💡Okay, so I have this problem about points on a coordinate plane. Let me try to understand it step by step. We have a set of 2000 points, each with coordinates (x_i, y_i). The conditions are that each x_i is between 0 and 83, each y_i is between 0 and 1, and all the x_i are unique. So, no two points share the same x-coordinate. That means each x-coordinate from 0 to 83 is used exactly once, but since there are 2000 points and only 84 possible integer x-coordinates (from 0 to 83 inclusive), wait, that can't be right. Wait, hold on, the x_i don't have to be integers, right? The problem just says 0 ≤ x_i ≤ 83, so they can be any real numbers in that interval, but all distinct. So, we have 2000 distinct x-coordinates spread out between 0 and 83, and each has a y-coordinate between 0 and 1.The question is asking for the largest positive integer n such that, no matter how these 2000 points are arranged (as long as they satisfy the given conditions), there will always be some unit square that contains exactly n points on its interior or boundary.Hmm, okay. So, I need to find the maximum n that is guaranteed to exist in any such configuration. That is, regardless of how someone places these 2000 points with unique x-coordinates between 0 and 83, and y-coordinates between 0 and 1, there will always be at least one unit square that contains exactly n points.I think this is related to the pigeonhole principle. Since we have 2000 points and a certain number of possible unit squares, we can figure out the minimum number of points that must lie within some unit square.But first, let me think about the grid. Since the x-coordinates range from 0 to 83 and the y-coordinates from 0 to 1, the entire area is 83 units wide and 1 unit tall. If we consider unit squares, each unit square will cover a 1x1 area. So, in the x-direction, we can have 83 such unit squares side by side, each covering an interval of 1 unit in x and the full 1 unit in y.Wait, but actually, since the y-coordinate only goes up to 1, each unit square in the y-direction would span from y=0 to y=1, right? So, each unit square is actually a rectangle of 1x1, but in our case, the height is fixed at 1, so each unit square is just a vertical strip of width 1 and height 1.But wait, no. A unit square has both width and height of 1. So, in this case, since the y-coordinate only goes up to 1, a unit square would have to span from some x to x+1 and from some y to y+1. But since y can't exceed 1, the maximum y for a unit square is 1. So, the unit squares can only be placed such that their bottom side is at y=0 and top side at y=1, or shifted up, but since y can't go beyond 1, the only possible unit squares are those that span from y=0 to y=1 and some x to x+1.Wait, that doesn't make sense because if the unit square has to have both width and height of 1, but our y-coordinate only goes up to 1, then the unit square can only be placed such that its bottom side is at y=0 and top side at y=1, and its left side is at some x and right side at x+1. So, in effect, each unit square is a vertical strip of width 1 and height 1, starting at x=0 to x=1, x=1 to x=2, etc., up to x=82 to x=83.So, there are 83 such unit squares in the x-direction, each covering a width of 1 and the full height of 1. Since we have 2000 points and 83 unit squares, by the pigeonhole principle, at least one of these unit squares must contain at least ⎡2000/83⎤ points.Calculating that: 2000 divided by 83 is approximately 24.096. So, the ceiling of that is 25. So, at least one unit square must contain at least 25 points.But the question is asking for the largest n such that some unit square contains exactly n points. So, is 25 the answer? Because in any configuration, there must be at least one unit square with at least 25 points, but could there be a configuration where all unit squares have at most 25 points? Or is it possible that in some configurations, some unit squares have more than 25 points, but we need the largest n that is guaranteed to exist.Wait, the question is asking for the largest n such that, for any good set, there exists a unit square with exactly n points. So, it's not about the minimum maximum, but rather the maximum n that is always achievable, regardless of the configuration.So, even if in some configurations, some unit squares have more than 25 points, we need the largest n that is guaranteed to be present in any configuration.So, if we can show that in any configuration, there must be at least one unit square with exactly 25 points, then 25 is the answer. But if it's possible to have a configuration where all unit squares have at most 24 points, then 24 would be the answer.But wait, from the pigeonhole principle, we know that at least one unit square must have at least 25 points. So, in any configuration, there must be at least one unit square with at least 25 points. But does that mean that there must be a unit square with exactly 25 points?Not necessarily. It could be that some unit squares have 25, 26, etc., but the question is about the existence of a unit square with exactly n points. So, if in some configurations, all unit squares have more than 25 points, then n would be higher. But we need the largest n that is guaranteed, regardless of the configuration.Wait, no. Because if in some configurations, the maximum number of points in a unit square is 25, then n can't be higher than 25, because in that case, there is no unit square with more than 25 points. But in other configurations, there might be unit squares with more points.But the question is asking for the largest n such that, for any good set, there exists a unit square with exactly n points. So, n has to be such that no matter how you arrange the points, you will always have at least one unit square with exactly n points.So, if in some configurations, the maximum number of points in a unit square is 25, and in others, it's higher, then n can't be higher than 25, because in the first case, there is no unit square with more than 25 points. Therefore, n must be 25.Wait, but let me think again. Suppose that in some configurations, all unit squares have exactly 24 or 25 points, but in others, some have 25, some have 26, etc. Then, the minimal maximum is 25, but the exact number of points in some unit square could vary.But the question is about the existence of a unit square with exactly n points. So, if in any configuration, there must be at least one unit square with exactly 25 points, then n is 25. But if it's possible to have a configuration where all unit squares have either 24 or 25 points, but not exactly 25, then n would be 24.Wait, but the pigeonhole principle tells us that at least one unit square has at least 25 points. So, in any configuration, there is at least one unit square with at least 25 points. But does that imply that there is a unit square with exactly 25 points? Not necessarily. It could be that all unit squares have more than 25 points, but that's not possible because 83 unit squares times 25 points would be 2075 points, but we only have 2000 points. So, 83*25=2075, which is more than 2000, so it's impossible for all unit squares to have at least 25 points. Therefore, the maximum number of points in a unit square must be at least 25, but not all can be 25.Wait, let me calculate: 83 unit squares, each with at least 25 points, would require 83*25=2075 points, but we only have 2000. So, 2000 - (83*24)=2000 - 1992=8 points. So, 8 unit squares would have 25 points, and the rest would have 24. So, in any configuration, there must be at least 8 unit squares with 25 points.Therefore, in any configuration, there are at least 8 unit squares with exactly 25 points. Therefore, the largest n such that in any good set, there exists a unit square with exactly n points is 25.Wait, but let me make sure. Suppose someone arranges the points such that as many unit squares as possible have 24 points, and the remaining points are distributed to make some unit squares have 25. Since 2000=83*24 + 8, so 8 unit squares would have 25 points, and the rest 75 would have 24. So, in any configuration, there must be at least 8 unit squares with 25 points. Therefore, n=25 is achievable, and it's the maximum because in some configurations, you can't have more than 25.Wait, but the question is about the interior and boundary of some unit square. So, does that include the boundary? Yes, it says "interior or boundary." So, points on the boundary are counted as well.But in our case, since the unit squares are aligned with the grid, and the points have unique x-coordinates, the points on the boundary would be those with x_i exactly at the edges of the unit squares. But since x_i are unique and can be anywhere, it's possible that some points lie exactly on the boundaries.But in terms of counting, whether a point is on the boundary or interior doesn't change the count, because the unit square includes both. So, the count remains the same.Therefore, the answer should be 25.Wait, but let me think again. Suppose someone arranges the points such that in each unit square, the number of points is as even as possible. So, 2000 points over 83 unit squares. 2000 divided by 83 is approximately 24.096. So, 24 points per unit square with some having 25.Specifically, 83*24=1992, so 2000-1992=8 points left, so 8 unit squares would have 25 points each. Therefore, in any arrangement, there must be at least 8 unit squares with 25 points. Therefore, n=25 is the minimal maximum, but since the question is about the existence of a unit square with exactly n points, and since in any arrangement, there are unit squares with exactly 25 points, then n=25 is the answer.I think that's it. So, the largest positive integer n is 25.

💡Alright, so I have this geometry problem here, and I need to prove that points A, P, and Q are collinear. Let me try to visualize the setup first.We have two circles, Γ₁ and Γ₂, intersecting at points A and B. There's a line passing through B that intersects Γ₁ at C and Γ₂ at D. So, I can imagine points C and D lying on this line, with B in between them. Now, on Γ₁, we choose a point E such that CE equals CB. Similarly, on Γ₂, we choose a point F such that BD equals DF. Importantly, E and F are not the same as B. Then, BF intersects Γ₁ again at P, and BE intersects Γ₂ again at Q. We need to show that A, P, and Q lie on a straight line.Hmm, okay. Let me try to draw this out mentally. Points A and B are common to both circles. The line through B cuts Γ₁ at C and Γ₂ at D. So, BC and BD are segments of this line. Then, E is on Γ₁ such that CE = CB, meaning E is another point on Γ₁ where the length from C to E is equal to CB. Similarly, F is on Γ₂ such that BD = DF, so F is another point on Γ₂ where the length from D to F is equal to BD.Now, BF intersects Γ₁ again at P, so starting from B, going through F, and intersecting Γ₁ at P. Similarly, BE intersects Γ₂ again at Q, so starting from B, going through E, and intersecting Γ₂ at Q. We need to show that A, P, and Q are collinear.I think I'll need to use some properties of circles, maybe power of a point, similar triangles, or cyclic quadrilaterals. Let me think about the given equal lengths: CE = CB and BD = DF. These might imply some isosceles triangles or equal angles.First, let's consider triangle CEB. Since CE = CB, triangle CEB is isosceles with base angles at E and B equal. Similarly, triangle DFB is isosceles with DF = BD, so angles at F and B are equal.Maybe I can find some angle equalities that can help me relate points P and Q to A. Since A is the other intersection point of Γ₁ and Γ₂, perhaps there are some cyclic quadrilateral properties I can use involving A.Let me try to look at angles involving point A. Since A is on both circles, angles subtended by AB in both circles might be useful. For example, angles at A subtended by chords BP and BQ.Wait, maybe I can use the Power of a Point theorem. For point B with respect to Γ₁ and Γ₂, the power would relate the lengths of segments from B to the points of intersection.For Γ₁, the power of B is BC * BB (but BB is zero, so that might not help). Wait, actually, the power of B with respect to Γ₁ is BC * BB', where BB' is the other intersection, but since B is on Γ₁, the power is zero. Hmm, maybe that's not the way to go.Alternatively, since E is on Γ₁ and CE = CB, maybe there's a reflection or rotation that maps C to E or something like that. Similarly, F is on Γ₂ with DF = BD, so maybe a similar transformation applies.Let me think about angles at points E and F. Since CE = CB, angle CEB equals angle CBE. Similarly, since DF = BD, angle DFB equals angle DBF.Maybe I can relate these angles to angles at points P and Q. Since P is on Γ₁, angles involving P might relate to angles at C or E. Similarly, Q is on Γ₂, so angles involving Q might relate to angles at D or F.Wait, perhaps using the cyclic quadrilateral properties. For example, in Γ₁, points B, C, E, P are on the circle, so opposite angles sum to 180 degrees. Similarly, in Γ₂, points B, D, F, Q are on the circle.Let me try to write down some angle equalities. In Γ₁, angle CEB equals angle CBE because triangle CEB is isosceles. Also, since E is on Γ₁, angle CEB equals angle CPB because they subtend the same arc CB. Wait, is that correct? Actually, angle CEB and angle CPB both subtend arc CB, so they should be equal.Similarly, in Γ₂, angle DFB equals angle DBF because triangle DFB is isosceles. Also, since F is on Γ₂, angle DFB equals angle DQB because they subtend the same arc DB.Hmm, so angle CPB equals angle CEB, and angle DQB equals angle DFB. Maybe I can relate these angles to angles at A.Since A is on both circles, maybe angles at A involving P and Q can be related to these angles. For example, angle APB might relate to angle ACB, and angle AQB might relate to angle ADB.Wait, let me think about the cyclic quadrilaterals. In Γ₁, quadrilateral BCPE is cyclic, so angle BPC equals angle BEC. Similarly, in Γ₂, quadrilateral BDQF is cyclic, so angle BQD equals angle BFD.But we know that angle BEC equals angle CBE, and angle BFD equals angle DBF. So, angle BPC equals angle CBE, and angle BQD equals angle DBF.Now, if I can relate angle CBE and angle DBF to angles at A, that might help. Since A is the other intersection point, maybe angles at A can be connected through some cyclic properties.Alternatively, maybe I can use the concept of spiral similarity or some kind of transformation that maps one triangle to another.Wait, another approach could be to use Menelaus' theorem or Ceva's theorem, but I'm not sure how to apply them here directly.Let me try to look at triangle BPQ and see if A lies on it. If I can show that A lies on line PQ, that would prove the collinearity.Alternatively, maybe I can show that angles PAQ and something else are equal, making A lie on PQ.Wait, perhaps using harmonic division or projective geometry concepts, but that might be too advanced for this problem.Let me go back to the angle equalities. Since angle BPC equals angle CBE, and angle BQD equals angle DBF, maybe I can relate these angles to angles at A.Since A is on both circles, angles at A subtended by BP and BQ might be related to these angles.Wait, in Γ₁, angle BAP equals angle BCP because they subtend the same arc BP. Similarly, in Γ₂, angle BAQ equals angle BDQ because they subtend the same arc BQ.But angle BCP is equal to angle CBE, and angle BDQ is equal to angle DBF. So, angle BAP equals angle CBE, and angle BAQ equals angle DBF.Now, if I can show that angle CBE equals angle DBF, then angle BAP would equal angle BAQ, implying that A lies on PQ.Wait, is angle CBE equal to angle DBF? Let me check.From the given, CE = CB, so triangle CEB is isosceles with angle CEB = angle CBE.Similarly, DF = BD, so triangle DFB is isosceles with angle DFB = angle DBF.But I don't see a direct relationship between angles CBE and DBF yet.Wait, maybe considering the line through B intersecting Γ₁ at C and Γ₂ at D, so line BCBD is a straight line. Therefore, angles at B related to C and D might be supplementary or something.Wait, actually, since C and D are on a straight line through B, angles CBE and DBF might be related through some vertical angles or something.Wait, no, because E is on Γ₁ and F is on Γ₂, so their positions are determined by the equal lengths.Hmm, maybe I need to consider the power of point B with respect to both circles.Wait, the power of B with respect to Γ₁ is BC * BB (but BB is zero, so that's not helpful). Similarly, for Γ₂, it's BD * BB, which is also zero. Hmm, maybe not.Alternatively, maybe using the fact that CE = CB and DF = BD, so E and F are reflections or something.Wait, if CE = CB, then E is the reflection of B over the perpendicular bisector of CB. Similarly, F is the reflection of B over the perpendicular bisector of BD.But I'm not sure if that helps directly.Wait, maybe considering triangles CEB and DFB. Both are isosceles, so maybe there's a similarity or congruence between them.But without knowing more about the lengths or angles, it's hard to say.Wait, perhaps using the fact that angles at E and F are equal to angles at B, as we established earlier.So, angle CEB = angle CBE, and angle DFB = angle DBF.If I can relate these angles to angles at P and Q, maybe I can find a relationship between P and Q that involves A.Wait, in Γ₁, angle BPC = angle BEC = angle CBE.Similarly, in Γ₂, angle BQD = angle BFD = angle DBF.So, angle BPC = angle CBE, and angle BQD = angle DBF.If I can show that angle CBE = angle DBF, then angle BPC = angle BQD, which might imply some relationship between P and Q.But how can I show that angle CBE = angle DBF?Wait, since CE = CB and DF = BD, maybe triangles CEB and DFB are similar.Wait, triangle CEB has sides CE = CB, and triangle DFB has sides DF = DB. So, both are isosceles triangles with two equal sides.But without knowing the included angles, it's hard to say if they are similar.Alternatively, maybe the angles at B are equal because of some symmetry.Wait, since C and D are on a straight line through B, and E and F are constructed similarly on each circle, maybe there's some symmetry that makes angles CBE and DBF equal.Alternatively, maybe considering the arcs subtended by these angles.Wait, in Γ₁, angle CBE subtends arc CE, and in Γ₂, angle DBF subtends arc DF.But since CE = CB and DF = BD, maybe the arcs are equal, leading to equal angles.Wait, in Γ₁, arc CE equals arc CB because CE = CB, so angle CBE, which subtends arc CE, equals angle CEB, which subtends arc CB.Similarly, in Γ₂, arc DF equals arc DB because DF = DB, so angle DBF, which subtends arc DF, equals angle DFB, which subtends arc DB.But I'm not sure if this directly leads to angle CBE = angle DBF.Wait, maybe considering the power of point E with respect to Γ₂ or something like that.Alternatively, maybe using inversion. But that might be too complicated.Wait, another idea: since E is on Γ₁ and CE = CB, maybe BE is the angle bisector of angle CBA or something.Similarly, since F is on Γ₂ and DF = BD, maybe BF is the angle bisector of angle DBA.But I'm not sure.Wait, let me think about the angles at A. Since A is on both circles, angles at A might be related to angles at P and Q.In Γ₁, angle BAP equals angle BCP because they subtend the same arc BP.Similarly, in Γ₂, angle BAQ equals angle BDQ because they subtend the same arc BQ.But angle BCP is equal to angle CBE, and angle BDQ is equal to angle DBF.So, angle BAP = angle CBE, and angle BAQ = angle DBF.If I can show that angle CBE = angle DBF, then angle BAP = angle BAQ, which would imply that A lies on PQ, making A, P, Q collinear.So, how can I show that angle CBE = angle DBF?Wait, since CE = CB and DF = BD, maybe triangles CEB and DFB are congruent?But for that, we need more information. We know two sides are equal, but we don't know the included angles.Alternatively, maybe using the fact that both triangles are isosceles and some angles are equal.Wait, in Γ₁, angle CEB = angle CBE, and in Γ₂, angle DFB = angle DBF.If I can relate angle CEB to angle DFB, maybe through some other angles.Wait, since points C, B, D are colinear, maybe angles involving E and F can be related through some transversal.Wait, perhaps considering the angles at E and F with respect to line CD.But I'm not sure.Wait, another approach: since CE = CB and DF = BD, maybe points E and F are such that BE and BF are symmedians or something.Alternatively, maybe using the fact that E and F are reflections of B over the midpoints of CB and BD, respectively.Wait, if CE = CB, then E is the reflection of B over the midpoint of CB. Similarly, F is the reflection of B over the midpoint of BD.So, if I consider midpoints M of CB and N of BD, then E is the reflection of B over M, and F is the reflection over N.This might imply that BE = 2BM and BF = 2BN.But I'm not sure if that helps directly.Wait, but if E and F are reflections, then lines BE and BF are related to the midpoints.Hmm.Alternatively, maybe considering vectors or coordinate geometry, but that might be too involved.Wait, going back to the angle approach. If I can show that angle CBE = angle DBF, then angle BAP = angle BAQ, implying A lies on PQ.So, let's try to find a relationship between angle CBE and angle DBF.Since CE = CB, triangle CEB is isosceles, so angle CEB = angle CBE.Similarly, DF = BD, so triangle DFB is isosceles, angle DFB = angle DBF.Now, in Γ₁, angle CEB = angle CBE = angle CPB (since they subtend the same arc CB).Similarly, in Γ₂, angle DFB = angle DBF = angle DQB (since they subtend the same arc DB).So, angle CPB = angle CBE and angle DQB = angle DBF.If I can relate angle CPB and angle DQB, maybe through some other angles.Wait, since points P and Q are intersections of BF and BE with the circles, maybe there's a relationship between arcs BP and BQ.Alternatively, maybe considering the angles at A.Wait, in Γ₁, angle BAP = angle BCP = angle CBE.In Γ₂, angle BAQ = angle BDQ = angle DBF.So, angle BAP = angle CBE and angle BAQ = angle DBF.If angle CBE = angle DBF, then angle BAP = angle BAQ, meaning that A lies on PQ.So, the key is to show that angle CBE = angle DBF.How can I show that?Wait, since CE = CB and DF = BD, maybe the triangles CEB and DFB are similar.Triangles CEB and DFB:- CE = CB- DF = DB- Angle at B: angle CBE and angle DBFIf I can show that the ratio of sides is equal and the included angles are equal, then the triangles would be similar.But CE/CB = 1, DF/DB = 1, so the ratios are equal.But what about the included angles? The included angles are angle CBE and angle DBF.Wait, if I can show that angle CBE = angle DBF, then by SAS similarity, triangles CEB and DFB would be similar.But that's what I'm trying to prove, so I can't assume that.Alternatively, maybe using the Law of Sines in triangles CEB and DFB.In triangle CEB:CE = CB, so it's isosceles, so angle CEB = angle CBE.By Law of Sines:CE / sin(angle CBE) = CB / sin(angle CEB)But since CE = CB and angle CBE = angle CEB, this holds.Similarly, in triangle DFB:DF = DB, so it's isosceles, angle DFB = angle DBF.By Law of Sines:DF / sin(angle DBF) = DB / sin(angle DFB)Again, since DF = DB and angle DBF = angle DFB, this holds.But I don't see how this helps me relate angle CBE and angle DBF.Wait, maybe considering the power of point E with respect to Γ₂.Power of E with respect to Γ₂ is EC * EB = ED * EF.But I don't know if that's helpful.Alternatively, power of F with respect to Γ₁ is FB * FD = FC * FE.Hmm, not sure.Wait, another idea: since E is on Γ₁ and F is on Γ₂, maybe lines BE and BF have some harmonic properties.Alternatively, maybe considering the cyclic quadrilaterals involving A.Wait, since A is on both circles, maybe quadrilateral APBQ is cyclic, but I don't know.Wait, actually, if A, P, Q are collinear, then APQ is a straight line, so it's not a quadrilateral.Wait, maybe using the radical axis theorem. The radical axis of Γ₁ and Γ₂ is line AB. So, any point on AB has equal power with respect to both circles.But I'm not sure how that helps here.Wait, maybe considering the power of point P with respect to Γ₂.Power of P with respect to Γ₂ is PB * PF = PD * PC.But I don't know if that helps.Wait, another approach: since E is such that CE = CB, and F is such that DF = BD, maybe points E and F are midpoints of arcs or something.Wait, in Γ₁, since CE = CB, point E might be the midpoint of arc CB that doesn't contain A.Similarly, in Γ₂, since DF = BD, point F might be the midpoint of arc DB that doesn't contain A.If that's the case, then lines BE and BF would be angle bisectors or something.But I'm not sure.Wait, if E is the midpoint of arc CB in Γ₁, then BE would bisect angle CBA.Similarly, if F is the midpoint of arc DB in Γ₂, then BF would bisect angle DBA.If that's the case, then angles CBE and DBF would be equal, each being half of angle CBA and DBA, respectively.But since points C, B, D are colinear, angle CBA and DBA are supplementary.Wait, no, because C and D are on a straight line through B, so angle CBA + angle DBA = 180 degrees.If BE and BF are bisecting these angles, then angle CBE = angle CBA / 2 and angle DBF = angle DBA / 2.But since angle CBA + angle DBA = 180, then angle CBE + angle DBF = 90 degrees.Hmm, but that doesn't necessarily make them equal.Wait, unless angle CBA = angle DBA, which would only happen if BC = BD, but we don't know that.So, maybe E and F are not necessarily midpoints of arcs.Hmm, this is getting complicated. Maybe I need to try a different approach.Wait, let's consider the spiral similarity that maps Γ₁ to Γ₂. Since A and B are common points, maybe there's a spiral similarity centered at A that maps Γ₁ to Γ₂.If such a similarity exists, it would map points on Γ₁ to points on Γ₂, preserving angles and ratios.So, maybe point C on Γ₁ maps to some point on Γ₂, and similarly for E and P.But I'm not sure how to apply this directly.Wait, another idea: since CE = CB and DF = BD, maybe triangles CEB and DFB are congruent.But for congruence, we need more than just two sides equal; we need either the included angle or another side.We know CE = CB and DF = DB, but without knowing the included angles or another side, we can't conclude congruence.Wait, but if we consider the circles, maybe the arcs subtended by CE and DF are equal, leading to equal angles.Wait, in Γ₁, arc CE equals arc CB because CE = CB.Similarly, in Γ₂, arc DF equals arc DB because DF = DB.So, the arcs are equal in their respective circles.But since the circles might have different radii, the angles subtended by these arcs at the centers might not be equal, but the angles at the circumference would be equal within each circle.Wait, but how does that help me relate angles in different circles?Hmm.Wait, maybe using the fact that angles subtended by equal arcs are equal.But since the circles are different, the arcs might not correspond.Wait, unless the arcs correspond through some transformation.I'm stuck here. Maybe I need to look for another property or theorem that can help me relate these points.Wait, perhaps using the concept of isogonal conjugates. If BE and BF are isogonal, then their reflections might relate to each other.But I'm not sure.Wait, another idea: since E is on Γ₁ and F is on Γ₂, maybe lines BE and BF are symmedians or something.Alternatively, maybe using the fact that E and F are such that BE and BF are reflections over some axis.Wait, I'm not making progress here. Maybe I need to try to write down all the angle equalities and see if something cancels out.So, let's recap:1. In Γ₁, triangle CEB is isosceles with CE = CB, so angle CEB = angle CBE.2. In Γ₂, triangle DFB is isosceles with DF = DB, so angle DFB = angle DBF.3. In Γ₁, angle BPC = angle BEC = angle CBE.4. In Γ₂, angle BQD = angle BFD = angle DBF.5. In Γ₁, angle BAP = angle BCP = angle CBE.6. In Γ₂, angle BAQ = angle BDQ = angle DBF.So, angle BAP = angle CBE and angle BAQ = angle DBF.If I can show that angle CBE = angle DBF, then angle BAP = angle BAQ, implying that A lies on PQ.So, how can I show that angle CBE = angle DBF?Wait, since points C, B, D are colinear, maybe angles involving E and F can be related through some transversal.Wait, perhaps considering the angles at E and F with respect to line CD.But I'm not sure.Wait, another idea: since CE = CB and DF = BD, maybe triangles CEB and DFB are congruent by SAS if angle CBE = angle DBF.But that's what I'm trying to prove, so I can't assume that.Alternatively, maybe using the Law of Cosines in triangles CEB and DFB.In triangle CEB:CE = CB, so let's denote CE = CB = x.Then, by Law of Cosines:EB² = CE² + CB² - 2 * CE * CB * cos(angle ECB)But angle ECB = angle CEB = angle CBE.Wait, this might not help directly.Similarly, in triangle DFB:DF = DB = y.FB² = DF² + DB² - 2 * DF * DB * cos(angle DFB)But angle DFB = angle DBF.Again, not sure.Wait, maybe considering the power of point E with respect to Γ₂.Power of E: EC * EB = ED * EF.But I don't know the lengths, so this might not help.Similarly, power of F with respect to Γ₁: FB * FD = FC * FE.Again, not helpful without more information.Wait, maybe considering the triangles EBC and FBD.They have sides CE = CB and DF = DB, but without knowing the included angles or another side, it's hard to relate them.Wait, unless the triangles are similar in some way.Wait, if angle CBE = angle DBF, then triangles CEB and DFB would be similar by SAS.But again, that's what I'm trying to prove.Wait, maybe using the fact that angles at E and F are equal to angles at B, as established earlier.So, angle CEB = angle CBE and angle DFB = angle DBF.If I can relate angle CEB to angle DFB, maybe through some other angles.Wait, since E is on Γ₁ and F is on Γ₂, maybe angles subtended by BE and BF can be related.Wait, in Γ₁, angle BEC = angle BPC.In Γ₂, angle BFD = angle BQD.So, angle BPC = angle CBE and angle BQD = angle DBF.If I can relate angle BPC and angle BQD, maybe through some other angles.Wait, since points P and Q are on Γ₁ and Γ₂ respectively, maybe there's a relationship between arcs BP and BQ.Alternatively, maybe considering the angles at A.Wait, in Γ₁, angle BAP = angle BCP = angle CBE.In Γ₂, angle BAQ = angle BDQ = angle DBF.So, angle BAP = angle CBE and angle BAQ = angle DBF.If I can show that angle CBE = angle DBF, then angle BAP = angle BAQ, implying that A lies on PQ.So, how can I show that angle CBE = angle DBF?Wait, maybe considering the cyclic quadrilaterals involving E and F.Wait, in Γ₁, quadrilateral BCPE is cyclic, so angle BEC = angle BPC.Similarly, in Γ₂, quadrilateral BDQF is cyclic, so angle BFD = angle BQD.But I already used that.Wait, another idea: since CE = CB and DF = BD, maybe points E and F are such that BE and BF are equal in some way.Wait, but without knowing the radii of the circles, I can't say that.Wait, maybe considering the inversion with respect to point B.If I invert the figure with respect to B, maybe some properties will become clearer.But inversion is a bit advanced, and I'm not sure.Wait, another idea: since CE = CB, point E is the reflection of B over the midpoint of CB.Similarly, F is the reflection of B over the midpoint of BD.So, if I consider midpoints M of CB and N of BD, then E = reflection of B over M, and F = reflection of B over N.So, BE = 2 BM and BF = 2 BN.But since M and N are midpoints, BM = MC and BN = ND.So, BE = 2 BM = 2 MC and BF = 2 BN = 2 ND.But I don't see how this helps.Wait, maybe considering the homothety that maps Γ₁ to Γ₂.If there's a homothety centered at A that maps Γ₁ to Γ₂, then points C and D would correspond, as well as E and F.But I'm not sure.Wait, another approach: since E is on Γ₁ and CE = CB, maybe arc CE equals arc CB in Γ₁.Similarly, in Γ₂, arc DF equals arc DB.So, the arcs are equal in their respective circles.But since the circles might have different radii, the angles subtended by these arcs at the centers might not be equal, but the angles at the circumference would be equal within each circle.Wait, but how does that help me relate angles in different circles?Hmm.Wait, maybe considering the angles at E and F with respect to line AB.But I'm not sure.Wait, another idea: since A is the other intersection point of Γ₁ and Γ₂, maybe lines AP and AQ are related through some symmetry.Wait, if I can show that AP and AQ are isogonal conjugates, then A would lie on PQ.But I'm not sure.Wait, maybe using the fact that angles BAP and BAQ are equal, as we saw earlier.If angle BAP = angle BAQ, then A lies on the angle bisector of angle PAQ, but since we need collinearity, that might not be enough.Wait, actually, if angle BAP = angle BAQ, then points P and Q lie on a circle such that A lies on the angle bisector, but I'm not sure.Wait, maybe considering triangle APQ and point B.But I'm not making progress.Wait, let me try to summarize what I have:- In Γ₁, angle BPC = angle CBE.- In Γ₂, angle BQD = angle DBF.- In Γ₁, angle BAP = angle CBE.- In Γ₂, angle BAQ = angle DBF.So, angle BAP = angle CBE and angle BAQ = angle DBF.If I can show that angle CBE = angle DBF, then angle BAP = angle BAQ, implying that A lies on PQ.So, the crux is to show angle CBE = angle DBF.Wait, maybe considering the triangle formed by points C, B, D.Since C, B, D are colinear, maybe some external angles can be related.Wait, in triangle CEB, angle CEB = angle CBE.In triangle DFB, angle DFB = angle DBF.If I can relate these angles through some other triangles or lines.Wait, maybe considering the lines BE and BF and their intersections with the circles.Wait, another idea: since E is on Γ₁ and F is on Γ₂, maybe lines BE and BF intersect the circles again at P and Q, which are related through some property.Wait, maybe using the fact that angles at P and Q are equal to angles at E and F, which are equal to angles at B.But I'm stuck.Wait, maybe considering the cyclic quadrilaterals involving A.Wait, since A is on both circles, maybe quadrilateral APBQ is cyclic, but I don't know.Wait, if I can show that angle PAQ = 180 degrees, then A, P, Q would be collinear.But how?Wait, angle PAQ is the sum of angle BAP and angle BAQ.If angle BAP = angle BAQ, then angle PAQ = 2 angle BAP.But I need angle PAQ = 180 degrees.Wait, unless angle BAP = 90 degrees, but that's not necessarily the case.Wait, maybe considering that angle BAP + angle BAQ = 180 degrees.But I don't know.Wait, another idea: since points P and Q are defined as the second intersections of BF and BE with the circles, maybe there's a relationship between arcs BP and BQ.Wait, in Γ₁, arc BP is equal to arc BE because angles subtended by them are equal.Similarly, in Γ₂, arc BQ is equal to arc BF.But I don't see how that helps.Wait, maybe considering the power of point A with respect to both circles.Power of A with respect to Γ₁: AB * AA' = AC * AE, but A is on Γ₁, so power is zero.Similarly, for Γ₂: AB * AA' = AD * AF, but again, power is zero.Hmm.Wait, maybe considering the radical axis. The radical axis of Γ₁ and Γ₂ is line AB, so any point on AB has equal power with respect to both circles.But I'm not sure.Wait, another idea: since E is on Γ₁ and F is on Γ₂, maybe lines AE and AF have some relationship.But I don't know.Wait, maybe considering the angles at A.In Γ₁, angle BAP = angle BCP = angle CBE.In Γ₂, angle BAQ = angle BDQ = angle DBF.If I can show that angle CBE + angle DBF = 180 degrees, then angle BAP + angle BAQ = 180 degrees, implying that A lies on PQ.But is angle CBE + angle DBF = 180 degrees?Wait, since points C, B, D are colinear, angle CBE + angle DBF = angle CBA + angle DBA.But angle CBA + angle DBA = 180 degrees because C, B, D are colinear.So, angle CBE + angle DBF = 180 degrees.Wait, but angle CBE and angle DBF are angles at B, so their sum is 180 degrees.But earlier, we saw that angle BAP = angle CBE and angle BAQ = angle DBF.So, angle BAP + angle BAQ = angle CBE + angle DBF = 180 degrees.Therefore, angle PAQ = 180 degrees, meaning that points A, P, Q are collinear.Yes! That makes sense.So, to summarize:1. In Γ₁, angle BAP = angle BCP = angle CBE.2. In Γ₂, angle BAQ = angle BDQ = angle DBF.3. Since points C, B, D are colinear, angle CBE + angle DBF = 180 degrees.4. Therefore, angle BAP + angle BAQ = 180 degrees.5. Hence, points A, P, Q are collinear.I think that's the solution.

💡Okay, so I have this problem with three equations involving variables x, y, and z. I need to find the value of xyz. Let me try to figure this out step by step.First, let me write down the given equations to have them clear in my mind:1. (x + y + z)(xy + xz + yz) = 362. x²(y + z) + y²(x + z) + z²(x + y) = 123. (x + y + z)² = x² + y² + z² + 12Hmm, okay. So, three equations. I need to find xyz. Let me think about how these equations relate to each other.Starting with the first equation: (x + y + z)(xy + xz + yz) = 36. I remember that when you multiply (x + y + z) by (xy + xz + yz), you get terms like x²y, x²z, xy², y²z, xz², yz², and also 3xyz. So, expanding this might help.Let me try expanding it:(x + y + z)(xy + xz + yz) = x(xy + xz + yz) + y(xy + xz + yz) + z(xy + xz + yz)Calculating each part:- x(xy + xz + yz) = x²y + x²z + xyz- y(xy + xz + yz) = xy² + xyz + y²z- z(xy + xz + yz) = xyz + xz² + yz²Adding all these together:x²y + x²z + xyz + xy² + xyz + y²z + xyz + xz² + yz²Combine like terms:x²y + x²z + xy² + y²z + xz² + yz² + 3xyzSo, the first equation becomes:x²y + x²z + xy² + y²z + xz² + yz² + 3xyz = 36Now, looking at the second equation: x²(y + z) + y²(x + z) + z²(x + y) = 12Let me expand this as well:x²(y + z) = x²y + x²zy²(x + z) = xy² + y²zz²(x + y) = xz² + yz²Adding them together:x²y + x²z + xy² + y²z + xz² + yz² = 12Wait a minute, this is exactly the same as the terms in the first equation except for the 3xyz. So, in the first equation, we have:(x²y + x²z + xy² + y²z + xz² + yz²) + 3xyz = 36But from the second equation, we know that (x²y + x²z + xy² + y²z + xz² + yz²) = 12So, substituting that into the first equation:12 + 3xyz = 36Subtract 12 from both sides:3xyz = 24Divide both sides by 3:xyz = 8Wait, so is it that straightforward? I mean, I used the first two equations and substituted the second into the first, and I got xyz = 8. That seems too easy, but let me check if I missed something.Looking back, the first equation expands to the sum of all those terms plus 3xyz, and the second equation is exactly the sum of those terms. So, substituting makes sense. Therefore, 3xyz = 36 - 12 = 24, so xyz = 8.But just to make sure, let me see if the third equation gives me any additional information or if it's consistent with what I found.The third equation is: (x + y + z)² = x² + y² + z² + 12I know that (x + y + z)² expands to x² + y² + z² + 2(xy + yz + xz). So, substituting that in:x² + y² + z² + 2(xy + yz + xz) = x² + y² + z² + 12Subtract x² + y² + z² from both sides:2(xy + yz + xz) = 12Divide both sides by 2:xy + yz + xz = 6So, the sum of the pairwise products is 6.Wait, in the first equation, we had (x + y + z)(xy + xz + yz) = 36. If I let S = x + y + z and P = xy + yz + xz, then the first equation is S * P = 36.From the third equation, we found that P = 6. So, S * 6 = 36, which means S = 6.So, x + y + z = 6, and xy + yz + xz = 6.Now, if I think about symmetric equations, these are the elementary symmetric sums for three variables. So, if I consider a cubic equation whose roots are x, y, z, it would be t³ - S t² + P t - Q = 0, where Q = xyz.So, the cubic equation would be t³ - 6t² + 6t - Q = 0.But I don't know if that helps me directly, but it's good to note.But from earlier, I found that xyz = 8. Let me see if that's consistent.If I have x + y + z = 6, xy + yz + xz = 6, and xyz = 8, then the cubic equation would be t³ - 6t² + 6t - 8 = 0.I can try to factor this to see if it has real roots, just to check consistency.Let me try t = 2: 8 - 24 + 12 - 8 = -12 ≠ 0t = 1: 1 - 6 + 6 - 8 = -7 ≠ 0t = 4: 64 - 96 + 24 - 8 = -16 ≠ 0Hmm, maybe t = something else. Alternatively, perhaps it's not necessary to factor it since the problem only asks for xyz, which I already found as 8.But just to make sure, let me recap:From the first equation: (x + y + z)(xy + xz + yz) = 36From the third equation: (x + y + z)² = x² + y² + z² + 12Which gave me that xy + yz + xz = 6So, substituting back into the first equation: (x + y + z)*6 = 36 => x + y + z = 6Then, from the second equation, I found that x²(y + z) + y²(x + z) + z²(x + y) = 12, which is equal to (x²y + x²z + y²x + y²z + z²x + z²y) = 12And from the expansion of the first equation, that sum is equal to 36 - 3xyz, so 12 = 36 - 3xyz => 3xyz = 24 => xyz = 8Yes, that seems consistent.So, I think I did it correctly. The value of xyz is 8.**Final Answer**The value of ( xyz ) is boxed{8}.

💡Okay, so I have this problem about two triangles, ABC and PQR. The conditions given are that angle A is equal to angle P, and the absolute difference between angles B and C is less than the absolute difference between angles Q and R. I need to show that the sum of the sines of angles A, B, and C is greater than the sum of the sines of angles P, Q, and R. Then, I also have to figure out what angles A, B, and C would maximize the sum of their sines.Alright, let's start by understanding the given conditions. We have two triangles, so each has three angles that add up to 180 degrees. The first condition is that angle A equals angle P. That means in both triangles, the angle at A and P are the same. The second condition is about the differences between the other two angles. Specifically, the absolute difference between angles B and C in triangle ABC is less than the absolute difference between angles Q and R in triangle PQR. So, in triangle ABC, angles B and C are closer to each other in measure compared to angles Q and R in triangle PQR.I need to show that sin A + sin B + sin C is greater than sin P + sin Q + sin R. Since angle A equals angle P, their sines are equal. So, the comparison really comes down to comparing sin B + sin C with sin Q + sin R.Hmm, maybe I can use some trigonometric identities here. I remember that the sum of sines can be expressed using product-to-sum formulas. Let me recall: sin B + sin C can be written as 2 sin((B + C)/2) cos((B - C)/2). Similarly, sin Q + sin R would be 2 sin((Q + R)/2) cos((Q - R)/2).Given that angles B + C in triangle ABC are equal to 180 - A, and angles Q + R in triangle PQR are equal to 180 - P. But since angle A equals angle P, that means B + C equals Q + R. So, the first part of the sine terms in both expressions, sin((B + C)/2) and sin((Q + R)/2), are actually equal because (B + C)/2 equals (Q + R)/2.That simplifies things a bit. So, sin B + sin C is equal to 2 sin((B + C)/2) cos((B - C)/2), and sin Q + sin R is equal to 2 sin((Q + R)/2) cos((Q - R)/2). Since sin((B + C)/2) equals sin((Q + R)/2), the comparison now depends on the cosine terms.The problem states that |B - C| is less than |Q - R|. Since cosine is a decreasing function in the interval [0, π], a smaller angle will result in a larger cosine value. Therefore, cos((B - C)/2) is greater than cos((Q - R)/2). Putting it all together, sin B + sin C is greater than sin Q + sin R because the cosine term in the first expression is larger, even though the sine terms are equal. Adding sin A to both sides, which is equal to sin P, we get that sin A + sin B + sin C is greater than sin P + sin Q + sin R. That seems to make sense.Now, moving on to the second part: what angles A, B, C maximize sin A + sin B + sin C? I know that the sine function reaches its maximum value of 1 at 90 degrees. However, in a triangle, the sum of angles is 180 degrees, so we can't have all three angles being 90 degrees. I think symmetry might play a role here. If all angles are equal, meaning the triangle is equilateral, each angle would be 60 degrees. The sine of 60 degrees is √3/2, so the sum would be 3*(√3/2) = (3√3)/2. Is this the maximum? Let me think. If I make one angle larger, say 90 degrees, then the other two angles would have to add up to 90 degrees. Let's say one is 45 degrees and the other is 45 degrees. Then, sin 90 + sin 45 + sin 45 = 1 + √2/2 + √2/2 = 1 + √2 ≈ 2.414. Comparing this to the equilateral case, which is approximately 2.598, the equilateral triangle gives a larger sum.What if I make one angle very small, say approaching 0 degrees, then the other two angles would approach 90 degrees each. Then, sin A would approach 0, and sin B and sin C would each approach 1. So, the sum would approach 2, which is less than the equilateral case.Alternatively, if I have an isosceles triangle with two angles equal, say 75 degrees and 30 degrees. Then, sin 75 + sin 75 + sin 30. Sin 75 is approximately 0.9659, so 2*0.9659 + 0.5 ≈ 2.4318, which is still less than 2.598.It seems like the maximum occurs when all angles are equal, i.e., the triangle is equilateral. Therefore, the angles that maximize sin A + sin B + sin C are all 60 degrees.Wait, let me verify this with calculus. Suppose we fix angle A and try to maximize sin B + sin C. Since B + C = 180 - A, we can express C as 180 - A - B. Then, sin B + sin C = sin B + sin(180 - A - B). Using the identity sin(180 - x) = sin x, this becomes sin B + sin(A + B). To maximize this, take the derivative with respect to B and set it to zero. The derivative is cos B + cos(A + B). Setting this equal to zero: cos B + cos(A + B) = 0. Using the identity cos(A + B) = cos A cos B - sin A sin B, we get cos B + cos A cos B - sin A sin B = 0. Factoring out cos B: cos B (1 + cos A) - sin A sin B = 0.This seems a bit complicated. Maybe instead, since B + C is fixed, the maximum of sin B + sin C occurs when B = C. Because the sine function is concave in [0, π], the maximum occurs at the midpoint. So, when B = C, sin B + sin C is maximized.Therefore, for fixed A, the maximum of sin B + sin C occurs when B = C. So, to maximize sin A + sin B + sin C, we should have all angles equal, meaning A = B = C = 60 degrees.Yeah, that makes sense. So, the maximum occurs when the triangle is equilateral.**Final Answer**The angles (A), (B), and (C) that maximize (sin A + sin B + sin C) are each (boxed{60^circ}).

💡Okay, so I have this problem about runners A, B, and C, and I need to figure out the length of the race based on how they beat each other. Let me try to break it down step by step.First, the problem says that runner A can beat runner B by 40 yards. That means when A finishes the race, B is still 40 yards behind. Similarly, runner B can beat runner C by 30 yards, so when B finishes the race, C is 30 yards behind. And finally, it says that runner A can beat runner C by 65 yards. So when A finishes the race, C is 65 yards behind. I need to find the length of the race.Hmm, okay. I think I need to relate their speeds somehow. Maybe I can set up some equations based on their speeds and the distances they cover.Let me denote the length of the race as ( d ) yards. Let's also denote the speeds of runners A, B, and C as ( a ), ( b ), and ( c ) respectively. Speed is distance over time, so if I can find the time it takes for each runner to finish the race, I can relate their speeds.When A beats B by 40 yards, that means when A finishes the race (which is ( d ) yards), B has only run ( d - 40 ) yards. So, the time it takes A to run ( d ) yards is the same as the time it takes B to run ( d - 40 ) yards. So, I can write the equation:[frac{d}{a} = frac{d - 40}{b}]Similarly, when B beats C by 30 yards, when B finishes the race (which is ( d ) yards), C has only run ( d - 30 ) yards. So, the time it takes B to run ( d ) yards is the same as the time it takes C to run ( d - 30 ) yards. That gives me another equation:[frac{d}{b} = frac{d - 30}{c}]And finally, when A beats C by 65 yards, when A finishes the race (which is ( d ) yards), C has only run ( d - 65 ) yards. So, the time it takes A to run ( d ) yards is the same as the time it takes C to run ( d - 65 ) yards. That gives me the third equation:[frac{d}{a} = frac{d - 65}{c}]Okay, so now I have three equations:1. ( frac{d}{a} = frac{d - 40}{b} )2. ( frac{d}{b} = frac{d - 30}{c} )3. ( frac{d}{a} = frac{d - 65}{c} )I need to solve these equations to find ( d ). Let me see how I can manipulate these equations to find ( d ).Starting with the first equation: ( frac{d}{a} = frac{d - 40}{b} ). I can rearrange this to express ( b ) in terms of ( a ):[b = frac{(d - 40)}{d} cdot a]So, ( b = frac{d - 40}{d} a ). That's equation 1 rearranged.Now, moving to the second equation: ( frac{d}{b} = frac{d - 30}{c} ). Similarly, I can rearrange this to express ( c ) in terms of ( b ):[c = frac{(d - 30)}{d} cdot b]So, ( c = frac{d - 30}{d} b ). That's equation 2 rearranged.But from equation 1, I already have ( b ) in terms of ( a ). So I can substitute that into this equation to express ( c ) in terms of ( a ).Substituting ( b = frac{d - 40}{d} a ) into ( c = frac{d - 30}{d} b ):[c = frac{d - 30}{d} cdot frac{d - 40}{d} a]Simplifying that:[c = frac{(d - 30)(d - 40)}{d^2} a]Okay, so now I have ( c ) in terms of ( a ). Let me note that down.Now, looking at the third equation: ( frac{d}{a} = frac{d - 65}{c} ). I can rearrange this to express ( c ) in terms of ( a ):[c = frac{(d - 65)}{d} a]So, ( c = frac{d - 65}{d} a ). That's equation 3 rearranged.Wait, now I have two expressions for ( c ):1. From equations 1 and 2: ( c = frac{(d - 30)(d - 40)}{d^2} a )2. From equation 3: ( c = frac{d - 65}{d} a )Since both expressions equal ( c ), I can set them equal to each other:[frac{(d - 30)(d - 40)}{d^2} a = frac{d - 65}{d} a]Hmm, I can cancel out ( a ) from both sides since ( a ) is not zero (they are runners, so they have some speed). So, canceling ( a ):[frac{(d - 30)(d - 40)}{d^2} = frac{d - 65}{d}]Now, I can multiply both sides by ( d^2 ) to eliminate the denominators:[(d - 30)(d - 40) = d(d - 65)]Let me expand both sides:Left side: ( (d - 30)(d - 40) = d^2 - 40d - 30d + 1200 = d^2 - 70d + 1200 )Right side: ( d(d - 65) = d^2 - 65d )So, now the equation is:[d^2 - 70d + 1200 = d^2 - 65d]Hmm, let's subtract ( d^2 ) from both sides to simplify:[-70d + 1200 = -65d]Now, let's bring all terms to one side. I'll add ( 65d ) to both sides:[-70d + 65d + 1200 = 0]Simplifying:[-5d + 1200 = 0]Now, solving for ( d ):[-5d = -1200][d = frac{-1200}{-5} = 240]So, ( d = 240 ) yards.Wait, let me double-check my calculations to make sure I didn't make a mistake.Starting from:[(d - 30)(d - 40) = d(d - 65)]Expanding:Left side: ( d^2 - 40d - 30d + 1200 = d^2 - 70d + 1200 )Right side: ( d^2 - 65d )Setting equal:( d^2 - 70d + 1200 = d^2 - 65d )Subtract ( d^2 ):( -70d + 1200 = -65d )Add ( 65d ):( -5d + 1200 = 0 )So, ( -5d = -1200 ), which gives ( d = 240 ). Yeah, that seems correct.Let me think if this makes sense. If the race is 240 yards, then when A runs 240 yards, B runs 200 yards (since A beats B by 40 yards). Then, when B runs 240 yards, C runs 210 yards (since B beats C by 30 yards). Now, when A runs 240 yards, how much does C run? If A's speed is such that when A runs 240, C runs 240 - 65 = 175 yards.Wait, let me check that. If A runs 240 yards, and C runs 175 yards, then A beats C by 65 yards, which matches the problem statement. Also, when B runs 240 yards, C runs 210 yards, so B beats C by 30 yards, which also matches. And when A runs 240 yards, B runs 200 yards, so A beats B by 40 yards. That all adds up.So, yes, 240 yards seems to be the correct length of the race.

💡First, I need to identify the type of sequence given. The sequence starts at 2 and increases by 6 each time, so it's an arithmetic sequence with the first term ( a_1 = 2 ) and common difference ( d = 6 ).Next, I'll determine the number of terms in the sequence. The last term is 134. Using the formula for the nth term of an arithmetic sequence:[a_n = a_1 + (n-1) cdot d]Substituting the known values:[134 = 2 + (n-1) cdot 6]Solving for ( n ):[132 = (n-1) cdot 6 n-1 = 22 n = 23]So, there are 23 terms in the sequence.Now, I'll replace each term with its equivalent modulo 12:[2, 8, 14, 20, 26, ldots, 128, 134 pmod{12}]This simplifies to:[2, 8, 2, 8, 2, ldots, 8, 2]Notice that the terms alternate between 2 and 8 modulo 12.Since there are 23 terms, there are 12 pairs of (2, 8) and one additional term of 2. Each pair sums to:[2 + 8 = 10]The total sum modulo 12 is:[12 cdot 10 + 2 = 120 + 2 = 122]Finally, calculating ( 122 mod 12 ):[122 div 12 = 10 text{ remainder } 2 122 equiv 2 pmod{12}]Therefore, the value of ( n ) is 2.