answer：First, I need to calculate the product of the numbers from 1 to 8, which is 40320. My goal is to find the largest five-digit integer where the product of its digits equals 40320.I'll start by factorizing 40320 into its prime factors: 2^7 * 3^2 * 5 * 7.Next, I'll look for the largest single-digit numbers that can be formed using these prime factors. The largest single-digit number possible is 9, which uses 3^2. Then, I can form an 8 using 2^3.Now, I have the remaining factors: 2^4 * 5 * 7. I can make another 8 from 2^3, leaving me with 2 * 5 * 7. These remaining factors can be used as the digits 2, 5, and 7.Finally, I'll arrange these digits in descending order to form the largest possible five-digit number: 98752.

answer：Okay, so I have this problem where I need to find the minimum value of the expression (a^2 + b^2 + c^2 + d^2 + ab + ac + ad + bc + bd + cd) given that (a), (b), (c), and (d) are positive numbers whose product is 1. The options are A: 0, B: 4, C: 8, D: 10. Hmm, let me think about how to approach this.First, since the product (abcd = 1), maybe I can use some inequality to find the minimum value. The expression given is a combination of squares and products of the variables. It looks kind of like a quadratic form. Maybe I can rewrite it in a way that makes it easier to apply inequalities like AM-GM or Cauchy-Schwarz.Let me write down the expression again:(a^2 + b^2 + c^2 + d^2 + ab + ac + ad + bc + bd + cd)Hmm, I notice that this expression includes all the pairwise products of (a), (b), (c), and (d), as well as their squares. That makes me think of the square of a sum. Specifically, the square of (a + b + c + d) is:((a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd))But in our expression, the coefficients of the pairwise products are 1 instead of 2. So maybe I can relate the given expression to this square somehow.Let me denote the given expression as (E):(E = a^2 + b^2 + c^2 + d^2 + ab + ac + ad + bc + bd + cd)Comparing this with the square of the sum, we have:((a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd))So, if I subtract (ab + ac + ad + bc + bd + cd) from both sides, I get:((a + b + c + d)^2 - (ab + ac + ad + bc + bd + cd) = a^2 + b^2 + c^2 + d^2 + ab + ac + ad + bc + bd + cd)Which means:(E = (a + b + c + d)^2 - (ab + ac + ad + bc + bd + cd))Hmm, not sure if that helps directly. Maybe another approach.Since the product (abcd = 1), perhaps I can use the AM-GM inequality, which states that for positive real numbers, the arithmetic mean is greater than or equal to the geometric mean.But the expression (E) is a bit complicated. Maybe I can group terms or find a substitution.Wait, another thought: if I let (a = b = c = d), since the product is 1, each variable would be 1, because (1 times 1 times 1 times 1 = 1). Let me compute (E) in that case.If (a = b = c = d = 1), then:(E = 1^2 + 1^2 + 1^2 + 1^2 + 1times1 + 1times1 + 1times1 + 1times1 + 1times1 + 1times1)Calculating that:(4 times 1 + 6 times 1 = 4 + 6 = 10)So, when all variables are 1, (E = 10). Hmm, that's one of the options, D: 10. But is this the minimum?I need to check if there's a way to make (E) smaller. Maybe by making some variables larger and others smaller, keeping the product 1.Let me try with two variables being larger and two smaller. For example, let me set (a = b = t) and (c = d = frac{1}{t}), so that (abcd = t times t times frac{1}{t} times frac{1}{t} = 1).Now, compute (E) in terms of (t):First, compute each term:(a^2 = t^2), same for (b^2), (c^2 = frac{1}{t^2}), same for (d^2).So, (a^2 + b^2 + c^2 + d^2 = 2t^2 + 2left(frac{1}{t^2}right))Now, the pairwise products:(ab = t times t = t^2)(ac = t times frac{1}{t} = 1), same for (ad), (bc), (bd)Wait, let's compute each:(ab = t^2)(ac = t times frac{1}{t} = 1)(ad = t times frac{1}{t} = 1)Similarly, (bc = t times frac{1}{t} = 1)(bd = t times frac{1}{t} = 1)(cd = frac{1}{t} times frac{1}{t} = frac{1}{t^2})So, the pairwise products sum up to:(ab + ac + ad + bc + bd + cd = t^2 + 1 + 1 + 1 + 1 + frac{1}{t^2} = t^2 + 4 + frac{1}{t^2})Therefore, the entire expression (E) is:(2t^2 + 2left(frac{1}{t^2}right) + t^2 + 4 + frac{1}{t^2})Simplify:(2t^2 + frac{2}{t^2} + t^2 + 4 + frac{1}{t^2} = (2t^2 + t^2) + left(frac{2}{t^2} + frac{1}{t^2}right) + 4 = 3t^2 + frac{3}{t^2} + 4)So, (E = 3t^2 + frac{3}{t^2} + 4). Now, to find the minimum of this expression with respect to (t > 0).Let me denote (f(t) = 3t^2 + frac{3}{t^2} + 4). To find its minimum, take the derivative and set it to zero.Compute (f'(t)):(f'(t) = 6t - frac{6}{t^3})Set (f'(t) = 0):(6t - frac{6}{t^3} = 0)Multiply both sides by (t^3):(6t^4 - 6 = 0)Simplify:(6t^4 = 6) => (t^4 = 1) => (t = 1) (since (t > 0))So, the minimum occurs at (t = 1). Plugging back into (f(t)):(f(1) = 3(1)^2 + frac{3}{(1)^2} + 4 = 3 + 3 + 4 = 10)So, even when I set (a = b = t) and (c = d = 1/t), the minimum value of (E) is still 10, achieved when (t = 1), meaning all variables are 1.Hmm, so maybe 10 is indeed the minimum. But let me check another case to be sure.What if I set three variables equal and the fourth different? For example, let (a = b = c = t) and (d = frac{1}{t^3}) so that (abcd = t times t times t times frac{1}{t^3} = 1).Compute (E) in terms of (t):First, squares:(a^2 = t^2), same for (b^2), (c^2), and (d^2 = frac{1}{t^6})So, (a^2 + b^2 + c^2 + d^2 = 3t^2 + frac{1}{t^6})Now, pairwise products:(ab = t^2), same for (ac), (bc)(ad = t times frac{1}{t^3} = frac{1}{t^2}), same for (bd), (cd)(cd = t times frac{1}{t^3} = frac{1}{t^2})Wait, actually, let me compute each pairwise product:(ab = t times t = t^2)(ac = t times t = t^2)(ad = t times frac{1}{t^3} = frac{1}{t^2})Similarly, (bc = t times t = t^2)(bd = t times frac{1}{t^3} = frac{1}{t^2})(cd = t times frac{1}{t^3} = frac{1}{t^2})So, the pairwise products sum up to:(ab + ac + ad + bc + bd + cd = t^2 + t^2 + frac{1}{t^2} + t^2 + frac{1}{t^2} + frac{1}{t^2})Simplify:(3t^2 + 3left(frac{1}{t^2}right))Therefore, the entire expression (E) is:(3t^2 + frac{1}{t^6} + 3t^2 + 3left(frac{1}{t^2}right))Wait, hold on, I think I made a mistake here. Let me re-express:The squares: (3t^2 + frac{1}{t^6})The pairwise products: (3t^2 + 3left(frac{1}{t^2}right))So, adding them together:(E = 3t^2 + frac{1}{t^6} + 3t^2 + 3left(frac{1}{t^2}right) = 6t^2 + frac{1}{t^6} + 3left(frac{1}{t^2}right))Hmm, that seems more complicated. Maybe taking the derivative here would be messy. Alternatively, perhaps this approach isn't the best.Alternatively, maybe I should consider symmetry. Since the expression is symmetric in all variables, the minimum might occur when all variables are equal. That is, when (a = b = c = d = 1), as I initially thought.But let me test another case where two variables are equal and the other two are different. For example, let (a = b = t) and (c = s), (d = frac{1}{t^2 s}), so that (abcd = t times t times s times frac{1}{t^2 s} = 1).Compute (E) in terms of (t) and (s):First, squares:(a^2 = t^2), same for (b^2), (c^2 = s^2), (d^2 = frac{1}{t^4 s^2})So, (a^2 + b^2 + c^2 + d^2 = 2t^2 + s^2 + frac{1}{t^4 s^2})Pairwise products:(ab = t^2)(ac = t s)(ad = t times frac{1}{t^2 s} = frac{1}{t s})Similarly, (bc = t s)(bd = t times frac{1}{t^2 s} = frac{1}{t s})(cd = s times frac{1}{t^2 s} = frac{1}{t^2})So, the pairwise products sum up to:(ab + ac + ad + bc + bd + cd = t^2 + t s + frac{1}{t s} + t s + frac{1}{t s} + frac{1}{t^2})Simplify:(t^2 + 2 t s + 2 times frac{1}{t s} + frac{1}{t^2})Therefore, the entire expression (E) is:(2t^2 + s^2 + frac{1}{t^4 s^2} + t^2 + 2 t s + 2 times frac{1}{t s} + frac{1}{t^2})Simplify:(3t^2 + s^2 + frac{1}{t^4 s^2} + 2 t s + 2 times frac{1}{t s} + frac{1}{t^2})This seems even more complicated. Maybe this approach isn't helpful.Alternatively, perhaps I should consider the expression (E) as a sum of squares plus the sum of pairwise products. Maybe I can write it in terms of ((a + b + c + d)^2) minus something.Wait, earlier I had:(E = (a + b + c + d)^2 - (ab + ac + ad + bc + bd + cd))But that didn't seem helpful. Alternatively, maybe I can consider the expression as:(E = frac{1}{2} left[ (a + b)^2 + (a + c)^2 + (a + d)^2 + (b + c)^2 + (b + d)^2 + (c + d)^2 right])But I'm not sure if that helps either.Wait, another idea: Maybe use the AM-GM inequality on the terms in (E). Since all variables are positive, AM-GM might give a lower bound.Let me count how many terms there are in (E). There are 4 square terms and 6 pairwise product terms, so 10 terms in total. Maybe I can apply AM-GM to all 10 terms.But AM-GM on 10 terms would give:(frac{a^2 + b^2 + c^2 + d^2 + ab + ac + ad + bc + bd + cd}{10} geq sqrt[10]{a^2 b^2 c^2 d^2 times (ab)(ac)(ad)(bc)(bd)(cd)})Wait, let's compute the geometric mean.First, the product of all terms:(a^2 times b^2 times c^2 times d^2 times ab times ac times ad times bc times bd times cd)Let me compute the exponents for each variable:For (a): 2 (from (a^2)) + 1 (from (ab)) + 1 (from (ac)) + 1 (from (ad)) = 5Similarly, for (b): 2 + 1 + 1 + 1 = 5For (c): 2 + 1 + 1 + 1 = 5For (d): 2 + 1 + 1 + 1 = 5So, the product is (a^5 b^5 c^5 d^5 = (abcd)^5 = 1^5 = 1)Therefore, the geometric mean is (sqrt[10]{1} = 1)So, by AM-GM:(frac{E}{10} geq 1) => (E geq 10)Equality holds when all terms are equal, i.e., when (a^2 = b^2 = c^2 = d^2 = ab = ac = ad = bc = bd = cd)Which implies that (a = b = c = d), since all the square terms and pairwise products are equal.Given that (abcd = 1), this implies (a = b = c = d = 1), as before.Therefore, the minimum value of (E) is 10, achieved when all variables are 1.So, the answer should be D: 10.**Final Answer**boxed{D}

answer：First, I need to determine the integer part of the square root of 80, which means finding the largest integer less than or equal to √80.I know that 8 squared is 64 and 9 squared is 81. Since 80 is between 64 and 81, it follows that √80 is between 8 and 9.Therefore, the largest integer less than √80 is 8.

answer：First, I need to determine how many trees the forester planted on Monday. Let's denote the number of trees planted on Monday as ( M ).On Tuesday, the forester planted a third of what was planted on Monday, which is ( frac{1}{3}M ).The total number of trees planted over the two days is given as 80. Therefore, I can set up the equation:[M + frac{1}{3}M = 80]Combining like terms, this simplifies to:[frac{4}{3}M = 80]To solve for ( M ), I multiply both sides of the equation by the reciprocal of ( frac{4}{3} ), which is ( frac{3}{4} ):[M = 80 times frac{3}{4} = 60]So, the forester planted 60 trees on Monday.Next, I calculate the total number of trees in the forest after Monday by adding the initial 30 native trees to the 60 trees planted:[30 + 60 = 90]Finally, to find the ratio of the total number of trees after Monday to the initial number of native trees, I divide 90 by 30:[frac{90}{30} = 3]Therefore, the ratio is ( 3:1 ).

answer：Alright, so I'm trying to solve this geometry problem where I have triangle ABC, and A1, B1, C1 are midpoints of sides BC, AC, and AB respectively. Points E and F are on the midlines C1B1 and A1B1 such that BE is the angle bisector of angle AEB1 and BF is the angle bisector of angle CFB1. I need to prove that the bisectors of angles ABC and FBE coincide.First, I should probably draw a diagram to visualize the problem. Let me sketch triangle ABC, mark the midpoints A1, B1, C1. Then, the midlines C1B1 and A1B1 are drawn. Points E and F are somewhere on these midlines. BE bisects angle AEB1, and BF bisects angle CFB1.Since A1, B1, C1 are midpoints, the midlines C1B1 and A1B1 are parallel to sides AB and BC respectively, and each is half their length. So, triangle A1B1C1 is similar to triangle ABC with a ratio of 1:2.Now, points E and F are on these midlines. Let me think about the properties of angle bisectors. If BE is the angle bisector of angle AEB1, then it divides angle AEB1 into two equal parts. Similarly, BF divides angle CFB1 into two equal parts.I wonder if there's a way to use similarity or congruence here. Since the midlines are parallel to the sides, maybe some triangles are similar.Let me consider triangle AEB1. Since C1B1 is a midline, EB1 is half of AB, right? Wait, no, EB1 is part of the midline, so it's half the length of AB. Similarly, EA is part of the midline C1B1, so it's half the length of CB.Wait, maybe I should assign coordinates to the points to make this more concrete. Let me place triangle ABC in a coordinate system. Let me set point B at the origin (0,0), point C at (2c, 0), and point A at (2a, 2b). Then, the midpoints would be:- A1 is the midpoint of BC: ((2c + 0)/2, (0 + 0)/2) = (c, 0)- B1 is the midpoint of AC: ((2a + 2c)/2, (2b + 0)/2) = (a + c, b)- C1 is the midpoint of AB: ((2a + 0)/2, (2b + 0)/2) = (a, b)So, midline C1B1 connects points C1(a, b) and B1(a + c, b). So, it's a horizontal line at y = b from x = a to x = a + c.Similarly, midline A1B1 connects A1(c, 0) and B1(a + c, b). The slope of A1B1 is (b - 0)/(a + c - c) = b/a.Now, points E and F are on C1B1 and A1B1 respectively. Let me parameterize these points.For point E on C1B1: since it's a horizontal line from (a, b) to (a + c, b), I can write E as (a + t, b) where t is between 0 and c.Similarly, for point F on A1B1: since A1 is (c, 0) and B1 is (a + c, b), the parametric equations for A1B1 can be written as x = c + s*a, y = 0 + s*b, where s ranges from 0 to 1. So, F is (c + s*a, s*b).Now, BE is the angle bisector of angle AEB1. Let me find the coordinates of E and then express the condition that BE bisects angle AEB1.Point E is (a + t, b). Point B is (0,0). So, vector BE is from B(0,0) to E(a + t, b). The angle at E is angle AEB1. So, points A, E, B1.Wait, point A is (2a, 2b), E is (a + t, b), and B1 is (a + c, b). So, angle AEB1 is the angle at E between points A, E, and B1.Similarly, BF is the angle bisector of angle CFB1. Point F is (c + s*a, s*b). So, angle at F between points C, F, and B1.This is getting a bit complicated. Maybe I can use vector methods or coordinate geometry to express the angle bisector conditions.Alternatively, maybe using mass point geometry or Ceva's theorem could help since we're dealing with angle bisectors.Wait, Ceva's theorem relates to concurrent cevians, but here we have two angle bisectors. Maybe not directly applicable.Alternatively, maybe using the angle bisector theorem. For triangle AEB1, BE is the angle bisector, so the ratio of the adjacent sides is equal to the ratio of the opposite sides.In triangle AEB1, BE bisects angle AEB1, so by the angle bisector theorem:(AE)/(EB1) = (AB)/(B1B)Wait, let me clarify. In triangle AEB1, the angle bisector at E is BE. So, the angle bisector theorem states that:(AE)/(EB1) = (AB)/(B1B)Wait, AB is a side of the original triangle, but in triangle AEB1, the sides are AE, EB1, and AB1.Wait, maybe I need to express the lengths in terms of coordinates.Point A is (2a, 2b), E is (a + t, b), B1 is (a + c, b).So, length AE is the distance between A(2a, 2b) and E(a + t, b):AE = sqrt[(2a - (a + t))^2 + (2b - b)^2] = sqrt[(a - t)^2 + b^2]Similarly, EB1 is the distance between E(a + t, b) and B1(a + c, b):EB1 = sqrt[(a + c - (a + t))^2 + (b - b)^2] = sqrt[(c - t)^2] = |c - t|Since t is between 0 and c, c - t is positive, so EB1 = c - t.By the angle bisector theorem in triangle AEB1, we have:(AE)/(EB1) = (AB)/(B1B)Wait, AB is the length from A(2a, 2b) to B(0,0):AB = sqrt[(2a)^2 + (2b)^2] = 2*sqrt(a^2 + b^2)B1B is the length from B1(a + c, b) to B(0,0):B1B = sqrt[(a + c)^2 + b^2]So, according to the angle bisector theorem:sqrt[(a - t)^2 + b^2] / (c - t) = 2*sqrt(a^2 + b^2) / sqrt[(a + c)^2 + b^2]This seems complicated, but maybe we can square both sides to eliminate the square roots:[(a - t)^2 + b^2] / (c - t)^2 = [4(a^2 + b^2)] / [(a + c)^2 + b^2]Cross-multiplying:[(a - t)^2 + b^2] * [(a + c)^2 + b^2] = 4(a^2 + b^2)(c - t)^2This is a quadratic equation in t. It might be messy, but perhaps we can find a value of t that satisfies this.Similarly, for point F on A1B1, we can apply the angle bisector theorem in triangle CFB1.Point F is (c + s*a, s*b). So, in triangle CFB1, BF is the angle bisector of angle CFB1.Applying the angle bisector theorem:(CF)/(FB1) = (CB)/(B1B)Wait, CB is the length from C(2c, 0) to B(0,0):CB = 2cB1B is sqrt[(a + c)^2 + b^2] as before.CF is the distance from C(2c, 0) to F(c + s*a, s*b):CF = sqrt[(2c - (c + s*a))^2 + (0 - s*b)^2] = sqrt[(c - s*a)^2 + (s*b)^2]FB1 is the distance from F(c + s*a, s*b) to B1(a + c, b):FB1 = sqrt[(a + c - (c + s*a))^2 + (b - s*b)^2] = sqrt[(a - s*a)^2 + (b(1 - s))^2] = sqrt[(a(1 - s))^2 + (b(1 - s))^2] = |1 - s| * sqrt(a^2 + b^2)Since s is between 0 and 1, 1 - s is positive, so FB1 = (1 - s)*sqrt(a^2 + b^2)By the angle bisector theorem:sqrt[(c - s*a)^2 + (s*b)^2] / [(1 - s)*sqrt(a^2 + b^2)] = 2c / sqrt[(a + c)^2 + b^2]Again, squaring both sides:[(c - s*a)^2 + (s*b)^2] / [(1 - s)^2(a^2 + b^2)] = [4c^2] / [(a + c)^2 + b^2]Cross-multiplying:[(c - s*a)^2 + (s*b)^2] * [(a + c)^2 + b^2] = 4c^2(1 - s)^2(a^2 + b^2)This is another quadratic equation in s.At this point, I'm getting bogged down with algebra. Maybe there's a better approach.Let me think about the properties of midlines and angle bisectors. Since E and F are on midlines, which are midsegments, maybe there's some symmetry or homothety involved.A homothety is a transformation that enlarges or reduces a figure by a scale factor relative to a center point. Since A1, B1, C1 are midpoints, a homothety with center B and factor 2 would map the midlines to the sides of the triangle.Wait, if I consider a homothety centered at B with factor 2, it would map the midline C1B1 to AB, and midline A1B1 to BC.Under this homothety, point E on C1B1 would map to a point on AB, and point F on A1B1 would map to a point on BC.But how does this help with the angle bisectors?Alternatively, maybe reflecting the triangle over the angle bisector of angle ABC could help. Reflections preserve angles and midpoints.If I reflect triangle ABC over the angle bisector of angle ABC, then point C maps to a point on AB, and point A maps to a point on BC. Maybe this reflection swaps E and F or relates them somehow.Alternatively, since BE and BF are angle bisectors, maybe they are symmetric with respect to the angle bisector of angle ABC.Wait, if I can show that the angle bisector of angle ABC is also the angle bisector of angle FBE, then they coincide.To do this, maybe I can show that the angle bisector of angle ABC divides angle FBE into two equal parts.Alternatively, maybe using trigonometric identities or vector methods to show that the direction vectors of the bisectors are the same.Another approach: since E and F are defined via angle bisectors on the midlines, maybe their positions are such that when connected to B, they form angles that are symmetric with respect to the angle bisector of angle ABC.Wait, perhaps using Ceva's theorem in triangle ABC with point B as the vertex. But Ceva's theorem involves concurrent cevians, and here we have two cevians BE and BF. Maybe not directly applicable.Alternatively, maybe using trigonometric Ceva's theorem, which relates to the sines of angles.Wait, trigonometric Ceva's theorem states that for concurrent cevians from a vertex, the product of the sines of the angles is equal. But I'm not sure if that applies here.Alternatively, maybe considering the incenter. The angle bisector of angle ABC is the incenter's direction. If I can show that the incenter lies on both BE and BF, but that seems unlikely since E and F are on midlines.Wait, maybe not the incenter, but some other center.Alternatively, perhaps using the fact that midlines create smaller similar triangles, and the angle bisectors in the smaller triangles correspond to those in the larger triangle.Wait, triangle A1B1C1 is similar to triangle ABC with a ratio of 1:2. So, any angle bisectors in triangle A1B1C1 would correspond to angle bisectors in triangle ABC scaled by 2.But points E and F are on midlines, not necessarily on the sides of triangle A1B1C1.Hmm, this is tricky. Maybe I need to think differently.Let me consider the angle bisector of angle ABC. Let's call this bisector l. I need to show that l is also the bisector of angle FBE.So, if I can show that l divides angle FBE into two equal angles, then l is the bisector.Alternatively, maybe showing that the reflection of F over l lies on BE, or something like that.Wait, reflection might be a good approach. If I reflect F over the angle bisector l, maybe the reflection lies on BE.Alternatively, since BE and BF are angle bisectors in their respective triangles, maybe they are symmetric with respect to l.Wait, another idea: since E and F are defined via angle bisectors on midlines, which are midsegments, maybe their positions are such that BE and BF are symmetric with respect to the angle bisector of angle ABC.Therefore, the angle between BE and the bisector is equal to the angle between BF and the bisector, implying that the bisector of angle FBE coincides with the bisector of angle ABC.Alternatively, maybe using the fact that the midlines are parallel to the sides, so the angles formed by BE and BF with the midlines are equal to those with the sides, leading to the bisectors coinciding.Wait, perhaps using the properties of parallel lines and corresponding angles.Since C1B1 is parallel to AB, the angle between BE and C1B1 is equal to the angle between BE and AB. Similarly, since A1B1 is parallel to BC, the angle between BF and A1B1 is equal to the angle between BF and BC.Given that BE and BF are angle bisectors in their respective triangles, maybe the angles they make with the midlines correspond to the angles they make with the sides, leading to the overall angle bisector of angle ABC.This is getting a bit abstract. Maybe I need to consider specific coordinates again.Let me assign specific coordinates to simplify. Let me set B at (0,0), C at (2,0), and A at (0,2). So, triangle ABC is a right triangle with legs of length 2.Then, midpoints:- A1 is midpoint of BC: (1,0)- B1 is midpoint of AC: (1,1)- C1 is midpoint of AB: (0,1)Midline C1B1 connects (0,1) to (1,1), which is the horizontal line y=1 from x=0 to x=1.Midline A1B1 connects (1,0) to (1,1), which is the vertical line x=1 from y=0 to y=1.Wait, in this specific case, midline C1B1 is horizontal, and midline A1B1 is vertical.So, point E is on C1B1: (t,1) where t is between 0 and1.Point F is on A1B1: (1, s) where s is between 0 and1.Now, BE is the angle bisector of angle AEB1.Point A is (0,2), E is (t,1), B1 is (1,1).So, angle AEB1 is the angle at E between points A, E, B1.Similarly, BF is the angle bisector of angle CFB1.Point C is (2,0), F is (1, s), B1 is (1,1).So, angle CFB1 is the angle at F between points C, F, B1.Let me first find the coordinates of E such that BE bisects angle AEB1.Using the angle bisector theorem in triangle AEB1.In triangle AEB1, BE is the angle bisector of angle AEB1.So, by the angle bisector theorem:(AE)/(EB1) = (AB)/(B1B)Compute lengths:AE: distance from A(0,2) to E(t,1):AE = sqrt[(t - 0)^2 + (1 - 2)^2] = sqrt[t^2 + 1]EB1: distance from E(t,1) to B1(1,1):EB1 = |1 - t|AB: distance from A(0,2) to B(0,0):AB = 2B1B: distance from B1(1,1) to B(0,0):B1B = sqrt[1^2 + 1^2] = sqrt(2)So, by angle bisector theorem:sqrt(t^2 + 1) / |1 - t| = 2 / sqrt(2) = sqrt(2)So,sqrt(t^2 + 1) = sqrt(2) * |1 - t|Square both sides:t^2 + 1 = 2*(1 - 2t + t^2)Simplify:t^2 + 1 = 2 - 4t + 2t^2Bring all terms to one side:0 = 2 - 4t + 2t^2 - t^2 -1Simplify:0 = t^2 - 4t +1Solve quadratic equation:t = [4 ± sqrt(16 - 4)] / 2 = [4 ± sqrt(12)] / 2 = [4 ± 2*sqrt(3)] / 2 = 2 ± sqrt(3)Since t is between 0 and1, we take t = 2 - sqrt(3) ≈ 2 - 1.732 ≈ 0.2679So, point E is at (2 - sqrt(3), 1)Similarly, let's find point F on A1B1 such that BF bisects angle CFB1.Point F is (1, s). So, in triangle CFB1, BF is the angle bisector of angle CFB1.By angle bisector theorem:(CF)/(FB1) = (CB)/(B1B)Compute lengths:CF: distance from C(2,0) to F(1,s):CF = sqrt[(2 -1)^2 + (0 - s)^2] = sqrt[1 + s^2]FB1: distance from F(1,s) to B1(1,1):FB1 = |1 - s|CB: distance from C(2,0) to B(0,0):CB = 2B1B: distance from B1(1,1) to B(0,0):B1B = sqrt(2) as beforeSo, by angle bisector theorem:sqrt(1 + s^2) / |1 - s| = 2 / sqrt(2) = sqrt(2)Thus,sqrt(1 + s^2) = sqrt(2)*|1 - s|Square both sides:1 + s^2 = 2*(1 - 2s + s^2)Simplify:1 + s^2 = 2 - 4s + 2s^2Bring all terms to one side:0 = 2 - 4s + 2s^2 -1 -s^2Simplify:0 = s^2 -4s +1Solve quadratic equation:s = [4 ± sqrt(16 -4)] / 2 = [4 ± sqrt(12)] / 2 = 2 ± sqrt(3)Since s is between 0 and1, s = 2 - sqrt(3) ≈ 0.2679So, point F is at (1, 2 - sqrt(3))Now, we have points E(2 - sqrt(3),1) and F(1, 2 - sqrt(3)).We need to find the bisector of angle FBE and show it coincides with the bisector of angle ABC.First, let's find the angle bisector of angle ABC.In triangle ABC, angle at B is the right angle since ABC is a right triangle with legs AB and BC. Wait, no, in my coordinate system, ABC is a right triangle with right angle at B? Wait, no, in my coordinate system, A is at (0,2), B at (0,0), C at (2,0). So, AB is vertical, BC is horizontal, and AC is the hypotenuse. So, angle at B is the right angle, 90 degrees.Wait, but in the problem statement, we need to find the bisector of angle ABC, which is 90 degrees, so its bisector would be the line that splits the right angle into two 45-degree angles.In this coordinate system, the angle bisector of angle ABC (the right angle) would be the line y = x, since it makes equal angles with the x-axis and y-axis.Wait, but in our coordinate system, AB is along the y-axis from (0,0) to (0,2), and BC is along the x-axis from (0,0) to (2,0). So, the angle bisector would indeed be the line y = x, which makes 45 degrees with both axes.Now, we need to find the bisector of angle FBE and show it coincides with y = x.First, let's find the coordinates of points F, B, E.Point F is (1, 2 - sqrt(3)) ≈ (1, 0.2679)Point B is (0,0)Point E is (2 - sqrt(3),1) ≈ (0.2679,1)So, angle FBE is the angle at B between points F, B, E.We need to find the bisector of this angle.First, let's find the vectors BF and BE.Vector BF is from B(0,0) to F(1, 2 - sqrt(3)): (1, 2 - sqrt(3))Vector BE is from B(0,0) to E(2 - sqrt(3),1): (2 - sqrt(3),1)We can find the angle bisector by finding a vector that is a linear combination of the unit vectors in the directions of BF and BE.First, find the unit vectors.Length of BF: sqrt(1^2 + (2 - sqrt(3))^2) = sqrt(1 + (4 - 4*sqrt(3) + 3)) = sqrt(8 - 4*sqrt(3))Similarly, length of BE: sqrt((2 - sqrt(3))^2 +1^2) = sqrt(4 -4*sqrt(3) +3 +1) = sqrt(8 -4*sqrt(3))Interestingly, both BF and BE have the same length, sqrt(8 -4*sqrt(3)).So, the unit vectors are:For BF: (1, 2 - sqrt(3)) / sqrt(8 -4*sqrt(3))For BE: (2 - sqrt(3),1) / sqrt(8 -4*sqrt(3))The angle bisector direction vector would be the sum of these unit vectors.Let me compute the sum:(1 + (2 - sqrt(3)), (2 - sqrt(3)) +1) / sqrt(8 -4*sqrt(3))Simplify numerator:x-component: 1 + 2 - sqrt(3) = 3 - sqrt(3)y-component: (2 - sqrt(3)) +1 = 3 - sqrt(3)So, the direction vector is (3 - sqrt(3), 3 - sqrt(3)) / sqrt(8 -4*sqrt(3))This simplifies to (1,1) scaled by (3 - sqrt(3))/sqrt(8 -4*sqrt(3))But (3 - sqrt(3))/sqrt(8 -4*sqrt(3)) can be simplified.Let me compute sqrt(8 -4*sqrt(3)).Let me write 8 -4*sqrt(3) as (sqrt(6) - sqrt(2))^2.Compute (sqrt(6) - sqrt(2))^2 = 6 + 2 - 2*sqrt(12) = 8 - 4*sqrt(3). Yes, that's correct.So, sqrt(8 -4*sqrt(3)) = sqrt(6) - sqrt(2)Therefore, the direction vector is (1,1) scaled by (3 - sqrt(3))/(sqrt(6) - sqrt(2))Let me rationalize the denominator:(3 - sqrt(3))/(sqrt(6) - sqrt(2)) * (sqrt(6) + sqrt(2))/(sqrt(6) + sqrt(2)) = [ (3 - sqrt(3))(sqrt(6) + sqrt(2)) ] / (6 - 2) = [ (3 - sqrt(3))(sqrt(6) + sqrt(2)) ] /4Let me compute the numerator:3*sqrt(6) + 3*sqrt(2) - sqrt(3)*sqrt(6) - sqrt(3)*sqrt(2)Simplify each term:3*sqrt(6) remains3*sqrt(2) remainssqrt(3)*sqrt(6) = sqrt(18) = 3*sqrt(2)sqrt(3)*sqrt(2) = sqrt(6)So, numerator becomes:3*sqrt(6) + 3*sqrt(2) - 3*sqrt(2) - sqrt(6) = (3*sqrt(6) - sqrt(6)) + (3*sqrt(2) - 3*sqrt(2)) = 2*sqrt(6) + 0 = 2*sqrt(6)Therefore, the direction vector is (1,1) scaled by (2*sqrt(6))/4 = (sqrt(6)/2)So, the direction vector is (sqrt(6)/2, sqrt(6)/2), which is just a scalar multiple of (1,1). Therefore, the angle bisector of angle FBE is along the line y = x, which is the same as the angle bisector of angle ABC.Hence, the bisectors coincide.**Final Answer**boxed{text{The bisectors of angles } ABC text{ and } FBE text{ coincide.}}

answer：Alright, so I've got this problem here: "Bryson bought 2 pair of shoes during the thanksgiving sale. Find the total number of new shoe he has now?" Okay, let's break this down step by step. First, I need to understand what exactly is being asked. It seems straightforward—Bryson bought some shoes, and we need to find out how many new shoes he has now.Alright, so the key information here is that Bryson bought 2 pair of shoes. Now, the word "pair" is important here. In everyday language, when someone buys shoes, they usually buy them in pairs, meaning one for the left foot and one for the right foot. So, one pair equals two shoes. That seems pretty standard.But wait, the problem says "2 pair of shoes." Hmm, that's interesting. It uses "pair" without the 's' at the end, but I think it's just a way of saying two pairs. So, two pairs of shoes. Okay, so if one pair is two shoes, then two pairs would be four shoes in total. That makes sense.But let me double-check to make sure I'm not missing anything. The problem mentions "during the thanksgiving sale," which might imply that there are some discounts or special offers, but I don't think that affects the number of shoes Bryson bought. It's just providing context about when he bought them.Now, the problem asks for the total number of new shoes he has now. So, we're assuming that these are new shoes, and we need to find out how many new shoes he has in total after this purchase.So, if one pair is two shoes, then two pairs would be two times two, which equals four shoes. That seems pretty straightforward. But let me think about it another way to make sure I'm not oversimplifying.Suppose Bryson already had some shoes before this purchase. The problem doesn't specify, so I think we can assume that we're only concerned with the number of new shoes he bought during the sale. So, regardless of how many shoes he had before, we're focusing on the two pairs he bought now.But wait, the problem says "the total number of new shoe he has now." It uses "shoe" instead of "shoes." That might be a typo, but I think it's safe to assume they meant "shoes." So, it's asking for the total number of new shoes he has now.Okay, so if he bought two pairs, that's four shoes. So, the total number of new shoes he has now is four.But let me think about possible exceptions or alternative interpretations. Could "pair" mean something different here? For example, in some contexts, a pair could refer to a matching set, but in the case of shoes, it's almost always one for each foot. So, I think it's safe to stick with the standard definition.Also, could there be any hidden information or assumptions I'm missing? The problem doesn't mention anything about Bryson discarding old shoes or receiving any additional shoes as part of the sale. So, I think it's just a straightforward calculation.In summary, Bryson bought two pairs of shoes, and each pair consists of two shoes. Therefore, two pairs equal four shoes. So, the total number of new shoes he has now is four.But just to be thorough, let's consider if the problem meant something else. Suppose "pair" was being used differently, like a set of shoes for a specific purpose, but that doesn't seem to fit here. The context is a thanksgiving sale, which is a general shopping event, not specialized.Also, if we consider the possibility that "pair" could mean something else, like a pair of socks or something, but no, the problem specifically mentions shoes. So, it's definitely about shoes.Another angle: could the problem be trying to trick me into thinking about something else, like the number of pairs instead of individual shoes? But the question asks for the total number of new shoes, not pairs. So, it's clear that they want the count of individual shoes.Alright, I think I've covered all bases here. The problem is straightforward: two pairs of shoes equal four individual shoes. So, Bryson has four new shoes now.