) * ( (b - (3/5)t ) / ( (48 -4b)/3 + (4/5)t ) ) = -1This is very complicated, but maybe we can substitute t from earlier.From t= sqrt(a² +b²)=sqrt( ((48 -7b)/3 )² +b² )But this seems too messy. Maybe instead, let's assume t=6 and see if it satisfies all equations.If t=6, then from equation (1):7b +3a=48And t= sqrt(a² +b²)=6So,a² +b²=36From equation (1):3a=48 -7b => a=(48 -7b)/3Substitute into a² +b²=36:( (48 -7b)/3 )² +b²=36Compute:(2304 - 672b +49b²)/9 +b²=36Multiply all terms by 9:2304 -672b +49b² +9b²=324Combine like terms:58b² -672b +2304 -324=058b² -672b +1980=0Divide all terms by 2:29b² -336b +990=0Now, solve for b:b=(336 ±sqrt(336² -4*29*990))/(2*29)Compute discriminant:336²=1128964*29*990=4*29*990=4*28710=114840So,sqrt(112896 -114840)=sqrt(-1944)Wait, that's negative, which is impossible. So, t=6 is not a solution. Hmm, that contradicts our earlier conclusion.Wait, this means that t=6 is not valid, which suggests that my earlier approach was wrong.Wait, but earlier when I used similar triangles, I got t=6, but when trying to verify with coordinates, it leads to a contradiction. So, where is the mistake?Wait, in the similar triangles approach, I assumed that the base of the smaller triangle is (5/4)t, but maybe that's incorrect.Let me re-examine that.The length of PQ along DF is t, but the actual Euclidean distance between P and Q is t*sqrt(1 + (slope)^2). Wait, no, the Euclidean distance between P and Q is t, but along DF, which has a slope, the distance along DF is longer.Wait, actually, the length of PQ along DF is t / cos(theta), where theta is the angle between DF and the x-axis. Since the slope of DF is -3/4, cos(theta)=4/5. So, the length along DF is t / (4/5)= (5/4)t.Therefore, the base of the smaller triangle is (5/4)t, and the height is h - t=48/5 -t.Using similar triangles:(5/4)t /20 = (48/5 -t)/ (48/5)Simplify:(5t/4)/20 = (48/5 -t)/(48/5)Multiply both sides by 20*(48/5):(5t/4)*(48/5) = (48/5 -t)*20Simplify:(5t/4)*(48/5)= (48/5 -t)*20The 5s cancel:(t/4)*48= (48/5 -t)*20Simplify:12t= (48/5 -t)*20Multiply out the right side:12t= (48/5)*20 -20t12t= 192 -20tAdd 20t to both sides:32t=192t=6So, t=6 is correct. But when I tried to verify with coordinates, I got a contradiction. So, maybe I made a mistake in the coordinate approach.Wait, let's try t=6 and see if it works.From equation (1):7b +3a=48From t= sqrt(a² +b²)=6So,a² +b²=36From equation (1):a=(48 -7b)/3Substitute into a² +b²=36:( (48 -7b)/3 )² +b²=36Compute:(2304 - 672b +49b²)/9 +b²=36Multiply all terms by 9:2304 -672b +49b² +9b²=324Combine like terms:58b² -672b +2304 -324=058b² -672b +1980=0Divide by 2:29b² -336b +990=0Now, discriminant D=336² -4*29*990=112896 -114840= -1944Negative discriminant, which is impossible. So, t=6 is not possible.Wait, this is a contradiction. So, where is the mistake?Wait, in the similar triangles approach, I assumed that the base of the smaller triangle is (5/4)t, but maybe that's incorrect because the base of the smaller triangle is not the projection of the square onto DF, but rather the segment from P to Q on DF, which is t units in length along DF. But the actual Euclidean distance between P and Q is t*sqrt(1 + (slope)^2)=t*sqrt(1 + (9/16))=t*sqrt(25/16)= (5/4)t.Wait, no, that's the length along DF. Wait, I'm confused.Wait, the length along DF between P and Q is t, but the Euclidean distance between P and Q is t*sqrt(1 + (slope)^2). Wait, no, the Euclidean distance is t, and the length along DF is t / cos(theta)= (5/4)t.Wait, I think I'm mixing up the concepts. Let me clarify.The side PQ of the square is t, which is the Euclidean distance between P and Q. Since P and Q are on DF, which has a slope of -3/4, the distance between them along DF is longer than t. Specifically, the distance along DF is t / cos(theta), where theta is the angle between DF and the x-axis.Since the slope is -3/4, tan(theta)=3/4, so cos(theta)=4/5. Therefore, the distance along DF between P and Q is t / (4/5)= (5/4)t.Therefore, the base of the smaller triangle is (5/4)t, and the height is h - t=48/5 -t.Using similar triangles:(5/4)t /20 = (48/5 -t)/(48/5)Simplify:(5t/4)/20 = (48/5 -t)/(48/5)Multiply both sides by 20*(48/5):(5t/4)*(48/5)= (48/5 -t)*20Simplify:(5t/4)*(48/5)= (48/5 -t)*20The 5s cancel:(t/4)*48= (48/5 -t)*20Simplify:12t= (48/5 -t)*20Multiply out the right side:12t= (48/5)*20 -20t12t= 192 -20tAdd 20t to both sides:32t=192t=6So, t=6 is correct. But when I tried to verify with coordinates, I got a contradiction. So, maybe my coordinate approach was wrong.Wait, in the coordinate approach, I assumed that the vector RS is (b, -a), and then vector SP is (a, b). But maybe that's not correct because the rotation might be clockwise instead of counterclockwise.Wait, if I rotate RS=(b, -a) by 90 degrees clockwise, the vector becomes (a, b). But if I rotate it counterclockwise, it becomes (-a, b). So, depending on the orientation, it could be either.Let me try rotating RS=(b, -a) by 90 degrees clockwise, which gives (a, b). So, vector SP=(a, b). Therefore, point P=S + SP=(b +a, 0 +b)=(a +b, b).Since P lies on DF: y= (-3/4)x +12So,b= (-3/4)(a +b) +12Multiply by 4:4b= -3(a +b) +484b= -3a -3b +487b +3a=48Which is equation (1).Now, since PQRS is a square, the length of RS is equal to the length of SP, which is t.So,sqrt(b² +a²)=tAlso, the distance between P and Q is t. Since P and Q are on DF, which is a straight line, the distance between them is t. But the coordinates of Q can be found by moving t units from P along DF.The direction vector of DF is (16, -12), which is (4, -3). The unit vector is (4/5, -3/5). So, moving t units from P along DF gives Q:x_Q= x_P + (4/5)ty_Q= y_P - (3/5)tBut Q is also a vertex of the square, so the vector from Q to R should be equal to the vector from S to P rotated 90 degrees.Wait, vector QR is from Q to R: (0 - x_Q, a - y_Q)= (-x_Q, a - y_Q)Vector SP is from S to P: (a, b)Since QR is a rotation of SP by 90 degrees, their dot product is zero:(-x_Q)(a) + (a - y_Q)(b)=0Also, their magnitudes are equal:sqrt(x_Q² + (a - y_Q)² )=sqrt(a² +b² )But this is getting too involved. Let me try substituting t=6 and see if it works.If t=6, then from equation (1):7b +3a=48And from t= sqrt(a² +b²)=6So,a² +b²=36From equation (1):a=(48 -7b)/3Substitute into a² +b²=36:( (48 -7b)/3 )² +b²=36Compute:(2304 - 672b +49b²)/9 +b²=36Multiply all terms by 9:2304 -672b +49b² +9b²=324Combine like terms:58b² -672b +2304 -324=058b² -672b +1980=0Divide by 2:29b² -336b +990=0Now, discriminant D=336² -4*29*990=112896 -114840= -1944Negative discriminant, which is impossible. So, t=6 is not a solution.Wait, this is a contradiction. So, my similar triangles approach must be wrong.Wait, maybe the base of the smaller triangle is not (5/4)t, but rather t. Let me try that.If the base of the smaller triangle is t, then:t /20 = (h - t)/hSo,t /20 = (48/5 -t)/(48/5)Multiply both sides by 20*(48/5):t*(48/5)=20*(48/5 -t)Simplify:(48/5)t=192 -20tMultiply both sides by 5:48t=960 -100t148t=960t=960/148=240/37≈6.486Which is the same as my first approach. But when I tried to verify with coordinates, I got a contradiction.Wait, maybe the base of the smaller triangle is t, not (5/4)t. So, the similar triangles approach gives t=240/37≈6.486.But earlier, when I used the projection, I got t=6, which led to a contradiction in coordinates.So, which one is correct?Wait, let's think about the similar triangles. The smaller triangle above the square is similar to DEF. The base of DEF is DF=20, and the base of the smaller triangle is the segment from P to Q on DF, which is t units in length along DF. But since DF is sloped, the actual Euclidean distance between P and Q is t*sqrt(1 + (slope)^2)=t*sqrt(1 + (9/16))=t*sqrt(25/16)= (5/4)t.Wait, so the base of the smaller triangle is (5/4)t, and the height is h - t=48/5 -t.Therefore, the ratio of the bases is (5/4)t /20, and the ratio of the heights is (48/5 -t)/(48/5).Setting them equal:(5/4)t /20 = (48/5 -t)/(48/5)Simplify:(5t/4)/20 = (48/5 -t)/(48/5)Multiply both sides by 20*(48/5):(5t/4)*(48/5)= (48/5 -t)*20Simplify:(5t/4)*(48/5)= (48/5 -t)*20The 5s cancel:(t/4)*48= (48/5 -t)*20Simplify:12t= (48/5 -t)*20Multiply out the right side:12t= (48/5)*20 -20t12t= 192 -20tAdd 20t to both sides:32t=192t=6So, t=6 is correct. But when I tried to verify with coordinates, I got a contradiction. So, maybe my coordinate approach was wrong.Wait, perhaps the error is in assuming that the vector SP is (a, b). Maybe it's actually (-a, b) or something else.Alternatively, maybe the square is not oriented as I thought. Maybe the side PQ is not along DF, but rather, the square is standing on DF with two vertices on DF, but the sides are not aligned with the axes.Wait, in the coordinate system, if P and Q are on DF, and R is on DE, S is on EF, then the square is rotated with respect to the axes.Therefore, the vectors RS and SP are not aligned with the axes, so their components are not simply (b, -a) and (a, b). Instead, they have components based on the slope of DF.This makes the coordinate approach more complicated because the vectors are not axis-aligned.Given the time I've spent and the contradictions arising from the coordinate approach, I think the similar triangles approach is more reliable, leading to t=6.Therefore, the side length of the square is 6.But wait, earlier when I tried t=6, the coordinate approach led to a contradiction. So, maybe I made a mistake in the coordinate approach.Alternatively, perhaps the correct answer is 240/37≈6.486, but I'm not sure.Wait, let me try to solve the equations again.From equation (1):7b +3a=48From t= sqrt(a² +b²)From the coordinates of P: (a + b, b) lies on DF: y= (-3/4)x +12So,b= (-3/4)(a + b) +12Multiply by 4:4b= -3a -3b +487b +3a=48Which is equation (1). So, no new info.Now, from the coordinates of Q, which is t units away from P along DF.The direction vector of DF is (4, -3), so unit vector is (4/5, -3/5).So, Q= P + t*(4/5, -3/5)So,x_Q= a + b + (4t)/5y_Q= b - (3t)/5But Q is also a vertex of the square, so the vector from Q to R should be equal to the vector from S to P rotated 90 degrees.Vector QR= R - Q= (0 - x_Q, a - y_Q)= (-x_Q, a - y_Q)Vector SP= P - S= (a + b - b, b -0)= (a, b)Since QR is a rotation of SP by 90 degrees, their dot product is zero:(-x_Q)(a) + (a - y_Q)(b)=0Also, their magnitudes are equal:sqrt(x_Q² + (a - y_Q)² )=sqrt(a² +b² )Let me substitute x_Q and y_Q:x_Q= a + b + (4t)/5y_Q= b - (3t)/5So,(- (a + b + (4t)/5 ))(a) + (a - (b - (3t)/5 ))(b)=0Simplify:- a(a + b + (4t)/5 ) + b(a - b + (3t)/5 )=0Expand:- a² -ab - (4a t)/5 + ab - b² + (3b t)/5=0Combine like terms:- a² -ab - (4a t)/5 + ab - b² + (3b t)/5=0Simplify:- a² - b² - (4a t)/5 + (3b t)/5=0Multiply through by 5 to eliminate denominators:-5a² -5b² -4a t +3b t=0Now, from equation (1):7b +3a=48 => 3a=48 -7b => a=(48 -7b)/3From t= sqrt(a² +b² )Let me substitute a=(48 -7b)/3 into the equation:-5a² -5b² -4a t +3b t=0First, compute a²:a²= ((48 -7b)/3 )²= (2304 - 672b +49b²)/9So,-5*(2304 -672b +49b²)/9 -5b² -4*(48 -7b)/3 *t +3b t=0Multiply all terms by 9 to eliminate denominators:-5*(2304 -672b +49b²) -45b² -12*(48 -7b)*t +27b t=0Expand:-11520 +3360b -245b² -45b² -576t +84b t +27b t=0Combine like terms:-11520 +3360b -290b² -576t +111b t=0Now, from t= sqrt(a² +b² )=sqrt( ((48 -7b)/3 )² +b² )=sqrt( (2304 -672b +49b²)/9 +b² )=sqrt( (2304 -672b +49b² +9b²)/9 )=sqrt( (2304 -672b +58b²)/9 )=(1/3)sqrt(58b² -672b +2304)This is very complicated. Let me try to express t in terms of b and substitute.Let me denote t=(1/3)sqrt(58b² -672b +2304)Now, substitute into the equation:-11520 +3360b -290b² -576*(1/3)sqrt(58b² -672b +2304) +111b*(1/3)sqrt(58b² -672b +2304)=0Simplify:-11520 +3360b -290b² -192sqrt(58b² -672b +2304) +37b sqrt(58b² -672b +2304)=0Factor out sqrt(58b² -672b +2304):-11520 +3360b -290b² + (-192 +37b)sqrt(58b² -672b +2304)=0This is a very complicated equation to solve analytically. Maybe I can try to find a value of b that satisfies this equation.Let me assume t=240/37≈6.486Then, from equation (1):7b +3a=48From t= sqrt(a² +b² )=240/37So,a² +b²=(240/37)²=57600/1369≈42.07From equation (1):a=(48 -7b)/3Substitute into a² +b²=57600/1369:( (48 -7b)/3 )² +b²=57600/1369Compute:(2304 -672b +49b²)/9 +b²=57600/1369Multiply all terms by 9:2304 -672b +49b² +9b²= (57600/1369)*9=518400/1369≈379.3So,58b² -672b +2304=518400/1369Multiply both sides by 1369:58b²*1369 -672b*1369 +2304*1369=518400This is a quadratic in b, but it's very messy. Maybe instead, I can try to solve numerically.Alternatively, maybe the correct answer is t=240/37≈6.486, which is approximately 6.486, but the exact value is 240/37.Wait, but earlier, using similar triangles, I got t=6, but that led to a contradiction in coordinates. So, perhaps the correct answer is 240/37.But I'm confused because the similar triangles approach seems logical, but the coordinate approach contradicts it.Wait, maybe the error is in assuming that the base of the smaller triangle is (5/4)t. Perhaps it's actually t, not (5/4)t.Wait, no, because the base of the smaller triangle is the projection of the square onto DF, which is longer than t. So, it should be (5/4)t.Therefore, the similar triangles approach leading to t=6 is correct, and the coordinate approach must have an error.Alternatively, maybe the side length is indeed 240/37, and the similar triangles approach was incorrectly applied.Given the time I've spent, I think the correct answer is t=240/37, which is approximately 6.486.But wait, when I used the similar triangles approach correctly, considering the projection, I got t=6, but that led to a contradiction in coordinates. So, perhaps the correct answer is 240/37.Wait, let me check online for similar problems. I recall that in a right triangle with legs a and b, the side length of the inscribed square is (a*b)/(a +b). But in this case, the legs are 12 and 16, so t=12*16/(12+16)=192/28=48/7≈6.857, which is different from both 6 and 240/37≈6.486.Wait, that formula is for a square with one side along the legs, but in this problem, the square has two vertices on the hypotenuse, so the formula doesn't apply.Therefore, I think the correct approach is the similar triangles one, leading to t=6.But given the contradictions in the coordinate approach, I'm not entirely sure.Wait, let me try to find the correct answer by solving the quadratic equation.From earlier, we have:29b² -336b +990=0Discriminant D=336² -4*29*990=112896 -114840= -1944Negative discriminant, so no real solution. Therefore, t=6 is not possible.Thus, the correct answer must be t=240/37≈6.486.Therefore, the side length of the square is 240/37.But wait, let me check the similar triangles approach again.If the base of the smaller triangle is t, then:t /20 = (h - t)/hSo,t /20 = (48/5 -t)/(48/5)Multiply both sides by 20*(48/5):t*(48/5)=20*(48/5 -t)Simplify:(48/5)t=192 -20tMultiply both sides by 5:48t=960 -100t148t=960t=960/148=240/37≈6.486Therefore, t=240/37 is correct.So, the side length of the square is 240/37.

💡Okay, so I have this problem about a circle A and a fixed point B. The circle A is given by the equation x² + y² + 2x - 15 = 0, and point B is at (1, 0). Then, for any point M on circle A, we're supposed to find the perpendicular bisector of segment MB and see where it intersects MA at point N. The trajectory of point N is called curve C.Part (I) asks for the equation of curve C. Hmm, okay. Let me try to visualize this. Circle A has the equation x² + y² + 2x - 15 = 0. Maybe I should rewrite this in standard form to find its center and radius. Completing the square for the x-terms:x² + 2x + y² = 15 (x + 1)² + y² = 16So, circle A is centered at (-1, 0) with a radius of 4. Point B is at (1, 0), which is on the x-axis, 2 units away from the center of circle A.Now, for any point M on circle A, we need to construct the perpendicular bisector of MB. The perpendicular bisector of a segment passes through the midpoint and is perpendicular to the segment. So, the midpoint of MB would be ((x + 1)/2, y/2) if M is (x, y). Then, the slope of MB is (y - 0)/(x - 1) = y/(x - 1), so the slope of the perpendicular bisector would be the negative reciprocal, which is -(x - 1)/y.But wait, M is on circle A, so it must satisfy (x + 1)² + y² = 16. So, any point M can be parameterized as (-1 + 4 cos θ, 0 + 4 sin θ), right? Using parametric equations for a circle.So, M is (-1 + 4 cos θ, 4 sin θ). Then, point B is (1, 0). So, the midpoint of MB is [(-1 + 4 cos θ + 1)/2, (4 sin θ + 0)/2] = (2 cos θ, 2 sin θ). The slope of MB is (4 sin θ - 0)/( (-1 + 4 cos θ) - 1) = (4 sin θ)/(4 cos θ - 2) = (2 sin θ)/(2 cos θ - 1). Therefore, the slope of the perpendicular bisector is the negative reciprocal, which is -(2 cos θ - 1)/(2 sin θ).So, the equation of the perpendicular bisector is y - 2 sin θ = [-(2 cos θ - 1)/(2 sin θ)](x - 2 cos θ).Now, we need to find where this perpendicular bisector intersects MA. Point A is (-1, 0). So, the line MA connects point M (-1 + 4 cos θ, 4 sin θ) to A (-1, 0). Let me find the parametric equations for MA.Parametric equations for MA can be written as:x = -1 + t[(-1 + 4 cos θ) - (-1)] = -1 + t(4 cos θ) y = 0 + t(4 sin θ - 0) = 4 t sin θwhere t ranges from 0 to 1.So, any point on MA is (-1 + 4 t cos θ, 4 t sin θ). Now, we need to find the intersection point N between the perpendicular bisector and MA.So, substitute x and y from MA into the equation of the perpendicular bisector.So, plugging into the perpendicular bisector equation:4 t sin θ - 2 sin θ = [-(2 cos θ - 1)/(2 sin θ)] [ (-1 + 4 t cos θ) - 2 cos θ ]Simplify the right-hand side:First, compute the x-component inside the brackets:(-1 + 4 t cos θ) - 2 cos θ = -1 + 4 t cos θ - 2 cos θ = -1 + (4 t - 2) cos θSo, the right-hand side becomes:[-(2 cos θ - 1)/(2 sin θ)] * (-1 + (4 t - 2) cos θ )Multiply numerator:= [ (2 cos θ - 1)(1 - (4 t - 2) cos θ ) ] / (2 sin θ )So, the equation is:4 t sin θ - 2 sin θ = [ (2 cos θ - 1)(1 - (4 t - 2) cos θ ) ] / (2 sin θ )Multiply both sides by 2 sin θ to eliminate the denominator:2 sin θ (4 t sin θ - 2 sin θ) = (2 cos θ - 1)(1 - (4 t - 2) cos θ )Simplify left side:2 sin θ * 4 t sin θ = 8 t sin² θ 2 sin θ * (-2 sin θ) = -4 sin² θ So, left side is 8 t sin² θ - 4 sin² θRight side:Expand (2 cos θ - 1)(1 - (4 t - 2) cos θ )= 2 cos θ * 1 + 2 cos θ * (- (4 t - 2) cos θ ) -1 * 1 -1 * (- (4 t - 2) cos θ )= 2 cos θ - 2(4 t - 2) cos² θ - 1 + (4 t - 2) cos θCombine like terms:= [2 cos θ + (4 t - 2) cos θ] - 2(4 t - 2) cos² θ - 1 = [ (2 + 4 t - 2) cos θ ] - 2(4 t - 2) cos² θ - 1 = 4 t cos θ - 2(4 t - 2) cos² θ - 1So, right side is 4 t cos θ - 2(4 t - 2) cos² θ - 1So, putting it all together:Left side: 8 t sin² θ - 4 sin² θ Right side: 4 t cos θ - 2(4 t - 2) cos² θ - 1Bring all terms to left side:8 t sin² θ - 4 sin² θ - 4 t cos θ + 2(4 t - 2) cos² θ + 1 = 0Simplify term by term:First term: 8 t sin² θ Second term: -4 sin² θ Third term: -4 t cos θ Fourth term: 2(4 t - 2) cos² θ = 8 t cos² θ - 4 cos² θ Fifth term: +1So, combine all:8 t sin² θ - 4 sin² θ - 4 t cos θ + 8 t cos² θ - 4 cos² θ + 1 = 0Now, group like terms:Terms with t:8 t sin² θ - 4 t cos θ + 8 t cos² θ = t (8 sin² θ - 4 cos θ + 8 cos² θ )Constant terms:-4 sin² θ - 4 cos² θ + 1So, equation becomes:t (8 sin² θ - 4 cos θ + 8 cos² θ ) + (-4 sin² θ - 4 cos² θ + 1 ) = 0Let me factor out 4 from the t terms:t [ 4(2 sin² θ + 2 cos² θ ) - 4 cos θ ] + (-4 sin² θ - 4 cos² θ + 1 ) = 0Note that 2 sin² θ + 2 cos² θ = 2(sin² θ + cos² θ ) = 2(1) = 2.So, t [4*2 - 4 cos θ ] + (-4(sin² θ + cos² θ ) + 1 ) = 0 = t [8 - 4 cos θ ] + (-4*1 + 1 ) = 0 = t (8 - 4 cos θ ) - 3 = 0So, solving for t:t = 3 / (8 - 4 cos θ ) = 3 / [4(2 - cos θ ) ] = (3/4) / (2 - cos θ )So, t = 3/(4(2 - cos θ )).Now, recall that point N is on MA, so its coordinates are:x = -1 + 4 t cos θ y = 4 t sin θSo, plug t into x and y:x = -1 + 4 * [3/(4(2 - cos θ ))] * cos θ = -1 + [3 cos θ ] / (2 - cos θ )Similarly,y = 4 * [3/(4(2 - cos θ ))] * sin θ = [3 sin θ ] / (2 - cos θ )So, we have:x = -1 + (3 cos θ ) / (2 - cos θ ) y = (3 sin θ ) / (2 - cos θ )Let me simplify x:x = (-1)(2 - cos θ ) / (2 - cos θ ) + (3 cos θ ) / (2 - cos θ ) = [ -2 + cos θ + 3 cos θ ] / (2 - cos θ ) = [ -2 + 4 cos θ ] / (2 - cos θ )So, x = (4 cos θ - 2 ) / (2 - cos θ )Similarly, y = (3 sin θ ) / (2 - cos θ )Now, to find the equation of curve C, we need to eliminate θ from these equations.Let me denote:Let’s set u = 2 - cos θ, so cos θ = 2 - u.But maybe another approach is better. Let me express x and y in terms of cos θ and sin θ.Let me write:Let’s denote:Let’s set:Let’s let’s denote s = cos θ, t = sin θ.Then, we have:x = (4s - 2)/(2 - s) y = 3t/(2 - s)Also, since s² + t² = 1.So, from x:x = (4s - 2)/(2 - s) Multiply both sides by (2 - s):x(2 - s) = 4s - 2 2x - x s = 4s - 2 Bring all terms to left:2x - x s - 4s + 2 = 0 Factor s:2x + 2 - s(x + 4) = 0 So,s(x + 4) = 2x + 2 Thus,s = (2x + 2)/(x + 4)Similarly, from y:y = 3t/(2 - s) But 2 - s = (2 - s) = denominator from x.From x, we have:x = (4s - 2)/(2 - s) So, 2 - s = (4s - 2)/x Thus,2 - s = (4s - 2)/x Multiply both sides by x:2x - x s = 4s - 2 Which is the same as before.But let me express t in terms of y:From y = 3t/(2 - s), so t = y(2 - s)/3But s = (2x + 2)/(x + 4), so 2 - s = 2 - (2x + 2)/(x + 4) = [2(x + 4) - (2x + 2)] / (x + 4) = [2x + 8 - 2x - 2]/(x + 4) = 6/(x + 4)So, t = y * [6/(x + 4)] / 3 = y * [2/(x + 4)]Thus, t = 2y/(x + 4)So, now, since s² + t² = 1,[(2x + 2)/(x + 4)]² + [2y/(x + 4)]² = 1Let me compute this:[(2x + 2)² + (2y)² ] / (x + 4)² = 1 Multiply both sides by (x + 4)²:(4x² + 8x + 4) + 4y² = (x + 4)² Expand (x + 4)²:x² + 8x + 16So, equation becomes:4x² + 8x + 4 + 4y² = x² + 8x + 16Bring all terms to left:4x² + 8x + 4 + 4y² - x² - 8x - 16 = 0 Simplify:3x² + 4y² - 12 = 0 Divide both sides by 12:x²/4 + y²/3 = 1So, the equation of curve C is x²/4 + y²/3 = 1, which is an ellipse.Okay, that seems to make sense. So, part (I) is done.Now, part (II): If the line y = k(x - 1) intersects curve C at points P and Q, is there a fixed point R on the x-axis such that angle ORP = angle ORQ regardless of the value of k? If it exists, find R; otherwise, explain why.Hmm, so we have line y = k(x - 1) intersecting ellipse C at P and Q. We need to find if there's a fixed point R on x-axis such that angles ORP and ORQ are equal for any k.Angles ORP and ORQ being equal suggests that R lies on the angle bisector of angle PRQ. But since this has to hold for any k, R must be such that it's the exis of symmetry for all such lines.Alternatively, perhaps R is such that it's the exis of symmetry for the ellipse, but the ellipse is symmetric about both axes. Wait, but the ellipse is x²/4 + y²/3 = 1, which is centered at the origin, major axis along x-axis.Wait, but the line y = k(x - 1) passes through (1, 0), which is point B. So, all lines pass through B(1, 0). So, points P and Q are intersections of line through B with ellipse C.So, the problem is, for any line through B(1, 0) intersecting ellipse C at P and Q, is there a fixed point R on x-axis such that angles ORP and ORQ are equal.Angles ORP and ORQ being equal suggests that R is such that OR is the angle bisector of angle PRQ. So, for any chord PQ through B, R is such that OR bisects angle PRQ.This is reminiscent of the concept of the "optical property" of ellipses, but I'm not sure.Alternatively, perhaps R is the other focus of the ellipse. The ellipse has foci at (-1, 0) and (1, 0). Wait, because the ellipse equation is x²/4 + y²/3 = 1, so a² = 4, b² = 3, so c² = a² - b² = 1, so c = 1. So, foci at (-1, 0) and (1, 0). So, point B is (1, 0), which is one focus.Wait, so if we have a point R on x-axis, perhaps the other focus at (-1, 0). But let's see.Wait, but in the problem, the line passes through B(1, 0), which is a focus. So, for any chord through a focus, is there a point R such that angles ORP and ORQ are equal?Alternatively, maybe R is the other focus, (-1, 0). Let me test.Suppose R is (-1, 0). Then, for any chord PQ through B(1, 0), does OR bisect angle PRQ?Wait, but OR is the line from origin to (-1, 0), which is just the x-axis from (0,0) to (-1, 0). So, the angle between OR and PR, and OR and QR.But since the ellipse is symmetric about x-axis, perhaps this holds.Wait, but let me think differently.Alternatively, perhaps R is (4, 0). Because in part (I), the ellipse has major axis length 4, so vertices at (±2, 0). Wait, but 4 is beyond that. Hmm.Alternatively, maybe it's related to the directrix.Wait, the directrix of the ellipse is at x = ±a²/c = ±4/1 = ±4. So, directrices at x = 4 and x = -4.So, maybe R is at (4, 0). Let me test.Suppose R is (4, 0). Then, for any chord PQ through B(1, 0), does OR bisect angle PRQ?Alternatively, perhaps using the concept of pole and polar.Alternatively, perhaps using coordinates.Let me set up the problem.Given ellipse C: x²/4 + y²/3 = 1.Line y = k(x - 1) intersects ellipse at P and Q.We need to find R(t, 0) on x-axis such that angles ORP and ORQ are equal.Angles ORP and ORQ being equal implies that R lies on the angle bisector of angle PRQ.But since this has to hold for any k, R must be such that for any chord PQ through B(1, 0), OR bisects angle PRQ.Alternatively, perhaps using the concept of isogonal conjugates.Alternatively, perhaps using coordinates.Let me find the coordinates of P and Q.Given line y = k(x - 1) intersects ellipse x²/4 + y²/3 = 1.Substitute y = k(x - 1) into ellipse:x²/4 + [k²(x - 1)²]/3 = 1Multiply both sides by 12 to eliminate denominators:3x² + 4k²(x² - 2x + 1) = 12 3x² + 4k²x² - 8k²x + 4k² = 12 (3 + 4k²)x² - 8k²x + (4k² - 12) = 0Let me denote this quadratic as:(4k² + 3)x² - 8k²x + (4k² - 12) = 0Let the roots be x1 and x2, corresponding to points P and Q.By Vieta's formula:x1 + x2 = (8k²)/(4k² + 3) x1 x2 = (4k² - 12)/(4k² + 3)Similarly, y1 = k(x1 - 1), y2 = k(x2 - 1)Now, we need to find R(t, 0) such that angles ORP and ORQ are equal.This condition can be translated into the slopes of lines RP and RQ being such that their angles with OR are equal.Alternatively, using the concept that the ratio of distances from R to the lines PR and QR is equal.But perhaps a better approach is to use the condition that R lies on the angle bisector, which can be expressed using the formula for angle bisector in terms of coordinates.Alternatively, using the condition that the reflection of R over the line OR lies on the circumcircle of triangle PRQ, but that might be complicated.Alternatively, perhaps using the condition that the product of the slopes of PR and QR is -1, but that would mean they are perpendicular, which is not necessarily the case.Wait, but the condition is that angles ORP and ORQ are equal. So, the angle between OR and PR is equal to the angle between OR and QR.This is equivalent to saying that R lies on the angle bisector of angle PRQ.But since this has to hold for any chord PQ through B, R must be such that it's the exis of symmetry for all such chords.Alternatively, perhaps using the concept of harmonic conjugate.Alternatively, perhaps using coordinates.Let me denote R(t, 0). Then, the condition is that the angles ORP and ORQ are equal.This can be expressed using the tangent of the angles, or using the slopes.Alternatively, using vectors.But perhaps a better approach is to use the condition that the reflection of R over the line OR lies on the circumcircle of triangle PRQ, but that might be too involved.Alternatively, perhaps using the condition that the ratio of distances from R to the lines PR and QR is equal.Wait, perhaps using the formula for the angle bisector.The condition that R lies on the angle bisector of angle PRQ can be expressed as:(RP / RQ) = (PP' / PQ'), where P' and Q' are the projections, but that might not be straightforward.Alternatively, perhaps using the formula that if R lies on the angle bisector, then (x1 - t)(x2 - t) + y1 y2 = 0, but I'm not sure.Wait, perhaps using the condition that the slopes of PR and QR satisfy a certain relation.Let me denote the slopes of PR and QR as m1 and m2.Then, the condition that angles ORP and ORQ are equal can be expressed as:(m1 - m_OR)/(1 + m1 m_OR) = (m2 - m_OR)/(1 + m2 m_OR)Where m_OR is the slope of OR, which is 0 since OR is along the x-axis from (0,0) to (t, 0). So, m_OR = 0.Thus, the condition simplifies to:(m1 - 0)/(1 + 0) = (m2 - 0)/(1 + 0) => m1 = m2But that would imply that PR and QR have the same slope, which is only possible if P and Q are symmetric with respect to the x-axis, which is not necessarily the case.Wait, that can't be right. Maybe I made a mistake.Alternatively, perhaps using the condition that the angles are equal in terms of direction vectors.Alternatively, perhaps using the condition that the reflection of OR over PR equals the reflection over QR, but that's too vague.Alternatively, perhaps using the condition that the product of the slopes of PR and QR is -1, but that would mean they are perpendicular, which is not necessarily the case.Wait, perhaps a better approach is to use the concept of isogonal conjugates.In triangle PRQ, if OR is the angle bisector, then R is the isogonal conjugate of some point.But I'm not sure.Alternatively, perhaps using coordinates.Let me denote R(t, 0). Then, the condition is that the angles ORP and ORQ are equal.This can be expressed using the formula for the tangent of the angles.The tangent of angle ORP is |(m_PR - m_OR)/(1 + m_PR m_OR)|, but since m_OR = 0, it's |m_PR|.Similarly, tangent of angle ORQ is |m_QR|.But since angles are equal, |m_PR| = |m_QR|.But that would mean that the absolute values of the slopes of PR and QR are equal.So, |(y1)/(x1 - t)| = |(y2)/(x2 - t)|Which implies that (y1)/(x1 - t) = ±(y2)/(x2 - t)But since this has to hold for any k, perhaps the positive case.So, (y1)/(x1 - t) = (y2)/(x2 - t)Cross-multiplying:y1 (x2 - t) = y2 (x1 - t)But y1 = k(x1 - 1), y2 = k(x2 - 1)So,k(x1 - 1)(x2 - t) = k(x2 - 1)(x1 - t)Assuming k ≠ 0, we can divide both sides by k:(x1 - 1)(x2 - t) = (x2 - 1)(x1 - t)Expand both sides:x1 x2 - t x1 - x2 + t = x1 x2 - t x2 - x1 + tSimplify:Left side: x1 x2 - t x1 - x2 + t Right side: x1 x2 - t x2 - x1 + tSubtract right side from left side:(-t x1 - x2 + t) - (-t x2 - x1 + t) = 0 = -t x1 - x2 + t + t x2 + x1 - t = 0 = (-t x1 + x1) + (-x2 + t x2) + (t - t) = 0 = x1(1 - t) + x2(t - 1) = 0 = (x1 - x2)(1 - t) = 0So, either x1 = x2 or 1 - t = 0.But x1 ≠ x2 because P and Q are distinct points (unless the line is tangent, but we are considering intersections, so generally x1 ≠ x2).Thus, 1 - t = 0 => t = 1.But wait, t = 1 is point B itself. But in the problem, R is a fixed point on x-axis, different from B? Or can it be B?Wait, but if R is at (1, 0), which is point B, then angles ORP and ORQ would be angles between OB and PR, QR. But since B is on the line PQ, perhaps this is trivial.But the problem says "regardless of the value of k", so R must be fixed for all k. If R is at (1, 0), then for any line through B, the angles ORP and ORQ would be equal because R is on the line PQ. So, the angles would be zero or something, which is trivial.But the problem probably expects a non-trivial R.Wait, perhaps I made a mistake in the condition.I assumed that |m_PR| = |m_QR|, but perhaps the correct condition is that the angles are equal, which could mean that the slopes satisfy a certain relation, not necessarily equal in absolute value.Alternatively, perhaps the condition is that the angles between OR and PR, and OR and QR are equal, which can be expressed as:tan(angle ORP) = tan(angle ORQ)Which would mean that the slopes satisfy:(y1)/(x1 - t) = (y2)/(x2 - t)But as above, leading to t = 1.But that seems trivial.Alternatively, perhaps the condition is that the angles are equal in direction, so the slopes are equal or negative reciprocals, but not necessarily equal in magnitude.Wait, perhaps the condition is that the angles are equal in measure, so their tangents are equal in magnitude but sign can vary.But in that case, we have:(y1)/(x1 - t) = ±(y2)/(x2 - t)But for this to hold for all k, the ± must be consistent.But if we take the negative sign, then:(y1)/(x1 - t) = - (y2)/(x2 - t)Which would imply:y1 (x2 - t) + y2 (x1 - t) = 0Substituting y1 = k(x1 - 1), y2 = k(x2 - 1):k(x1 - 1)(x2 - t) + k(x2 - 1)(x1 - t) = 0Divide by k (assuming k ≠ 0):(x1 - 1)(x2 - t) + (x2 - 1)(x1 - t) = 0Expand:x1 x2 - t x1 - x2 + t + x1 x2 - t x2 - x1 + t = 0Combine like terms:2 x1 x2 - t(x1 + x2) - (x1 + x2) + 2 t = 0Factor:2 x1 x2 - (t + 1)(x1 + x2) + 2 t = 0Now, using Vieta's formulas from earlier:x1 + x2 = (8k²)/(4k² + 3) x1 x2 = (4k² - 12)/(4k² + 3)Substitute into the equation:2*(4k² - 12)/(4k² + 3) - (t + 1)*(8k²)/(4k² + 3) + 2 t = 0Multiply all terms by (4k² + 3) to eliminate denominators:2*(4k² - 12) - (t + 1)*8k² + 2 t*(4k² + 3) = 0Expand:8k² - 24 - 8k²(t + 1) + 8 t k² + 6 t = 0Simplify term by term:8k² -24 -8 t k² -8k² +8 t k² +6 t =0Combine like terms:(8k² -8k²) + (-8 t k² +8 t k²) + (-24 +6 t )=0 = 0 + 0 + (-24 +6 t )=0 So,-24 +6 t =0 6 t =24 t=4So, t=4.Thus, R is at (4, 0).Therefore, there exists a fixed point R(4, 0) on the x-axis such that angles ORP and ORQ are equal for any k.So, the answer is R(4, 0).

💡Alright, so I have this problem where I'm given that a real number α satisfies cos α = tan α. I need to find the value of 1/sin α + cos⁴ α. Hmm, okay, let me try to figure this out step by step.First, I know that tan α is equal to sin α over cos α. So, if cos α equals tan α, I can write that as:cos α = sin α / cos αHmm, that looks like an equation I can manipulate. Maybe I can multiply both sides by cos α to get rid of the denominator. Let's try that:cos α * cos α = sin αWhich simplifies to:cos² α = sin αOkay, so now I have cos squared alpha equals sin alpha. I remember that there's a Pythagorean identity that relates sin and cos: sin² α + cos² α = 1. Maybe I can use that here. Since I have cos² α in terms of sin α, I can substitute that into the identity.So, substituting cos² α with sin α, the identity becomes:sin² α + sin α = 1Now, that's a quadratic equation in terms of sin α. Let me write it as:sin² α + sin α - 1 = 0Alright, to solve this quadratic equation, I can use the quadratic formula. The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a). In this case, a = 1, b = 1, and c = -1. Plugging those values in:sin α = [-1 ± sqrt(1² - 4*1*(-1))]/(2*1)sin α = [-1 ± sqrt(1 + 4)]/2sin α = [-1 ± sqrt(5)]/2So, sin α can be either (-1 + sqrt(5))/2 or (-1 - sqrt(5))/2. But wait, the sine of an angle can only be between -1 and 1. Let's check both solutions:First solution: (-1 + sqrt(5))/2. Since sqrt(5) is approximately 2.236, this becomes (-1 + 2.236)/2 ≈ 1.236/2 ≈ 0.618. That's within the valid range.Second solution: (-1 - sqrt(5))/2. That would be (-1 - 2.236)/2 ≈ (-3.236)/2 ≈ -1.618. That's less than -1, which isn't possible for sin α. So, we discard the second solution.Therefore, sin α = (-1 + sqrt(5))/2.Now, I need to find 1/sin α + cos⁴ α. Let's tackle each part separately.First, 1/sin α. Since sin α is (-1 + sqrt(5))/2, then 1/sin α is 2/(-1 + sqrt(5)). Hmm, that denominator looks a bit messy. Maybe I can rationalize it. To rationalize, I'll multiply numerator and denominator by the conjugate of the denominator, which is (-1 - sqrt(5)):1/sin α = 2/(-1 + sqrt(5)) * (-1 - sqrt(5))/(-1 - sqrt(5))Multiplying out the numerator: 2*(-1 - sqrt(5)) = -2 - 2sqrt(5)Multiplying out the denominator: (-1 + sqrt(5))*(-1 - sqrt(5)) = (-1)^2 - (sqrt(5))^2 = 1 - 5 = -4So, 1/sin α = (-2 - 2sqrt(5))/(-4) = (2 + 2sqrt(5))/4 = (1 + sqrt(5))/2Okay, that simplifies nicely. So, 1/sin α is (1 + sqrt(5))/2.Next, let's find cos⁴ α. Earlier, we found that cos² α = sin α, which is (-1 + sqrt(5))/2. So, cos⁴ α is (cos² α)^2, which is [(-1 + sqrt(5))/2]^2.Let's compute that:[(-1 + sqrt(5))/2]^2 = [(-1)^2 + 2*(-1)*(sqrt(5)) + (sqrt(5))^2]/4= [1 - 2sqrt(5) + 5]/4= (6 - 2sqrt(5))/4= (3 - sqrt(5))/2So, cos⁴ α is (3 - sqrt(5))/2.Now, let's add the two results together:1/sin α + cos⁴ α = (1 + sqrt(5))/2 + (3 - sqrt(5))/2Since the denominators are the same, we can add the numerators:= [1 + sqrt(5) + 3 - sqrt(5)]/2= (1 + 3 + sqrt(5) - sqrt(5))/2= 4/2= 2Wait, that's interesting. The sqrt(5) terms cancel out, and we're left with 4/2, which is 2. So, the value of 1/sin α + cos⁴ α is 2.Let me just double-check my steps to make sure I didn't make any mistakes.1. Started with cos α = tan α, which led to cos² α = sin α.2. Substituted into the Pythagorean identity to get a quadratic in sin α.3. Solved the quadratic and found sin α = (-1 + sqrt(5))/2.4. Calculated 1/sin α by rationalizing the denominator and got (1 + sqrt(5))/2.5. Found cos⁴ α by squaring cos² α, which was sin α, resulting in (3 - sqrt(5))/2.6. Added the two results together, and the sqrt(5) terms canceled out, leaving 2.Everything seems to check out. So, I think the final answer is 2.**Final Answer**boxed{2}

💡Okay, so I have this equation: ( x(x - 2)^8 = a_0 + a_1(x - 1) + a_2(x - 1)^2 + ldots + a_9(x - 1)^9 ). I need to find the value of ( a_6 ). Hmm, let me think about how to approach this.First, I remember that when you have a polynomial expressed in terms of ( (x - 1) ), it's like shifting the variable. Maybe I can make a substitution to simplify things. Let me set ( t = x - 1 ). That means ( x = t + 1 ). So, substituting this into the equation, I get:( (t + 1)(t - 1)^8 = a_0 + a_1 t + a_2 t^2 + ldots + a_9 t^9 ).Okay, so now I have everything in terms of ( t ). This should make it easier to find the coefficients. I need to find ( a_6 ), which is the coefficient of ( t^6 ) in the expansion of ( (t + 1)(t - 1)^8 ).Let me recall the binomial theorem. The expansion of ( (t - 1)^8 ) can be written as ( sum_{k=0}^{8} binom{8}{k} t^{k} (-1)^{8 - k} ). But since I'm multiplying this by ( (t + 1) ), I need to consider how that affects each term.Specifically, when I multiply ( (t + 1) ) by ( (t - 1)^8 ), each term in the expansion of ( (t - 1)^8 ) will be multiplied by both ( t ) and ( 1 ). So, the coefficient of ( t^6 ) in the product will come from two places:1. The term where ( t ) is multiplied by the ( t^5 ) term in ( (t - 1)^8 ).2. The term where ( 1 ) is multiplied by the ( t^6 ) term in ( (t - 1)^8 ).So, I need to find the coefficients of ( t^5 ) and ( t^6 ) in ( (t - 1)^8 ) and then combine them accordingly.Using the binomial theorem, the coefficient of ( t^k ) in ( (t - 1)^8 ) is ( binom{8}{k} (-1)^{8 - k} ).Let me calculate the coefficient of ( t^5 ):( binom{8}{5} (-1)^{8 - 5} = binom{8}{5} (-1)^3 ).Calculating ( binom{8}{5} ):( binom{8}{5} = frac{8!}{5!(8 - 5)!} = frac{8 times 7 times 6}{3 times 2 times 1} = 56 ).So, the coefficient of ( t^5 ) is ( 56 times (-1)^3 = -56 ).Now, the coefficient of ( t^6 ):( binom{8}{6} (-1)^{8 - 6} = binom{8}{6} (-1)^2 ).Calculating ( binom{8}{6} ):( binom{8}{6} = frac{8!}{6!(8 - 6)!} = frac{8 times 7}{2 times 1} = 28 ).So, the coefficient of ( t^6 ) is ( 28 times (-1)^2 = 28 ).Now, going back to the original expression ( (t + 1)(t - 1)^8 ), the coefficient of ( t^6 ) will be the sum of:1. The coefficient of ( t^5 ) multiplied by ( t ) (which gives ( t^6 )).2. The coefficient of ( t^6 ) multiplied by ( 1 ).So, that's ( (-56) times 1 + 28 times 1 = -56 + 28 = -28 ).Wait, but looking back at the answer choices, -28 is option A. But I thought maybe it's one of the other options. Let me double-check my calculations.First, ( binom{8}{5} = 56 ), correct. Then, ( (-1)^3 = -1 ), so ( 56 times (-1) = -56 ). That seems right.Next, ( binom{8}{6} = 28 ), correct. ( (-1)^2 = 1 ), so ( 28 times 1 = 28 ). That's also correct.Adding them together: ( -56 + 28 = -28 ). Hmm, that's consistent. So, maybe the answer is indeed -28, which is option A.But wait, I remember sometimes when dealing with expansions, especially with shifts, there might be a factor I'm missing. Let me think again.I set ( t = x - 1 ), so ( x = t + 1 ). Then, substituting into ( x(x - 2)^8 ), I get ( (t + 1)(t - 1)^8 ). That seems correct.Expanding ( (t - 1)^8 ), the coefficients are as I calculated. Then, multiplying by ( (t + 1) ), the ( t^5 ) term becomes ( t^6 ) when multiplied by ( t ), and the ( t^6 ) term remains as ( t^6 ) when multiplied by 1. So, adding those coefficients is the right approach.Therefore, I think my calculation is correct, and ( a_6 = -28 ). So, the answer should be option A.

💡Okay, so I have this problem where I need to factor the polynomial ( x^2 + 6x + 9 - 64x^4 ) into two quadratic polynomials with integer coefficients. The final answer should be in the form ( (ax^2 + bx + c)(dx^2 + ex + f) ) with ( a < d ). Alright, let me try to figure this out step by step.First, I notice that the polynomial has a mix of terms with different degrees: ( x^4 ), ( x^2 ), ( x ), and a constant. It might help to rearrange the terms in descending order of degree. So, let me rewrite the polynomial:( -64x^4 + x^2 + 6x + 9 )Hmm, factoring a quartic polynomial can be tricky, but maybe I can find a way to express it as a product of two quadratics. Let me think about the structure of the polynomial. I see that ( x^2 + 6x + 9 ) is a perfect square because ( (x + 3)^2 = x^2 + 6x + 9 ). So, I can rewrite the polynomial as:( -64x^4 + (x + 3)^2 )Now, this looks like a difference of squares because ( (x + 3)^2 ) is a square and ( 64x^4 ) is also a square since ( (8x^2)^2 = 64x^4 ). So, I can express the polynomial as:( (x + 3)^2 - (8x^2)^2 )Ah, yes! This is a difference of squares, which has the form ( a^2 - b^2 = (a - b)(a + b) ). Applying this formula, where ( a = x + 3 ) and ( b = 8x^2 ), I get:( (x + 3 - 8x^2)(x + 3 + 8x^2) )Let me write that out:( (-8x^2 + x + 3)(8x^2 + x + 3) )Wait, the problem asks for two quadratic polynomials with integer coefficients, and I have that here. Each factor is a quadratic, and all coefficients are integers. Also, the leading coefficients are -8 and 8, so if I want to present them in the form ( (ax^2 + bx + c)(dx^2 + ex + f) ) with ( a < d ), I need to ensure that the first quadratic has a smaller leading coefficient than the second.Looking at the factors, the first one is ( -8x^2 + x + 3 ) and the second is ( 8x^2 + x + 3 ). Since -8 is less than 8, this satisfies the condition ( a < d ). So, I think this is the correct factorization.Just to double-check, I'll multiply the two quadratics to see if I get back the original polynomial:First, multiply ( -8x^2 ) by each term in the second quadratic:( -8x^2 times 8x^2 = -64x^4 )( -8x^2 times x = -8x^3 )( -8x^2 times 3 = -24x^2 )Next, multiply ( x ) by each term in the second quadratic:( x times 8x^2 = 8x^3 )( x times x = x^2 )( x times 3 = 3x )Finally, multiply 3 by each term in the second quadratic:( 3 times 8x^2 = 24x^2 )( 3 times x = 3x )( 3 times 3 = 9 )Now, let's add all these terms together:-64x^4 -8x^3 -24x^2 + 8x^3 + x^2 + 3x + 24x^2 + 3x + 9Combine like terms:-64x^4 + (-8x^3 + 8x^3) + (-24x^2 + x^2 + 24x^2) + (3x + 3x) + 9Simplify each group:-64x^4 + 0x^3 + ( -24x^2 + x^2 + 24x^2 ) + 6x + 9Calculating the x^2 terms:-24x^2 + x^2 = -23x^2-23x^2 + 24x^2 = x^2So, combining everything:-64x^4 + x^2 + 6x + 9Which is exactly the original polynomial. Great, so the factorization is correct.Therefore, the polynomial ( x^2 + 6x + 9 - 64x^4 ) factors into ( (-8x^2 + x + 3)(8x^2 + x + 3) ), and since -8 < 8, this satisfies the condition ( a < d ).**Final Answer**The given expression factors as boxed{(-8x^2 + x + 3)(8x^2 + x + 3)}.

💡Okay, so I have this problem where I need to find the equation of the tangent line to the function f(x) = x³ - 2 ln x at the point (1, f(1)). Hmm, I remember that to find a tangent line, I need two things: the slope of the tangent at that point and a point through which the tangent passes. Since the point is given as (1, f(1)), I already have the point, so I just need to find the slope.First, I should probably find f(1) to make sure I know the exact point. Let me plug in x = 1 into the function. So, f(1) = (1)³ - 2 ln(1). I know that ln(1) is 0 because e⁰ = 1. So, f(1) = 1 - 0 = 1. Okay, so the point is (1, 1).Now, to find the slope of the tangent line at x = 1, I need to find the derivative of f(x). The derivative f’(x) will give me the slope at any point x. Let me recall how to differentiate functions like this. The function f(x) is x³ minus 2 ln x. I can differentiate each term separately.The derivative of x³ is straightforward. Using the power rule, the derivative of xⁿ is n xⁿ⁻¹. So, the derivative of x³ is 3x². Good.Next, the derivative of 2 ln x. I remember that the derivative of ln x is 1/x. So, the derivative of 2 ln x should be 2*(1/x) = 2/x. But since it's minus 2 ln x, the derivative will be -2/x.Putting it all together, the derivative f’(x) = 3x² - 2/x.Now, I need to evaluate this derivative at x = 1 to find the slope of the tangent line at that specific point. So, f’(1) = 3*(1)² - 2/(1). Calculating that, 3*(1) is 3, and 2/(1) is 2. So, 3 - 2 = 1. Therefore, the slope of the tangent line at x = 1 is 1.Alright, so now I have the slope (which is 1) and the point (1, 1). I can use the point-slope form of a line to write the equation of the tangent line. The point-slope formula is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point.Plugging in the values, we get y - 1 = 1*(x - 1). Simplifying this, y - 1 = x - 1. If I add 1 to both sides, I get y = x. So, the equation of the tangent line is y = x.Wait, let me double-check my steps to make sure I didn't make any mistakes. First, f(1) = 1, that's correct. Then, the derivative f’(x) = 3x² - 2/x, which seems right. Plugging in x = 1 gives f’(1) = 3 - 2 = 1, so the slope is 1. Using point-slope form with (1,1) gives y - 1 = 1*(x - 1), which simplifies to y = x. That seems correct.I guess another way to write the equation is in standard form, which would be x - y = 0. Both forms are correct, but sometimes standard form is preferred. So, depending on what's required, both y = x and x - y = 0 are acceptable.Let me visualize this for a moment. The function f(x) = x³ - 2 ln x. At x = 1, it's equal to 1. The tangent line at that point has a slope of 1, so it's a straight line passing through (1,1) with a 45-degree angle. That makes sense because the slope is 1, which corresponds to a 45-degree angle.Just to be thorough, maybe I can check the equation of the tangent line by plugging in x = 1 into both the function and the tangent line equation. For the function, f(1) = 1, and for the tangent line, y = 1 when x = 1, so they match. If I pick another point close to x = 1, say x = 1.1, I can approximate the function and the tangent line to see if they are close.Calculating f(1.1): (1.1)³ - 2 ln(1.1). Let's compute that. (1.1)³ is approximately 1.331. ln(1.1) is approximately 0.09531. So, 2 ln(1.1) is about 0.19062. Therefore, f(1.1) ≈ 1.331 - 0.19062 ≈ 1.14038.Now, the tangent line at x = 1.1 is y = 1.1. So, the function value is approximately 1.14038, and the tangent line gives 1.1. The difference is about 0.04038, which seems reasonable for a linear approximation near x = 1.If I go a bit further, say x = 2, just to see how the tangent line behaves. f(2) = 8 - 2 ln 2 ≈ 8 - 1.386 ≈ 6.614. The tangent line at x = 2 would be y = 2. So, the function is much higher there, as expected, since the function is cubic and grows faster than the linear tangent line.This makes sense because the tangent line is only a good approximation near the point of tangency, which is x = 1 in this case. Farther away, the function and the tangent line diverge.I think I've covered all the steps and double-checked my work. I don't see any mistakes in my calculations or reasoning. The equation of the tangent line is indeed y = x or x - y = 0.