💡Okay, so I have this problem here: there's a sphere O with an inscribed cube, and the surface area of the sphere is 36π. I need to find the length of the edge of the cube. Hmm, let me think about how to approach this.First, I remember that when a cube is inscribed in a sphere, all the cube's vertices touch the sphere. That means the sphere's diameter is equal to the space diagonal of the cube. So, if I can find the sphere's radius from the surface area, I can then relate that to the cube's edge length.The surface area of a sphere is given by the formula 4πr², where r is the radius. The problem says the surface area is 36π, so I can set up the equation:4πr² = 36πTo find r, I can divide both sides by 4π:r² = 36π / 4π = 9So, r² = 9, which means r = 3. Therefore, the radius of the sphere is 3 units.Since the diameter of the sphere is twice the radius, the diameter is 2 * 3 = 6 units. This diameter is equal to the space diagonal of the cube.Now, I need to relate the space diagonal of the cube to its edge length. For a cube with edge length x, the space diagonal can be found using the formula derived from the Pythagorean theorem in three dimensions. The space diagonal d is:d = x√3We know that the space diagonal d is equal to the sphere's diameter, which is 6. So:x√3 = 6To solve for x, I can divide both sides by √3:x = 6 / √3But it's not nice to have a square root in the denominator, so I can rationalize it by multiplying numerator and denominator by √3:x = (6√3) / (√3 * √3) = (6√3) / 3 = 2√3So, the edge length of the cube is 2√3 units.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the surface area, I found the radius correctly, then the diameter, which is the space diagonal. Then, using the space diagonal formula for a cube, I solved for x. Everything seems to add up. I think I did it right.Just to recap:1. Surface area of sphere = 36π2. 4πr² = 36π ⇒ r² = 9 ⇒ r = 33. Diameter = 2r = 64. Space diagonal of cube = 65. Space diagonal formula: x√3 = 6 ⇒ x = 6/√3 = 2√3Yep, that all checks out. So, the edge length of the cube is 2√3.**Final Answer**The length of the edge of the cube is boxed{2sqrt{3}}.

💡Okay, so I've got this problem about a 4x4 grid with a token starting on a gray cell. The token moves according to the arrows in each cell, and after each move, the arrow in the cell it just left flips direction. I need to figure out from which cell the token will exit the grid.First, I need to understand the setup better. There's a 4x4 grid, so rows are labeled A to D, and columns are 1 to 4. The token starts on a gray cell, but the problem doesn't specify which one. Hmm, that's a bit confusing. Maybe it's implied that the gray cell is a specific one, like C2 or something? I'll have to assume that.Assuming the token starts on cell C2, as in the previous example. The arrows in each cell direct the token to move in a certain direction—up, down, left, or right. After each move, the arrow in the cell the token was just on flips to the opposite direction. So if it was pointing right, it now points left, and so on.I think the best way to approach this is to map out each move step by step, keeping track of the token's position and the direction of the arrows as they change. Let's start with the initial position.1. **Starting Position:** Cell C2. Let's assume the arrow here points right.2. **First Move:** From C2, the token moves right to C3. After moving, the arrow in C2 flips to the opposite direction, so now it points left.3. **Second Move:** Now, the token is on C3. Let's assume the arrow here points up. So it moves up to B3. The arrow in C3 flips to point down.4. **Third Move:** On B3, assuming the arrow points up, it moves to A3. The arrow in B3 flips to point down.5. **Fourth Move:** On A3, assuming the arrow points right, it moves to A4. The arrow in A3 flips to point left.6. **Fifth Move:** On A4, assuming the arrow points down, it moves to B4. The arrow in A4 flips to point up.7. **Sixth Move:** On B4, assuming the arrow points down, it moves to C4. The arrow in B4 flips to point up.8. **Seventh Move:** On C4, assuming the arrow points left, it moves to C3. The arrow in C4 flips to point right.9. **Eighth Move:** On C3, the arrow was previously flipped to point down. So it moves down to D3. The arrow in C3 flips to point up.10. **Ninth Move:** On D3, assuming the arrow points left, it moves to D2. The arrow in D3 flips to point right.11. **Tenth Move:** On D2, assuming the arrow points up, it moves to C2. The arrow in D2 flips to point down.12. **Eleventh Move:** On C2, the arrow was flipped to point left earlier. So it moves left to C1. The arrow in C2 flips to point right.13. **Twelfth Move:** On C1, assuming the arrow points up, it moves to B1. The arrow in C1 flips to point down.14. **Thirteenth Move:** On B1, assuming the arrow points up, it moves to A1. The arrow in B1 flips to point down.15. **Fourteenth Move:** On A1, assuming the arrow points right, it moves to A2. The arrow in A1 flips to point left.16. **Fifteenth Move:** On A2, assuming the arrow points right, it would move to A3. But wait, A3's arrow was flipped to point left earlier. So from A2, moving right to A3, but A3's arrow points left now. So the token would move back to A2. This seems like it's stuck in a loop between A2 and A3.Hmm, that's not good. Maybe I made a wrong assumption about the initial arrow directions. Let me double-check.Perhaps the initial arrow directions are different. Maybe not all arrows point right or up initially. Maybe they have a specific pattern. Without knowing the initial arrow directions, it's hard to predict the exact path. Maybe I need to assume a standard initial configuration.Alternatively, maybe the token doesn't get stuck and eventually exits the grid. Let me try a different approach.Let's consider that the token starts on C2, and the arrows are set in such a way that it moves in a specific pattern leading it to exit the grid. Maybe it's designed to exit from a particular cell.If I follow the previous steps, the token moves around the grid, flipping arrows as it goes. Eventually, it might reach a cell where the next move takes it outside the grid.Looking back at the steps, after moving to A2, the next move would be to A3, but A3's arrow points left, so it would go back to A2. This creates a loop. To break the loop, maybe the arrow directions change in a way that allows the token to exit.Alternatively, perhaps the token exits from a different cell earlier in the sequence. Let me check the earlier moves again.After moving to A4, the token moves to B4, then to C4, then to C3, then to D3, then to D2, then to C2, then to C1, then to B1, then to A1, then to A2.At A2, if the arrow points right, it would go to A3, but A3's arrow points left, so it goes back to A2. If the arrow at A2 points left, it would go to A1, but A1's arrow points right, so it goes back to A2. This seems like an endless loop.Maybe the token exits from A2 when it tries to move beyond the grid. If the arrow at A2 points up, it would try to move to a non-existent cell above A2, thus exiting the grid.Wait, A2 is on the top row, so moving up from A2 would take it out of the grid. So if the arrow at A2 points up, the token would exit from A2.But in my earlier steps, the arrow at A2 was pointing right or left, causing a loop. Maybe the arrow directions change in such a way that eventually, the arrow at A2 points up.Let me see. After moving to A2, the arrow at A1 flips to point left. If the token is on A2, and the arrow there points up, it would exit. But how does the arrow at A2 get set to up?Looking back, when the token was on A3, it moved to A4, flipping A3's arrow to left. Then from A4 to B4, flipping A4's arrow to up. From B4 to C4, flipping B4's arrow to up. From C4 to C3, flipping C4's arrow to right. From C3 to D3, flipping C3's arrow to up. From D3 to D2, flipping D3's arrow to right. From D2 to C2, flipping D2's arrow to down. From C2 to C1, flipping C2's arrow to right. From C1 to B1, flipping C1's arrow to down. From B1 to A1, flipping B1's arrow to down. From A1 to A2, flipping A1's arrow to left.So, at this point, the arrow at A2 hasn't been flipped yet. It still points right or left, depending on its initial direction. If it points right, it goes to A3, which points left back to A2. If it points left, it goes to A1, which points right back to A2. So it's stuck.Unless, somehow, the arrow at A2 gets flipped to point up. But how? The only way the arrow at A2 gets flipped is if the token moves from A2 to another cell, flipping A2's arrow. But if it's stuck in a loop between A2 and A3 or A2 and A1, it never leaves A2 to flip its arrow.This suggests that the token might not exit the grid in this configuration, which contradicts the problem statement that it does exit. Maybe my initial assumption about the starting position or arrow directions is wrong.Alternatively, perhaps the token starts on a different cell. Let me try assuming it starts on C3 instead.1. **Starting Position:** Cell C3, arrow points right.2. **First Move:** C3 to C4, arrow in C3 flips to left.3. **Second Move:** C4, arrow points down, moves to D4, arrow in C4 flips to up.4. **Third Move:** D4, arrow points left, moves to D3, arrow in D4 flips to right.5. **Fourth Move:** D3, arrow points left, moves to D2, arrow in D3 flips to right.6. **Fifth Move:** D2, arrow points up, moves to C2, arrow in D2 flips to down.7. **Sixth Move:** C2, arrow points right, moves to C3, arrow in C2 flips to left.8. **Seventh Move:** C3, arrow was flipped to left, moves to C2, arrow in C3 flips to right.9. **Eighth Move:** C2, arrow was flipped to left, moves to C1, arrow in C2 flips to right.10. **Ninth Move:** C1, arrow points up, moves to B1, arrow in C1 flips to down.11. **Tenth Move:** B1, arrow points up, moves to A1, arrow in B1 flips to down.12. **Eleventh Move:** A1, arrow points right, moves to A2, arrow in A1 flips to left.13. **Twelfth Move:** A2, arrow points right, moves to A3, arrow in A2 flips to left.14. **Thirteenth Move:** A3, arrow points right, moves to A4, arrow in A3 flips to left.15. **Fourteenth Move:** A4, arrow points down, moves to B4, arrow in A4 flips to up.16. **Fifteenth Move:** B4, arrow points down, moves to C4, arrow in B4 flips to up.17. **Sixteenth Move:** C4, arrow was flipped to up, moves to B4, arrow in C4 flips to down.18. **Seventeenth Move:** B4, arrow was flipped to up, moves to A4, arrow in B4 flips to down.19. **Eighteenth Move:** A4, arrow was flipped to up, moves to A3, arrow in A4 flips to down.20. **Nineteenth Move:** A3, arrow was flipped to left, moves to A2, arrow in A3 flips to right.21. **Twentieth Move:** A2, arrow was flipped to left, moves to A1, arrow in A2 flips to right.22. **Twenty-First Move:** A1, arrow was flipped to left, moves to A2, arrow in A1 flips to right.23. **Twenty-Second Move:** A2, arrow was flipped to right, moves to A3, arrow in A2 flips to left.This seems like another loop between A2, A3, and A1. It doesn't exit the grid.Maybe the starting position is different. Let's try starting on B2.1. **Starting Position:** Cell B2, arrow points right.2. **First Move:** B2 to B3, arrow in B2 flips to left.3. **Second Move:** B3, arrow points up, moves to A3, arrow in B3 flips to down.4. **Third Move:** A3, arrow points right, moves to A4, arrow in A3 flips to left.5. **Fourth Move:** A4, arrow points down, moves to B4, arrow in A4 flips to up.6. **Fifth Move:** B4, arrow points down, moves to C4, arrow in B4 flips to up.7. **Sixth Move:** C4, arrow points left, moves to C3, arrow in C4 flips to right.8. **Seventh Move:** C3, arrow points down, moves to D3, arrow in C3 flips to up.9. **Eighth Move:** D3, arrow points left, moves to D2, arrow in D3 flips to right.10. **Ninth Move:** D2, arrow points up, moves to C2, arrow in D2 flips to down.11. **Tenth Move:** C2, arrow points right, moves to C3, arrow in C2 flips to left.12. **Eleventh Move:** C3, arrow was flipped to up, moves to B3, arrow in C3 flips to down.13. **Twelfth Move:** B3, arrow was flipped to down, moves to C3, arrow in B3 flips to up.14. **Thirteenth Move:** C3, arrow was flipped to down, moves to D3, arrow in C3 flips to up.15. **Fourteenth Move:** D3, arrow was flipped to right, moves to D4, arrow in D3 flips to left.16. **Fifteenth Move:** D4, arrow points up, moves to C4, arrow in D4 flips to down.17. **Sixteenth Move:** C4, arrow was flipped to right, moves to D4, arrow in C4 flips to left.18. **Seventeenth Move:** D4, arrow was flipped to down, moves to C4, arrow in D4 flips to up.19. **Eighteenth Move:** C4, arrow was flipped to left, moves to C3, arrow in C4 flips to right.20. **Nineteenth Move:** C3, arrow was flipped to up, moves to B3, arrow in C3 flips to down.21. **Twentieth Move:** B3, arrow was flipped to up, moves to A3, arrow in B3 flips to down.22. **Twenty-First Move:** A3, arrow was flipped to left, moves to A2, arrow in A3 flips to right.23. **Twenty-Second Move:** A2, arrow points right, moves to A3, arrow in A2 flips to left.24. **Twenty-Third Move:** A3, arrow was flipped to right, moves to A4, arrow in A3 flips to left.25. **Twenty-Fourth Move:** A4, arrow was flipped to up, moves to A3, arrow in A4 flips to down.This is getting too long, and it seems like the token is still not exiting the grid. Maybe I'm approaching this wrong. Perhaps there's a pattern or a specific sequence of moves that leads to the token exiting.Alternatively, maybe the token exits from a corner cell, like A4 or D1, depending on the path. But without knowing the initial arrow directions, it's hard to be certain.Wait, maybe the problem assumes that all arrows initially point in a specific direction, like all right or all down, and then they flip accordingly. If that's the case, the path might be more predictable.Assuming all arrows initially point right:1. **Starting Position:** C2, arrow right.2. **First Move:** C2 to C3, arrow in C2 flips to left.3. **Second Move:** C3, arrow right, moves to C4, arrow in C3 flips to left.4. **Third Move:** C4, arrow right, moves to C5, which is outside the grid. So it exits from C4.But wait, the grid is only 4x4, so column 4 is the last column. Moving right from C4 would exit the grid. So the token exits from C4.But the problem asks for the cell from which it exits, so that would be C4, which is Row C, Column 4.But in the previous example, the token didn't exit from C4 because it was stuck in a loop. Maybe the initial assumption about all arrows pointing right is incorrect.Alternatively, if all arrows initially point down:1. **Starting Position:** C2, arrow down.2. **First Move:** C2 to D2, arrow in C2 flips to up.3. **Second Move:** D2, arrow down, moves to D3, arrow in D2 flips to up.4. **Third Move:** D3, arrow down, moves to D4, arrow in D3 flips to up.5. **Fourth Move:** D4, arrow down, moves to D5, which is outside the grid. So it exits from D4.But again, the grid is only 4x4, so row D is the last row. Moving down from D4 would exit the grid. So the token exits from D4.But the problem doesn't specify the initial arrow directions, so it's unclear.Wait, maybe the initial arrow directions are such that the token moves in a specific path leading to a particular exit cell. Perhaps it's designed to exit from A2 or D4 based on the initial setup.Given the ambiguity, I think the most logical assumption is that the token exits from A2, as in the previous example, when the arrow at A2 points up, causing it to exit from A2.But I'm not entirely sure. Maybe I need to look for a pattern or a property of the grid and arrow flipping that ensures the token exits from a specific cell.Alternatively, perhaps the token exits from the cell it starts on, but that doesn't make sense because it needs to move out of the grid.Wait, another thought: since the token flips the arrow in the cell it leaves, it's changing the grid's state as it moves. This could lead to the token eventually reaching a cell where the next move takes it out of the grid.Given that, and considering the previous steps, the token might exit from A2 when the arrow there points up.So, after several moves, the token reaches A2, and due to the flipping of arrows, the arrow at A2 points up, causing it to exit from A2.Therefore, the token exits from cell A2, which is Row A, Column 2.

💡Alright, so I've got this problem here: Let x be a non-zero real number such that x + 1/x is an integer. I need to show that for all natural numbers n, x^n + 1/x^n is also an integer. Hmm, okay. Let me try to wrap my head around this.First off, I know that x is a real number and not zero, which is good because dividing by zero would be problematic. The key here is that x + 1/x is an integer. Let's call that integer k for simplicity. So, x + 1/x = k, where k is some integer. Now, I need to show that for any natural number n, x^n + 1/x^n is also an integer. That sounds like it might involve some sort of recursive relationship or maybe using mathematical induction. Induction seems like a good approach here because the problem is about showing something holds for all natural numbers.Let me recall how mathematical induction works. To prove a statement for all natural numbers, I need to show two things: the base case is true, and if the statement holds for some arbitrary natural number n, then it also holds for n + 1. If both of these are satisfied, then the statement is true for all natural numbers.So, let's start with the base case. For n = 1, x^1 + 1/x^1 is just x + 1/x, which is given to be an integer. So, the base case holds.Now, for the inductive step, I need to assume that for some arbitrary natural number n, x^n + 1/x^n is an integer, and then show that x^(n+1) + 1/x^(n+1) is also an integer. Hmm, how can I relate x^(n+1) + 1/x^(n+1) to x^n + 1/x^n?I remember that there's a recursive formula for expressions like these. Maybe I can use the fact that x^(n+1) + 1/x^(n+1) can be expressed in terms of x^n + 1/x^n and x^(n-1) + 1/x^(n-1). Let me try to recall or derive that.Let's consider multiplying x + 1/x by x^n + 1/x^n. So, (x + 1/x)(x^n + 1/x^n) = x^(n+1) + x^(n-1) + 1/x^(n-1) + 1/x^(n+1). Hmm, that's interesting. So, that product equals x^(n+1) + 1/x^(n+1) + x^(n-1) + 1/x^(n-1).Wait, so if I rearrange that, I get x^(n+1) + 1/x^(n+1) = (x + 1/x)(x^n + 1/x^n) - (x^(n-1) + 1/x^(n-1)). That seems useful. So, if I denote a_n = x^n + 1/x^n, then the above equation becomes a_{n+1} = (x + 1/x)a_n - a_{n-1}. Since x + 1/x is an integer (let's call it k), and if I assume that a_n and a_{n-1} are integers (which is the inductive hypothesis), then a_{n+1} would be k*a_n - a_{n-1}, which is an integer because it's a linear combination of integers with integer coefficients.Therefore, if a_n and a_{n-1} are integers, then a_{n+1} is also an integer. But wait, in the standard induction, we usually assume a_n is true and then prove a_{n+1}. Here, it seems like I need to assume both a_n and a_{n-1} are true to prove a_{n+1}. Does that mean I need to use strong induction or something?Actually, in this case, since the recursive formula relates a_{n+1} to a_n and a_{n-1}, I might need to use a form of induction that accounts for two previous steps. So, perhaps I should verify the base cases for n = 1 and n = 2, and then assume it holds for n and n-1 to prove for n+1.Let me check n = 2. For n = 2, x^2 + 1/x^2. I can compute this using the identity (x + 1/x)^2 = x^2 + 2 + 1/x^2. So, x^2 + 1/x^2 = (x + 1/x)^2 - 2. Since x + 1/x is an integer, its square is also an integer, and subtracting 2 keeps it an integer. So, a_2 is indeed an integer.Okay, so now I have a_1 and a_2 as integers. Let's assume that for some n >= 2, both a_n and a_{n-1} are integers. Then, using the recursive formula, a_{n+1} = k*a_n - a_{n-1}, which is an integer because k, a_n, and a_{n-1} are all integers. Therefore, by induction, a_{n+1} is an integer.This seems to cover all natural numbers. So, starting from n = 1 and n = 2, and using the recursive relationship, we can show that a_n is an integer for all n in N.Wait, but what about n = 0? The problem states n is a natural number, which typically starts at 1, but sometimes people consider 0 as well. If n = 0, then x^0 + 1/x^0 = 1 + 1 = 2, which is an integer. So, even if we consider n = 0, it still holds. But since the problem specifies n in N, which usually starts at 1, maybe we don't need to worry about n = 0.But just to be thorough, let's make sure that the induction covers all cases. We've shown that if a_n and a_{n-1} are integers, then a_{n+1} is an integer. We've established the base cases for n = 1 and n = 2. So, by induction, it should hold for all n >= 1.Is there another way to think about this? Maybe using properties of polynomials or roots of equations. Since x + 1/x is an integer, perhaps x satisfies a quadratic equation. Let me explore that.If x + 1/x = k, then multiplying both sides by x gives x^2 + 1 = kx, which rearranges to x^2 - kx + 1 = 0. So, x is a root of the quadratic equation x^2 - kx + 1 = 0. The roots of this equation are x = [k ± sqrt(k^2 - 4)] / 2. Since k is an integer, the discriminant k^2 - 4 must be non-negative for real roots, so k^2 >= 4, meaning |k| >= 2.So, x is either [k + sqrt(k^2 - 4)] / 2 or [k - sqrt(k^2 - 4)] / 2. These are real numbers because the discriminant is non-negative.Now, if x is a root of x^2 - kx + 1 = 0, then we can use the recurrence relation for the powers of x. Specifically, x^n satisfies a linear recurrence relation based on the characteristic equation.Given the quadratic equation x^2 = kx - 1, we can express higher powers of x in terms of lower powers. For example, x^3 = x*(x^2) = x*(kx - 1) = kx^2 - x = k(kx - 1) - x = k^2x - k - x = (k^2 - 1)x - k.Similarly, x^4 = x*(x^3) = x*((k^2 - 1)x - k) = (k^2 - 1)x^2 - kx = (k^2 - 1)(kx - 1) - kx = (k^3 - k)x - (k^2 - 1) - kx = (k^3 - 2k)x - (k^2 - 1).Hmm, this seems a bit messy, but I notice a pattern. Each power of x can be expressed as a linear combination of x and 1, with coefficients that are integers because k is an integer. Therefore, x^n can be written as a_n x + b_n, where a_n and b_n are integers.Similarly, since 1/x is also a root of the same quadratic equation, we can express 1/x^n in terms of 1/x and 1, with the same coefficients. Therefore, 1/x^n = a_n (1/x) + b_n.Adding x^n + 1/x^n, we get (a_n x + b_n) + (a_n (1/x) + b_n) = a_n (x + 1/x) + 2b_n. Since x + 1/x = k is an integer, and a_n and b_n are integers, the entire expression is an integer.Wait, that seems like a different approach but arrives at the same conclusion. So, whether I use induction or the properties of the quadratic equation, I end up showing that x^n + 1/x^n is an integer for all natural numbers n.Let me summarize my thoughts:1. Given x + 1/x is an integer, denote it as k.2. Show that x^n + 1/x^n is an integer for all n in N.3. Use induction: base cases n=1 and n=2 are integers.4. Assume for n and n-1, then show for n+1 using the recursive formula.5. Alternatively, use the quadratic equation satisfied by x and express x^n in terms of x and 1, leading to x^n + 1/x^n being an integer.Both approaches seem valid and lead to the conclusion that x^n + 1/x^n is indeed an integer for all natural numbers n.I think I've covered all the necessary steps and verified the logic. I don't see any gaps or errors in the reasoning. It's a bit abstract, but breaking it down into these steps helps solidify the proof.

💡First, I recognize that when a positive integer ( x ) is divided by another positive integer ( y ), it results in a quotient and a remainder. In this case, the quotient is 96 and the remainder is 9.I can express this relationship mathematically as:[x = 96y + 9]Additionally, the problem states that ( frac{x}{y} = 96.45 ). This means:[frac{x}{y} = 96 + 0.45]Substituting the expression for ( x ) from the division equation into this, we get:[frac{96y + 9}{y} = 96.45]Simplifying the left side:[96 + frac{9}{y} = 96.45]Subtracting 96 from both sides:[frac{9}{y} = 0.45]To solve for ( y ), I invert both sides:[frac{y}{9} = frac{1}{0.45}]Calculating the right side:[frac{1}{0.45} = frac{20}{9}]Therefore:[frac{y}{9} = frac{20}{9}]Multiplying both sides by 9:[y = 20]So, the value of ( y ) is 20.

💡Okay, so I have this problem about triangle ABC. The circumcircle of triangle ABC has a diameter of (4√3)/3. The sides opposite angles A, B, and C are a, b, and c respectively, and angle C is 60 degrees. There are two parts to the problem.First, I need to find the value of (a + b + c)/(sin A + sin B + sin C). Hmm, okay. I remember that in a triangle, the Law of Sines relates the sides and the sines of the opposite angles. The Law of Sines says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumcircle. Wait, the problem gives the diameter, which is 2R, so R is half of that. The diameter is (4√3)/3, so R must be (2√3)/3. So, 2R is (4√3)/3. That means each ratio a/sin A, b/sin B, c/sin C is equal to (4√3)/3. So, if I think about the numerator a + b + c, that's the sum of the sides. The denominator is sin A + sin B + sin C. If I factor out the common term from the numerator and denominator, it's like (a + b + c)/(sin A + sin B + sin C) = (a/sin A + b/sin B + c/sin C)/(sin A + sin B + sin C) * (sin A + sin B + sin C). Wait, that might not be the right way to think about it. Alternatively, since each a = 2R sin A, b = 2R sin B, c = 2R sin C, so a + b + c = 2R (sin A + sin B + sin C). Therefore, (a + b + c)/(sin A + sin B + sin C) = 2R. Since 2R is given as the diameter, which is (4√3)/3, that should be the value. So, the first part is straightforward once I recall the Law of Sines.Now, moving on to the second part. If a + b = ab, find the area of triangle ABC. Hmm, okay. So, given that a + b = ab, and angle C is 60 degrees. I need to find the area. First, let's recall that the area of a triangle can be found using (1/2)ab sin C, where a and b are sides and C is the included angle. Since angle C is 60 degrees, sin C is √3/2. So, if I can find the product ab, I can compute the area.But I need to find ab. I know that a + b = ab. So, ab = a + b. Maybe I can use the Law of Cosines here. The Law of Cosines states that c² = a² + b² - 2ab cos C. Since angle C is 60 degrees, cos C is 0.5. So, c² = a² + b² - ab.But I don't know c yet. Wait, maybe I can find c using the Law of Sines. From the first part, we know that c = 2R sin C. R is (2√3)/3, so 2R is (4√3)/3. So, c = (4√3)/3 * sin 60°. Sin 60° is √3/2, so c = (4√3)/3 * (√3)/2 = (4*3)/6 = 2. So, c is 2.So, c² = 4. Therefore, from the Law of Cosines: 4 = a² + b² - ab. But we also know that a + b = ab. Let me denote ab as x. Then, a + b = x. Also, from the equation above, a² + b² - ab = 4. I can express a² + b² in terms of (a + b)². Since (a + b)² = a² + 2ab + b², so a² + b² = (a + b)² - 2ab. Substituting into the equation: (a + b)² - 2ab - ab = 4. So, (a + b)² - 3ab = 4. But since a + b = ab, which is x, then x² - 3x = 4. So, x² - 3x - 4 = 0.Solving this quadratic equation: x² - 3x - 4 = 0. The discriminant is 9 + 16 = 25. So, x = [3 ± 5]/2. Since x is ab, which is a positive quantity, we take the positive root: x = (3 + 5)/2 = 4. So, ab = 4.Therefore, the area is (1/2)ab sin C = (1/2)*4*(√3/2) = (1/2)*4*(√3)/2 = (2)*(√3)/2 = √3. So, the area is √3.Wait, let me double-check. If ab = 4, then a + b = 4 as well. So, a and b are roots of the equation t² - 4t + 4 = 0, which factors as (t - 2)² = 0. So, a = b = 2. So, triangle ABC has sides a = 2, b = 2, c = 2. Wait, that's an equilateral triangle? But angle C is 60 degrees, which is consistent with an equilateral triangle. So, all sides are equal, and all angles are 60 degrees. So, the area is indeed (√3/4)*2² = √3. That matches.So, both parts are solved. The first part is (4√3)/3, and the second part is √3.**Final Answer**1. boxed{dfrac{4sqrt{3}}{3}}2. boxed{sqrt{3}}

💡Okay, so I have this problem where the radius of the inscribed circle of a triangle is 1, and the lengths of the altitudes are integers. I need to prove that the triangle is equilateral. Hmm, let me try to figure this out step by step.First, I remember that the radius of the inscribed circle (inradius) of a triangle is related to the area and the semiperimeter. The formula is ( r = frac{S}{s} ), where ( S ) is the area of the triangle and ( s ) is the semiperimeter. Since the inradius ( r ) is given as 1, that means ( S = s ). So, the area of the triangle is equal to its semiperimeter.Next, I know that the area of a triangle can also be expressed using any of its altitudes. For example, if ( h_a ) is the altitude corresponding to side ( a ), then ( S = frac{1}{2} a h_a ). Similarly, ( S = frac{1}{2} b h_b ) and ( S = frac{1}{2} c h_c ) for the other sides and their corresponding altitudes.Since the problem states that the altitudes are integers, ( h_a ), ( h_b ), and ( h_c ) are all integers. Also, because the inradius is 1, the area ( S ) is equal to the semiperimeter ( s ). So, ( s = frac{1}{2} a h_a = frac{1}{2} b h_b = frac{1}{2} c h_c ).Let me write that down:1. ( s = frac{1}{2} a h_a )2. ( s = frac{1}{2} b h_b )3. ( s = frac{1}{2} c h_c )Since all these expressions equal ( s ), they must equal each other. So,( frac{1}{2} a h_a = frac{1}{2} b h_b = frac{1}{2} c h_c )Which simplifies to:( a h_a = b h_b = c h_c )Let me denote this common value as ( k ). So, ( a h_a = b h_b = c h_c = k ).Now, since ( s = frac{a + b + c}{2} ), and ( s = frac{1}{2} a h_a ), substituting ( s ) gives:( frac{a + b + c}{2} = frac{1}{2} a h_a )Multiplying both sides by 2:( a + b + c = a h_a )Similarly, since ( a h_a = b h_b = c h_c = k ), we can write:( a + b + c = k )But ( k = a h_a ), so:( a + b + c = a h_a )Which implies:( b + c = a (h_a - 1) )Similarly, we can write:( a + c = b (h_b - 1) )( a + b = c (h_c - 1) )Hmm, this is getting a bit complicated. Maybe I should consider specific cases or look for some inequalities.I know that in any triangle, the sum of any two sides must be greater than the third side. So, ( a + b > c ), ( a + c > b ), and ( b + c > a ).Given that ( b + c = a (h_a - 1) ), and ( a + b > c ), substituting ( c ) from the first equation:( a + b > a (h_a - 1) - b )Simplify:( a + b > a h_a - a - b )Bring all terms to one side:( a + b + a + b - a h_a > 0 )Combine like terms:( 2a + 2b - a h_a > 0 )Factor out ( a ):( a (2 - h_a) + 2b > 0 )Since ( h_a ) is an integer greater than or equal to 1, but in a triangle, the altitude must be greater than the inradius times 2, right? Because the inradius is the distance from the center to the side, so the altitude must be more than twice that. So, ( h_a > 2r = 2 times 1 = 2 ). Therefore, ( h_a geq 3 ).So, ( h_a geq 3 ). Plugging that into the inequality:( a (2 - h_a) + 2b > 0 )Since ( h_a geq 3 ), ( 2 - h_a leq -1 ). Therefore, ( a (negative) + 2b > 0 ).Which implies:( -a + 2b > 0 )So,( 2b > a )Similarly, from the other equations:From ( a + c = b (h_b - 1) ), we can derive ( 2c > b ).And from ( a + b = c (h_c - 1) ), we can derive ( 2a > c ).So, we have:1. ( 2b > a )2. ( 2c > b )3. ( 2a > c )Let me see if I can combine these inequalities.From 1: ( a < 2b )From 2: ( b < 2c )From 3: ( c < 2a )So, combining 1 and 2: ( a < 2b < 4c )From 3: ( c < 2a ), so ( 4c < 8a )Therefore, ( a < 8a ), which is always true, but not very helpful.Maybe I can express all variables in terms of one variable.Let me assume that the triangle is equilateral. Then, all sides are equal, say ( a = b = c ). Then, the semiperimeter ( s = frac{3a}{2} ). The area ( S = frac{sqrt{3}}{4} a^2 ). The inradius ( r = frac{S}{s} = frac{frac{sqrt{3}}{4} a^2}{frac{3a}{2}} = frac{sqrt{3}}{6} a ). Given ( r = 1 ), so ( frac{sqrt{3}}{6} a = 1 implies a = frac{6}{sqrt{3}} = 2sqrt{3} ).Then, the altitude ( h = frac{sqrt{3}}{2} a = frac{sqrt{3}}{2} times 2sqrt{3} = 3 ). So, the altitude is 3, which is an integer. So, in the equilateral case, it works.But the problem is to prove that the triangle must be equilateral given the conditions. So, I need to show that no other triangle can satisfy these conditions.Suppose the triangle is not equilateral. Then, without loss of generality, assume ( a > b geq c ). Then, from the inequalities above, ( 2b > a ), so ( a < 2b ). Also, ( 2c > b ), so ( c > frac{b}{2} ). And ( 2a > c ), so ( c < 2a ).But since ( a < 2b ), and ( c > frac{b}{2} ), combining these, ( frac{b}{2} < c < 2a < 4b ). Hmm, not sure if that helps.Let me go back to the expressions for the area.We have ( S = s = frac{a + b + c}{2} ).Also, ( S = frac{1}{2} a h_a ), so ( frac{a + b + c}{2} = frac{1}{2} a h_a implies a + b + c = a h_a ).Similarly, ( a + b + c = b h_b ) and ( a + b + c = c h_c ).So, ( a h_a = b h_b = c h_c = a + b + c ).Let me denote ( P = a + b + c ). Then, ( a h_a = P ), ( b h_b = P ), ( c h_c = P ).So, ( h_a = frac{P}{a} ), ( h_b = frac{P}{b} ), ( h_c = frac{P}{c} ).Since ( h_a ), ( h_b ), ( h_c ) are integers, ( frac{P}{a} ), ( frac{P}{b} ), ( frac{P}{c} ) are integers.Therefore, ( a ), ( b ), ( c ) must be divisors of ( P ).Also, since ( a ), ( b ), ( c ) are sides of a triangle, they must satisfy the triangle inequality.Let me think about the possible values of ( a ), ( b ), ( c ).Since ( h_a = frac{P}{a} geq 3 ), as established earlier, ( P geq 3a ).Similarly, ( P geq 3b ) and ( P geq 3c ).But ( P = a + b + c ), so ( a + b + c geq 3a implies b + c geq 2a ).But from triangle inequality, ( b + c > a ). So, ( 2a leq b + c ).Similarly, ( 2b leq a + c ) and ( 2c leq a + b ).Wait, this seems like the triangle is such that each side is at least half the sum of the other two sides.This is only possible if all sides are equal, right? Because if ( a ) is the largest side, then ( b + c > a ), but if ( b + c geq 2a ), then ( a ) can't be too large compared to ( b ) and ( c ).Let me test this.Suppose ( a ) is the largest side. Then, ( b + c geq 2a ).But from triangle inequality, ( b + c > a ). So, combining these, ( 2a leq b + c ).But ( b + c = P - a ). So, ( 2a leq P - a implies 3a leq P ).But ( P = a + b + c ), so ( 3a leq a + b + c implies 2a leq b + c ).Which is the same as before.So, if ( a ) is the largest side, ( b + c geq 2a ).But in a triangle, the largest side must be less than the sum of the other two sides. So, ( a < b + c ).Combining with ( b + c geq 2a ), we have ( 2a leq b + c < 2a + something ). Wait, no, actually, ( b + c ) is just greater than ( a ), but we have ( b + c geq 2a ).This seems contradictory unless ( a = b = c ).Because if ( a ) is the largest side, and ( b + c geq 2a ), but ( a < b + c ), so ( 2a leq b + c < 2a + something ). Wait, actually, ( b + c ) can be as large as needed, but in our case, ( P = a + b + c ), and ( h_a = frac{P}{a} geq 3 ).So, ( P geq 3a implies a + b + c geq 3a implies b + c geq 2a ).Similarly, ( b + c < 2a + something ), but actually, ( b + c ) can be much larger, but in our case, since ( h_a ) is an integer, ( P ) must be a multiple of ( a ), ( b ), and ( c ).Wait, maybe I should consider specific integer values for ( h_a ), ( h_b ), ( h_c ).Since ( h_a geq 3 ), let's try ( h_a = 3 ).Then, ( P = 3a ).Similarly, ( h_b = frac{P}{b} = frac{3a}{b} ) must be an integer.Similarly, ( h_c = frac{3a}{c} ) must be an integer.So, ( b ) must divide ( 3a ), and ( c ) must divide ( 3a ).Also, since ( P = 3a = a + b + c implies 2a = b + c ).So, ( b + c = 2a ).But from triangle inequality, ( b + c > a ), which is satisfied since ( 2a > a ).Also, ( |b - c| < a ).Given that ( b + c = 2a ), and ( |b - c| < a ), let me see what this implies.Let me denote ( b = a - d ) and ( c = a + d ), where ( d ) is a positive number less than ( a ).But wait, since ( b + c = 2a ), if ( b = a - d ), then ( c = a + d ).But ( b ) and ( c ) must be positive, so ( d < a ).Also, from triangle inequality, ( b + c > a ), which is ( 2a > a ), which is true.Also, ( a + b > c implies a + (a - d) > a + d implies 2a - d > a + d implies a > 2d ).Similarly, ( a + c > b implies a + (a + d) > a - d implies 2a + d > a - d implies a + 2d > 0 ), which is always true.So, the main constraint is ( a > 2d ).Now, since ( h_b = frac{3a}{b} = frac{3a}{a - d} ) must be an integer.Similarly, ( h_c = frac{3a}{c} = frac{3a}{a + d} ) must be an integer.So, both ( frac{3a}{a - d} ) and ( frac{3a}{a + d} ) must be integers.Let me denote ( k = frac{3a}{a - d} ) and ( m = frac{3a}{a + d} ), where ( k ) and ( m ) are integers.So,( k = frac{3a}{a - d} implies k(a - d) = 3a implies ka - kd = 3a implies a(k - 3) = kd implies d = frac{a(k - 3)}{k} )Similarly,( m = frac{3a}{a + d} implies m(a + d) = 3a implies ma + md = 3a implies a(m - 3) = -md implies d = frac{a(3 - m)}{m} )So, from both expressions for ( d ):( frac{a(k - 3)}{k} = frac{a(3 - m)}{m} )Cancel ( a ) (assuming ( a neq 0 )):( frac{k - 3}{k} = frac{3 - m}{m} )Cross-multiplying:( m(k - 3) = k(3 - m) )Expanding both sides:( mk - 3m = 3k - km )Bring all terms to one side:( mk - 3m - 3k + km = 0 )Combine like terms:( 2mk - 3m - 3k = 0 )Factor:( mk(2) - 3(m + k) = 0 )Hmm, not sure if that helps. Let me rearrange:( 2mk = 3(m + k) )So,( 2mk - 3m - 3k = 0 )Let me add 9 to both sides:( 2mk - 3m - 3k + 9 = 9 )Factor:( (2m - 3)(2k - 3) = 9 )Wait, let me check that.Let me try to factor ( 2mk - 3m - 3k ).Let me write it as:( 2mk - 3m - 3k = 0 )Add 9/2 to both sides:( 2mk - 3m - 3k + frac{9}{2} = frac{9}{2} )But that might not help. Alternatively, let me factor:Let me factor out m and k:( m(2k - 3) - 3k = 0 )Hmm, not helpful.Alternatively, let me write it as:( 2mk = 3m + 3k )Divide both sides by mk:( 2 = frac{3}{k} + frac{3}{m} )So,( frac{3}{k} + frac{3}{m} = 2 )Divide both sides by 3:( frac{1}{k} + frac{1}{m} = frac{2}{3} )So, we have ( frac{1}{k} + frac{1}{m} = frac{2}{3} ), where ( k ) and ( m ) are integers greater than or equal to 3 (since ( h_a geq 3 ), and ( h_b ) and ( h_c ) are also integers).Let me find integer solutions for ( k ) and ( m ) such that ( frac{1}{k} + frac{1}{m} = frac{2}{3} ).Let me denote ( k leq m ) without loss of generality.So, ( frac{1}{k} + frac{1}{m} = frac{2}{3} )Let me try ( k = 3 ):Then, ( frac{1}{3} + frac{1}{m} = frac{2}{3} implies frac{1}{m} = frac{1}{3} implies m = 3 ).So, ( k = 3 ), ( m = 3 ).Check if this works.So, ( k = 3 ), ( m = 3 ).From earlier, ( d = frac{a(k - 3)}{k} = frac{a(0)}{3} = 0 ).But ( d = 0 ) implies ( b = a ) and ( c = a ), so the triangle is equilateral.So, that's one solution.What about ( k = 4 ):Then, ( frac{1}{4} + frac{1}{m} = frac{2}{3} implies frac{1}{m} = frac{2}{3} - frac{1}{4} = frac{8}{12} - frac{3}{12} = frac{5}{12} implies m = frac{12}{5} ), which is not an integer.Similarly, ( k = 5 ):( frac{1}{5} + frac{1}{m} = frac{2}{3} implies frac{1}{m} = frac{2}{3} - frac{1}{5} = frac{10}{15} - frac{3}{15} = frac{7}{15} implies m = frac{15}{7} ), not integer.( k = 6 ):( frac{1}{6} + frac{1}{m} = frac{2}{3} implies frac{1}{m} = frac{2}{3} - frac{1}{6} = frac{4}{6} - frac{1}{6} = frac{3}{6} = frac{1}{2} implies m = 2 ).But ( m = 2 ) is less than 3, which contradicts our earlier conclusion that ( h_b geq 3 ).So, ( k = 6 ), ( m = 2 ) is invalid.Similarly, ( k = 2 ):But ( k geq 3 ), so we can't consider ( k = 2 ).What about ( k = 4 ), ( m = 12/5 ) is invalid.So, the only integer solution is ( k = 3 ), ( m = 3 ), which leads to ( d = 0 ), meaning the triangle is equilateral.Therefore, the only possible triangle under these conditions is an equilateral triangle.So, I think this proves that the triangle must be equilateral if the inradius is 1 and all altitudes are integers.