answer：Okay, so I have these two geometry problems to solve about convex polyhedrons. Let me take them one at a time. Starting with part a: Prove that in any convex polyhedron, there is either a triangular face or a trihedral angle. Hmm, okay. So, a convex polyhedron is a three-dimensional shape where all its interior angles are less than 180 degrees, and every face is a convex polygon. A triangular face is a face that's a triangle, and a trihedral angle is a point where three edges meet, right? So, the problem is saying that in any such polyhedron, you can't avoid both triangular faces and trihedral angles. At least one of them has to exist.I remember that Euler's formula relates the number of vertices, edges, and faces in a convex polyhedron. Euler's formula is V - E + F = 2, where V is vertices, E is edges, and F is faces. Maybe that can help here. Let me think. If I can show that if there are no triangular faces, then there must be a trihedral angle, or vice versa. So, suppose, for contradiction, that there are no triangular faces and no trihedral angles. Then, every face must have at least four edges, and every vertex must have at least four edges meeting there. Wait, but in a polyhedron, each edge is shared by two faces, and each vertex is where at least three edges meet. So, if every face has at least four edges, then the total number of edges can be calculated as (sum of all edges around each face)/2. Similarly, each vertex has at least four edges, so the total number of edges can also be calculated as (sum of all edges around each vertex)/2.Let me write that down. Let F be the number of faces, each with at least four edges, so the total number of edges E is at least (4F)/2 = 2F. Similarly, let V be the number of vertices, each with at least four edges, so the total number of edges E is at least (4V)/2 = 2V. So, E >= 2F and E >= 2V.But from Euler's formula, V - E + F = 2. If E >= 2F and E >= 2V, then substituting into Euler's formula, we have V - E + F <= V - 2F + F = V - F. But Euler's formula says this equals 2, so V - F >= 2. Similarly, V - E + F <= V - 2V + F = -V + F. But this is also equal to 2, so -V + F >= 2. So, combining these, V - F >= 2 and F - V >= 2. But that can't happen because if V - F >= 2, then F - V <= -2, which contradicts F - V >= 2. Therefore, our initial assumption that there are no triangular faces and no trihedral angles must be wrong. So, there must be at least one triangular face or one trihedral angle. Wait, does that cover all cases? Let me double-check. If a polyhedron has no triangular faces, then every face has at least four edges, so E >= 2F. Similarly, if it has no trihedral angles, then every vertex has at least four edges, so E >= 2V. Then, plugging into Euler's formula, V - E + F = 2, but if E >= 2F and E >= 2V, then V - E + F <= V - 2F + F = V - F, which must be >= 2. Similarly, V - E + F <= -V + F, which must be >= 2. So, V - F >= 2 and F - V >= 2, which is impossible. Therefore, our assumption is wrong, so there must be at least one triangular face or one trihedral angle. Okay, that seems solid.Moving on to part b: Prove that for any convex polyhedron, the sum of the number of triangular faces and the number of trihedral angles is at least 8. So, let T be the number of triangular faces, and S be the number of trihedral angles. We need to show that T + S >= 8.Hmm, okay. Maybe I can use similar reasoning as in part a, but more detailed. Let's denote:- Let F be the total number of faces.- Let V be the total number of vertices.- Let E be the total number of edges.We know Euler's formula: V - E + F = 2.Also, let me denote:- Let T be the number of triangular faces.- Let S be the number of trihedral angles (vertices where three edges meet).- Let Q be the number of quadrilateral faces.- Let P be the number of pentagonal faces.And so on. Similarly, for vertices, let me denote:- Let S be the number of trihedral angles.- Let Qv be the number of vertices where four edges meet.- Let Pv be the number of vertices where five edges meet.And so on.So, each face contributes a certain number of edges, and each vertex contributes a certain number of edges. Since each edge is shared by two faces and two vertices, we can write:Sum over all faces of the number of edges per face = 2E.Similarly, sum over all vertices of the number of edges per vertex = 2E.So, for faces:3T + 4Q + 5P + ... = 2E.For vertices:3S + 4Qv + 5Pv + ... = 2E.So, both sums equal 2E.Now, our goal is to relate T and S. Let me try to find inequalities involving T and S.First, note that if a face is not triangular, it has at least four edges. Similarly, if a vertex is not trihedral, it has at least four edges.So, let me write:Sum over faces: 3T + 4(F - T) <= 2E.Wait, no. If T is the number of triangular faces, then the remaining F - T faces have at least four edges each. So, the total number of edges is at least 3T + 4(F - T). But since each edge is counted twice, 2E >= 3T + 4(F - T). So, 2E >= 4F - T.Similarly, for vertices: each vertex is either trihedral or has at least four edges. So, the total number of edges is at least 3S + 4(V - S). So, 2E >= 3S + 4(V - S) = 4V - S.So, we have:2E >= 4F - T,and2E >= 4V - S.Therefore, combining these, 4F - T <= 2E,and 4V - S <= 2E.So, 4F - T <= 2E,and 4V - S <= 2E.So, adding these two inequalities:4F - T + 4V - S <= 4E.But from Euler's formula, V - E + F = 2, so V + F = E + 2.Therefore, 4V + 4F = 4E + 8.So, 4V + 4F = 4E + 8.But from the previous inequality, 4F - T + 4V - S <= 4E.So, substituting 4V + 4F = 4E + 8, we have:(4E + 8) - T - S <= 4E.Subtracting 4E from both sides:8 - T - S <= 0.Which simplifies to:T + S >= 8.So, that's exactly what we needed to prove. Therefore, the sum of the number of triangular faces and trihedral angles is at least 8.Wait, let me make sure I didn't make a mistake in the algebra. So, starting from:4F - T <= 2E,and4V - S <= 2E.Adding them:4F + 4V - T - S <= 4E.But from Euler's formula, V + F = E + 2, so 4V + 4F = 4E + 8.Therefore, substituting into the inequality:(4E + 8) - T - S <= 4E.Subtract 4E:8 - T - S <= 0,which gives T + S >= 8.Yes, that seems correct. So, part b is proved.I think that's solid. So, in summary, for part a, assuming no triangular faces and no trihedral angles leads to a contradiction via Euler's formula, so at least one must exist. For part b, using the same approach but with more detailed inequalities, we find that the sum of triangular faces and trihedral angles must be at least 8.**Final Answer**a) boxed{text{Proven}} b) boxed{8}

answer：Okay, so I have this problem here where I need to find the ordered pair (a, b) given that one of the roots of the cubic equation (x^3 + ax^2 - x + b = 0) is (1 - 2i). Hmm, let me think about how to approach this.First, I remember that for polynomials with real coefficients, if there's a complex root, its complex conjugate must also be a root. That's because the coefficients are real, so the imaginary parts have to cancel out. So, if (1 - 2i) is a root, then (1 + 2i) must also be a root. That gives me two roots already: (1 - 2i) and (1 + 2i).Since it's a cubic equation, there must be a third root, which I'll call (r), that's real because the other two are complex conjugates. So, the three roots are (1 - 2i), (1 + 2i), and (r).Now, I can use Vieta's formulas to relate the coefficients of the polynomial to the sums and products of its roots. Vieta's formulas for a cubic equation (x^3 + px^2 + qx + c = 0) tell us that:1. The sum of the roots is (-p).2. The sum of the products of the roots taken two at a time is (q).3. The product of the roots is (-c).In our case, the equation is (x^3 + ax^2 - x + b = 0), so comparing to the standard form, (p = a), (q = -1), and (c = b).Let me write down the Vieta's formulas for this specific equation:1. Sum of roots: ((1 - 2i) + (1 + 2i) + r = -a)2. Sum of products two at a time: ((1 - 2i)(1 + 2i) + (1 - 2i)r + (1 + 2i)r = -1)3. Product of roots: ((1 - 2i)(1 + 2i)r = -b)Let me compute each of these step by step.Starting with the sum of the roots:((1 - 2i) + (1 + 2i) + r = (1 + 1) + (-2i + 2i) + r = 2 + 0i + r = 2 + r)According to Vieta, this equals (-a), so:(2 + r = -a) --> Equation (1)Next, the sum of the products two at a time:First, compute ((1 - 2i)(1 + 2i)). That's a difference of squares, so:((1)^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5)Then, compute ((1 - 2i)r + (1 + 2i)r). Factor out the r:(r[(1 - 2i) + (1 + 2i)] = r[2 + 0i] = 2r)So, the sum of the products two at a time is (5 + 2r). According to Vieta, this equals (-1):(5 + 2r = -1) --> Equation (2)Now, let me solve Equation (2) for r:(5 + 2r = -1)Subtract 5 from both sides:(2r = -6)Divide both sides by 2:(r = -3)Okay, so the third root is (-3). Now, plug this back into Equation (1):(2 + (-3) = -a)Simplify:(-1 = -a)Multiply both sides by -1:(a = 1)Alright, so a is 1. Now, let's find b using the product of the roots.The product of the roots is ((1 - 2i)(1 + 2i)(-3)). We already computed ((1 - 2i)(1 + 2i)) earlier as 5, so:(5 times (-3) = -15)According to Vieta, the product of the roots is (-b), so:(-15 = -b)Multiply both sides by -1:(b = 15)So, putting it all together, (a = 1) and (b = 15). Therefore, the ordered pair is ((1, 15)).Wait, let me double-check my calculations to make sure I didn't make a mistake.First, sum of roots: 1 - 2i + 1 + 2i + (-3) = (1 + 1 - 3) + (-2i + 2i) = (-1) + 0i = -1. So, -a = -1, which means a = 1. That seems correct.Sum of products two at a time: (1 - 2i)(1 + 2i) = 5, and then (1 - 2i)(-3) + (1 + 2i)(-3) = -3 + 6i -3 -6i = -6. Wait, hold on, that doesn't match with my earlier calculation.Wait, no, hold on. Wait, no, when I calculated the sum of products two at a time, I had:((1 - 2i)(1 + 2i) + (1 - 2i)r + (1 + 2i)r = 5 + 2r). But when I plug in r = -3, that becomes 5 + 2*(-3) = 5 - 6 = -1, which matches Vieta's formula because it's supposed to be equal to -1. So that's correct.But when I just thought about computing (1 - 2i)(-3) + (1 + 2i)(-3), that's equal to -3(1 - 2i + 1 + 2i) = -3*(2) = -6. Wait, so why is that different from 5 + 2r?Wait, no, actually, the sum of the products two at a time includes all possible pairs. So, it's (1 - 2i)(1 + 2i) + (1 - 2i)(r) + (1 + 2i)(r). So, that's 5 + r*(1 - 2i + 1 + 2i) = 5 + r*2. So, that's 5 + 2r, which when r = -3, becomes 5 - 6 = -1. So that's correct.But when I just thought about computing (1 - 2i)(-3) + (1 + 2i)(-3), that's only two terms, but actually, the sum of products two at a time includes three terms: (1 - 2i)(1 + 2i), (1 - 2i)(-3), and (1 + 2i)(-3). So, that's why it's 5 + (-3)(1 - 2i + 1 + 2i) = 5 + (-3)(2) = 5 - 6 = -1. So, that's consistent.So, my calculation is correct. So, a = 1 and b = 15.Just to be thorough, let me plug the root (1 - 2i) into the equation and see if it satisfies it with a = 1 and b = 15.So, the equation is (x^3 + x^2 - x + 15 = 0). Let's compute ( (1 - 2i)^3 + (1 - 2i)^2 - (1 - 2i) + 15 ).First, compute ( (1 - 2i)^2 ):( (1 - 2i)^2 = 1^2 - 2*1*(2i) + (2i)^2 = 1 - 4i + 4i^2 = 1 - 4i - 4 = -3 - 4i )Then, compute ( (1 - 2i)^3 ):We can write it as ( (1 - 2i)*(1 - 2i)^2 = (1 - 2i)*(-3 - 4i) ).Multiply this out:(1*(-3) + 1*(-4i) - 2i*(-3) - 2i*(-4i))= (-3 - 4i + 6i + 8i^2)= (-3 + 2i + 8*(-1))= (-3 + 2i - 8)= (-11 + 2i)Now, plug into the equation:( (1 - 2i)^3 + (1 - 2i)^2 - (1 - 2i) + 15 )= ((-11 + 2i) + (-3 - 4i) - (1 - 2i) + 15)Compute each term:First term: -11 + 2iSecond term: -3 - 4iThird term: -1 + 2i (because subtracting (1 - 2i) is the same as adding (-1 + 2i))Fourth term: +15Now, add them all together:Real parts: -11 - 3 - 1 + 15 = (-15) + 15 = 0Imaginary parts: 2i - 4i + 2i = 0iSo, total is 0 + 0i = 0. Perfect, it satisfies the equation. So, my values for a and b are correct.Therefore, the ordered pair is (1, 15).**Final Answer**The ordered pair is boxed{(1, 15)}.

answer：First, determine the total distance Peter needs to walk, which is 2.5 miles. He has already walked 1 mile, so the remaining distance is 2.5 minus 1, which equals 1.5 miles.Next, calculate the time it takes to walk the remaining distance. Since Peter takes 20 minutes to walk one mile, the time for 1.5 miles is 20 multiplied by 1.5, resulting in 30 minutes.Therefore, Peter needs to walk for 30 more minutes to reach the grocery store.

answer：First, I need to determine the total number of jellybeans accounted for by the blue, purple, and orange colors. There are 14 blue jellybeans, 26 purple jellybeans, and 40 orange jellybeans. Adding these together gives:14 + 26 + 40 = 80 jellybeans.The total number of jellybeans in the jar is 200. To find out how many are red, I subtract the number of non-red jellybeans from the total:200 - 80 = 120 jellybeans.Therefore, there are 120 red jellybeans in the jar.

answer：First, I need to determine how many rolls of toilet paper Stella uses each day. Since there are 6 bathrooms and she stocks 1 roll per bathroom per day, she uses 6 rolls daily.Next, I'll calculate the total number of rolls used over 4 weeks. There are 7 days in a week, so over 4 weeks, that's 4 multiplied by 7, which equals 28 days. Multiplying the daily usage by the number of days gives 6 rolls/day multiplied by 28 days, totaling 168 rolls.Each pack contains 1 dozen rolls, which is 12 rolls per pack. To find out how many packs Stella needs to buy, I'll divide the total number of rolls by the number of rolls per pack: 168 rolls divided by 12 rolls per pack equals 14 packs.Therefore, Stella needs to buy 14 packs of toilet paper.

answer：Alright, so I've got this function here: f(x) = 1/(3^x + 1). And I need to find the value of f(log₂3) + f(log₄(1/9)). Hmm, okay. Let me break this down step by step.First off, I need to figure out what f(log₂3) is. That means I have to plug log base 2 of 3 into the function f(x). So, substituting x with log₂3, the function becomes f(log₂3) = 1/(3^{log₂3} + 1). Hmm, that exponent looks a bit tricky. Let me think about how to simplify 3^{log₂3}.I remember that a^{log_b c} can be rewritten using the change of base formula. Specifically, a^{log_b c} = c^{log_b a}. So, applying that here, 3^{log₂3} would be equal to 3^{log₂3}... Wait, that doesn't seem helpful. Maybe I should use natural logarithms to simplify this.Let me recall that a^{log_b c} = e^{(ln a)(log_b c)}. Alternatively, I can express log₂3 in terms of natural logs: log₂3 = ln3 / ln2. So, 3^{log₂3} = 3^{(ln3 / ln2)}. Hmm, that's 3 raised to the power of (ln3 / ln2). Maybe I can write this as e^{(ln3)(ln3 / ln2)}? Wait, that might not be the right approach.Alternatively, I remember that 3^{log₂3} can be expressed as 2^{log₂3 * log_2 3}? Hmm, no, that doesn't seem right either. Maybe I should think about it differently. Let me denote y = 3^{log₂3}. Taking log base 2 of both sides: log₂y = log₂(3^{log₂3}) = log₂3 * log₂3 = (log₂3)^2. Hmm, but that doesn't seem to help me directly.Wait, maybe I can express 3^{log₂3} as 2^{log₂3 * log_2 3}... Wait, no, that's not correct. Let me try a different approach. Let me recall that 3 = 2^{log₂3}, so 3^{log₂3} = (2^{log₂3})^{log₂3} = 2^{(log₂3)^2}. Hmm, that seems complicated, but maybe it's useful.Alternatively, perhaps I can use the identity that a^{log_b c} = c^{log_b a}. So, 3^{log₂3} = 3^{log₂3} = 3^{log₂3}. Wait, that's just restating it. Maybe I need to think numerically. Let me compute log₂3 first. I know that log₂3 is approximately 1.58496. So, 3^{1.58496} is roughly 3^{1.58496}. Let me compute that. 3^1 is 3, 3^2 is 9, so 3^1.58496 is somewhere between 3 and 9. Maybe around 5 or 6? Hmm, not sure if that helps.Wait, maybe I can use the fact that 3^{log₂3} = 2^{log₂3 * log_2 3}... No, that seems messy. Maybe I should instead consider the expression 3^{log₂3} + 1. Is there a way to relate this to something else?Wait, perhaps if I let x = log₂3, then 3^x = 3^{log₂3}. But 3^x = 3^{log₂3} = (2^{log₂3})^{log₂3} = 2^{(log₂3)^2}. Hmm, not helpful.Wait, maybe I can use the identity that 3^{log₂3} = 2^{log₂3 * log_2 3}? No, that's not correct. Alternatively, perhaps I can write 3^{log₂3} as e^{(ln3)(log₂3)}. Let's see, log₂3 is ln3 / ln2, so 3^{log₂3} = e^{(ln3)(ln3 / ln2)} = e^{(ln3)^2 / ln2}. Hmm, that's an expression, but I don't know if that helps me compute it.Wait, maybe I can think about the function f(x) = 1/(3^x + 1). If I can find a relationship between f(x) and f(y) where x and y are related in some way, maybe I can find a symmetry or something.Looking at the second term, f(log₄(1/9)). Let me compute log₄(1/9) first. Since 1/9 is 9^{-1}, which is (3^2)^{-1} = 3^{-2}. So, log₄(3^{-2}) = -2 * log₄3. Now, log₄3 can be expressed as log₂3 / log₂4 = log₂3 / 2. So, log₄(1/9) = -2 * (log₂3 / 2) = -log₂3. So, f(log₄(1/9)) = f(-log₂3).So, now I have f(log₂3) + f(-log₂3). Let me denote x = log₂3, so the expression becomes f(x) + f(-x). So, f(x) = 1/(3^x + 1), and f(-x) = 1/(3^{-x} + 1). Let me compute f(x) + f(-x):f(x) + f(-x) = 1/(3^x + 1) + 1/(3^{-x} + 1).Let me simplify the second term: 1/(3^{-x} + 1) = 1/(1/3^x + 1) = 1/( (1 + 3^x)/3^x ) = 3^x / (1 + 3^x).So, f(x) + f(-x) = 1/(3^x + 1) + 3^x/(3^x + 1) = (1 + 3^x)/(3^x + 1) = 1.Wow, that's neat! So, f(x) + f(-x) = 1. Therefore, f(log₂3) + f(-log₂3) = 1.But wait, in the original problem, the second term is f(log₄(1/9)), which we found to be f(-log₂3). So, indeed, f(log₂3) + f(log₄(1/9)) = f(x) + f(-x) = 1.So, the value is 1.Wait, that seems too straightforward. Let me double-check my steps.1. I started by trying to compute f(log₂3) and f(log₄(1/9)) separately, but that led me to a complicated exponent. Instead, I noticed that log₄(1/9) simplifies to -log₂3, so f(log₄(1/9)) becomes f(-log₂3).2. Then, I set x = log₂3, so the expression becomes f(x) + f(-x).3. I expanded f(x) and f(-x): f(x) = 1/(3^x + 1) f(-x) = 1/(3^{-x} + 1) = 3^x / (1 + 3^x)4. Adding them together: 1/(3^x + 1) + 3^x/(3^x + 1) = (1 + 3^x)/(3^x + 1) = 1.Yes, that seems correct. So, regardless of the value of x, as long as we're adding f(x) and f(-x), we get 1. Therefore, the answer is 1.I think that's solid. I don't see any mistakes in the reasoning. So, the value of f(log₂3) + f(log₄(1/9)) is 1.**Final Answer**boxed{1}