💡First, I need to determine how many goals The Kickers scored in each period. They scored 2 goals in the first period and twice that amount in the second period, which means they scored 4 goals in the second period.Next, I'll calculate The Spiders' goals. They scored half the amount that The Kickers scored in the first period, which is 1 goal. Additionally, The Spiders scored twice the amount that The Kickers scored in their second period, which is 8 goals.Now, I'll add up all the goals scored by both teams. The Kickers scored a total of 6 goals (2 in the first period and 4 in the second period). The Spiders scored a total of 9 goals (1 goal from the first period and 8 goals from the second period).Finally, by adding the goals from both teams, the total number of goals scored in the match is 15.

💡First, I need to determine the slopes of both lines to find the value of ( a ).For the first line ( ax + 3y - 1 = 0 ), I'll rearrange it into the slope-intercept form ( y = mx + b ):[3y = -ax + 1 implies y = -frac{a}{3}x + frac{1}{3}]So, the slope ( m_1 ) is ( -frac{a}{3} ).For the second line ( 3x - y + 2 = 0 ), I'll also rearrange it into the slope-intercept form:[-y = -3x - 2 implies y = 3x + 2]So, the slope ( m_2 ) is ( 3 ).Since the lines are perpendicular, the product of their slopes should be ( -1 ):[m_1 times m_2 = -1 implies -frac{a}{3} times 3 = -1]Simplifying this equation:[-a = -1 implies a = 1]Therefore, the value of ( a ) is ( 1 ).

💡First, I need to find the sum of (frac{1}{3}) and (frac{1}{4}). To do this, I'll determine a common denominator for the two fractions. The least common multiple of 3 and 4 is 12.Next, I'll convert each fraction to have this common denominator:[frac{1}{3} = frac{4}{12} quad text{and} quad frac{1}{4} = frac{3}{12}]Now, I'll add the two fractions:[frac{4}{12} + frac{3}{12} = frac{7}{12}]Finally, to find the reciprocal of (frac{7}{12}), I'll invert the fraction:[text{Reciprocal of } frac{7}{12} = frac{12}{7}]

💡Okay, so I have this problem here about a parabola. Let me try to figure it out step by step. The equation of the parabola is given as ( y^2 = 4x ). I remember that this is a standard form of a parabola that opens to the right. The general form is ( y^2 = 4ax ), where 'a' is the distance from the vertex to the focus. So in this case, comparing ( y^2 = 4x ) with the general form, I can see that ( 4a = 4 ), which means ( a = 1 ). So, the vertex of this parabola is at the origin (0,0), and the focus is 1 unit to the right of the vertex along the x-axis. That means the focus is at (1, 0). I also recall that the directrix of a parabola is a vertical line on the opposite side of the vertex from the focus. Since the parabola opens to the right, the directrix should be a vertical line to the left of the vertex. For this parabola, the directrix is ( x = -1 ).Now, the problem says that the distance from a point P on the parabola to the y-axis is 2. Hmm, the distance from a point to the y-axis is just the absolute value of its x-coordinate. So if the distance is 2, that means the x-coordinate of point P is either 2 or -2. But wait, since the parabola ( y^2 = 4x ) only includes points where x is non-negative because ( y^2 ) is always non-negative, and 4x must also be non-negative. So x can't be negative here. Therefore, the x-coordinate of point P must be 2.So, point P is somewhere on the parabola with an x-coordinate of 2. Let me find the corresponding y-coordinate(s). Plugging x = 2 into the equation ( y^2 = 4x ), we get ( y^2 = 8 ). Taking the square root of both sides, y can be either ( sqrt{8} ) or ( -sqrt{8} ). Simplifying, ( sqrt{8} ) is ( 2sqrt{2} ), so y is ( 2sqrt{2} ) or ( -2sqrt{2} ). Therefore, point P can be either (2, ( 2sqrt{2} )) or (2, ( -2sqrt{2} )).Now, the question asks for the distance from point P to the focus of the parabola. The focus is at (1, 0), as I found earlier. So, I need to calculate the distance between point P, which is (2, ( 2sqrt{2} )) or (2, ( -2sqrt{2} )), and the focus at (1, 0).I can use the distance formula for this. The distance between two points (x1, y1) and (x2, y2) is given by ( sqrt{(x2 - x1)^2 + (y2 - y1)^2} ). Let's apply this formula.First, let's take point P as (2, ( 2sqrt{2} )). The distance to the focus (1, 0) would be:( sqrt{(1 - 2)^2 + (0 - 2sqrt{2})^2} )Calculating each part:( (1 - 2)^2 = (-1)^2 = 1 )( (0 - 2sqrt{2})^2 = (-2sqrt{2})^2 = (2sqrt{2})^2 = 4 * 2 = 8 )Adding these together: 1 + 8 = 9Taking the square root: ( sqrt{9} = 3 )So, the distance is 3.Just to be thorough, let's check the other point P, which is (2, ( -2sqrt{2} )). The distance to the focus (1, 0) would be:( sqrt{(1 - 2)^2 + (0 - (-2sqrt{2}))^2} )Simplify:( (1 - 2)^2 = (-1)^2 = 1 )( (0 - (-2sqrt{2}))^2 = (2sqrt{2})^2 = 8 )Adding them: 1 + 8 = 9Square root: ( sqrt{9} = 3 )Same result. So regardless of whether P is (2, ( 2sqrt{2} )) or (2, ( -2sqrt{2} )), the distance to the focus is 3.Wait, but I remember something about parabolas where the distance from any point on the parabola to the focus is equal to the distance from that point to the directrix. Maybe I can use that property instead of calculating the distance directly. Let me think.The definition of a parabola is the set of all points equidistant from the focus and the directrix. So, if I can find the distance from point P to the directrix, that should be equal to the distance from P to the focus.The directrix is the line ( x = -1 ). The distance from a point (x, y) to the directrix is the horizontal distance from that point to the line ( x = -1 ). Since the directrix is a vertical line, the distance is just the absolute difference in the x-coordinates.So, for point P at (2, ( 2sqrt{2} )), the distance to the directrix is |2 - (-1)| = |3| = 3. Therefore, the distance from P to the focus should also be 3, which matches what I found earlier.This makes me more confident that the answer is indeed 3.Alternatively, I could think about the definition of the parabola. Since the distance from P to the focus equals the distance from P to the directrix, and since we know the distance from P to the directrix is 3, then the distance from P to the focus is also 3.So, whether I calculate it directly using the distance formula or use the definition of the parabola, I arrive at the same answer.Therefore, the distance from point P to the focus is 3.**Final Answer**boxed{C}

💡Okay, so I have this problem about an arithmetic sequence. It says that the sum of the first n terms is S_n, and we're given two conditions: S_4 equals three times S_2, and the seventh term, a_7, is 15. We need to find the common difference of the sequence. Hmm, let me think about how to approach this.First, I remember that in an arithmetic sequence, each term is obtained by adding a constant difference, which we call 'd', to the previous term. The general formula for the nth term of an arithmetic sequence is a_n = a_1 + (n-1)d, where a_1 is the first term. Now, the sum of the first n terms, S_n, can be calculated using the formula S_n = n/2 * [2a_1 + (n-1)d]. Alternatively, it can also be written as S_n = n*(a_1 + a_n)/2. Maybe both forms will be useful here.Given that S_4 = 3S_2, let me write down the expressions for S_4 and S_2 using the sum formula.So, S_4 = 4/2 * [2a_1 + (4-1)d] = 2*(2a_1 + 3d) = 4a_1 + 6d.Similarly, S_2 = 2/2 * [2a_1 + (2-1)d] = 1*(2a_1 + d) = 2a_1 + d.According to the problem, S_4 = 3S_2. So, substituting the expressions we have:4a_1 + 6d = 3*(2a_1 + d)Let me expand the right side:4a_1 + 6d = 6a_1 + 3dNow, let's bring all terms to one side to solve for a_1 and d:4a_1 + 6d - 6a_1 - 3d = 0Simplify:-2a_1 + 3d = 0So, -2a_1 + 3d = 0. Let me write this as equation (1):-2a_1 + 3d = 0Now, the other condition given is a_7 = 15. Using the general term formula:a_7 = a_1 + (7-1)d = a_1 + 6d = 15So, a_1 + 6d = 15. Let's call this equation (2):a_1 + 6d = 15Now, we have two equations:1) -2a_1 + 3d = 02) a_1 + 6d = 15I can solve this system of equations to find a_1 and d. Let's start with equation (1):-2a_1 + 3d = 0Let me solve for a_1:-2a_1 = -3dDivide both sides by -2:a_1 = (3d)/2So, a_1 is (3/2)d. Now, let's substitute this into equation (2):a_1 + 6d = 15Substituting a_1:(3d)/2 + 6d = 15Let me combine the terms. To do that, I'll express 6d as (12d)/2 so that both terms have the same denominator:(3d)/2 + (12d)/2 = (15d)/2So, (15d)/2 = 15Now, solve for d:Multiply both sides by 2:15d = 30Divide both sides by 15:d = 2So, the common difference d is 2. Let me check if this makes sense.If d = 2, then from equation (1), a_1 = (3*2)/2 = 3. So, a_1 is 3.Let me verify the conditions:First, S_4 should be 3 times S_2.Calculate S_2:S_2 = 2a_1 + d = 2*3 + 2 = 6 + 2 = 8S_4 = 4a_1 + 6d = 4*3 + 6*2 = 12 + 12 = 24Is 24 equal to 3 times 8? Yes, 3*8=24. So that condition is satisfied.Now, check a_7:a_7 = a_1 + 6d = 3 + 6*2 = 3 + 12 = 15. That's correct as per the given condition.Therefore, the common difference d is indeed 2.**Final Answer**The common difference of the arithmetic sequence is boxed{2}.

💡Alright, so I have this problem about a trapezoid and some points defined by intersections with circumcircles. Let me try to unpack it step by step.First, the problem states: Let ABCD be a non-isosceles trapezoid. Define a point A_1 as the intersection of the circumcircle of triangle BCD and the line AC (choosing A_1 distinct from C). Similarly, define points B_1, C_1, and D_1 in the same way. We need to prove that A_1B_1C_1D_1 is also a trapezoid.Okay, so I need to visualize this. Let me draw a rough sketch in my mind. ABCD is a trapezoid, so it has one pair of sides parallel. Since it's non-isosceles, the non-parallel sides are of different lengths. Let's assume AB is parallel to CD for simplicity.Now, for point A_1, we're taking the circumcircle of triangle BCD and intersecting it with line AC. Since C is already on both the circumcircle and the line AC, we choose the other intersection point as A_1. Similarly, B_1 is the other intersection of the circumcircle of triangle ACD with line BD, and so on.Hmm, okay. So each new point is constructed by intersecting a circumcircle of a triangle formed by three of the original trapezoid's vertices with a diagonal of the trapezoid.I think I need to recall some properties of cyclic quadrilaterals and trapezoids. Since ABCD is a trapezoid, AB parallel CD. Maybe this parallelism can help us show that A_1B_1C_1D_1 is also a trapezoid by showing that some sides are parallel.Let me think about the circumcircle of triangle BCD. Point A_1 lies on this circle, so angles subtended by the same chord should be equal. Specifically, angle BA_1D = angle BCD because they both subtend arc BD in the circumcircle of BCD.Wait, is that right? Let me double-check. In the circumcircle of BCD, point A_1 is on the circle, so angles subtended by chord BD at points A_1 and C should be equal. So, angle BA_1D = angle BCD.Similarly, in the circumcircle of ACD, point B_1 is on this circle, so angle AB_1D = angle ACD.Hmm, interesting. Maybe I can use these angle equalities to find some similar triangles or something.Alternatively, maybe I can use power of a point. Since A_1 is on the circumcircle of BCD and on line AC, the power of point A with respect to the circumcircle of BCD should relate AA_1 and AC.Wait, the power of point A with respect to the circumcircle of BCD is AB cdot AD = AA_1 cdot AC. Is that correct? Let me recall the power of a point formula: for a point P outside a circle, the power is PA cdot PB where PA and PB are lengths of segments from P to the points of intersection with the circle.So, in this case, point A lies outside the circumcircle of BCD, and line AC intersects the circle at C and A_1. Therefore, the power of A is AC cdot AA_1.Similarly, the power can also be expressed as AB cdot AD because AB and AD are the lengths of the two sides from A to the circle. Wait, is that accurate? Actually, the power of point A with respect to the circle is equal to AB cdot AD only if AB and AD are tangent segments, but in this case, AB and AD are not necessarily tangents.Hmm, maybe I need a different approach. Let me think about harmonic division or projective geometry, but that might be too advanced.Alternatively, maybe I can use Menelaus' theorem or Ceva's theorem. Since we're dealing with lines intersecting sides of triangles, these theorems might come in handy.Wait, let's consider triangle BCD and the line AC intersecting it at C and A_1. Maybe Menelaus' theorem can relate the ratios of the segments.But Menelaus' theorem applies to a transversal cutting through the sides of a triangle, but in this case, AC is not necessarily cutting through all three sides. Hmm.Alternatively, maybe Ceva's theorem, which involves concurrent lines, but I don't see an immediate application here.Wait, perhaps I should consider the cyclic quadrilaterals. Since A_1 is on the circumcircle of BCD, quadrilateral BCDA_1 is cyclic. Similarly, quadrilateral ACD B_1 is cyclic, and so on.So, in cyclic quadrilateral BCDA_1, we have that angle BA_1D = angle BCD as I thought earlier. Similarly, in cyclic quadrilateral ACD B_1, we have angle AB_1D = angle ACD.Wait, so both angle BA_1D and angle AB_1D are equal to angles in the original trapezoid. Maybe I can relate these angles to show that sides of A_1B_1C_1D_1 are parallel.Alternatively, maybe I can use the concept of spiral similarity or some kind of similarity transformation.Wait, another idea: since AB parallel CD, the angles at A and D are supplementary with the angles at B and C. Maybe these properties can help in showing that the new quadrilateral has a pair of parallel sides.Alternatively, maybe I can use coordinates. Assign coordinates to the trapezoid and compute the coordinates of A_1, B_1, etc., then check if the sides are parallel.Let me try that approach. Let's place trapezoid ABCD on a coordinate system with AB and CD parallel to the x-axis. Let me assign coordinates:Let’s set point A at (0, 0), point B at (b, 0), point C at (c, h), and point D at (d, h), where h is the height of the trapezoid. Since it's a trapezoid, AB is parallel to CD, so both have slope 0.Now, let's find the coordinates of A_1. A_1 is the intersection of line AC and the circumcircle of triangle BCD (other than C).First, let's find the equation of line AC. Points A(0,0) and C(c, h). The slope is h/c, so the equation is y = (h/c)x.Next, find the equation of the circumcircle of triangle BCD. Points B(b, 0), C(c, h), and D(d, h). Let's find the circumcircle equation.The general equation of a circle is x^2 + y^2 + Dx + Ey + F = 0. Plugging in the three points:For B(b, 0): b^2 + 0 + Db + E*0 + F = 0 Rightarrow b^2 + Db + F = 0.For C(c, h): c^2 + h^2 + Dc + Eh + F = 0.For D(d, h): d^2 + h^2 + Dd + Eh + F = 0.Now, subtract the equation for C from the equation for D:(d^2 + h^2 + Dd + Eh + F) - (c^2 + h^2 + Dc + Eh + F) = 0Simplify: d^2 - c^2 + D(d - c) = 0 Rightarrow (d - c)(d + c) + D(d - c) = 0Factor out (d - c): (d - c)(d + c + D) = 0Since d neq c (as CD is a side of the trapezoid and not degenerate), we have d + c + D = 0 Rightarrow D = - (c + d).Now, from the equation for B: b^2 + Db + F = 0 Rightarrow b^2 - (c + d)b + F = 0 Rightarrow F = (c + d)b - b^2.Now, substitute D and F into the equation for C:c^2 + h^2 + Dc + Eh + F = 0Substitute D = - (c + d) and F = (c + d)b - b^2:c^2 + h^2 - (c + d)c + Eh + (c + d)b - b^2 = 0Simplify:c^2 + h^2 - c^2 - cd + Eh + cb + db - b^2 = 0Simplify further:h^2 - cd + Eh + cb + db - b^2 = 0Group terms:Eh + (cb + db - cd - b^2 + h^2) = 0Solve for E:Eh = - (cb + db - cd - b^2 + h^2)E = frac{ - (cb + db - cd - b^2 + h^2) }{ h }So, now we have D, E, and F in terms of b, c, d, and h.Now, the equation of the circumcircle of BCD is:x^2 + y^2 - (c + d)x + Ey + F = 0We can write it as:x^2 + y^2 - (c + d)x + left( frac{ - (cb + db - cd - b^2 + h^2) }{ h } right) y + ( (c + d)b - b^2 ) = 0Now, we need to find the intersection of this circle with line AC, which is y = (h/c)x.Substitute y = (h/c)x into the circle equation:x^2 + left( frac{h}{c}x right)^2 - (c + d)x + left( frac{ - (cb + db - cd - b^2 + h^2) }{ h } right) left( frac{h}{c}x right ) + ( (c + d)b - b^2 ) = 0Simplify term by term:1. x^22. left( frac{h^2}{c^2} right) x^23. - (c + d)x4. left( frac{ - (cb + db - cd - b^2 + h^2) }{ h } right) left( frac{h}{c}x right ) = frac{ - (cb + db - cd - b^2 + h^2) }{ c } x5. ( (c + d)b - b^2 )Combine all terms:x^2 + frac{h^2}{c^2}x^2 - (c + d)x - frac{ (cb + db - cd - b^2 + h^2) }{ c } x + ( (c + d)b - b^2 ) = 0Factor out x^2 and x:left( 1 + frac{h^2}{c^2} right) x^2 - left( c + d + frac{ cb + db - cd - b^2 + h^2 }{ c } right) x + ( (c + d)b - b^2 ) = 0Simplify coefficients:First coefficient: 1 + frac{h^2}{c^2} = frac{c^2 + h^2}{c^2}Second coefficient:c + d + frac{ cb + db - cd - b^2 + h^2 }{ c } = c + d + b + frac{ db - cd - b^2 + h^2 }{ c }Wait, let me compute that step by step:frac{ cb + db - cd - b^2 + h^2 }{ c } = frac{ cb }{ c } + frac{ db }{ c } - frac{ cd }{ c } - frac{ b^2 }{ c } + frac{ h^2 }{ c } = b + frac{ db }{ c } - d - frac{ b^2 }{ c } + frac{ h^2 }{ c }So, the second coefficient becomes:c + d + b + frac{ db }{ c } - d - frac{ b^2 }{ c } + frac{ h^2 }{ c } = c + b + frac{ db - b^2 + h^2 }{ c }So, putting it all together, the equation is:frac{c^2 + h^2}{c^2} x^2 - left( c + b + frac{ db - b^2 + h^2 }{ c } right ) x + ( (c + d)b - b^2 ) = 0Multiply through by c^2 to eliminate denominators:(c^2 + h^2) x^2 - left( c(c + b) + db - b^2 + h^2 right ) x + c^2 ( (c + d)b - b^2 ) = 0Let me expand each term:First term: (c^2 + h^2) x^2Second term: - [ c^2 + cb + db - b^2 + h^2 ] xThird term: c^2 ( cb + db - b^2 ) = c^3 b + c^2 d b - c^2 b^2So, the quadratic equation is:(c^2 + h^2) x^2 - (c^2 + cb + db - b^2 + h^2) x + (c^3 b + c^2 d b - c^2 b^2) = 0This seems complicated, but maybe we can factor it or find roots.We know that x = c is a root because point C is on both the circle and the line AC. Let's verify:Plug x = c into the equation:(c^2 + h^2)c^2 - (c^2 + cb + db - b^2 + h^2)c + (c^3 b + c^2 d b - c^2 b^2) = 0Compute each term:1. (c^2 + h^2)c^2 = c^4 + c^2 h^22. - (c^2 + cb + db - b^2 + h^2)c = -c^3 - c^2 b - c d b + c b^2 - c h^23. c^3 b + c^2 d b - c^2 b^2Add them together:c^4 + c^2 h^2 - c^3 - c^2 b - c d b + c b^2 - c h^2 + c^3 b + c^2 d b - c^2 b^2Simplify term by term:- c^4- c^2 h^2 - c h^2 = c h^2 (c - 1)- - c^3- - c^2 b + c^3 b = c^2 b ( -1 + c )- - c d b + c^2 d b = c d b ( -1 + c )- c b^2 - c^2 b^2 = c b^2 (1 - c )Hmm, this doesn't seem to simplify to zero unless specific conditions are met. Maybe I made a mistake in the substitution or calculation.Alternatively, perhaps instead of trying to find the exact coordinates, I can use parametric equations or another method.Wait, maybe instead of coordinates, I can use projective geometry or look for homothety.Given that AB parallel CD, maybe the transformation that maps ABCD to A_1B_1C_1D_1 is a homothety, which preserves parallelism.But I need to find a center of homothety or something. Alternatively, maybe the new quadrilateral is similar to the original one, but scaled.Alternatively, perhaps the new quadrilateral is also a trapezoid because the construction preserves the parallelism.Wait, another idea: since A_1 is on the circumcircle of BCD, and AB parallel CD, maybe the angles at A_1 and B_1 are such that A_1B_1 is parallel to C_1D_1.Alternatively, maybe I can use the concept of similar triangles.Wait, let's consider triangles ABA_1 and CDA_1. Since AB parallel CD, maybe these triangles are similar.But I'm not sure. Let me think again.Wait, in the circumcircle of BCD, point A_1 is such that angle BA_1D = angle BCD. Similarly, in the circumcircle of ACD, point B_1 is such that angle AB_1D = angle ACD.Since AB parallel CD, angle BCD = angle ABC (alternate interior angles). Similarly, angle ACD = angle BAD.So, angle BA_1D = angle ABC and angle AB_1D = angle BAD.Hmm, interesting. So, maybe triangles BA_1D and ABC are similar because they have equal angles. Similarly, triangles AB_1D and BAD are similar.If that's the case, then the ratios of sides would be preserved, leading to some proportionalities.Wait, if triangles BA_1D and ABC are similar, then:frac{BA_1}{BA} = frac{BD}{BC}Similarly, if triangles AB_1D and BAD are similar, then:frac{AB_1}{AB} = frac{AD}{BD}But I'm not sure if this directly helps in showing that A_1B_1C_1D_1 is a trapezoid.Wait, maybe I can consider the cross ratio or something related to projective geometry, but that might be overcomplicating.Alternatively, maybe I can use the concept of harmonic conjugates.Wait, another approach: since A_1 is the second intersection of AC with the circumcircle of BCD, and similarly for B_1, C_1, D_1, maybe the quadrilateral A_1B_1C_1D_1 is the image of ABCD under some inversion or projective transformation that preserves parallelism.But I'm not sure about that.Wait, maybe I can use the fact that in a trapezoid, the intersection point of the diagonals lies on the line at infinity if it's a parallelogram, but since it's a trapezoid, the diagonals intersect at a finite point.Wait, let me recall that in a trapezoid, the intersection point of the diagonals, say O, divides the diagonals proportionally. That is, frac{AO}{OC} = frac{BO}{OD}.Maybe this ratio can help in showing that A_1B_1C_1D_1 is also a trapezoid.Wait, let's denote O as the intersection of diagonals AC and BD in trapezoid ABCD. Then, frac{AO}{OC} = frac{AB}{CD}.Similarly, since A_1 is on AC, maybe we can express A_1 in terms of O and some ratio.Wait, let's consider the power of point O with respect to the circumcircle of BCD. The power is OB cdot OD = OA_1 cdot OC.Similarly, the power of O with respect to the circumcircle of ACD is OA cdot OC_1 = OB cdot OD.Wait, so from the first power, OB cdot OD = OA_1 cdot OC.From the second power, OA cdot OC_1 = OB cdot OD.Therefore, OA_1 cdot OC = OA cdot OC_1.Which implies frac{OA_1}{OA} = frac{OC_1}{OC}.Similarly, for points B_1 and D_1, we can derive similar ratios.Wait, so if frac{OA_1}{OA} = frac{OC_1}{OC}, then points A_1 and C_1 are related by a homothety centered at O.Similarly, points B_1 and D_1 are related by the same homothety.Therefore, the quadrilateral A_1B_1C_1D_1 is the image of ABCD under a homothety centered at O.Since homothety preserves parallelism, if AB parallel CD, then A_1B_1 parallel C_1D_1.Therefore, A_1B_1C_1D_1 is also a trapezoid.Wait, that seems promising. Let me recap:1. Let O be the intersection of diagonals AC and BD in trapezoid ABCD.2. The power of O with respect to the circumcircle of BCD is OB cdot OD = OA_1 cdot OC.3. Similarly, the power of O with respect to the circumcircle of ACD is OA cdot OC_1 = OB cdot OD.4. Equating the two, we get OA_1 cdot OC = OA cdot OC_1, leading to frac{OA_1}{OA} = frac{OC_1}{OC}.5. This implies that A_1 and C_1 are images of A and C under a homothety centered at O with ratio frac{OA_1}{OA}.6. Similarly, for B_1 and D_1, we can show that they are images of B and D under the same homothety.7. Since homothety preserves parallelism, and AB parallel CD, their images A_1B_1 and C_1D_1 are also parallel.8. Therefore, A_1B_1C_1D_1 is a trapezoid.Yes, that makes sense. So, the key idea is that the points A_1, B_1, C_1, D_1 are images of A, B, C, D under a homothety centered at the intersection of the diagonals, which preserves parallelism, hence the new quadrilateral is also a trapezoid.