answer：Okay, so I need to prove that the quartic polynomial (x^4 + 2x^2 + 2x + 2) can't be factored into two quadratic trinomials with integer coefficients. Hmm, quartic means degree 4, and quadratic trinomials are degree 2. So, if it factors into two quadratics, each of those quadratics would have to be of the form (x^2 + ax + b) and (x^2 + cx + d), where (a), (b), (c), and (d) are integers. Alright, let me start by assuming that it can be factored like that. So, let's say:[(x^2 + ax + b)(x^2 + cx + d) = x^4 + 2x^2 + 2x + 2]Now, I need to expand the left side and then equate the coefficients to the right side. Let's do that step by step.First, expand the left side:[(x^2 + ax + b)(x^2 + cx + d) = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd]Now, let's write down the coefficients from the original polynomial:- The coefficient of (x^4) is 1.- The coefficient of (x^3) is 0 (since there's no (x^3) term).- The coefficient of (x^2) is 2.- The coefficient of (x) is 2.- The constant term is 2.So, setting the coefficients equal, we get the following system of equations:1. (a + c = 0) (from the (x^3) term)2. (ac + b + d = 2) (from the (x^2) term)3. (ad + bc = 2) (from the (x) term)4. (bd = 2) (from the constant term)Alright, let's try to solve this system step by step.From equation 1: (a + c = 0), so (c = -a). That's helpful because it reduces the number of variables.Now, let's substitute (c = -a) into the other equations.Equation 2 becomes:[a(-a) + b + d = 2 implies -a^2 + b + d = 2]Equation 3 becomes:[a d + b(-a) = 2 implies a d - a b = 2 implies a(d - b) = 2]Equation 4 is still:[b d = 2]So, now we have:1. (c = -a)2. (-a^2 + b + d = 2)3. (a(d - b) = 2)4. (b d = 2)Now, let's look at equation 4: (b d = 2). Since (b) and (d) are integers, their possible pairs are limited. The integer pairs that multiply to 2 are:- (1, 2)- (2, 1)- (-1, -2)- (-2, -1)So, we can consider each of these pairs and see if they satisfy the other equations.Let's start with the first pair: (b = 1), (d = 2).Substituting into equation 3: (a(2 - 1) = 2 implies a(1) = 2 implies a = 2).Then, from equation 1, (c = -a = -2).Now, substitute (a = 2), (c = -2), (b = 1), (d = 2) into equation 2:[-2^2 + 1 + 2 = -4 + 3 = -1 neq 2]That's not equal to 2, so this pair doesn't work.Next, let's try (b = 2), (d = 1).Substituting into equation 3: (a(1 - 2) = 2 implies a(-1) = 2 implies a = -2).Then, (c = -a = 2).Now, substitute (a = -2), (c = 2), (b = 2), (d = 1) into equation 2:[-(-2)^2 + 2 + 1 = -4 + 3 = -1 neq 2]Again, not equal to 2. So, this pair doesn't work either.Now, let's try the negative pairs. First, (b = -1), (d = -2).Substituting into equation 3: (a(-2 - (-1)) = 2 implies a(-1) = 2 implies a = -2).Then, (c = -a = 2).Substitute into equation 2:[-(-2)^2 + (-1) + (-2) = -4 - 1 - 2 = -7 neq 2]Not equal to 2. Doesn't work.Next, (b = -2), (d = -1).Substituting into equation 3: (a(-1 - (-2)) = 2 implies a(1) = 2 implies a = 2).Then, (c = -a = -2).Substitute into equation 2:[-(2)^2 + (-2) + (-1) = -4 - 2 - 1 = -7 neq 2]Still not equal to 2. So, none of the integer pairs for (b) and (d) satisfy all the equations.Wait, did I miss any pairs? Let me double-check. The pairs are (1,2), (2,1), (-1,-2), (-2,-1). I think I covered all of them.So, in all cases, substituting the possible integer pairs for (b) and (d) leads to a contradiction in equation 2. Therefore, there are no integer solutions for (a), (b), (c), and (d) that satisfy all the equations.Hence, the quartic polynomial (x^4 + 2x^2 + 2x + 2) cannot be factored into the product of two quadratic trinomials with integer coefficients.

answer：Alright, so I'm trying to understand this geometry problem. It says that if there's a convex quadrilateral (ABCD) with both an incircle (a circle tangent to all four sides) and an excircle (a circle tangent to the extensions of all four sides), then the diagonals of this quadrilateral must be perpendicular. Hmm, that's interesting. I need to figure out why that's the case.First, let me recall what a convex quadrilateral is. It's a four-sided figure where all interior angles are less than 180 degrees, and the sides don't cross each other. Now, having an incircle means that there's a circle inside the quadrilateral that touches all four sides. Such quadrilaterals are called tangential quadrilaterals. I remember that in a tangential quadrilateral, the sums of the lengths of opposite sides are equal. So, (AB + CD = BC + AD). That's a useful property.Now, the problem also mentions an excircle. An excircle is a circle outside the quadrilateral that is tangent to the extensions of all four sides. I'm not as familiar with excircles in quadrilaterals, but I think they exist for certain types of quadrilaterals. Maybe this quadrilateral is bicentric, meaning it has both an incircle and a circumcircle. Wait, no, a circumcircle would pass through all four vertices, which is different from an excircle tangent to the extensions of the sides.I need to visualize this. Let me try to sketch a convex quadrilateral with an incircle and an excircle. The incircle touches all four sides from the inside, while the excircle touches the extensions of the sides from the outside. It's like having two circles, one inside and one outside, both related to the quadrilateral in a specific way.Now, the key is to show that the diagonals (AC) and (BD) are perpendicular. I know that in some quadrilaterals, like rhombuses and squares, the diagonals are perpendicular, but those are special cases. Here, we're dealing with a general convex quadrilateral with specific circle tangency properties.Maybe I can use properties of tangents to circles. In a tangential quadrilateral, the lengths from a vertex to the points of tangency are equal. For example, if the incircle touches side (AB) at point (P), side (BC) at (Q), side (CD) at (R), and side (DA) at (S), then (AP = AS), (BP = BQ), (CQ = CR), and (DR = DS). These equal tangent lengths might help in establishing some relationships between the sides and angles of the quadrilateral.Similarly, for the excircle, the points of tangency on the extensions of the sides will also have equal tangent lengths. Let's denote the excircle touching the extension of side (AB) at (P'), the extension of side (BC) at (Q'), the extension of side (CD) at (R'), and the extension of side (DA) at (S'). Then, the tangent lengths from each vertex to these points should also be equal. For example, (AP' = AS'), (BP' = BQ'), (CQ' = CR'), and (DR' = DS').I wonder if these equal tangent lengths can help me relate the sides of the quadrilateral in a way that forces the diagonals to be perpendicular. Maybe by setting up equations based on these tangent lengths and using the fact that the quadrilateral is convex, I can derive some relationships between the angles or sides that lead to the conclusion about the diagonals.Another approach could be to use coordinates. If I assign coordinates to the vertices of the quadrilateral, I might be able to express the conditions for the existence of an incircle and an excircle, and then derive the equations for the diagonals to show that they are perpendicular. This might involve some complex algebra, but it could be a systematic way to approach the problem.Alternatively, maybe there's a theorem or property that directly links the existence of an incircle and an excircle to the perpendicularity of the diagonals in a convex quadrilateral. I'm not sure if I remember such a theorem, but it's possible that this is a known result in geometry.Let me think about the properties of bicentric quadrilaterals. Wait, bicentric quadrilaterals have both an incircle and a circumcircle, but in this problem, we're dealing with an incircle and an excircle. So, it's not exactly the same as bicentric quadrilaterals. Maybe there's a similar property or a related theorem that applies here.I also recall that in some cases, the existence of both an incircle and an excircle can impose strong constraints on the shape of the quadrilateral, potentially forcing it to have perpendicular diagonals. Perhaps by analyzing the symmetry or the angle relationships imposed by these circles, I can show that the diagonals must intersect at right angles.Another idea is to consider the incenter and excenter of the quadrilateral. The incenter is the center of the incircle, and the excenter is the center of the excircle. These points might have specific relationships with the diagonals, such as lying on them or being equidistant from them, which could lead to the diagonals being perpendicular.I'm also thinking about the concept of harmonic division or projective geometry, but I'm not sure if that's applicable here. Maybe using vector geometry could help, by expressing the sides and diagonals as vectors and showing that their dot product is zero, which would imply perpendicularity.Wait, let's get back to the basics. In a tangential quadrilateral, the incenter is equidistant from all sides. Similarly, the excenter is equidistant from the extensions of all sides. These distances are the radii of the incircle and excircle, respectively. Maybe by relating these radii to the sides and angles of the quadrilateral, I can find a relationship that leads to the diagonals being perpendicular.I also remember that in some quadrilaterals, like kites, the diagonals are perpendicular, but kites have two distinct pairs of adjacent sides equal. I don't know if the quadrilateral in question has such properties, but maybe the presence of both an incircle and an excircle imposes similar constraints.Perhaps I can use trigonometry. If I can express the angles of the quadrilateral in terms of the radii of the incircle and excircle, I might be able to show that the angles between the diagonals are 90 degrees.Another thought: maybe the quadrilateral is orthodiagonal (having perpendicular diagonals) if and only if the sum of the squares of two opposite sides equals the sum of the squares of the other two opposite sides. I think that's a property of orthodiagonal quadrilaterals. So, if I can show that (AB^2 + CD^2 = BC^2 + AD^2), then the diagonals are perpendicular.Given that the quadrilateral has both an incircle and an excircle, perhaps I can derive such a relationship between the sides. Since we know that in a tangential quadrilateral, (AB + CD = BC + AD), and maybe combining this with properties from the excircle could lead to the necessary condition for perpendicular diagonals.I'm also considering the areas of the quadrilateral. The area can be expressed in terms of the inradius and the semiperimeter, and maybe also in terms of the exradius. Perhaps equating these expressions or relating them could provide some insight into the relationship between the sides and the diagonals.Wait, let's think about the exradius. For a tangential quadrilateral, the inradius (r) is related to the area (K) and the semiperimeter (s) by (K = r cdot s). For an excircle, there's also a relationship involving the exradius (R), but I'm not sure exactly what it is. Maybe it's similar, but with a different semiperimeter or something.I need to look up or recall the exact formula for the exradius in a quadrilateral. I think it might be related to the area and some other parameter, but I'm not certain. If I can express both the inradius and exradius in terms of the sides and area, maybe I can find a relationship that ties them together and leads to the conclusion about the diagonals.Another approach could be to use the Pitot theorem, which states that for a convex quadrilateral to be tangential, the sum of the lengths of the two opposite sides must be equal. We already have that (AB + CD = BC + AD). Maybe combining this with properties from the excircle could give us additional equations that relate the sides in a way that forces the diagonals to be perpendicular.I'm also thinking about the Newton's theorem, which relates the midpoints of the diagonals in a quadrilateral, but I'm not sure if that's directly applicable here.Perhaps I can consider the specific case of a quadrilateral that has both an incircle and an excircle and see if it must be a kite or some other quadrilateral with perpendicular diagonals. If I can show that the quadrilateral must satisfy the properties of a kite or another orthodiagonal quadrilateral, then the conclusion follows.Alternatively, maybe I can use coordinate geometry. Let's assign coordinates to the vertices of the quadrilateral and use the conditions for the existence of an incircle and an excircle to derive equations that the coordinates must satisfy. Then, by calculating the slopes of the diagonals, I can show that their product is (-1), which would imply that they are perpendicular.This might involve setting up a coordinate system where one vertex is at the origin, and then expressing the other vertices in terms of variables. Then, using the conditions for tangency to both circles, I can set up equations and solve for the relationships between the coordinates. Finally, by calculating the slopes of the diagonals, I can verify if they are perpendicular.I think this could be a viable approach, although it might be algebraically intensive. Let's try to outline the steps:1. Assign coordinates to the vertices (A), (B), (C), and (D).2. Express the conditions for the existence of an incircle and an excircle in terms of these coordinates.3. Derive equations based on these conditions.4. Solve these equations to find relationships between the coordinates.5. Calculate the slopes of the diagonals (AC) and (BD).6. Show that the product of these slopes is (-1), indicating perpendicularity.This seems like a systematic way to approach the problem, even if it's a bit involved. I might need to use the formula for the distance from a point to a line to express the conditions for tangency.Alternatively, maybe there's a more geometric approach that avoids heavy algebra. Perhaps by considering the properties of the incircle and excircle and their centers, I can find a relationship between the diagonals.I recall that in a tangential quadrilateral, the incenter lies at the intersection of the angle bisectors. Similarly, the excenter lies at the intersection of the external angle bisectors. Maybe the line connecting the incenter and excenter has some special property related to the diagonals.If I can show that this line is perpendicular to one of the diagonals, or that it bisects the angle between the diagonals, it might help in proving that the diagonals are perpendicular.Another idea is to consider the midpoints of the diagonals. In some quadrilaterals, like parallelograms, the midpoints coincide, but in general quadrilaterals, they don't. Maybe the presence of both an incircle and an excircle imposes some condition on these midpoints that leads to the diagonals being perpendicular.I'm also thinking about the concept of harmonic conjugates and projective geometry, but I'm not sure if that's necessary here. Maybe a simpler approach using basic geometry principles would suffice.Wait, let's think about the properties of the excircle. Since it's tangent to the extensions of the sides, it's like an ex-tangential quadrilateral. I think there's a term for quadrilaterals that have both an incircle and an excircle, but I'm not sure. Maybe they're called bitangential quadrilaterals?If I can find some properties of bitangential quadrilaterals, perhaps I can use them to show that the diagonals are perpendicular. I recall that in some cases, bitangential quadrilaterals have perpendicular diagonals, but I need to verify if that's always true.Alternatively, maybe I can use the fact that both circles are tangent to the sides (or their extensions) to establish some congruent triangles or similar triangles within the quadrilateral, which could lead to the conclusion about the diagonals.I'm also considering the possibility of using the Pythagorean theorem. If I can show that the triangles formed by the diagonals satisfy the Pythagorean theorem, then the diagonals must be perpendicular.Another thought: perhaps the existence of both an incircle and an excircle implies that the quadrilateral is both tangential and ex-tangential, and that these properties together force the diagonals to be perpendicular. Maybe there's a theorem that states this directly.I think I need to look up or recall if there's a specific theorem that connects the existence of both an incircle and an excircle to the perpendicularity of the diagonals in a convex quadrilateral. If such a theorem exists, it would make the proof straightforward.Alternatively, if no such theorem comes to mind, I'll need to proceed with a more detailed geometric analysis, possibly using coordinate geometry or vector methods, to derive the necessary relationships.In summary, my approach will be:1. Recall properties of tangential and ex-tangential quadrilaterals.2. Consider the implications of having both an incircle and an excircle.3. Explore geometric properties, such as equal tangent lengths, incenter and excenter relationships, and potential congruence or similarity of triangles.4. Use coordinate geometry or vector methods if necessary to establish the required relationships.5. Conclude that the diagonals must be perpendicular based on the derived relationships.I think starting with the properties of tangential quadrilaterals and then incorporating the properties of the excircle will be the most straightforward path. Let's proceed with that.

answer：To determine the values of ( m ) and ( n ) such that points ( A(-3, m) ) and ( B(n, 2) ) are symmetric with respect to the origin, we can use the property of symmetry about the origin. When two points are symmetric with respect to the origin, their coordinates are negatives of each other. This means:- The x-coordinate of point ( A ) is the negative of the x-coordinate of point ( B ).- The y-coordinate of point ( A ) is the negative of the y-coordinate of point ( B ).Starting with the x-coordinates:[-3 = -n implies n = 3]Next, for the y-coordinates:[m = -2]Adding these values together:[m + n = -2 + 3 = 1]

answer：First, identify the distinct marbles and the identical marbles. Tom has four unique marbles (red, green, blue, purple) and two sets of identical marbles: four yellow and three orange.Next, calculate the number of ways to choose two identical marbles:- For the yellow marbles, since they are identical, there is only 1 way to choose two.- Similarly, for the orange marbles, there is only 1 way to choose two.Then, calculate the number of ways to choose two marbles of different colors:- There are 4 unique marbles, and the number of ways to choose two different colors is C(4,2) = 6.Additionally, calculate the number of ways to pair each unique marble with the identical marbles:- Each unique marble can be paired with 4 yellow marbles, resulting in 4 × 4 = 16 pairs.- Each unique marble can be paired with 3 orange marbles, resulting in 4 × 3 = 12 pairs.Finally, add up all the possible combinations:- 1 (two yellow) + 1 (two orange) + 6 (two different colors) + 16 (one unique and one yellow) + 12 (one unique and one orange) = 36.Therefore, the total number of distinct pairs of marbles Tom can choose is 36.

answer：First, I need to determine the work rates of individuals ( A ), ( B ), and ( C ). Let ( r_A ), ( r_B ), and ( r_C ) represent their respective work rates in jobs per day.From the given information:1. ( A ) and ( B ) together can complete the job in 3 days: [ 3(r_A + r_B) = 1 implies r_A + r_B = frac{1}{3} ]2. ( B ) and ( C ) together can complete the job in 2 days: [ 2(r_B + r_C) = 1 implies r_B + r_C = frac{1}{2} ]3. ( A ) and ( C ) together can complete the job in 3 days: [ 3(r_A + r_C) = 1 implies r_A + r_C = frac{1}{3} ]Next, I'll add all three equations to find the combined work rate of ( A ), ( B ), and ( C ):[(r_A + r_B) + (r_B + r_C) + (r_A + r_C) = frac{1}{3} + frac{1}{2} + frac{1}{3}][2(r_A + r_B + r_C) = frac{7}{6}][r_A + r_B + r_C = frac{7}{12}]Now, I'll subtract the second equation (( r_B + r_C = frac{1}{2} )) from the combined work rate to find ( r_A ):[r_A = frac{7}{12} - frac{1}{2} = frac{7}{12} - frac{6}{12} = frac{1}{12}]Finally, the number of days required for ( A ) to complete the job alone is the reciprocal of ( r_A ):[text{Days required by } A = frac{1}{r_A} = frac{1}{frac{1}{12}} = 12 text{ days}]

answer：First, I need to understand what elevation means. Elevation is the vertical height of a point above or below sea level.Mount Everest is the highest mountain peak in the world, and it is located above sea level. To represent this elevation using a positive number, I will denote it as +8844 meters.On the other hand, the Mariana Trench is the lowest point on Earth and is located below sea level. To represent this elevation using a negative number, I will denote it as -11034 meters.By assigning positive and negative signs to these elevations, I can clearly distinguish between points above and below sea level.