💡Okay, so I have this problem here: Given that x=1 and x=3 are two adjacent extreme points of the function f(x) = sin(ωx + φ) where ω > 0. Also, the derivative of f(x) at x = 3/2 is negative, meaning f’(3/2) < 0. I need to find f(1/3). Hmm, let me try to figure this out step by step.First, I remember that for a sine function of the form sin(ωx + φ), the extreme points (maxima and minima) occur where the derivative is zero. The derivative of sin(ωx + φ) is ωcos(ωx + φ). So, setting this equal to zero gives cos(ωx + φ) = 0. The solutions to this equation are where ωx + φ = π/2 + kπ, where k is any integer.Given that x=1 and x=3 are two adjacent extreme points, the distance between them should be half the period of the sine function because between a maximum and a minimum, the function completes half a period. Wait, actually, adjacent extreme points could be a maximum and a minimum, which are half a period apart, or two maxima or two minima, which would be a full period apart. Hmm, so I need to clarify that.But in this case, since they are adjacent, it's likely that they are a maximum and a minimum. So, the distance between x=1 and x=3 is 2 units. If this is half the period, then the full period T would be 4 units. The period of the sine function is given by T = 2π / ω. So, if T = 4, then ω = 2π / 4 = π/2. Okay, that seems right.So, ω is π/2. Now, let's write the function as f(x) = sin((π/2)x + φ). Now, I need to find φ. To do that, I can use the information about the derivative at x = 3/2 being negative.First, let's find the derivative f’(x). As I mentioned earlier, f’(x) = ωcos(ωx + φ). So, plugging in ω = π/2, we get f’(x) = (π/2)cos((π/2)x + φ). At x = 3/2, this derivative is negative. So, (π/2)cos((π/2)(3/2) + φ) < 0. Since π/2 is positive, the cosine term must be negative. Therefore, cos((3π/4) + φ) < 0.Now, let's think about the extreme points at x=1 and x=3. At x=1, f(x) is either a maximum or a minimum. Similarly, at x=3, it's the opposite. Since the derivative at x=3/2 is negative, which is between x=1 and x=3, that tells us something about the behavior of the function in that interval.Let me calculate the value of the function at x=1. f(1) = sin((π/2)(1) + φ) = sin(π/2 + φ). Similarly, at x=3, f(3) = sin((π/2)(3) + φ) = sin(3π/2 + φ).Now, since x=1 and x=3 are adjacent extreme points, one of them is a maximum and the other is a minimum. Let's assume that at x=1, the function is at a maximum. Then, f(1) = 1, which would mean sin(π/2 + φ) = 1. The sine function equals 1 at π/2 + 2πk, where k is an integer. So, π/2 + φ = π/2 + 2πk, which implies φ = 2πk. Since sine is periodic with period 2π, we can take φ = 0 without loss of generality.Alternatively, if x=1 were a minimum, then f(1) = -1, which would mean sin(π/2 + φ) = -1. That would imply π/2 + φ = 3π/2 + 2πk, so φ = π + 2πk. But let's see if that fits with the derivative condition.Wait, let's go back to the derivative condition. We have cos(3π/4 + φ) < 0. Let's plug in φ = 0. Then, cos(3π/4) = -√2/2, which is indeed negative. So, that works. If φ were π, then cos(3π/4 + π) = cos(7π/4) = √2/2, which is positive, contradicting the derivative condition. Therefore, φ must be 0.So, φ = 0. Therefore, the function is f(x) = sin((π/2)x). Now, we need to find f(1/3). Let's compute that.f(1/3) = sin((π/2)(1/3)) = sin(π/6). I remember that sin(π/6) is 1/2. So, f(1/3) = 1/2.Wait, let me double-check everything to make sure I didn't make a mistake. The period was calculated as 4, which gives ω = π/2. The extreme points at x=1 and x=3, with the derivative negative at x=3/2, led us to conclude that φ = 0. Plugging back in, f(x) = sin((π/2)x). Then, f(1/3) = sin(π/6) = 1/2. That seems correct.I think that's the answer. f(1/3) is 1/2.

💡Alright, so I have this geometry problem here about an acute-angled triangle ABC. The problem says that the internal angle bisectors of angles A, B, and C intersect the circumcircle of triangle ABC at points A1, B1, and C1 respectively. Then, the intersections of AA1 with B1C1, BB1 with C1A1, and CC1 with A1B1 are denoted as A0, B0, and C0 respectively. I need to prove that triangles ABC and A0B0C0 have the same incenter.Hmm, okay. Let me start by recalling some basic concepts. The incenter of a triangle is the point where the internal angle bisectors meet, and it's also the center of the incircle, which touches all three sides of the triangle. So, for triangle ABC, the incenter is where AA1, BB1, and CC1 meet.Now, the problem mentions that these angle bisectors intersect the circumcircle again at A1, B1, and C1. So, A1 is the second intersection point of the angle bisector of angle A with the circumcircle, and similarly for B1 and C1.Next, A0 is the intersection of AA1 with B1C1, B0 is the intersection of BB1 with C1A1, and C0 is the intersection of CC1 with A1B1. So, A0, B0, and C0 are points inside the triangle where these lines intersect.I need to show that the incenter of triangle ABC is also the incenter of triangle A0B0C0. That means I need to show that the internal angle bisectors of triangle A0B0C0 meet at the same point I, which is the incenter of ABC.Maybe I can start by considering the properties of the circumcircle and the angle bisectors. Since A1, B1, and C1 lie on the circumcircle, perhaps there are some cyclic quadrilateral properties or arc relationships that could be useful.Also, since A0 is the intersection of AA1 and B1C1, maybe I can find some similar triangles or use properties of intersecting chords in the circle.Wait, another thought: since AA1 is an angle bisector, and A1 is on the circumcircle, perhaps there are some known properties about the angle bisector intersecting the circumcircle. Maybe something related to the arc lengths or angles subtended by those points.Let me try to draw a diagram in my mind. Triangle ABC with circumcircle passing through A, B, C. The internal bisectors of angles A, B, C meet the circumcircle again at A1, B1, C1. Then, lines AA1, BB1, CC1 are drawn, and they intersect the opposite sides (or their extensions) at A0, B0, C0.Hmm, perhaps I can use Ceva's Theorem here? Ceva's Theorem relates the concurrency of cevians in a triangle. But in this case, the cevians are AA1, BB1, CC1, which already concur at the incenter I. But we are looking at their intersections with the opposite sides (B1C1, C1A1, A1B1) to get A0, B0, C0.Wait, maybe not Ceva, but perhaps some properties of harmonic division or projective geometry? Alternatively, maybe using power of a point with respect to the circumcircle.Another idea: since A1 is on the circumcircle and on the angle bisector, maybe the arc lengths from A to A1 correspond to half the angle at A? Let me think. The angle bisector divides angle A into two equal parts. Since A1 is on the circumcircle, the arc BA1 and A1C should be related to the angles at B and C.Wait, actually, in a triangle, the angle bisector from A meets the circumcircle at the midpoint of the arc BC that doesn't contain A. Is that correct? Let me recall: yes, in a triangle, the angle bisector of angle A passes through the midpoint of the arc BC that doesn't contain A. So, A1 is the midpoint of arc BC not containing A.Similarly, B1 is the midpoint of arc AC not containing B, and C1 is the midpoint of arc AB not containing C.That seems important. So, A1, B1, C1 are midpoints of arcs opposite to A, B, C respectively.Therefore, points A1, B1, C1 are the ex-points or something? Wait, no, they are midpoints of arcs. So, perhaps they are the circumcenters of some related triangles?Wait, actually, in triangle ABC, the midpoint of arc BC is equidistant from B and C, and lies on the angle bisector of angle A. So, A1 is equidistant from B and C, and lies on the circumcircle.Similarly for B1 and C1.So, perhaps triangle A1B1C1 is the medial triangle of the circumcircle? Or maybe it's related to the circumcevian midarc triangle.Hmm, not sure. Maybe I can consider the properties of triangle A1B1C1.Alternatively, perhaps I can consider the triangle A0B0C0 and find its incenter.Wait, the incenter is the intersection of the angle bisectors. So, if I can show that the angle bisectors of triangle A0B0C0 pass through I, the incenter of ABC, then I is also the incenter of A0B0C0.Alternatively, maybe I can show that I is equidistant from the sides of triangle A0B0C0, which would imply it's the incenter.But perhaps a better approach is to use properties of cyclic quadrilaterals and angle bisectors.Let me consider point A0. A0 is the intersection of AA1 and B1C1. Since A1 is the midpoint of arc BC, and B1 and C1 are midpoints of arcs AC and AB respectively.Wait, so B1C1 is a line connecting midpoints of arcs AC and AB. So, B1C1 is a chord of the circumcircle connecting midpoints of arcs AC and AB.Similarly, AA1 is the angle bisector of angle A, passing through A1.So, A0 is the intersection of AA1 and B1C1.Similarly for B0 and C0.Hmm, perhaps I can consider the angles at A0, B0, C0 and see if they relate to the incenter.Alternatively, maybe I can use inversion. But that might be complicated.Wait, another idea: since A1, B1, C1 are midpoints of arcs, perhaps triangle A1B1C1 is the circumcevian midarc triangle, and triangle A0B0C0 is its cevian triangle with respect to I.But I'm not sure.Wait, maybe I can consider that since A1 is the midpoint of arc BC, then the line B1C1 is the perpendicular bisector of BC? No, not necessarily.Wait, in the circumcircle, the midpoint of arc BC is equidistant from B and C, so the line from A1 to the midpoint of BC is perpendicular to BC. But B1C1 is a different line.Wait, perhaps I can consider the angles at A0.Let me try to compute some angles.Since A1 is the midpoint of arc BC, the angles at A1 with respect to BC are equal.Similarly, B1 and C1 are midpoints of their respective arcs.So, perhaps angles involving A0 can be related to the angles of ABC.Wait, maybe I can use the fact that AA1 is the angle bisector, so angle BAA1 = angle CA A1 = (1/2) angle BAC.Similarly, since A1 is the midpoint of arc BC, angle BA1C = angle BAC / 2.Wait, no, actually, angle BA1C is equal to half the measure of arc BC, which is equal to (1/2) angle BAC because A1 is the midpoint.Wait, no, actually, in the circumcircle, the measure of an arc is twice the measure of the inscribed angle subtended by it. So, if A1 is the midpoint of arc BC, then arc BA1 = arc A1C = (1/2) arc BC.But arc BC is equal to 2 angle BAC, because in the circumcircle, the central angle is twice the inscribed angle. Wait, no, actually, the central angle is twice the inscribed angle subtended by the same arc.So, angle BAC is the inscribed angle subtended by arc BC, so the central angle for arc BC is 2 angle BAC.Therefore, since A1 is the midpoint of arc BC, arc BA1 = arc A1C = angle BAC.Similarly, arc AC is 2 angle ABC, so midpoint B1 divides arc AC into two arcs each of measure angle ABC.Similarly, arc AB is 2 angle ACB, so midpoint C1 divides arc AB into two arcs each of measure angle ACB.Okay, so now, considering point A0, which is the intersection of AA1 and B1C1.Let me try to find some angle relationships at A0.Since A0 lies on AA1, which is the angle bisector, and on B1C1.Maybe I can consider triangle A0B1C1.Wait, since B1 and C1 are midpoints of arcs AC and AB, respectively, perhaps lines B1C1 have some symmetric properties.Alternatively, maybe I can use power of a point.Wait, point A0 lies on AA1 and B1C1. So, perhaps I can compute the power of A0 with respect to the circumcircle.But since A0 is inside the circle, its power would be negative, equal to OA0^2 - R^2, but I'm not sure if that helps.Alternatively, maybe I can use Menelaus' Theorem on triangle ABC with transversal B1C1.Wait, Menelaus' Theorem relates the ratios of lengths when a transversal crosses the sides of a triangle. But in this case, B1C1 is a line cutting through the circumcircle, not necessarily the sides of ABC.Hmm, perhaps not directly applicable.Wait, another idea: since A1 is the midpoint of arc BC, then the line AA1 is the angle bisector and also the symmedian in some contexts. Wait, no, the symmedian is different.Wait, but maybe I can use properties of symmedians or isogonal conjugates.Alternatively, perhaps considering that A0 is the intersection of AA1 and B1C1, and since B1 and C1 are midpoints of arcs, perhaps A0 has some symmetrical properties.Wait, maybe I can consider that triangle A0B0C0 is similar to triangle ABC or something like that.Alternatively, maybe I can consider that the incenter I lies on the angle bisectors of triangle A0B0C0.Wait, since I is the incenter of ABC, it lies on AA1, BB1, CC1. So, if I can show that I also lies on the angle bisectors of A0B0C0, then I would be the incenter.Alternatively, maybe I can show that the internal angle bisectors of A0B0C0 pass through I.Wait, perhaps I can compute the angles at A0, B0, C0 and show that the bisectors meet at I.Alternatively, maybe I can use trigonometric Ceva's Theorem for concurrency.Wait, I'm getting a bit stuck. Maybe I can try to look for some known theorem or lemma that relates the incenter with these points.Wait, I recall that in some cases, the incenter can be the orthocenter or centroid of another triangle, but I'm not sure if that applies here.Wait, another idea: since A1, B1, C1 are midpoints of arcs, perhaps triangle A1B1C1 is called the circumcevian midarc triangle, and triangle A0B0C0 is its cevian triangle with respect to I.Wait, if that's the case, then perhaps triangle A0B0C0 is similar to ABC or has some relation that preserves the incenter.Alternatively, maybe I can consider that I is the incenter of both triangles because of some reflection properties.Wait, perhaps I can consider that the lines AA1, BB1, CC1 are concurrent at I, and their intersections with B1C1, C1A1, A1B1 are A0, B0, C0. So, maybe triangle A0B0C0 is the cevian triangle of I with respect to triangle A1B1C1.If that's the case, then perhaps the incenter of A0B0C0 is the same as the incenter of ABC.Wait, but I'm not sure about that. Maybe I need to explore more.Alternatively, perhaps I can use coordinates. Assign coordinates to triangle ABC, compute the coordinates of A1, B1, C1, then find A0, B0, C0, and then compute the incenter.But that might be too involved, especially since it's an acute triangle, so coordinates could be messy.Wait, maybe I can use barycentric coordinates with respect to triangle ABC.In barycentric coordinates, the incenter has coordinates proportional to the lengths of the sides.But I'm not sure if that would help directly.Wait, another thought: since A1 is the midpoint of arc BC, then the line B1C1 is the perpendicular bisector of BC? Wait, no, because B1 and C1 are midpoints of arcs AC and AB, not necessarily related to BC.Wait, but in the circumcircle, the midpoint of arc BC is equidistant from B and C, so the line from A1 to the midpoint of BC is perpendicular to BC. But B1C1 is a different line.Wait, perhaps I can consider that B1C1 is the angle bisector of angle BAC or something.Wait, no, B1 is the midpoint of arc AC, so the line from B1 to C1 would relate to angles at B1 and C1.Wait, maybe I can compute some angles at A0.Since A0 is on AA1 and B1C1, perhaps I can find the angles at A0 with respect to B1 and C1.Wait, let me try to compute angle BA0C.Wait, no, maybe angle B0A0C0.Wait, this is getting too vague.Wait, perhaps I can use the fact that A1, B1, C1 are midpoints of arcs, so the lines B1C1, C1A1, A1B1 are the perpendicular bisectors of the sides of triangle ABC.Wait, no, the perpendicular bisectors of the sides pass through the circumcenter, not necessarily related to B1C1.Wait, but in this case, since B1 and C1 are midpoints of arcs, perhaps lines B1C1 are related to the internal angle bisectors.Wait, maybe I can consider that lines B1C1 are parallel to some sides or have some proportional relationships.Wait, I'm not making progress here. Maybe I need to look for some properties or lemmas that connect these points.Wait, I recall that in some cases, the incenter lies on the circumcircle of another triangle, but I don't think that's directly applicable here.Wait, another idea: since A1 is the midpoint of arc BC, then the line AA1 is the angle bisector and also the symmedian. Wait, no, the symmedian is the reflection of the median over the angle bisector.Wait, but maybe I can use properties of symmedians here.Alternatively, perhaps I can use the fact that A0 lies on AA1 and B1C1, so maybe A0 is the midpoint of some segment or has some harmonic properties.Wait, maybe I can use harmonic division. If AA1 intersects B1C1 at A0, perhaps there is a harmonic bundle here.Alternatively, maybe I can use the fact that A1 is the midpoint of arc BC, so the polar of A1 with respect to the circumcircle passes through the midpoint of BC.Wait, but I'm not sure how that helps.Wait, perhaps I can consider the polar of I with respect to the circumcircle. Since I is the incenter, its polar might have some relation to the sides or something.Wait, I'm getting too abstract here. Maybe I need to take a different approach.Wait, let me consider the incenter I of ABC. I need to show that I is also the incenter of A0B0C0. So, I need to show that I lies on the angle bisectors of A0B0C0.Alternatively, I can show that I is equidistant from the sides of A0B0C0.But how?Wait, maybe I can consider the distances from I to the sides of A0B0C0.Since I is the incenter of ABC, it is equidistant from the sides of ABC. But how does that relate to the sides of A0B0C0?Alternatively, maybe I can show that the internal angle bisectors of A0B0C0 pass through I.Wait, perhaps I can compute the angles at A0, B0, C0 and show that the bisectors meet at I.Wait, let me try to compute angle B0A0C0.Since A0 is the intersection of AA1 and B1C1, and B0 is the intersection of BB1 and C1A1, perhaps I can find some relationships between these angles.Wait, maybe I can use cyclic quadrilaterals. For example, since A1, B1, C1 are on the circumcircle, perhaps quadrilaterals A0B1C1A1 are cyclic or something.Wait, but A0 is on AA1 and B1C1, so maybe quadrilateral A0B1C1A1 is cyclic? Let me check.If A0 lies on AA1 and B1C1, then points A0, A1, B1, C1 lie on a circle only if certain angle conditions are met.Wait, but I don't know if that's the case.Alternatively, maybe I can use power of a point for A0 with respect to the circumcircle.Power of A0: A0A * A0A1 = A0B1 * A0C1.But since A0 lies on AA1 and B1C1, this might give some relation.Wait, but I'm not sure how to use this to find the incenter.Wait, maybe I can consider that since I is the incenter, it lies on AA1, BB1, CC1, and perhaps also on the angle bisectors of A0B0C0.Wait, but how?Alternatively, maybe I can consider that triangle A0B0C0 is the incentroid triangle or something similar.Wait, another idea: since A1, B1, C1 are midpoints of arcs, perhaps lines AA1, BB1, CC1 are concurrent at I, and their intersections with B1C1, C1A1, A1B1 are A0, B0, C0. So, maybe triangle A0B0C0 is the cevian triangle of I with respect to triangle A1B1C1.If that's the case, then perhaps the incenter of A0B0C0 is the same as the incenter of ABC.But I'm not sure about that.Wait, maybe I can use the fact that the cevian triangle of the incenter has the same incenter.Wait, but I need to think more carefully.Alternatively, maybe I can use trigonometric Ceva's Theorem to show that the cevians of A0B0C0 concur at I.Wait, but I'm not sure.Wait, another approach: since A1, B1, C1 are midpoints of arcs, perhaps triangle A1B1C1 is the medial triangle of the circumcircle, and triangle A0B0C0 is its cevian triangle with respect to I.In that case, perhaps the incenter of A0B0C0 is I.But I need to verify this.Alternatively, maybe I can use the fact that I is the incenter of ABC and lies on AA1, BB1, CC1, and then show that it also lies on the angle bisectors of A0B0C0.Wait, perhaps I can compute the angles at A0, B0, C0 and show that the bisectors meet at I.Wait, let me try to compute angle B0A0C0.Since A0 is the intersection of AA1 and B1C1, and B0 is the intersection of BB1 and C1A1, perhaps I can find some relationships between these angles.Wait, maybe I can use the fact that A1 is the midpoint of arc BC, so angle BA1C = angle BAC / 2.Similarly, angle CB1A = angle ABC / 2, and angle AC1B = angle ACB / 2.Wait, but how does that help with angle B0A0C0?Alternatively, maybe I can use the fact that A0 lies on B1C1, so angle B0A0C0 is equal to angle B1A0C1.Wait, but I'm not sure.Wait, perhaps I can use the fact that A1 is the midpoint of arc BC, so the line B1C1 is the perpendicular bisector of BC.Wait, no, that's not necessarily true.Wait, but since B1 and C1 are midpoints of arcs AC and AB, respectively, perhaps the line B1C1 is related to the angle bisector of angle BAC.Wait, maybe I can consider that B1C1 is parallel to BC or something.Wait, no, that's not necessarily the case.Wait, perhaps I can consider that since B1 and C1 are midpoints of arcs AC and AB, the line B1C1 is the angle bisector of angle BAC.Wait, but I'm not sure.Wait, another idea: since A1 is the midpoint of arc BC, the line AA1 is the angle bisector, and A0 is the intersection of AA1 with B1C1.So, maybe A0 is the midpoint of AA1 or something.Wait, but I don't think that's necessarily true.Wait, perhaps I can use mass point geometry or something like that.Wait, but I'm not sure.Wait, another idea: since A1 is the midpoint of arc BC, the line B1C1 is the perpendicular bisector of BC.Wait, no, that's not necessarily true.Wait, but in the circumcircle, the midpoint of arc BC is equidistant from B and C, so the line from A1 to the midpoint of BC is perpendicular to BC.But B1C1 is a different line.Wait, perhaps I can consider that B1C1 is the perpendicular bisector of BC.Wait, no, because B1 and C1 are midpoints of arcs AC and AB, not BC.Wait, perhaps I can consider that B1C1 is the angle bisector of angle BAC.Wait, but I'm not sure.Wait, maybe I can compute the angles at A0.Since A0 is on AA1, which is the angle bisector, and on B1C1, perhaps I can find some angle equalities.Wait, let me try to compute angle BA0C.Wait, no, maybe angle B0A0C0.Wait, this is getting too vague.Wait, maybe I can use the fact that A1, B1, C1 are midpoints of arcs, so the triangle A1B1C1 is equilateral or something.Wait, no, unless ABC is equilateral, which it's not necessarily.Wait, but in any case, triangle A1B1C1 has some special properties.Wait, another idea: since A1, B1, C1 are midpoints of arcs, the triangle A1B1C1 is called the circumcevian midarc triangle, and it's known that the incenter of ABC is the orthocenter of A1B1C1.Wait, is that true? Let me think.Yes, actually, in triangle ABC, the midpoints of the arcs A1, B1, C1 form a triangle called the circumcevian midarc triangle, and the incenter of ABC is the orthocenter of this triangle.So, I is the orthocenter of triangle A1B1C1.Therefore, the altitudes of triangle A1B1C1 meet at I.But how does that help with triangle A0B0C0?Wait, since A0 is the intersection of AA1 and B1C1, and similarly for B0 and C0, then triangle A0B0C0 is the orthic triangle of triangle A1B1C1.Because the orthic triangle is formed by the feet of the altitudes, but in this case, A0 is the intersection of AA1 (which is an altitude) with B1C1.Wait, no, actually, in triangle A1B1C1, the altitudes are from A1, B1, C1 perpendicular to the opposite sides.But in our case, A0 is the intersection of AA1 (which is an altitude) with B1C1.Wait, so A0 is the foot of the altitude from A1 onto B1C1.Similarly, B0 is the foot from B1 onto C1A1, and C0 is the foot from C1 onto A1B1.Therefore, triangle A0B0C0 is the orthic triangle of triangle A1B1C1.Since I is the orthocenter of A1B1C1, then the orthic triangle A0B0C0 has I as its incenter.Wait, is that a known property? That the orthocenter of the original triangle is the incenter of its orthic triangle.Wait, no, actually, in general, the orthocenter of the original triangle is the incenter of its orthic triangle only in specific cases, like in acute triangles.Wait, but in our case, since ABC is acute, A1B1C1 is also acute, so its orthic triangle A0B0C0 would have I as its incenter.Therefore, I is the incenter of A0B0C0.Therefore, the incenter of ABC is the same as the incenter of A0B0C0.So, that's the proof.**Final Answer**boxed{triangle ABC ; text{and} ; triangle A_0B_0C_0 text{ have the same incenter.}}

💡First, I'll define the variables to represent the number of calls and agents in each team. Let ( C ) be the number of calls processed by each member of team B, and ( N ) be the number of agents in team B.Since each member of team A processes ( frac{7}{5} ) times the number of calls compared to team B, each member of team A processes ( frac{7}{5}C ) calls.Team A has ( frac{5}{8} ) of the number of agents that team B has, so team A has ( frac{5}{8}N ) agents.Next, I'll calculate the total number of calls processed by each team. For team A, it's the product of the number of calls per agent and the number of agents:[text{Total calls by team A} = left(frac{7}{5}Cright) times left(frac{5}{8}Nright) = frac{7}{8}CN]For team B, the total number of calls is:[text{Total calls by team B} = C times N = CN]Now, I'll find the total number of calls processed by both teams combined:[text{Total calls} = text{Total calls by team A} + text{Total calls by team B} = frac{7}{8}CN + CN = frac{15}{8}CN]Finally, to determine the fraction of the total calls processed by team B, I'll divide the total calls by team B by the total calls processed by both teams:[text{Fraction by team B} = frac{CN}{frac{15}{8}CN} = frac{8}{15}]Therefore, team B processed ( frac{8}{15} ) of the total calls.

💡Okay, so I have this problem where I need to find the smallest integer x greater than 1 such that x squared is one more than a multiple of 7. Hmm, let me try to understand what this means. So, mathematically, I think this translates to finding x where x² ≡ 1 mod 7. That means when I divide x squared by 7, the remainder should be 1. Got it. So, I need to find the smallest integer x > 1 that satisfies this condition.Alright, let's start by checking the squares of numbers modulo 7. Maybe if I list out the squares of numbers from 2 upwards and see when the remainder is 1 when divided by 7.Starting with x = 2:2² = 4. 4 divided by 7 is 0 with a remainder of 4. So, 4 mod 7 is 4, which isn't 1. So, x=2 doesn't work.Next, x=3:3² = 9. 9 divided by 7 is 1 with a remainder of 2. So, 9 mod 7 is 2. Not 1 either.Moving on to x=4:4² = 16. 16 divided by 7 is 2 with a remainder of 2. So, 16 mod 7 is 2. Still not 1.How about x=5:5² = 25. 25 divided by 7 is 3 with a remainder of 4. So, 25 mod 7 is 4. Not 1.Next up, x=6:6² = 36. 36 divided by 7 is 5 with a remainder of 1. So, 36 mod 7 is 1. Ah, that's what we're looking for!Wait, so x=6 works because 6 squared is 36, and 36 divided by 7 leaves a remainder of 1. So, 36 is one more than a multiple of 7 (since 35 is 5*7, and 36 is 35+1). That makes sense.Just to make sure I didn't miss anything, let me check x=7:7² = 49. 49 mod 7 is 0, which isn't 1. So, x=7 doesn't work. But since we're looking for the smallest integer greater than 1, and x=6 works, we don't need to go further.So, yeah, x=6 is the smallest integer greater than 1 such that x squared is one more than a multiple of 7. I think that's the answer.**Final Answer**The smallest integer ( x ) is boxed{6}.

💡First, calculate the total length of wire Bonnie uses. She uses 12 pieces of wire, each 4 inches long, so the total length is 12 multiplied by 4, which equals 48 inches.Next, determine the volume of Bonnie's cube. Since each edge of the cube is 4 inches long, the volume is 4 cubed, which is 64 cubic inches.Roark creates unit cubes, each with a volume of 1 cubic inch. To match Bonnie's volume, he needs 64 unit cubes.Each unit cube requires 12 pieces of 1-inch wire, totaling 12 inches of wire per cube. Therefore, for 64 cubes, Roark uses 64 multiplied by 12, which equals 768 inches of wire.Finally, find the ratio of Bonnie's total wire length to Roark's total wire length. This is 48 divided by 768, which simplifies to 1/16.

💡Alright, so I have this problem about lines in a plane. Let me try to understand it step by step. The problem says:"In the plane, there are ( n ) lines such that every pair of lines intersects, but no four lines pass through a single point. There are a total of 16 intersection points, and 6 of these points are intersections of three lines each. Find the value of ( n )."Okay, so let's break this down. We have ( n ) lines, and every pair of lines intersects. That means if I pick any two lines, they will meet at exactly one point. But it also says that no four lines pass through a single point. So, while multiple lines can intersect at the same point, it's limited to at most three lines intersecting at a single point.The total number of intersection points is 16, and out of these, 6 points are where three lines intersect each. So, these 6 points are special because they have three lines meeting there, instead of just two.I need to find ( n ), the number of lines. Hmm, okay. Let's think about how to approach this.First, if every pair of lines intersects, the maximum number of intersection points without any three lines intersecting at the same point would be ( binom{n}{2} ), which is the number of ways to choose two lines from ( n ) lines. That's because each pair of lines intersects at exactly one point.But in this problem, some points have three lines intersecting, which affects the total count of intersection points. Specifically, each point where three lines intersect would have been counted multiple times if we just used ( binom{n}{2} ). So, we need to adjust our count accordingly.Let me recall that if three lines intersect at a single point, instead of having three separate intersection points (each pair of lines intersecting), we have just one point. So, for each such point, we're actually reducing the total number of intersection points by 2. Because instead of 3 points, we have 1, so the difference is 2.Given that there are 6 such points where three lines intersect, each of these points reduces the total number of intersection points by 2. So, the total reduction is ( 6 times 2 = 12 ).Therefore, the actual number of intersection points, considering these overlaps, is ( binom{n}{2} - 12 ). But the problem states that the total number of intersection points is 16. So, we can set up the equation:[binom{n}{2} - 12 = 16]Simplifying this, we get:[binom{n}{2} = 28]Now, ( binom{n}{2} ) is equal to ( frac{n(n-1)}{2} ). So, substituting that in:[frac{n(n-1)}{2} = 28]Multiplying both sides by 2 to eliminate the denominator:[n(n - 1) = 56]Now, we need to find an integer ( n ) such that ( n(n - 1) = 56 ). Let's try to factor 56 to find such an ( n ).56 can be factored as:- 7 × 8 = 56So, if ( n = 8 ), then ( n - 1 = 7 ), and indeed, ( 8 times 7 = 56 ).Therefore, ( n = 8 ) is the solution.Wait a second, let me double-check to make sure I didn't make a mistake. If ( n = 8 ), then the number of intersection points without any three lines intersecting would be ( binom{8}{2} = 28 ). But since we have 6 points where three lines intersect, each of these reduces the total count by 2, so the total reduction is 12. Therefore, the actual number of intersection points is ( 28 - 12 = 16 ), which matches the problem statement. So, yes, ( n = 8 ) is correct.I think that makes sense. So, the value of ( n ) is 8.