💡Okay, so I've got this problem here with four positive integers a, b, c, and d. The equations given are:a² = c(d + 29) b² = c(d - 29)And I need to find the value of d. Hmm, let's see how to approach this.First, both a² and b² are expressed in terms of c and d. It seems like c is a common factor here. Maybe I can find a relationship between a and b by eliminating c or d. Let me try dividing the two equations to see if that helps.If I take a² / b², that should equal [c(d + 29)] / [c(d - 29)]. The c cancels out, so I get:(a / b)² = (d + 29) / (d - 29)Let me denote this ratio as k, so:k = (d + 29) / (d - 29)Which means:k = (d + 29) / (d - 29)I can rearrange this to solve for d. Let's cross-multiply:k(d - 29) = d + 29 kd - 29k = d + 29 kd - d = 29k + 29 d(k - 1) = 29(k + 1) d = [29(k + 1)] / (k - 1)Hmm, so d is expressed in terms of k, which is (a / b)². Since a and b are integers, k must be a rational number. Let me think about how k relates to a and b.Since k = (a / b)², let's say a = m * b, where m is some rational number. But since a and b are integers, m must be a rational number such that m² is also rational. To keep things simple, let's assume m is an integer. So, a = m * b.Wait, but if a and b are integers, and m is an integer, then a must be a multiple of b. Let me test this assumption.If a = m * b, then substituting back into the first equation:(m * b)² = c(d + 29) m² * b² = c(d + 29)But from the second equation, we have b² = c(d - 29). So, substituting c from the second equation into the first:m² * b² = [b² / (d - 29)] * (d + 29)Simplify:m² * b² = b² * (d + 29) / (d - 29)Divide both sides by b² (since b is positive, it's non-zero):m² = (d + 29) / (d - 29)Which is the same as our earlier equation for k. So, m² = k.Therefore, m must be an integer, and m² = (d + 29) / (d - 29). Let's denote m² as k again for simplicity.So, m² = (d + 29) / (d - 29)Let me rearrange this equation to solve for d:m²(d - 29) = d + 29 m²d - 29m² = d + 29 m²d - d = 29m² + 29 d(m² - 1) = 29(m² + 1) d = [29(m² + 1)] / (m² - 1)So, d is equal to 29 times (m² + 1) divided by (m² - 1). Since d must be an integer, the denominator (m² - 1) must divide the numerator 29(m² + 1).Let me write this as:d = 29 * [ (m² + 1) / (m² - 1) ]So, (m² - 1) must divide 29(m² + 1). Since 29 is a prime number, the factors of 29 are limited. Let's consider the possible values of m² - 1 that can divide 29(m² + 1).Let me denote m² - 1 as t. Then, t divides 29(m² + 1). But m² + 1 = t + 2, because m² + 1 = (m² - 1) + 2 = t + 2.So, t divides 29(t + 2). Therefore, t divides 29*2 = 58, because:29(t + 2) = 29t + 58So, t divides 58. Therefore, t must be a positive divisor of 58.The positive divisors of 58 are: 1, 2, 29, 58.So, m² - 1 can be 1, 2, 29, or 58.Let's test each case:1. If m² - 1 = 1: Then m² = 2. But m must be an integer, and 2 is not a perfect square. So, this case is invalid.2. If m² - 1 = 2: Then m² = 3. Again, 3 is not a perfect square. So, invalid.3. If m² - 1 = 29: Then m² = 30. 30 is not a perfect square. Invalid.4. If m² - 1 = 58: Then m² = 59. 59 is not a perfect square. Invalid.Hmm, none of these cases give m² as a perfect square. Did I make a mistake somewhere?Wait, maybe I need to consider that (m² - 1) doesn't necessarily have to be a divisor of 58, but rather that (m² - 1) divides 29(m² + 1). So, perhaps t divides 29(m² + 1), but t = m² - 1.Alternatively, maybe I should consider that t divides 58, but t can also be a factor that includes 29. Let me think differently.Let me write d as:d = 29 * [ (m² + 1) / (m² - 1) ]For d to be an integer, (m² - 1) must divide 29(m² + 1). Let me denote m² as x for simplicity.So, x - 1 divides 29(x + 1). Therefore, x - 1 divides 29(x + 1).Which implies that x - 1 divides 29(x + 1) - 29(x - 1) = 29*2 = 58.So, x - 1 divides 58. Therefore, x - 1 is a positive divisor of 58.The positive divisors of 58 are 1, 2, 29, 58.So, x - 1 can be 1, 2, 29, or 58.Therefore, x = 2, 3, 30, 59.But x = m², so m² must be 2, 3, 30, or 59. However, none of these are perfect squares except for 1, 4, 9, etc. So, none of these x values are perfect squares.Wait, that's a problem. It seems like there's no integer m that satisfies this condition. Did I do something wrong?Let me go back to the beginning. Maybe I made an incorrect assumption when I set a = m * b. Perhaps a and b are not necessarily multiples of each other, but share some common factor.Let me try a different approach. Let's denote the greatest common divisor (gcd) of a and b as g. So, a = g * a', and b = g * b', where gcd(a', b') = 1.Substituting into the original equations:(g * a')² = c(d + 29) (g * b')² = c(d - 29)So, g² * a'² = c(d + 29) g² * b'² = c(d - 29)Dividing the first equation by the second:(a'²) / (b'²) = (d + 29) / (d - 29)Let me denote this ratio as k again:k = (d + 29) / (d - 29) = (a' / b')²Since a' and b' are coprime, (a' / b')² must be in its simplest form. Therefore, (d + 29) and (d - 29) must be multiples of squares.Let me write:d + 29 = m² * t d - 29 = n² * tWhere t is some positive integer, and m and n are positive integers with gcd(m, n) = 1.Subtracting the two equations:(d + 29) - (d - 29) = m² * t - n² * t 58 = t(m² - n²)So, t(m² - n²) = 58.Since 58 factors into 2 * 29, and m and n are coprime, we need to find integers m, n, t such that t(m² - n²) = 58.Let me consider the possible factorizations of 58:1. t = 1, m² - n² = 58 2. t = 2, m² - n² = 29 3. t = 29, m² - n² = 2 4. t = 58, m² - n² = 1Let's analyze each case:1. t = 1, m² - n² = 58 We need to find integers m, n such that m² - n² = 58. This can be factored as (m - n)(m + n) = 58. The factor pairs of 58 are (1, 58) and (2, 29). Let's test these: - (m - n) = 1, (m + n) = 58 Adding: 2m = 59 => m = 29.5, not integer. - (m - n) = 2, (m + n) = 29 Adding: 2m = 31 => m = 15.5, not integer. So, no solution in this case.2. t = 2, m² - n² = 29 Factor pairs of 29: (1, 29) So, (m - n) = 1, (m + n) = 29 Adding: 2m = 30 => m = 15 Subtracting: 2n = 28 => n = 14 So, m = 15, n = 14. This works because 15² - 14² = 225 - 196 = 29.3. t = 29, m² - n² = 2 Factor pairs of 2: (1, 2) So, (m - n) = 1, (m + n) = 2 Adding: 2m = 3 => m = 1.5, not integer. No solution here.4. t = 58, m² - n² = 1 Factor pairs of 1: (1, 1) So, (m - n) = 1, (m + n) = 1 Adding: 2m = 2 => m = 1 Subtracting: 2n = 0 => n = 0, but n must be positive. So, no solution here.Therefore, the only valid case is case 2, where t = 2, m = 15, n = 14.Now, substituting back:d + 29 = m² * t = 15² * 2 = 225 * 2 = 450 d - 29 = n² * t = 14² * 2 = 196 * 2 = 392So, solving for d:From d + 29 = 450: d = 450 - 29 = 421Let me verify this:Check if a² = c(d + 29) and b² = c(d - 29) hold with d = 421.From earlier, a = g * a' = g * m = g * 15 b = g * b' = g * n = g * 14From the equations:a² = c(d + 29) (15g)² = c(421 + 29) 225g² = c * 450 c = 225g² / 450 = (225 / 450)g² = (1/2)g²Similarly, from b² = c(d - 29):(14g)² = c(421 - 29) 196g² = c * 392 c = 196g² / 392 = (196 / 392)g² = (1/2)g²So, c = (1/2)g². Since c must be an integer, g² must be even, which implies g is even. Let me set g = 2k, where k is a positive integer.Then, c = (1/2)(2k)² = (1/2)(4k²) = 2k²So, c is an integer as required.Therefore, the solution is valid for any positive integer k, but since we're only asked for d, which is 421, that's our answer.I think that makes sense. I went through different approaches, realized my initial assumption might have been too restrictive, then considered the gcd approach, which led me to the correct solution. It took a while, but breaking it down step by step helped.

💡First, calculate the cost price of the candy bars. The club buys candy bars at a rate of 8 bars for 3. Therefore, the cost per bar is:[frac{3 text{ dollars}}{8 text{ bars}} = 0.375 text{ dollars per bar}]Since they buy 1500 bars, the total cost is:[1500 text{ bars} times 0.375 text{ dollars per bar} = 562.5 text{ dollars}]Next, calculate the selling price of the candy bars. The club sells the candy bars at a rate of 3 bars for 2. Therefore, the selling price per bar is:[frac{2 text{ dollars}}{3 text{ bars}} = 0.6667 text{ dollars per bar}]Since they sell 1500 bars, the total revenue is:[1500 text{ bars} times 0.6667 text{ dollars per bar} = 1000 text{ dollars}]Finally, calculate the net profit by subtracting the total cost and the booth setup cost from the total revenue:[text{Net Profit} = 1000 text{ dollars} - 562.5 text{ dollars} - 50 text{ dollars} = 387.5 text{ dollars}]The school club's net profit is 387.5 dollars.

💡Okay, so I have these two algebra problems to solve, and I want to make sure I understand each step properly. Let me take them one at a time.Starting with the first problem: (ab)^6 ÷ (ab)^2 × (ab)^4. Hmm, that looks like a combination of exponents with the same base, which is "ab". I remember that when you divide exponents with the same base, you subtract the exponents, and when you multiply, you add them. So maybe I can simplify this by combining the exponents.Let me write it out step by step. First, (ab)^6 divided by (ab)^2 would be (ab)^(6-2), which is (ab)^4. Then, I need to multiply that result by (ab)^4. So, (ab)^4 × (ab)^4 would be (ab)^(4+4), which is (ab)^8. That seems straightforward. So, the first problem simplifies to (ab)^8.Now, moving on to the second problem: (3x^3)^2 × x^5 - (-x^2)^6 ÷ x. Okay, this one looks a bit more complicated because there are multiple operations and negative signs involved. Let me break it down.First, I'll tackle (3x^3)^2. I know that when you raise a product to a power, you raise each factor to that power. So, 3 squared is 9, and (x^3)^2 is x^(3×2) which is x^6. So, (3x^3)^2 becomes 9x^6.Next, I need to multiply that by x^5. So, 9x^6 × x^5. When multiplying like bases, I add the exponents. So, x^6 × x^5 is x^(6+5) which is x^11. Therefore, 9x^6 × x^5 is 9x^11.Now, moving on to the second part: (-x^2)^6 ÷ x. I need to be careful with the negative sign here. When raising a negative number to an even power, the result is positive. So, (-x^2)^6 is the same as (x^2)^6 because the negative becomes positive. Then, (x^2)^6 is x^(2×6) which is x^12.Now, I have to divide that by x. So, x^12 ÷ x is x^(12-1) which is x^11. So, (-x^2)^6 ÷ x simplifies to x^11.Putting it all together, the expression becomes 9x^11 - x^11. Subtracting these like terms, 9x^11 - x^11 is 8x^11.Wait, let me double-check to make sure I didn't make any mistakes. For the first problem, combining the exponents correctly: 6 - 2 + 4 equals 8, so (ab)^8 seems right. For the second problem, squaring (3x^3) gives 9x^6, multiplying by x^5 gives 9x^11. Then, (-x^2)^6 is x^12, dividing by x gives x^11. Subtracting x^11 from 9x^11 gives 8x^11. Yeah, that all checks out.I think I've got it!

💡First, I need to determine the number of different purchasing plans when buying at least one out of three books.I can approach this by considering all possible ways to choose one, two, or all three books.For choosing one book, there are 3 possible choices.For choosing two books, there are also 3 possible combinations.For choosing all three books, there is only 1 way.Adding these possibilities together, the total number of purchasing plans is 3 + 3 + 1, which equals 7.Therefore, the correct answer is 7.

💡First, I need to determine Jake's sister's current weight. Since the combined weight of Jake and his sister is 153 pounds, and Jake currently weighs 113 pounds, I can subtract Jake's weight from the total to find his sister's weight.Next, I want Jake's weight to be twice his sister's weight. Let’s denote the weight Jake needs to lose as ( x ). After losing ( x ) pounds, Jake's new weight will be ( 113 - x ) pounds.I set up the equation ( 113 - x = 2 times 40 ) to represent the condition where Jake's weight is twice his sister's weight. Solving for ( x ) gives me the number of pounds Jake needs to lose.Finally, by performing the calculations, I find that Jake needs to lose 33 pounds to achieve the desired weight ratio.

💡Okay, so I have this problem about a convex polyhedron and its road network. The problem says that the road network connects all the vertices without splitting the polyhedron into two parts, and I need to prove that this road system has at least 4 dead ends. Dead ends are vertices with only one road connected to them, right? In the figure mentioned, vertices A, B, C, and D are dead ends. First, I need to understand what a convex polyhedron is. It's a three-dimensional shape where all its faces are convex polygons, and every edge is shared by exactly two faces. The road network along the edges means that the roads are the edges of the polyhedron, connecting all the vertices. So, essentially, the road network is the graph formed by the vertices and edges of the polyhedron.The problem mentions that the road network connects all the vertices without splitting the polyhedron into two parts. That means the graph is connected, right? So, it's a connected planar graph because it's embedded on the surface of a convex polyhedron. Now, I need to prove that there are at least four dead ends, which are vertices of degree one. In graph theory terms, these are called pendant vertices. So, I need to show that in any connected planar graph representing the edges of a convex polyhedron, there must be at least four vertices of degree one.Wait, but hold on. In a convex polyhedron, each face is a polygon, and each edge is shared by two faces. So, the graph is 3-vertex-connected because it's a convex polyhedron, right? According to Steinitz's theorem, the graph of a convex polyhedron is 3-vertex-connected and planar. So, that gives me some properties to work with.But I'm not sure if that directly helps me with the number of pendant vertices. Maybe I should think about Euler's formula. Euler's formula for planar graphs says that V - E + F = 2, where V is the number of vertices, E is the number of edges, and F is the number of faces. If I can relate the number of edges and vertices, maybe I can find some constraints on the degrees of the vertices. Let's denote the number of vertices as V, edges as E, and faces as F. In a convex polyhedron, each face is a polygon, so each face has at least three edges. But since each edge is shared by two faces, the total number of edges can be related to the number of faces. Specifically, 3F ≤ 2E, which simplifies to F ≤ (2/3)E. Plugging this into Euler's formula: V - E + F = 2. Since F ≤ (2/3)E, we have V - E + (2/3)E ≥ 2, which simplifies to V - (1/3)E ≥ 2, or V ≥ (1/3)E + 2. But I'm not sure if this is directly helpful. Maybe I should think about the degrees of the vertices. The sum of the degrees of all vertices is equal to twice the number of edges, right? So, if I denote the degree of vertex v_i as d_i, then Σd_i = 2E.If I want to find the number of pendant vertices, which are vertices with degree one, maybe I can use some inequality or average degree argument. Let's suppose that there are k pendant vertices. Then, the sum of the degrees would be at least k*1 + (V - k)*2, since the other vertices must have at least degree two to keep the graph connected. Wait, but in a convex polyhedron, the graph is 3-vertex-connected, so actually, every vertex must have degree at least three. Is that right? Because in a 3-vertex-connected graph, you can't disconnect the graph by removing fewer than three vertices, which implies that each vertex has degree at least three. Hold on, that contradicts the idea of having pendant vertices, which have degree one. So, maybe I'm misunderstanding something. If the graph is 3-vertex-connected, then every vertex must have degree at least three, meaning there are no pendant vertices. But the problem says there are dead ends, which are pendant vertices. So, there must be something wrong with my reasoning.Wait, maybe the graph of the polyhedron itself is 3-vertex-connected, but the road network is a spanning subgraph that connects all the vertices without splitting the polyhedron into two parts. So, the road network is a connected spanning subgraph, but it's not necessarily the entire graph of the polyhedron. So, the road network is a connected graph embedded on the polyhedron, but it's not necessarily the complete graph of the polyhedron. Therefore, it can have vertices of degree one, which are the dead ends. Okay, that makes more sense. So, the road network is a connected spanning subgraph, and I need to show that it must have at least four vertices of degree one. How can I approach this? Maybe I can use some properties of planar graphs or consider the dual graph. Alternatively, I can think about the number of edges in the road network. Since the road network is connected and has V vertices, it must have at least V - 1 edges. But it's embedded on the polyhedron, which is a planar graph, so it must also satisfy planarity constraints. Wait, but the road network is a subgraph of the polyhedron's graph, which is planar. So, the road network is also planar. Therefore, it must satisfy Euler's formula: V - E + F = 2. But I'm not sure how to relate this to the number of pendant vertices. Maybe I can use the Handshaking Lemma, which says that the sum of degrees is equal to twice the number of edges. If I denote the number of pendant vertices as k, then the sum of degrees is at least k*1 + (V - k)*2, since the other vertices must have degree at least two. But the sum of degrees is also equal to 2E. So, 2E ≥ k + 2(V - k) = 2V - k. Therefore, 2E ≥ 2V - k, which implies that k ≥ 2V - 2E. But I need to find a lower bound for k, so I need to find an upper bound for 2V - 2E. Wait, from Euler's formula, V - E + F = 2, so E = V + F - 2. If I can bound F, the number of faces, then I can express E in terms of V and F. In the road network, each face must be bounded by at least three edges, but since it's a subgraph, some faces might be larger. However, each edge is shared by at most two faces. So, 3F ≤ 2E, which gives F ≤ (2/3)E. Plugging this into Euler's formula: V - E + (2/3)E ≥ 2, so V - (1/3)E ≥ 2, which implies that V ≥ (1/3)E + 2. Rearranging, E ≤ 3V - 6. So, the number of edges in the road network is at most 3V - 6. But the road network is connected, so it has at least V - 1 edges. Now, going back to the inequality k ≥ 2V - 2E. Since E ≤ 3V - 6, then 2E ≤ 6V - 12. Therefore, 2V - 2E ≥ 2V - (6V - 12) = -4V + 12. But this gives k ≥ -4V + 12, which isn't helpful because k can't be negative. Maybe I need a different approach. Let's think about the dual graph. The dual of a planar graph has a vertex for each face of the original graph, and edges connecting vertices whose corresponding faces share an edge. But I'm not sure if that helps directly. Maybe I should consider the number of vertices of degree one. In a connected planar graph, the number of vertices of degree one must be at least four. Is that a known result? I'm not sure, but maybe I can try to prove it.Suppose, for contradiction, that there are fewer than four pendant vertices. So, k ≤ 3. Then, the sum of degrees is 2E = Σd_i ≥ k*1 + (V - k)*2 = k + 2V - 2k = 2V - k. So, 2E ≥ 2V - k. But since k ≤ 3, 2E ≥ 2V - 3. From Euler's formula, E = V + F - 2. Also, since each face is bounded by at least three edges, 3F ≤ 2E, so F ≤ (2/3)E. Plugging into Euler's formula: V - E + (2/3)E ≥ 2 ⇒ V - (1/3)E ≥ 2 ⇒ V ≥ (1/3)E + 2 ⇒ E ≤ 3V - 6. So, E ≤ 3V - 6. From earlier, 2E ≥ 2V - 3 ⇒ E ≥ V - 1.5. But E must be an integer, so E ≥ V - 1. But E is also ≤ 3V - 6. So, combining these: V - 1 ≤ E ≤ 3V - 6. But I need to relate this to k. Maybe I can find a contradiction if k is too small.Wait, let's try plugging in k = 3. Then, 2E ≥ 2V - 3. But from Euler's formula, E = V + F - 2, and F ≤ (2/3)E. So, E = V + F - 2 ≤ V + (2/3)E - 2 ⇒ E - (2/3)E ≤ V - 2 ⇒ (1/3)E ≤ V - 2 ⇒ E ≤ 3V - 6, which is consistent with what we had before.But I need to find a contradiction when k is small. Maybe I can use the fact that in a planar graph, the number of edges is bounded, and the number of pendant vertices affects the degree sequence.Alternatively, maybe I can use induction on the number of vertices. Suppose the statement is true for all convex polyhedrons with fewer than V vertices. Now, consider a polyhedron with V vertices. If I can show that removing a vertex or an edge leads to a smaller polyhedron with at least four pendant vertices, then the statement holds.But I'm not sure if that's the right way to go. Maybe another approach is to consider that in a convex polyhedron, the dual graph is also a convex polyhedron, so it's 3-vertex-connected. Wait, but the road network is a connected spanning subgraph, not necessarily the dual. Alternatively, maybe I can think about the graph being 2-edge-connected or something like that. Wait, in a convex polyhedron, the graph is 3-edge-connected, right? Because it's 3-vertex-connected. So, the road network, being a connected spanning subgraph, must also be 2-edge-connected? Or maybe not necessarily.I'm getting a bit confused here. Maybe I should look for a different strategy. What if I consider the number of vertices of degree one in a connected planar graph. Is there a theorem that states that a connected planar graph must have at least four vertices of degree one? I don't recall such a theorem, but maybe it's related to the number of leaves in a spanning tree.Wait, in a spanning tree of a connected planar graph, the number of leaves is at least two. But in our case, the road network is a connected spanning subgraph, which might not be a tree. If it's a tree, then it has V - 1 edges and at least two leaves. But our road network can have more edges, so it can have more cycles, which might allow for more leaves.But we need to show that it has at least four leaves. Maybe I can use the fact that in a planar graph, the number of vertices of degree one is related to the number of edges and faces. Wait, let's try to use the Handshaking Lemma again. Let k be the number of pendant vertices (degree 1). Then, the sum of degrees is 2E = k*1 + Σ_{v not pendant} d_v. Since the graph is connected, the other vertices must have degree at least two. So, Σ_{v not pendant} d_v ≥ 2(V - k). Therefore, 2E ≥ k + 2(V - k) = 2V - k. So, 2E ≥ 2V - k ⇒ k ≥ 2V - 2E. But I need to find a lower bound for k, so I need an upper bound for 2V - 2E. From Euler's formula, V - E + F = 2 ⇒ E = V + F - 2. Also, since each face is bounded by at least three edges, 3F ≤ 2E ⇒ F ≤ (2/3)E. Plugging into E = V + F - 2, we get E ≤ V + (2/3)E - 2 ⇒ (1/3)E ≤ V - 2 ⇒ E ≤ 3V - 6. So, E ≤ 3V - 6. Therefore, 2E ≤ 6V - 12. So, 2V - 2E ≥ 2V - (6V - 12) = -4V + 12. But this gives k ≥ -4V + 12, which isn't helpful because k can't be negative. Hmm, maybe I need a different approach. Let's think about the average degree. The average degree d_avg = (2E)/V. From E ≤ 3V - 6, we have d_avg ≤ (6V - 12)/V = 6 - 12/V. So, the average degree is less than 6. But how does that help with the number of pendant vertices? Maybe I can use the fact that if there are few pendant vertices, the average degree would be higher. Suppose there are only three pendant vertices. Then, the sum of degrees is 2E = 3*1 + Σ_{v not pendant} d_v. The other V - 3 vertices must have degree at least two, so Σ_{v not pendant} d_v ≥ 2(V - 3). Therefore, 2E ≥ 3 + 2(V - 3) = 2V - 3. So, 2E ≥ 2V - 3 ⇒ E ≥ V - 1.5. But E must be an integer, so E ≥ V - 1. But from Euler's formula, E ≤ 3V - 6. So, V - 1 ≤ E ≤ 3V - 6. But I need to find a contradiction. Maybe if I assume k = 3, then the number of edges is constrained in a way that contradicts planarity or something else.Wait, let's consider the number of edges in terms of the number of faces. If there are k pendant vertices, then the number of edges is E = (Σd_i)/2 = (k + Σ_{v not pendant} d_v)/2. If k = 3, then E = (3 + Σ_{v not pendant} d_v)/2. But Σ_{v not pendant} d_v ≥ 2(V - 3), so E ≥ (3 + 2(V - 3))/2 = (2V - 3)/2 = V - 1.5. Since E must be an integer, E ≥ V - 1. But from Euler's formula, E = V + F - 2. So, V + F - 2 ≥ V - 1 ⇒ F ≥ 1. Which is true, but not helpful. Maybe I need to consider the number of faces in terms of edges. Each face is bounded by at least three edges, so 3F ≤ 2E ⇒ F ≤ (2/3)E. From Euler's formula, V - E + F = 2 ⇒ F = E - V + 2. So, E - V + 2 ≤ (2/3)E ⇒ E - (2/3)E ≤ V - 2 ⇒ (1/3)E ≤ V - 2 ⇒ E ≤ 3V - 6, which is consistent with what we had before.I'm going in circles here. Maybe I need to think differently. What if I consider the dual graph? The dual of a planar graph has a vertex for each face and edges connecting adjacent faces. But I'm not sure if that helps with the number of pendant vertices. Alternatively, maybe I can use the fact that in a convex polyhedron, the graph is 3-vertex-connected, so it's also 3-edge-connected. But the road network is a connected spanning subgraph, so it's at least 1-edge-connected. Wait, but if the road network had only three pendant vertices, then removing those three vertices would disconnect the graph, which contradicts the 3-vertex-connectedness of the polyhedron's graph. Wait, no, because the road network is a subgraph, not necessarily the entire graph. So, the road network could have pendant vertices even if the entire graph is 3-vertex-connected. But maybe the road network must preserve some connectivity properties. Wait, the problem says the road network connects all the vertices without splitting the polyhedron into two parts. So, it's a connected spanning subgraph, but it's not necessarily 3-vertex-connected. But the original polyhedron's graph is 3-vertex-connected, so any connected spanning subgraph must have certain properties. I'm not sure. Maybe I need to think about the number of edges in the road network. If the road network has too few edges, it might not be connected, but it's given that it's connected. Wait, maybe I can use the fact that in a planar graph, the number of edges is bounded, and the number of pendant vertices affects the number of edges. If there are fewer than four pendant vertices, then the number of edges would be too large or too small, contradicting planarity. But I'm not sure how to formalize that. Wait, let's try to assume that there are only three pendant vertices and see if that leads to a contradiction. Suppose k = 3. Then, the sum of degrees is 2E = 3 + Σ_{v not pendant} d_v. The other V - 3 vertices must have degree at least two, so Σ_{v not pendant} d_v ≥ 2(V - 3). Therefore, 2E ≥ 3 + 2(V - 3) = 2V - 3. So, E ≥ V - 1.5. But E must be an integer, so E ≥ V - 1. From Euler's formula, E = V + F - 2. So, V + F - 2 ≥ V - 1 ⇒ F ≥ 1. Which is true, but not helpful. But also, from the face-edge relationship, 3F ≤ 2E ⇒ F ≤ (2/3)E. From Euler's formula, F = E - V + 2. So, E - V + 2 ≤ (2/3)E ⇒ (1/3)E ≤ V - 2 ⇒ E ≤ 3V - 6. So, combining E ≥ V - 1 and E ≤ 3V - 6, we have V - 1 ≤ E ≤ 3V - 6. But I need to find a contradiction. Maybe if I consider the number of edges in terms of the number of faces. If E = V - 1, then F = (V - 1) - V + 2 = 1. So, there's only one face. But in a convex polyhedron, the road network must cover all the faces, but if there's only one face, that would mean the road network is a tree, which can't be because a tree has V - 1 edges and only one face (the outer face), but in a polyhedron, there are multiple faces. Wait, but the road network is a connected spanning subgraph, not necessarily the entire graph. So, it could have only one face if it's a tree, but in that case, it would have V - 1 edges and only one face. But in a convex polyhedron, the original graph has multiple faces, but the road network could be a tree with only one face. But the problem states that the road network connects all the vertices without splitting the polyhedron into two parts. So, it must be connected, but it doesn't necessarily have to cover all the faces. Wait, but if the road network is a tree, it would have V - 1 edges and only one face, but the original polyhedron has more faces. So, maybe the road network must cover all the faces? No, the road network is just a connected spanning subgraph, so it doesn't have to cover all the faces. It can be embedded on the polyhedron without covering all faces. But then, if it's a tree, it would have only one face, but the polyhedron has multiple faces. So, maybe the road network must have at least as many faces as the polyhedron? No, I don't think so. The road network is a subgraph, so it can have fewer faces. I'm getting stuck here. Maybe I need to think about the number of vertices of degree one in a planar graph. I recall that in a planar graph, the number of vertices of degree one is at least four. Is that true? Wait, no, that's not necessarily true. For example, a triangle has no vertices of degree one. A square has four vertices of degree two. But in our case, the road network is a connected spanning subgraph of a convex polyhedron's graph, which is 3-vertex-connected. Wait, maybe the road network must have at least four vertices of degree one because the original graph is 3-vertex-connected, and removing edges to form the road network would leave at least four vertices with degree one. But I'm not sure. Maybe I can think about it this way: in a 3-vertex-connected graph, every vertex has degree at least three. So, if I remove edges to form a connected spanning subgraph, the degrees of the vertices can only decrease. But to keep the subgraph connected, I can't remove all edges from a vertex, so each vertex must retain at least one edge. But if I remove edges from a vertex, its degree decreases. If I remove two edges from a vertex, its degree becomes one, making it a pendant vertex. Since the original graph is 3-vertex-connected, each vertex has degree at least three. So, to form a connected spanning subgraph, I can remove edges, but I have to ensure that the subgraph remains connected. If I remove two edges from a vertex, its degree becomes one, which is a pendant vertex. But how many such vertices can I have? If I remove two edges from four different vertices, I get four pendant vertices. But is it possible to have fewer than four pendant vertices? Suppose I remove only one edge from each of three vertices, making their degrees two, and leave the rest as is. Then, I would have no pendant vertices, which contradicts the problem statement. Wait, but the problem says that the road network connects all the vertices without splitting the polyhedron into two parts. So, it must be connected, but it doesn't necessarily have to have pendant vertices. But the problem states that it does have at least four dead ends, which are pendant vertices. So, maybe the way the edges are removed to form the road network necessarily creates at least four pendant vertices. But I'm not sure how to formalize that. Wait, maybe I can use the fact that in a 3-vertex-connected graph, any connected spanning subgraph must have at least four vertices of degree one. Is that a known result? I'm not sure, but maybe I can try to prove it. Suppose, for contradiction, that there are fewer than four pendant vertices. So, k ≤ 3. Then, the sum of degrees is 2E = k + Σ_{v not pendant} d_v. The other V - k vertices must have degree at least two, so Σ_{v not pendant} d_v ≥ 2(V - k). Therefore, 2E ≥ k + 2(V - k) = 2V - k. So, 2E ≥ 2V - k. But since k ≤ 3, 2E ≥ 2V - 3. From Euler's formula, E = V + F - 2. Also, since each face is bounded by at least three edges, 3F ≤ 2E ⇒ F ≤ (2/3)E. Plugging into Euler's formula: V - E + (2/3)E ≥ 2 ⇒ V - (1/3)E ≥ 2 ⇒ V ≥ (1/3)E + 2 ⇒ E ≤ 3V - 6. So, E ≤ 3V - 6. From 2E ≥ 2V - 3, we have E ≥ V - 1.5. But E must be an integer, so E ≥ V - 1. So, V - 1 ≤ E ≤ 3V - 6. But I need to find a contradiction. Maybe if I consider the number of edges in terms of the number of faces. If E = V - 1, then F = 1, which would mean the road network is a tree with only one face. But in a convex polyhedron, the road network must cover all the vertices, but it doesn't have to cover all the faces. Wait, but if the road network is a tree, it would have V - 1 edges and only one face, but the original polyhedron has more faces. So, maybe the road network must have at least as many faces as the polyhedron? No, the road network is a subgraph, so it can have fewer faces. I'm stuck again. Maybe I need to think about the dual graph. The dual graph of the road network would have a vertex for each face of the road network. Since the road network is connected, its dual is also connected. But I'm not sure how that helps with the number of pendant vertices. Wait, maybe I can use the fact that in a planar graph, the number of vertices of degree one is related to the number of faces. If there are k pendant vertices, then the number of faces is related to the number of edges and vertices. But I'm not sure. Wait, let's try to think about it differently. In a convex polyhedron, the graph is 3-vertex-connected, so it's also 3-edge-connected. If the road network is a connected spanning subgraph, then it must retain some of the connectivity properties. But if the road network had fewer than four pendant vertices, then removing those few vertices would disconnect the graph, which contradicts the 3-vertex-connectedness of the original graph. Wait, no, because the road network is a subgraph, not the entire graph. So, the road network could have pendant vertices even if the original graph is 3-vertex-connected. But maybe the road network must have at least four pendant vertices to maintain connectivity without splitting the polyhedron. I'm not sure. Maybe I need to think about specific examples. Take a cube, which is a convex polyhedron. It has 8 vertices and 12 edges. If I form a road network that connects all vertices without splitting the cube, how many pendant vertices would it have? If I take a spanning tree of the cube, it would have 7 edges and 8 - 1 = 7 vertices of degree one. But that's more than four. But the problem says at least four, so maybe in some cases, you can have more. Wait, but the problem is asking to prove that there are at least four, not exactly four. So, in the cube, any connected spanning subgraph must have at least four pendant vertices. But how? If I try to make a connected spanning subgraph of the cube with only three pendant vertices, is that possible? Let's see. The cube has 8 vertices. If I have three pendant vertices, then the other five vertices must have degree at least two. The sum of degrees would be 3*1 + 5*2 = 13, which is odd, but the sum of degrees must be even because it's twice the number of edges. So, that's impossible. Therefore, in the cube, any connected spanning subgraph must have an even number of pendant vertices. Wait, but the problem says at least four. So, in the cube, you can't have three pendant vertices because the sum of degrees would be odd, which is impossible. Therefore, you must have at least four pendant vertices. Ah, that's a good point. The sum of degrees must be even, so the number of pendant vertices must be even. Therefore, if you have fewer than four pendant vertices, you could have two or zero. But in a connected graph with more than two vertices, you can't have zero pendant vertices because that would mean all vertices have degree at least two, which would make the graph cyclic, but it's connected. Wait, no, a connected graph can have all vertices of degree at least two if it's a cycle or something more complex. But in our case, the road network is a connected spanning subgraph of a 3-vertex-connected graph. Wait, but in the cube example, we saw that having three pendant vertices leads to an odd sum of degrees, which is impossible. Therefore, the number of pendant vertices must be even. So, in the cube, the minimum number of pendant vertices in a connected spanning subgraph is four. Therefore, in general, for a convex polyhedron, any connected spanning subgraph must have at least four pendant vertices. So, that's the idea. The number of pendant vertices must be even, and having fewer than four would lead to either an impossible degree sequence or a disconnected graph. Therefore, the road system must have at least four dead ends.