🧠:Okay, so I have these four functions to expand into Taylor series and find their radii of convergence. Let me take them one by one. Starting with part a: expanding sin(3z + 2) in powers of z + 1. Hmm. Taylor series expansions usually involve expressing a function as a sum of terms involving (z - a)^n, where a is the point around which we're expanding. Here, it's in powers of z + 1, which means a = -1. So we need to center the expansion at z = -1. First, I remember that the Taylor series for sin(w) around w = 0 is the standard sum from n=0 to infinity of (-1)^n * w^(2n+1)/(2n+1)!). So maybe I can substitute w with something related to (z + 1). Let me see. The function is sin(3z + 2). Let's set w = z + 1. Then z = w - 1. Substituting into the argument of sine: 3z + 2 = 3(w - 1) + 2 = 3w - 3 + 2 = 3w - 1. So sin(3z + 2) becomes sin(3w - 1), where w = z + 1. So we need to expand sin(3w - 1) around w = 0. Alternatively, maybe I can write sin(3w - 1) as sin(3w - 1). That's like sin(a - b), but maybe using angle subtraction formula. Let's see: sin(A - B) = sin A cos B - cos A sin B. If I let A = 3w and B = 1, then sin(3w - 1) = sin(3w)cos(1) - cos(3w)sin(1). Then, I can expand sin(3w) and cos(3w) as their Taylor series around w=0. Yes, that might work. Let me recall that sin(x) = sum_{n=0}^infty (-1)^n x^{2n+1}/(2n+1)! and cos(x) = sum_{n=0}^infty (-1)^n x^{2n}/(2n)!. So substituting x with 3w, we get:sin(3w) = sum_{n=0}^infty (-1)^n (3w)^{2n+1}/(2n+1)!,cos(3w) = sum_{n=0}^infty (-1)^n (3w)^{2n}/(2n)!.So then sin(3w - 1) = sin(3w)cos(1) - cos(3w)sin(1) = cos(1) * sum_{n=0}^infty (-1)^n (3w)^{2n+1}/(2n+1)! - sin(1) * sum_{n=0}^infty (-1)^n (3w)^{2n}/(2n)!.Then substituting back w = z + 1, we get the series in terms of (z + 1). So each term will have (3(z + 1))^{2n+1} or (3(z + 1))^{2n}. Therefore, the expansion is valid as long as the original sine and cosine series converge, which is everywhere, but the radius of convergence for the Taylor series of sine and cosine is infinite. However, since we substituted 3w, but w is centered at 0, and we didn't have any singularities, the radius of convergence should still be infinite. Wait, but let me confirm. The substitution is linear, so the radius of convergence remains the same. Therefore, the radius of convergence for part a) is infinite.Moving on to part b: e^z in powers of 2z + 1. Hmm. Powers of 2z + 1. Let's see. Again, to expand around a certain point. Let me express 2z + 1 as the variable. Let me set w = 2z + 1. Then, solving for z, we have z = (w - 1)/2. So we need to express e^{z} = e^{(w - 1)/2} in powers of w. So expanding e^{(w - 1)/2} around w=0. Alternatively, we can write e^{( (w - 1)/2 )} as e^{-1/2} * e^{w/2}. Then, the Taylor series of e^{w/2} around w=0 is sum_{n=0}^infty (w/2)^n /n! , so multiplying by e^{-1/2}, we have e^{-1/2} sum_{n=0}^infty (w/2)^n /n! = sum_{n=0}^infty e^{-1/2} (1/2)^n w^n /n! . Then, substituting back w = 2z + 1, but wait, hold on. Wait, if we are expanding in powers of 2z + 1, that is, expressing the function as a series in terms of (2z + 1). So actually, the expansion variable is (2z + 1). Let me clarify. Wait, when the problem says "in powers of 2z + 1", it means that the expansion is centered at the point where 2z + 1 = 0, i.e., z = -1/2. Because in Taylor series, if you expand in powers of (z - a), then here, expanding in powers of (2z + 1) is equivalent to expanding in powers of (2(z + 1/2)). So the center is at z = -1/2. Therefore, the function e^z is being expanded around z = -1/2, with the series expressed in terms of (2z + 1). Alternatively, maybe the problem is expecting us to write the expansion as a series in (2z + 1), treating (2z + 1) as the variable. Let me check.Suppose we let t = 2z + 1. Then z = (t - 1)/2. So e^z = e^{(t - 1)/2} = e^{-1/2} e^{t/2}. Then, expanding e^{t/2} around t=0 gives the series sum_{n=0}^infty (t/2)^n /n! . So substituting back t = 2z + 1, we get e^{-1/2} sum_{n=0}^infty ( (2z + 1)/2 )^n /n! = e^{-1/2} sum_{n=0}^infty (2z + 1)^n / (2^n n! ). So the expansion is e^{-1/2} times sum_{n=0}^infty (2z + 1)^n / (2^n n! ). But in the problem statement, it says "in powers of 2z + 1", so perhaps this is correct. Now, the radius of convergence for the exponential function is infinite, so even after substitution, the radius in terms of t is infinite, but since t = 2z + 1, which is a linear substitution, the radius in terms of z would also be infinite. Wait, but radius of convergence is about the variable in which the series is expressed. Here, the series is in powers of (2z + 1), so the radius of convergence R in terms of t = 2z + 1 is infinity. Therefore, for the original variable z, since t can be any complex number, z can be any complex number as well. Therefore, the radius of convergence is infinite. So part b) also has radius of convergence infinity.Wait, but let me verify. The exponential function e^{t/2} has a Taylor series with infinite radius of convergence. Therefore, regardless of the substitution t = 2z + 1, which is just a linear shift and scaling, the radius remains infinite. So yes, R = infinity.Part c: (z + 1)/(z² - 3z + 2) in powers of z. So expanding around z = 0. First, let's factor the denominator. The denominator is z² - 3z + 2. Factoring: (z - 1)(z - 2). So the function is (z + 1)/[(z - 1)(z - 2)]. To expand this as a Taylor series around z = 0. Since the function is rational, we can perform partial fraction decomposition first. Let me try that.Let me write (z + 1)/[(z - 1)(z - 2)] = A/(z - 1) + B/(z - 2). Solving for A and B. Multiply both sides by (z - 1)(z - 2):z + 1 = A(z - 2) + B(z - 1).To find A and B, set up equations by substituting z = 1 and z = 2.For z = 1: 1 + 1 = A(1 - 2) + B(1 - 1) => 2 = -A + 0 => A = -2.For z = 2: 2 + 1 = A(2 - 2) + B(2 - 1) => 3 = 0 + B(1) => B = 3.So the function becomes -2/(z - 1) + 3/(z - 2). Now, we need to express each term as a Taylor series around z = 0. So let's handle each term separately.First term: -2/(z - 1) = 2/(1 - z). Since 1/(1 - z) is the sum from n=0 to infinity of z^n for |z| < 1. So 2/(1 - z) = 2 sum_{n=0}^infty z^n, radius of convergence 1.Second term: 3/(z - 2) = -3/(2 - z) = (-3/2)/(1 - z/2). Using the formula 1/(1 - w) = sum_{n=0}^infty w^n for |w| < 1. Here, w = z/2. So -3/2 * sum_{n=0}^infty (z/2)^n = -3/2 sum_{n=0}^infty z^n / 2^n = sum_{n=0}^infty (-3)/2^{n+1} z^n. The radius of convergence for this series is |z/2| < 1 => |z| < 2.Therefore, combining both series: 2 sum z^n + sum (-3)/2^{n+1} z^n. So the coefficients are 2 - 3/2^{n+1} for each z^n. So the entire series is sum_{n=0}^infty [2 - 3/(2^{n+1})] z^n. Now, the radius of convergence is the minimum of the radii from each series. Since the first term converges for |z| < 1 and the second term converges for |z| < 2, the overall radius is the smaller one, which is 1. However, we need to check if there are any singularities closer than 1. The original function has singularities at z = 1 and z = 2. The distance from the center (z=0) to the nearest singularity is 1. Therefore, the radius of convergence should be 1. So part c) has radius 1.Wait, but the partial fractions were each converted to series valid within their radii, and combined, the radius is the minimum, which is 1. So that seems correct.Part d: ln(3 - z) in powers of z. Expanding around z = 0. The natural logarithm function. Let me recall that ln(1 + w) has a Taylor series sum_{n=1}^infty (-1)^{n+1} w^n /n for |w| < 1. So maybe we can manipulate ln(3 - z) into a form similar to ln(1 + (some expression)).Let me write ln(3 - z) = ln(3(1 - z/3)) = ln(3) + ln(1 - z/3). Then, ln(1 - z/3) can be expanded using the series for ln(1 - w) which is -sum_{n=1}^infty w^n /n for |w| < 1. So substituting w = z/3:ln(3 - z) = ln(3) - sum_{n=1}^infty (z/3)^n /n = ln(3) - sum_{n=1}^infty z^n / (3^n n).Therefore, the expansion is ln(3) - sum_{n=1}^infty z^n / (3^n n). The radius of convergence is determined by the distance from the center z=0 to the nearest singularity. The function ln(3 - z) has a singularity at z = 3 (where the argument of the logarithm becomes zero, and logarithm has a branch point there). Therefore, the radius of convergence is the distance from 0 to 3, which is 3. Alternatively, in the series for ln(1 - z/3), the radius is |z/3| < 1 => |z| < 3. Hence, radius of convergence is 3.Wait, but let me confirm. The expansion of ln(1 - w) is valid for |w| < 1, so substituting w = z/3, we have |z/3| < 1 => |z| < 3. So radius of convergence 3. That's correct. Therefore, part d) has radius of convergence 3.Let me recap each part to ensure I didn't make a mistake.a) sin(3z + 2) around z = -1: we transformed it into sin(3w - 1) with w = z + 1, expanded using sine and cosine addition formulas, resulting in a series with infinite radius.b) e^z in powers of 2z + 1: set t = 2z + 1, so z = (t - 1)/2, expanded e^{(t - 1)/2} as e^{-1/2} * e^{t/2}, which is a series in t with infinite radius, so radius of convergence infinite.c) (z + 1)/(z² - 3z + 2): partial fractions into -2/(z - 1) + 3/(z - 2), expanded each around z=0, combined series with coefficients 2 - 3/2^{n+1}, radius of convergence 1 (distance to z=1).d) ln(3 - z): expressed as ln(3) + ln(1 - z/3), expanded the second term as a series, radius 3.All steps seem logical. Let me just check part a again. When expanding sin(3w - 1) using sin(a - b) formula, yes, that's valid. Then substituting the series for sin(3w) and cos(3w), which have infinite radii, so combining them, the series will also have infinite radius. Correct. So part a's radius is indeed infinity.Similarly, part b, substitution in exponential function which is entire, so radius remains infinite. Part c: the combined radius is 1, as the closest singularity is at z=1. Part d: closest singularity at z=3, so radius 3. All seems correct. I think these are the right answers.

✅:a) sin(3z + 2) in powers of z + 11. Substitution: Let w = z + 1 implies z = w - 1. Then: [ sin(3z + 2) = sin(3(w - 1) + 2) = sin(3w - 1). ]2. Angle Subtraction Formula: [ sin(3w - 1) = sin(3w)cos(1) - cos(3w)sin(1). ]3. Expand sin(3w) and cos(3w): [ sin(3w) = sum_{n=0}^infty frac{(-1)^n (3w)^{2n+1}}{(2n+1)!}, quad cos(3w) = sum_{n=0}^infty frac{(-1)^n (3w)^{2n}}{(2n)!}. ]4. Combine Series: [ sin(3w - 1) = cos(1)sum_{n=0}^infty frac{(-1)^n (3w)^{2n+1}}{(2n+1)!} - sin(1)sum_{n=0}^infty frac{(-1)^n (3w)^{2n}}{(2n)!}. ]5. Substitute w = z + 1 and simplify. The series converges everywhere (entire function).Radius of Convergence: infty.---b) e^z in powers of 2z + 11. Substitution: Let t = 2z + 1 implies z = frac{t - 1}{2}. Then: [ e^z = e^{frac{t - 1}{2}} = e^{-1/2}e^{t/2}. ]2. Expand e^{t/2}: [ e^{t/2} = sum_{n=0}^infty frac{(t/2)^n}{n!} implies e^{-1/2}sum_{n=0}^infty frac{(2z + 1)^n}{2^n n!}. ]3. Convergence: Exponential series converges everywhere.Radius of Convergence: infty.---c) frac{z+1}{z^2 - 3z + 2} in powers of z1. Partial Fractions: [ frac{z+1}{(z-1)(z-2)} = frac{-2}{z-1} + frac{3}{z-2}. ]2. Expand Each Term: - frac{-2}{z-1} = 2sum_{n=0}^infty z^n (valid for |z| < 1). - frac{3}{z-2} = -frac{3}{2}sum_{n=0}^infty left(frac{z}{2}right)^n (valid for |z| < 2).3. Combine Series: [ sum_{n=0}^infty left(2 - frac{3}{2^{n+1}}right)z^n. ]4. Convergence: Governed by the nearest singularity at z=1.Radius of Convergence: 1.---d) ln(3 - z) in powers of z1. Rewrite: [ ln(3 - z) = lnleft(3left(1 - frac{z}{3}right)right) = ln(3) + lnleft(1 - frac{z}{3}right). ]2. Expand ln(1 - w): [ lnleft(1 - frac{z}{3}right) = -sum_{n=1}^infty frac{z^n}{3^n n} quad (|z| < 3). ]3. Full Series: [ ln(3) - sum_{n=1}^infty frac{z^n}{3^n n}. ]4. Convergence: Nearest singularity at z=3.Radius of Convergence: 3.---Final Answers:- a) boxed{infty} - b) boxed{infty} - c) boxed{1} - d) boxed{3}

🔑:We will solve each of the parts step-by-step and calculate the Taylor series along with the radius of convergence.# Part (a)Let's find the Taylor series expansion of sin (3z + 2) around z = -1. 1. We know the general formula for the Taylor series expansion of sin(az + b) around z = c is given by: [ sin(az + b) = sum_{n=0}^{infty} frac{a^n}{n!} sin left(ac + b + frac{pi n}{2} right) (z - c)^n ]2. For our specific problem, we have a = 3, b = 2, and c = -1. Therefore, the Taylor series expansion becomes: [ sin(3z + 2) = sum_{n=0}^{infty} frac{3^n}{n!} sin left(3(-1) + 2 + frac{pi n}{2}right) (z + 1)^n ]3. Simplifying the argument of the sine function in the above series: [ 3(-1) + 2 + frac{pi n}{2} = -3 + 2 + frac{pi n}{2} = -1 + frac{pi n}{2} ] Hence, the expansion becomes: [ sin(3z + 2) = sum_{n=0}^{infty} frac{3^n}{n!} sin left(-1 + frac{pi n}{2}right) (z + 1)^n ]4. To find the radius of convergence, we use the ratio test (D'Alembert's ratio test): [ lim_{n to infty} left| frac{u_{n+1}(z)}{u_n(z)} right| = lim_{n to infty} left| frac{frac{3^{n+1}}{(n+1)!} sin left( -1 + frac{pi (n+1)}{2} right) (z + 1)^{n+1}}{frac{3^n}{n!} sin left( -1 + frac{pi n}{2} right) (z + 1)^n} right| ]5. This simplifies further as: [ = lim_{n to infty} frac{3 |z + 1|}{n + 1} left| cot left( -1 + frac{pi n}{2} right) right| = 0 ] Since for large n, the cotangent term oscillates between finite values.6. Thus, the series converges for all z and the radius of convergence is: [ R = +infty ]Conclusion for part (a):[boxed{sum_{n=0}^{infty} frac{3^n}{n!} sin left(-1 + frac{pi n}{2}right) (z + 1)^n, , R = +infty}]# Part (b)Now, we turn to finding the Taylor series expansion of e^z in powers of 2z+1.1. We start by recalling that the Taylor series expansion of e^z around z = 0 (Maclaurin series) is: [ e^z = sum_{n=0}^{infty} frac{z^n}{n!} ]2. We need to re-express e^z in terms of 2z+1. First, we substitute w = 2z+1 to write z = frac{w-1}{2}: [ e^z = e^{frac{w-1}{2}} ]3. Next, we rewrite the exponent using the property of exponentials: [ e^{frac{w-1}{2}} = e^{frac{-1}{2}} cdot e^{frac{w}{2}} = frac{1}{sqrt{e}} cdot e^{frac{w}{2}} ]4. We now expand e^{frac{w}{2}} as a series: [ e^{frac{w}{2}} = sum_{n=0}^{infty} frac{left(frac{w}{2}right)^n}{n!} = sum_{n=0}^{infty} frac{w^n}{2^n n!} ] Substituting back w = 2z + 1: [ e^{frac{w}{2}} = sum_{n=0}^{infty} frac{(2z + 1)^n}{2^n n!} ]5. Hence, the series expansion for e^z in terms of 2z+1 is: [ e^z = frac{1}{sqrt{e}} sum_{n=0}^{infty} frac{(2z + 1)^n}{2^n n!} ]6. To find the radius of convergence, we use the ratio test: [ lim_{n to infty} left| frac{u_{n+1}(z)}{u_n(z)} right| = lim_{n to infty} left| frac{frac{1}{sqrt{e}} frac{(2z+1)^{n+1}}{2^{n+1}(n+1)!}}{frac{1}{sqrt{e}} frac{(2z+1)^n}{2^n n!}} right| = lim_{n to infty} left| frac{2z + 1}{2(n + 1)} right| = 0 ]7. Thus, the series converges for all z and the radius of convergence is: [ R = +infty ]Conclusion for part (b):[boxed{frac{1}{sqrt{e}} sum_{n=0}^{infty} frac{(2z+1)^n}{2^n n!}, , R = +infty}]# Part (c)Next, we find the Taylor series expansion of frac{z+1}{z^2 - 3z + 2} in terms of powers of z.1. First, factorize the denominator: [ frac{z+1}{z^2 - 3z + 2} = frac{z+1}{(z-1)(z-2)} ]2. We perform partial fraction decomposition: [ frac{z+1}{(z-1)(z-2)} = frac{A}{z - 1} + frac{B}{z - 2} ]3. Solving for A and B: [ (z+1) = A(z-2) + B(z-1) ] Setting z = 1, we find: [ 1+1 = A(1-2) + B(1-1) implies 2 = -A implies A = -2 ] Setting z = 2, we find: [ 2+1 = A(2-2) + B(2-1) implies 3 = B implies B = 3 ] So: [ frac{z+1}{(z-1)(z-2)} = frac{-2}{z-1} + frac{3}{z-2} ]4. Express these terms as geometric series: [ frac{-2}{z-1} = -2 sum_{n=0}^{infty} z^n quad text{if} quad |z| < 1 ] [ frac{3}{z-2} = frac{-3}{2} sum_{n=0}^{infty} left( frac{z}{2} right)^n quad text{if} quad |z| < 2 ]5. Combining these series: [ frac{z+1}{(z-1)(z-2)} = -2 sum_{n=0}^{infty} z^n - frac{3}{2} sum_{n=0}^{infty} left( frac{z}{2} right)^n ]6. Simplifying: [ -2 sum_{n=0}^{infty} z^n + frac{3}{2} sum_{n=0}^{infty} left( frac{z}{2} right)^n ]7. The radius of convergence is the minimum of the radii of the individual series. The individual radii are 1 and 2 respectively, so the overall radius of convergence is: [ R = 1 ]Conclusion for part (c):[boxed{-2 sum_{n=0}^{infty} z^n - frac{3}{2} sum_{n=0}^{infty} left( frac{z}{2} right)^n, , R = 1}]# Part (d)Finally, find the Taylor series expansion of ln(3 - z) around z = 0.1. We start with the known series expansion: [ ln(1 - z) = -sum_{n=1}^{infty} frac{z^n}{n} ]2. We rewrite ln(3 - z): [ ln(3 - z) = lnleft(3 left( 1 - frac{z}{3} right) right) = ln 3 + lnleft(1 - frac{z}{3}right) ]3. Substituting the series for ln left( 1 - frac{z}{3} right): [ ln(3 - z) = ln 3 - sum_{n=1}^{infty} frac{1}{n} left( frac{z}{3} right)^n ]4. Simplifying the series: [ ln(3 - z) = ln 3 - sum_{n=1}^{infty} frac{z^n}{3^n n} ]5. To find the radius of convergence: [ lim_{n to infty} left| frac{u_{n+1}(z)}{u_n(z)} right| = lim_{n to infty} left| frac{frac{z^{n+1}}{3^{n+1} (n+1)}}{frac{z^n}{3^n n}} right| = frac{|z|}{3} ] The series converges if this is less than 1: [ frac{|z|}{3} < 1 implies |z| < 3 ]Conclusion for part (d):[boxed{ln 3 - sum_{n=1}^{infty} frac{z^n}{3^n n}, , R = 3}]

🧠:Okay, so I need to find the probability that when two dice are thrown, the sum is even and at least one die shows a six. Let me start by understanding the problem step by step.First, when two dice are thrown, each die has 6 faces, so there are a total of 6 x 6 = 36 possible outcomes. Each outcome is equally likely if the dice are fair. The problem has two conditions: the sum must be even, and at least one die must show a six. I need to find the number of outcomes that satisfy both conditions and then divide that by the total number of outcomes (36) to get the probability.Let me break down the problem into parts. First, let's consider the condition that the sum is even. The sum of two numbers is even if both numbers are even or both are odd. So, for two dice, each die can be even or odd. The possible even numbers on a die are 2, 4, 6, and the odd numbers are 1, 3, 5. So, for the sum to be even, both dice must be even, or both must be odd. Let's calculate the number of outcomes for each case.Case 1: Both dice are even. Each die has 3 even numbers, so 3 x 3 = 9 outcomes.Case 2: Both dice are odd. Similarly, each die has 3 odd numbers, so another 3 x 3 = 9 outcomes.Therefore, the total number of outcomes where the sum is even is 9 + 9 = 18. So, the probability of the sum being even is 18/36 = 1/2. That makes sense because the sum is equally likely to be odd or even.Now, the second condition is that at least one die shows a six. This is a bit trickier. The phrase "at least one" makes me think of using complementary counting. The complementary event of "at least one six" is "no sixes at all." So, the number of outcomes with at least one six is total outcomes minus outcomes with no sixes. Total outcomes are 36. Outcomes with no sixes: each die has 5 options (1-5), so 5 x 5 = 25. Therefore, outcomes with at least one six are 36 - 25 = 11. Wait, but let me verify this. If I list them, when the first die is 6, the second can be 1-6 (6 outcomes). When the second die is 6, the first can be 1-5 (since 6 was already counted in the first case), so 5 more outcomes. So total 6 + 5 = 11. Yes, that matches. So there are 11 outcomes where at least one die is a six.But now, the problem requires both conditions: sum is even AND at least one six. So I need the intersection of these two events. So I can't just multiply the probabilities because they might not be independent. Let me instead find the number of outcomes that satisfy both conditions.Let me list all outcomes where at least one die is a six and the sum is even. Let's think about how the presence of a six affects the parity of the sum.If one die is six (which is even), the other die must also be even for the sum to be even. Because even + even = even, and even + odd = odd. So, if exactly one die is six (even), the other die must be even. But if both dice are six, then both are even, so the sum is even.Wait, so let's break it down:Case 1: Exactly one die is six. Then, since six is even, the other die must be even for the sum to be even. The other die can be 2, 4, or 6. But wait, if exactly one die is six, the other die can be 2, 4, or 6? Wait, no. If exactly one die is six, the other die can be any number except six. But for the sum to be even, since one die is even (six), the other die must be even. So the other die has to be 2, 4, or 6. But if exactly one die is six, the other die can be 2, 4, or 6 (but not six). Wait, no. If exactly one die is six, then the other die must not be six. So, if one die is six (even), the other die must be even but not six. So the other die can be 2 or 4. Because 6 is excluded since we want exactly one six. Wait, this is confusing. Let me clarify.If exactly one die is six, then the other die must be even (so the sum is even), but not six. So the other die can be 2 or 4. Because 6 would make it two sixes, which is a separate case. Therefore, for exactly one six:First die is 6, second die is 2 or 4: that's 2 outcomes.Second die is 6, first die is 2 or 4: that's another 2 outcomes.So total of 4 outcomes where exactly one die is six and the other is even (2 or 4). Then, sum is even.Case 2: Both dice are six. Then, sum is 12, which is even. So that's 1 outcome: (6,6).Therefore, total outcomes where at least one die is six and sum is even is 4 + 1 = 5.Wait, but hold on. Let me check this again. If exactly one die is six, the other die must be even. But even numbers on a die are 2, 4, 6. But if we want exactly one six, then the other die can't be six. Therefore, the other die must be 2 or 4. So, first die 6 and second die 2 or 4: two outcomes. Similarly, second die 6 and first die 2 or 4: two outcomes. So 4 in total. Then both dice six: 1 outcome. So total 5.Alternatively, let's list all possible outcomes where at least one die is six and the sum is even.Possible outcomes with at least one six:(6,1), (6,2), (6,3), (6,4), (6,5), (6,6),(1,6), (2,6), (3,6), (4,6), (5,6).Total 11 outcomes as before.Now, from these 11, which have an even sum?Let's check each:(6,1): 6+1=7 (odd) -> exclude.(6,2): 6+2=8 (even) -> include.(6,3): 6+3=9 (odd) -> exclude.(6,4): 6+4=10 (even) -> include.(6,5): 6+5=11 (odd) -> exclude.(6,6): 6+6=12 (even) -> include.(1,6): 1+6=7 (odd) -> exclude.(2,6): 2+6=8 (even) -> include.(3,6): 3+6=9 (odd) -> exclude.(4,6): 4+6=10 (even) -> include.(5,6): 5+6=11 (odd) -> exclude.So the outcomes that qualify are: (6,2), (6,4), (6,6), (2,6), (4,6). That's 5 outcomes. So 5 out of 36.Therefore, the probability is 5/36. Wait, that seems straightforward. But let me confirm once more.Alternatively, using combinatorics. Let me think again.We need the number of outcomes where at least one die is six and the sum is even.Let’s consider two scenarios:1. Exactly one six: Then, as discussed, the other die must be even (since six is even, to make sum even, the other must be even). The other die can be 2 or 4 (since if it were 6, it would be two sixes). So two possibilities for each die. So two outcomes when the first die is six and second is 2 or 4, and two outcomes when the second die is six and first is 2 or 4. Total 4.2. Both dice are six: One outcome (6,6). Sum is 12, which is even.Thus, total of 4 +1 =5 outcomes. So probability 5/36.Therefore, the answer is 5/36.But let me check if there's another approach. For example, using conditional probability.We can think of the probability that the sum is even given that at least one die is six, multiplied by the probability that at least one die is six. But maybe that complicates it.Alternatively, since we need P(A and B) where A is sum even and B is at least one six. So P(A ∩ B) = number of favorable outcomes /36 =5/36.Alternatively, another way. Let me think about the possible combinations.If at least one die is six, then possible pairs are (6,x) and (x,6) where x is 1-6. But we have to subtract the duplicate (6,6). So total 11 as before.From these 11, sum is even. As we saw, 5 of them.So yes, 5/36.But wait, another way: let's use the principle of inclusion-exclusion.But maybe overcomplicating. Alternatively, the problem is straightforward when you list the outcomes.Alternatively, let's use parity. Since we need sum even and at least one six.Let me note that in order for the sum to be even, the two numbers must be both even or both odd.But at least one die is six (which is even). Therefore, since one die is even, the other die must also be even for the sum to be even. Because even + even = even, and even + odd = odd. Therefore, if at least one die is six (even), the other die must be even for the sum to be even.Therefore, the favorable outcomes are those where at least one die is six, and the other die is even.So, how many such outcomes?Case 1: Exactly one die is six and the other die is even (2 or 4). As before: 2 (for 6 on first die and 2 or 4 on second) + 2 (for 6 on second die and 2 or 4 on first) =4.Case 2: Both dice are six: 1.Total:5.Therefore, 5/36.Yes, that seems consistent.Alternatively, think in terms of variables. Let me denote the two dice as D1 and D2.We need D1 + D2 is even and (D1=6 or D2=6).Since D1=6 or D2=6, let's split into cases.Case 1: D1=6. Then D2 must be even for D1 + D2 to be even. So D2 ∈ {2,4,6}. So outcomes (6,2), (6,4), (6,6). Three outcomes.Case 2: D2=6. Then D1 must be even for D1 + D6 to be even. So D1 ∈ {2,4,6}. However, (6,6) is already counted in Case 1, so we subtract that. So outcomes (2,6), (4,6). Two outcomes.Total:3 +2=5. Hence, 5 outcomes.Yes, that also works. So that's another way to see it. So total 5, probability 5/36.Therefore, the probability is 5/36. I think that's the correct answer. Let me check once more.Alternatively, another method:The total number of outcomes with at least one six is 11. Of these, how many have even sum?As we listed earlier, 5. So 5/11 of the 11 outcomes, but wait, no. Wait, no. The probability is number of favorable outcomes over total possible outcomes, which is 5/36. Because probability is not conditional here. The problem is not asking for conditional probability, but the probability of both events happening. So it's 5/36.Yes, that's correct. If it had asked for the probability that the sum is even given that at least one die is six, then it would be 5/11. But here, it's the probability that both happen, so 5/36.Therefore, the final answer is 5/36.

✅:The probability that the sum of the points on two rolled dice is even and at least one die shows a six is calculated as follows:1. Total Outcomes: There are (6 times 6 = 36) possible outcomes when two dice are thrown.2. Favorable Outcomes: - At least one die shows a six: There are 11 such outcomes: ((6,1)) to ((6,6)) and ((1,6)) to ((5,6)). - Sum is even: For the sum to be even, if one die is six (even), the other die must also be even. This leads to: - Exactly one six with the other die being 2 or 4: ((6,2), (6,4), (2,6), (4,6)) (4 outcomes). - Both dice six: ((6,6)) (1 outcome). Total favorable outcomes: (4 + 1 = 5).3. Probability: (frac{text{Favorable Outcomes}}{text{Total Outcomes}} = frac{5}{36}).Answer: (boxed{dfrac{5}{36}})

🔑:To solve this problem, we need to consider the experiment of rolling two fair six-sided dice. Each die has faces numbered from 1 to 6. We are asked to find the probability of obtaining an even sum AND that at least one of the dice shows a 6.1. Calculate the total number of possible outcomes: - For two dice, each die has 6 faces. - Therefore, the total number of possible outcomes when both dice are rolled is: [ 6 times 6 = 36 ]2. Identify the favorable outcomes: - We are looking for outcomes where the sum is even and at least one die shows 6. - Let's break down the cases: - If one die shows a 6 (let’s consider the first die shows 6), then the second die must show an even number to ensure the sum is even. - The even numbers are 2, 4, and 6. - Accordingly, the combinations are: ( (6, 2), (6, 4), (6, 6) ) Now, let's consider the case where the second die shows a 6: - Applying the same logic, the combinations where the first die shows an even number are: ( (2, 6), (4, 6), (6, 6) ) - We must be careful to not double-count the outcome ( (6, 6) ). Therefore, the favorable outcomes are: [ (6, 2), (6, 4), (6, 6), (2, 6), (4, 6) ]3. Count the number of favorable outcomes: - From the previous step, we identified 5 unique favorable outcomes. - They are: ( (6, 2), (6, 4), (6, 6), (2, 6), (4, 6) ).4. Calculate the probability: - The probability ( P ) of an event occurring is given by the ratio of the number of favorable outcomes to the total number of possible outcomes. - Therefore: [ P = frac{text{Number of favorable outcomes}}{text{Total number of possible outcomes}} = frac{5}{36} ] Conclusion:The probability that the sum of the points on the two dice is even and at least one of the dice shows a 6 is:[boxed{frac{5}{36}}]

🧠:Okay, so I need to prove that the common chord of the two circles with diameters CM and BN passes through the orthocenter of triangle ABC. Let me start by recalling some definitions and properties related to orthocenters and common chords.First, the orthocenter of a triangle is the point where the three altitudes intersect. An altitude is a perpendicular line from a vertex to the opposite side. So, in triangle ABC, the orthocenter H is where the altitudes from A, B, and C meet.Next, the common chord of two circles is the line segment joining the points of intersection of the two circles. The common chord is perpendicular to the line joining the centers of the circles. So, to find the common chord, I need to find the intersections of the two circles and then show that this line passes through H.Given that the circles have diameters CM and BN, their centers will be the midpoints of CM and BN, respectively. Let me denote the centers as O1 and O2. Then, O1 is the midpoint of CM, and O2 is the midpoint of BN. The radii of the circles will be half of CM and BN, respectively.Now, let me try to visualize the problem. We have triangle ABC with points M on AB and N on AC. The circles with diameters CM and BN. Their common chord should pass through H, the orthocenter. I need to show that H lies on this common chord.One approach might be to show that H lies on both circles, but that might not necessarily be true. Alternatively, since the common chord is the set of points that are equidistant from both circles, maybe H satisfies that condition. Or perhaps there's a property related to orthocenters and circles with certain diameters.Another thought: maybe using coordinate geometry? Assign coordinates to points A, B, C, then find the equations of the circles and their common chord, then check if H lies on that line. But that might get messy. However, since the problem is general, coordinate geometry could work if I choose a coordinate system that simplifies calculations.Alternatively, synthetic geometry using properties of cyclic quadrilaterals, orthocenters, and midpoints. Let's see.First, recall that in a triangle, the orthocenter has several properties related to reflections and circles. For example, reflecting the orthocenter over a side of the triangle gives a point on the circumcircle. Not sure if that helps here.Wait, maybe considering the circles with diameters CM and BN. If a circle has diameter CM, then any angle subtended by CM is a right angle. So, any point on the circle with diameter CM will form a right angle with points C and M. Similarly for the circle with diameter BN, any point on it forms a right angle with B and N.So, if H is the orthocenter, then the altitudes from the vertices are perpendicular to the opposite sides. Let's denote H as the orthocenter. Then, for example, the altitude from B is BH, which is perpendicular to AC. Similarly, the altitude from C is CH, perpendicular to AB.Wait, since H is the orthocenter, then BH is perpendicular to AC, and CH is perpendicular to AB. Now, points M and N are on AB and AC, respectively. So, maybe there is a relation between H and these points M and N.Alternatively, since the circles have diameters CM and BN, then points C and M are endpoints of a diameter, so the circle with diameter CM includes all points P such that angle CPM is 90 degrees. Similarly, the circle with diameter BN includes all points Q such that angle BQN is 90 degrees.If H is on both circles, then angle CHM and BHN would be right angles. Let me check:Is angle CHM 90 degrees? H is the orthocenter, so CH is perpendicular to AB. Since M is on AB, then CH is perpendicular to AB, so HM is along AB. Wait, maybe not. Let me think.Wait, if H is the orthocenter, then the altitude from C is CH, which is perpendicular to AB. So, CH is perpendicular to AB, and M is a point on AB. So, HM is a segment from H to M on AB. Since CH is perpendicular to AB, then HM is along AB, but H is not on AB unless the triangle is right-angled. So, in general, H is outside AB unless it's a right triangle. Wait, no. If ABC is acute, the orthocenter is inside; if it's obtuse, it's outside. So, depending on the triangle, H can be inside or outside.But regardless, since CH is perpendicular to AB, then the line CH is an altitude. Similarly, BH is perpendicular to AC. So, the altitude from B is BH, which is perpendicular to AC. Since N is on AC, then BH is perpendicular to AC, so HN is along AC? Wait, no. If H is the orthocenter, then BH is perpendicular to AC, so the line BH is an altitude, meeting AC at some foot, say D. Then HD is part of BH, which is perpendicular to AC. But N is another point on AC. So, HN is not necessarily perpendicular unless N is D. But N is an arbitrary point on AC. Wait, but the problem doesn't specify where M and N are; they are just any points on AB and AC. So, maybe the result holds regardless of where M and N are. Interesting.So, regardless of where M and N are on AB and AC, the common chord of the two circles passes through the orthocenter. That suggests that H has a special property related to both circles.Perhaps H lies on both circles? Let me test that. If H lies on the circle with diameter CM, then angle CHM is 90 degrees. Similarly, if H lies on the circle with diameter BN, then angle BHN is 90 degrees. Are these angles necessarily right angles?Let me see. Since H is the orthocenter, then as mentioned earlier, CH is perpendicular to AB. But M is on AB, so HM is a segment from H to M on AB. So, angle CHM is the angle between CH and HM. But since CH is perpendicular to AB, and HM is on AB, then angle CHM is 90 degrees. Wait, yes! Because CH is perpendicular to AB, and HM is along AB (since M is on AB), so the angle between CH and HM is 90 degrees. Therefore, angle CHM is 90 degrees, which implies that H lies on the circle with diameter CM. Similarly, angle BHN: BH is perpendicular to AC, and N is on AC, so HN is along AC. Therefore, angle BHN is 90 degrees, so H lies on the circle with diameter BN. Therefore, H is a common point of both circles. Therefore, H is one of the intersection points of the two circles, hence the common chord passes through H. Therefore, the common chord is the line through H and the other intersection point. Therefore, it must pass through H.Wait, that seems straightforward. If H is on both circles, then it's one of the points of intersection, so the common chord, which is the line through both intersection points, must pass through H.But let me verify this step by step.First, to check if H is on the circle with diameter CM. For this, angle CHM must be 90 degrees. Since CH is the altitude from C to AB, which is perpendicular to AB. M is a point on AB. Then, HM is a segment from H to M on AB. Since AB is a line, and CH is perpendicular to AB, then the angle between CH and HM is 90 degrees. Therefore, triangle CHM has a right angle at H. Therefore, by Thales' theorem, H lies on the circle with diameter CM. Similarly, angle BHN is 90 degrees because BH is the altitude from B to AC, which is perpendicular to AC, and N is on AC. So, triangle BHN has a right angle at H, so H lies on the circle with diameter BN. Therefore, H is a common point of both circles. Therefore, the common chord, which is the line through all common points (in this case, H and another point), must pass through H. Hence, the common chord passes through H.Therefore, the proof is straightforward once we realize that H lies on both circles, making it lie on their common chord.Wait, but maybe the circles intersect at two points, so the common chord is the line through both points, and H is one of them. Therefore, regardless of the other point, the common chord must pass through H. Therefore, the conclusion follows.Is there a possibility that H is the only intersection point? That would mean the circles are tangent at H, but since H is inside the triangle if it's acute, but circles with diameters CM and BN might intersect at two points. Unless CM and BN are arranged in a way that the circles are tangent. But since M and N are arbitrary points on AB and AC, it's possible they could be tangent, but in general, they would intersect at two points, one of which is H. Therefore, the common chord is the line through both points, hence passing through H.Therefore, the key insight is recognizing that H lies on both circles, hence lies on their common chord.But let me double-check if angle CHM is indeed 90 degrees. Since CH is an altitude, it is perpendicular to AB. HM is a segment from H to M on AB. Therefore, HM is along AB, and CH is perpendicular to AB. Therefore, the angle between CH and HM is 90 degrees, so angle CHM is 90 degrees. Therefore, by the converse of Thales' theorem, H lies on the circle with diameter CM. Similarly for the other circle.Yes, that seems correct.Alternatively, if we use coordinate geometry, we could set coordinates for triangle ABC and compute the orthocenter, then check if it lies on both circles.Let me try that approach to confirm.Let’s place triangle ABC in a coordinate system. Let’s assume point A is at (0, 0), B is at (c, 0), and C is at (d, e). Then, AB is from (0,0) to (c,0), AC is from (0,0) to (d,e). Points M on AB and N on AC can be parameterized. Let’s let M be a point on AB: since AB is from (0,0) to (c,0), M can be represented as (m, 0) where 0 ≤ m ≤ c. Similarly, N is on AC, so AC is from (0,0) to (d,e), so N can be represented as (kd, ke) where 0 ≤ k ≤ 1.Now, the orthocenter H of triangle ABC can be found by solving the intersection of the altitudes. The altitude from B to AC: since AC has slope (e - 0)/(d - 0) = e/d, so the altitude from B is perpendicular to AC, so its slope is -d/e. It passes through B (c, 0), so its equation is y = (-d/e)(x - c).Similarly, the altitude from C to AB: AB is horizontal, so the altitude from C is vertical if AB is horizontal. Wait, AB is along the x-axis from (0,0) to (c,0), so its slope is 0. Therefore, the altitude from C is vertical, so it has an undefined slope, passing through C (d, e) and going straight down to AB. But since AB is horizontal, the foot of the altitude from C is (d, 0), so the altitude is the line x = d.Wait, but if AB is from (0,0) to (c,0), then unless d = c or something, the foot of the altitude from C might not be on AB. Wait, no. The altitude from C to AB is the perpendicular line from C to AB. Since AB is horizontal, the altitude is vertical, so it's x = d, but if d is not between 0 and c, then the foot of the altitude is outside AB. Wait, but in a triangle, the foot of the altitude must lie on the side or its extension. If the triangle is obtuse, the orthocenter lies outside.But regardless, let's proceed. The altitude from C is the vertical line x = d, which intersects AB at (d, 0). If d is between 0 and c, then it's on AB; otherwise, it's on the extension.Similarly, the altitude from B is y = (-d/e)(x - c). The orthocenter H is the intersection of these two altitudes. So, substituting x = d into the equation of the altitude from B gives y = (-d/e)(d - c). Therefore, H is at (d, (-d/e)(d - c)) = (d, (d(c - d))/e).Now, let's find the circles with diameters CM and BN. First, find the coordinates of M and N. Let’s take M as (m, 0) on AB, and N as (kd, ke) on AC.The circle with diameter CM: endpoints C (d, e) and M (m, 0). The center is the midpoint of CM: ((d + m)/2, (e + 0)/2) = ((d + m)/2, e/2). The radius is half the distance between C and M. The distance between C and M is sqrt((d - m)^2 + (e - 0)^2) = sqrt((d - m)^2 + e^2), so the radius is (1/2)sqrt((d - m)^2 + e^2).Similarly, the circle with diameter BN: endpoints B (c, 0) and N (kd, ke). The center is the midpoint of BN: ((c + kd)/2, (0 + ke)/2) = ((c + kd)/2, ke/2). The radius is half the distance between B and N: sqrt((c - kd)^2 + (0 - ke)^2) = sqrt((c - kd)^2 + (ke)^2), so radius is (1/2)sqrt((c - kd)^2 + (ke)^2).Now, we need to find the equation of both circles and find their common chord, then check if H lies on that line.First, equation of the circle with diameter CM:Center: ((d + m)/2, e/2)Radius: (1/2)sqrt((d - m)^2 + e^2)The equation is:(x - (d + m)/2)^2 + (y - e/2)^2 = [(1/2)sqrt((d - m)^2 + e^2)]^2Simplifying the right-hand side:(1/4)((d - m)^2 + e^2)So, equation is:(x - (d + m)/2)^2 + (y - e/2)^2 = (1/4)((d - m)^2 + e^2)Similarly, the equation of the circle with diameter BN:Center: ((c + kd)/2, ke/2)Radius: (1/2)sqrt((c - kd)^2 + (ke)^2)Equation:(x - (c + kd)/2)^2 + (y - ke/2)^2 = (1/4)((c - kd)^2 + (ke)^2)Now, to find the common chord, we can subtract the two equations to get the equation of the radical axis, which is the line containing the common chord.Let’s compute the difference between the two circle equations.First, expand both equations.First circle:(x - (d + m)/2)^2 + (y - e/2)^2 = (1/4)((d - m)^2 + e^2)Expanding left side:[x^2 - (d + m)x + ((d + m)/2)^2] + [y^2 - ey + (e/2)^2] = (1/4)((d - m)^2 + e^2)Similarly, second circle:(x - (c + kd)/2)^2 + (y - ke/2)^2 = (1/4)((c - kd)^2 + (ke)^2)Expanding left side:[x^2 - (c + kd)x + ((c + kd)/2)^2] + [y^2 - key + (ke/2)^2] = (1/4)((c - kd)^2 + (ke)^2)Subtract the second equation from the first:[x^2 - (d + m)x + ((d + m)/2)^2 + y^2 - ey + (e/2)^2] - [x^2 - (c + kd)x + ((c + kd)/2)^2 + y^2 - key + (ke/2)^2] = (1/4)((d - m)^2 + e^2) - (1/4)((c - kd)^2 + (ke)^2)Simplify left side:x^2 - (d + m)x + ((d + m)/2)^2 + y^2 - ey + (e/2)^2 - x^2 + (c + kd)x - ((c + kd)/2)^2 - y^2 + key - (ke/2)^2The x^2 and y^2 terms cancel out:- (d + m)x + ((d + m)/2)^2 - ey + (e/2)^2 + (c + kd)x - ((c + kd)/2)^2 + key - (ke/2)^2Combine like terms:[- (d + m)x + (c + kd)x] + [ - ey + key ] + [ ((d + m)/2)^2 + (e/2)^2 - ((c + kd)/2)^2 - (ke/2)^2 ]Factor x terms:x [ - (d + m) + (c + kd) ] + y [ - e + ke ] + [ ((d + m)^2 + e^2 - (c + kd)^2 - (ke)^2 ) / 4 ]Simplify coefficients:For x: - (d + m) + c + kd = c - d - m + kdFor y: - e + ke = e(-1 + k)So, the radical axis equation is:(c - d - m + kd)x + e(-1 + k)y + [ ( (d + m)^2 + e^2 - (c + kd)^2 - (ke)^2 ) / 4 ] = 0This seems complicated, but maybe substituting H into this equation will show that it satisfies the equation.Recall that H has coordinates (d, (d(c - d))/e ). Let's plug x = d and y = d(c - d)/e into the radical axis equation.First, compute each term:Left side:(c - d - m + kd)*d + e(-1 + k)*(d(c - d)/e ) + [ ( (d + m)^2 + e^2 - (c + kd)^2 - (ke)^2 ) / 4 ]Simplify term by term:First term: (c - d - m + kd)*d = d(c - d - m + kd) = d(c - d - m) + kd^2Second term: e*(-1 + k)*(d(c - d)/e ) = (-1 + k)*d(c - d)Third term: [ ( (d + m)^2 + e^2 - (c + kd)^2 - (ke)^2 ) / 4 ]So, combining first and second terms:d(c - d - m) + kd^2 + (-1 + k)*d(c - d)Factor d:d[ (c - d - m) + kd + (-1 + k)(c - d) ]Expand (-1 + k)(c - d):= - (c - d) + k(c - d)Thus, inside the brackets:(c - d - m) + kd - (c - d) + k(c - d)Simplify term by term:(c - d - m) - (c - d) = -mThen, kd + k(c - d) = k(d + c - d) = kcSo, total inside brackets: -m + kcTherefore, first two terms combine to d(-m + kc)Third term: [ ( (d + m)^2 + e^2 - (c + kd)^2 - (ke)^2 ) / 4 ]Let me compute numerator:(d + m)^2 + e^2 - (c + kd)^2 - (ke)^2Expand squares:= (d^2 + 2dm + m^2) + e^2 - (c^2 + 2ckd + k^2d^2) - k^2e^2= d^2 + 2dm + m^2 + e^2 - c^2 - 2ckd - k^2d^2 - k^2e^2Group like terms:= (d^2 - k^2d^2) + (2dm - 2ckd) + (m^2) + (e^2 - k^2e^2) - c^2= d^2(1 - k^2) + 2d(m - ck) + m^2 + e^2(1 - k^2) - c^2Factor:= (1 - k^2)(d^2 + e^2) + 2d(m - ck) + m^2 - c^2Hmm, not sure if this simplifies further.Therefore, the third term is [ (1 - k^2)(d^2 + e^2) + 2d(m - ck) + m^2 - c^2 ] / 4Therefore, the entire left side is:d(-m + kc) + [ (1 - k^2)(d^2 + e^2) + 2d(m - ck) + m^2 - c^2 ] / 4Now, we need to check if this equals zero when substituting H's coordinates.But this seems very involved. Maybe there's a mistake in the coordinate approach, or perhaps it's too tedious. Alternatively, maybe there's a pattern or substitution that can simplify this.Alternatively, let's assume specific coordinates for simplicity. Let me choose coordinates where ABC is a simple triangle, maybe a right-angled triangle to simplify calculations.Let’s assume ABC is a right-angled triangle at A. So, A is at (0,0), B at (1,0), C at (0,1). Then, the orthocenter H of a right-angled triangle is at the right angle vertex, which is A (0,0). Wait, but in a right-angled triangle, the orthocenter is at the right-angle vertex. So, H is at A (0,0). Wait, but in this case, the problem states that the common chord passes through H, which is A. Let me check with this example.Points M on AB and N on AC. Let’s choose M as (m, 0) on AB (from (0,0) to (1,0)), and N as (0, n) on AC (from (0,0) to (0,1)).Now, circles with diameters CM and BN. First, find points C (0,1), M (m,0), B (1,0), N (0,n).Circle with diameter CM: center at midpoint of CM, which is ((0 + m)/2, (1 + 0)/2) = (m/2, 1/2). Radius is half the distance between C and M: (1/2)sqrt((m - 0)^2 + (0 - 1)^2) = (1/2)sqrt(m^2 + 1).Equation: (x - m/2)^2 + (y - 1/2)^2 = (m^2 + 1)/4Similarly, circle with diameter BN: endpoints B (1,0) and N (0,n). Center at midpoint ((1 + 0)/2, (0 + n)/2) = (1/2, n/2). Radius is half the distance between B and N: (1/2)sqrt((1 - 0)^2 + (0 - n)^2) = (1/2)sqrt(1 + n^2).Equation: (x - 1/2)^2 + (y - n/2)^2 = (1 + n^2)/4Now, find the common chord. Subtract the two equations:(x - m/2)^2 + (y - 1/2)^2 - (x - 1/2)^2 - (y - n/2)^2 = (m^2 + 1)/4 - (1 + n^2)/4Expand both sides:Left side:[x^2 - m x + (m^2)/4 + y^2 - y + 1/4] - [x^2 - x + 1/4 + y^2 - n y + (n^2)/4]Simplify term by term:x^2 - m x + m²/4 + y² - y + 1/4 - x² + x - 1/4 - y² + n y - n²/4The x² and y² terms cancel:- m x + m²/4 - y + 1/4 + x - 1/4 + n y - n²/4Combine like terms:(-m x + x) + (- y + n y) + (m²/4 - n²/4)Factor:x(1 - m) + y(n - 1) + (m² - n²)/4Right side:(m² + 1 - 1 - n²)/4 = (m² - n²)/4Therefore, the equation becomes:x(1 - m) + y(n - 1) + (m² - n²)/4 = (m² - n²)/4Subtracting (m² - n²)/4 from both sides:x(1 - m) + y(n - 1) = 0Thus, the radical axis is the line x(1 - m) + y(n - 1) = 0.Now, check if H lies on this line. In our coordinate system, H is at A (0,0). Substituting x=0, y=0 into the equation:0*(1 - m) + 0*(n - 1) = 0, which is 0 = 0. True. Therefore, the common chord passes through H, which is (0,0) in this case.Therefore, in this specific case, the radical axis passes through H. So, this example supports the general proof.Moreover, in this case, since H is the orthocenter (at A in a right-angled triangle), and we see that regardless of m and n, the radical axis passes through H. Therefore, this gives more confidence in the general proof.Alternatively, in another example, let's take an acute triangle where H is inside.Suppose ABC is an equilateral triangle with coordinates A (0,0), B (2,0), C (1, √3). The orthocenter of an equilateral triangle is at the centroid, which is ( (0 + 2 + 1)/3, (0 + 0 + √3)/3 ) = (1, √3/3).Points M on AB and N on AC. Let’s choose M as the midpoint of AB: (1, 0). Let’s choose N as the midpoint of AC: (0.5, √3/2).Now, circles with diameters CM and BN.First, find points C (1, √3), M (1, 0). The circle with diameter CM: center at midpoint of CM: ( (1 + 1)/2, (√3 + 0)/2 ) = (1, √3/2). Radius is half the distance between C and M: distance CM is sqrt( (1 - 1)^2 + (√3 - 0)^2 ) = √3, so radius is √3/2.Equation: (x - 1)^2 + (y - √3/2)^2 = (3/4)Circle with diameter BN: endpoints B (2,0) and N (0.5, √3/2). Midpoint is ( (2 + 0.5)/2, (0 + √3/2)/2 ) = (1.25, √3/4). Distance BN: sqrt( (2 - 0.5)^2 + (0 - √3/2)^2 ) = sqrt( (1.5)^2 + ( (√3)/2 )^2 ) = sqrt(2.25 + 0.75 ) = sqrt(3), so radius is sqrt(3)/2.Equation: (x - 1.25)^2 + (y - √3/4)^2 = 3/4Now, find the radical axis. Subtract the two equations.First equation: (x - 1)^2 + (y - √3/2)^2 = 3/4Second equation: (x - 1.25)^2 + (y - √3/4)^2 = 3/4Subtract second from first:[(x - 1)^2 - (x - 1.25)^2] + [(y - √3/2)^2 - (y - √3/4)^2] = 0Expand each term.First term:(x - 1)^2 - (x - 1.25)^2= [x² - 2x + 1] - [x² - 2.5x + 1.5625]= x² - 2x + 1 - x² + 2.5x - 1.5625= 0.5x - 0.5625Second term:(y - √3/2)^2 - (y - √3/4)^2= [y² - √3 y + 3/4] - [y² - (√3/2)y + 3/16]= y² - √3 y + 3/4 - y² + (√3/2)y - 3/16= (-√3 y + (√3/2)y ) + (3/4 - 3/16 )= (-√3/2 y) + (12/16 - 3/16 )= (-√3/2 y) + (9/16 )Combine both terms:0.5x - 0.5625 - √3/2 y + 9/16 = 0Convert 0.5625 to 9/16:0.5x - 9/16 - √3/2 y + 9/16 = 0Simplify:0.5x - √3/2 y = 0Multiply both sides by 2:x - √3 y = 0So, radical axis is x = √3 y.Now, check if H (1, √3/3) lies on this line. Substitute x = 1, y = √3/3:1 = √3*(√3/3) = (3)/3 = 1. Yes, it satisfies. Therefore, the radical axis passes through H.Therefore, even in an acute triangle, the common chord passes through the orthocenter. This further confirms the general proof.Therefore, both the synthetic approach and coordinate examples support the conclusion that the common chord passes through the orthocenter because the orthocenter lies on both circles, making it a point on their radical axis (common chord).I think this solidifies the proof. The key insight is recognizing that the orthocenter lies on both circles due to the right angles formed with the diameters, and thus lies on their common chord.

✅:To prove that the common chord of the circles with diameters ( CM ) and ( BN ) passes through the orthocenter ( H ) of ( triangle ABC ), follow these steps:1. Identify Key Properties: - The orthocenter ( H ) is the intersection of the altitudes of ( triangle ABC ). - A circle with diameter ( XY ) implies that any point ( P ) on the circle satisfies ( angle XPY = 90^circ ).2. Show ( H ) Lies on Both Circles: - Circle with Diameter ( CM ): - ( CH ) is an altitude, so ( CH perp AB ). Since ( M ) is on ( AB ), ( HM ) lies on ( AB ). - Thus, ( angle CHM = 90^circ ), so ( H ) lies on the circle with diameter ( CM ) (by Thales' theorem). - Circle with Diameter ( BN ): - ( BH ) is an altitude, so ( BH perp AC ). Since ( N ) is on ( AC ), ( HN ) lies on ( AC ). - Thus, ( angle BHN = 90^circ ), so ( H ) lies on the circle with diameter ( BN ) (by Thales' theorem).3. Conclude Common Chord Passes Through ( H ): - The common chord of two circles is the radical axis, which consists of all points that lie on both circles. - Since ( H ) is a common point of both circles, it lies on their radical axis (common chord). - Therefore, the common chord must pass through ( H ).Final Answer:The common chord of the two circles with diameters ( CM ) and ( BN ) passes through the orthocenter of ( triangle ABC ) because the orthocenter ( H ) lies on both circles, making it a point on their radical axis. Thus, the common chord necessarily includes ( H ).(boxed{text{The common chord passes through the orthocenter of } triangle ABC.})

🔑:1. Introduction and Setup: Let’s consider a triangle ( triangle ABC ) with points ( M ) and ( N ) chosen such that ( M ) lies on line ( AB ) and ( N ) lies on line ( AC ). We need to demonstrate that the common chord of the circles with diameters ( CM ) and ( BN ) passes through the orthocenter of ( triangle ABC ).2. Construct Third Circle: To proceed with the proof, introduce a third circle with diameter ( BC ). This circle is significant because it connects the midpoints of segments ( B ) and ( C ) to the triangle's vertices.3. Identify Heights as Chords: The key observation is to note where the heights (altitudes) of ( triangle ABC ) intersect the corresponding sides: - The height from ( B ) to ( AC ) is a chord of the circle with diameter ( BC ). - The height from ( C ) to ( AB ) is also a chord of the circle with diameter ( BC ).4. Apply Known Result: Referring to the previously solved problem (Problem 187), it has been established that the common chord of the circles intersects at the orthocenter when the diameters are formed by the triangle's altitudes. Specifically: - The common chords of the first and third circles (with diameters ( CM ) and ( BC )) include the height from ( B ) to ( AC ). - Similarly, the common chords of the second and third circles (with diameters ( BN ) and ( BC )) include the height from ( C ) to ( AB ). By leveraging these known results, the common chord of the circles with diameters ( CM ) and ( BN ) must pass through the orthocenter of ( triangle ABC ).5. Final Assertion: This follows directly from the geometric properties demonstrated and applying the orthogonal intersections with the diameters yielding our desired results.Conclusion:Since all construction and results robustly affirm that the common chord of the circles defined by diameters ( CM ) and ( BN ) passes through the orthocenter of ( triangle ABC ):[ boxed{text{Common chord of the two circles passes through the orthocenter of } triangle ABC} ]

🧠:Okay, so I need to solve this geometry problem about a pyramid P-ABCD. Let me start by visualizing the pyramid. The base ABCD is a rectangle, and PA is perpendicular to the base. The given lengths are AB = √3, BC = 1, and PA = 2. Point E is the midpoint of PD. The problem has two parts: first, finding the cosine of the angle between lines AC and PB, and second, finding a point N in face PAB such that NE is perpendicular to the plane PAC, and then calculating the distances from N to AB and AP.Starting with part 1: the angle between AC and PB. Since we need the cosine of the angle between two lines, I think we can use the dot product formula. But first, I need coordinates for all the points to find the vectors AC and PB. Let me set up a coordinate system.Let’s place the base ABCD on the xy-plane with point A at the origin (0, 0, 0). Since ABCD is a rectangle, AB and AD are the sides. Given AB = √3 and BC = 1, so AB is along the x-axis from (0,0,0) to (√3, 0, 0), and BC is along the y-axis from (√3, 0, 0) to (√3, 1, 0). Then AD would be along the y-axis from (0,0,0) to (0, 1, 0), but wait, BC is the other side. Wait, ABCD is a rectangle, so AB and BC are adjacent sides. So if AB is length √3, BC is length 1, then the sides are AB and BC. So ABCD would have coordinates:A: (0, 0, 0)B: (√3, 0, 0)C: (√3, 1, 0)D: (0, 1, 0)Because moving from A to B along x-axis, then from B to C along y-axis, then from C to D along negative x-axis, and from D to A along negative y-axis. Wait, but in a rectangle, opposite sides are equal. So AB = CD = √3, and BC = AD = 1. That makes sense.PA is perpendicular to the base, so PA is along the z-axis. Since PA = 2, point P is at (0, 0, 2).Now, E is the midpoint of PD. Point D is (0, 1, 0), so PD goes from P(0,0,2) to D(0,1,0). The midpoint E would be the average of the coordinates: ((0+0)/2, (0+1)/2, (2+0)/2) = (0, 0.5, 1).Wait, let me check that again. PD is from P(0,0,2) to D(0,1,0). So midpoint E is ((0+0)/2, (0+1)/2, (2+0)/2) = (0, 0.5, 1). Yes, that's correct.Now, part 1: angle between AC and PB. Let me find vectors AC and PB.Vector AC is from A(0,0,0) to C(√3,1,0). So the vector is (√3, 1, 0).Vector PB is from P(0,0,2) to B(√3,0,0). So the vector is (√3 - 0, 0 - 0, 0 - 2) = (√3, 0, -2).The angle θ between AC and PB can be found using the dot product formula:cosθ = (AC · PB) / (|AC| |PB|)First compute the dot product:AC · PB = (√3)(√3) + (1)(0) + (0)(-2) = 3 + 0 + 0 = 3.Compute |AC|: sqrt((√3)^2 + 1^2 + 0^2) = sqrt(3 + 1) = sqrt(4) = 2.Compute |PB|: sqrt((√3)^2 + 0^2 + (-2)^2) = sqrt(3 + 0 + 4) = sqrt(7).Therefore, cosθ = 3 / (2 * sqrt(7)) = 3/(2√7). Rationalizing the denominator: (3√7)/(2*7) = (3√7)/14. But maybe they just want it as 3/(2√7). Wait, the question asks for the cosine of the angle, so either form is acceptable, but likely they prefer rationalized. Let me check:3/(2√7) is equivalent to 3√7/14. So either way. Maybe the answer is 3√7/14. Wait, let me compute that. 3 divided by 2√7 times √7/√7 is 3√7/14. So yes, that's correct. So the cosine is 3√7/14.Wait, but let me double-check the vectors. AC is from A to C: (√3,1,0). PB is from P to B: (√3,0,-2). The dot product is indeed √3*√3 + 1*0 + 0*(-2) = 3 + 0 + 0 = 3. The magnitudes are correct. So yes, cosθ = 3/(2*sqrt(7)) = 3√7/14. So part 1 answer is 3√7/14.Moving on to part 2: In face PAB, find a point N such that NE is perpendicular to plane PAC. Then find distances from N to AB and AP.First, let me understand the face PAB. Face PAB is the triangle connecting points P(0,0,2), A(0,0,0), and B(√3,0,0). So this is a vertical triangle in the x-z plane (since y=0 for all three points).Point N is somewhere on this face. So coordinates of N must satisfy the equation of the plane PAB. Since PAB is the plane y=0, but since all three points have y=0, so any point on PAB must have y=0. Wait, but in the problem statement, it's specified that N is in face PAB, so N must lie on the triangular face PAB. Therefore, coordinates of N can be parametrized. Let me think about parametrizing point N.Alternatively, since the face PAB is a triangle, we can express N in terms of parameters. Since PAB is a triangle with vertices at (0,0,2), (0,0,0), and (√3,0,0). So this is a triangle in the x-z plane (y=0). So any point N on PAB can be written as (x, 0, z), where x and z satisfy the plane equation of PAB. Wait, but PAB is a triangle, so the coordinates of N must lie within the triangle. So parametrizing using barycentric coordinates or using parameters along edges.Alternatively, express N as a combination of points. Let me think. Since N is in PAB, it can be expressed as a linear combination with weights that sum to 1. But since it's a face in 3D, maybe better to use coordinates.Alternatively, since the face PAB is the set of points that can be written as P + s*(A - P) + t*(B - P), where s and t are in [0,1] and s + t ≤ 1.But maybe easier to parametrize using two variables. Let's see. Let’s consider parametric equations for the face PAB.Since P is (0,0,2), A is (0,0,0), B is (√3,0,0). The vectors PA and PB are from P to A and P to B. Wait, but in the face PAB, any point can be written as P + u*(A - P) + v*(B - P), where u, v ≥ 0 and u + v ≤ 1.But since A - P is (0,0, -2) and B - P is (√3,0,-2). So a general point N on PAB would be (0 + √3*v, 0, 2 - 2u -2v). Wait, that seems a bit complicated. Maybe another approach.Alternatively, since the face PAB is a triangle with vertices at P(0,0,2), A(0,0,0), and B(√3,0,0). So in the x-z plane, y=0. So any point on PAB can be expressed as (x, 0, z) where x and z satisfy the plane equation of the triangle. Since the triangle is between points (0,0,2), (0,0,0), and (√3,0,0). So the plane is y=0, but within the triangle, x ranges from 0 to √3, z ranges from 0 to 2, but in such a way that the points lie on the triangle. So parametrizing in terms of x and z.Alternatively, using parameters. Let me consider moving from P to A to B. Let me parameterize N as follows: Since PA is vertical from (0,0,2) to (0,0,0), and PB is the line from (0,0,2) to (√3,0,0). So the face PAB can be parametrized by two parameters, say, moving along PA and PB.But perhaps a better approach is to use coordinates. Let me assume that point N has coordinates (x, 0, z) where y=0 (since it's on face PAB). Now, we need to find such a point N such that vector NE is perpendicular to plane PAC.First, let's find the equation of plane PAC. Plane PAC is the plane containing points P(0,0,2), A(0,0,0), and C(√3,1,0). Wait, point C is (√3,1,0). So three points: P(0,0,2), A(0,0,0), and C(√3,1,0). Let me find two vectors on this plane. Vector PA is A - P = (0,0,0) - (0,0,2) = (0,0,-2). Vector PC is C - P = (√3,1,0) - (0,0,2) = (√3,1,-2).The normal vector to plane PAC can be found by taking the cross product of PA and PC.PA = (0,0,-2)PC = (√3,1,-2)Cross product PA × PC:|i j k||0 0 -2||√3 1 -2|= i*(0*(-2) - (-2)*1) - j*(0*(-2) - (-2)*√3) + k*(0*1 - 0*√3)= i*(0 + 2) - j*(0 + 2√3) + k*(0 - 0)= 2i - 2√3 j + 0kSo the normal vector is (2, -2√3, 0). We can simplify by dividing by 2: (1, -√3, 0).Therefore, the equation of plane PAC is 1(x - 0) - √3(y - 0) + 0(z - 2) = 0, which simplifies to x - √3 y = 0.Wait, let me check. The general plane equation is a(x - x0) + b(y - y0) + c(z - z0) = 0, where (a,b,c) is the normal vector and (x0,y0,z0) is a point on the plane. Let's take point P(0,0,2). So plugging into the normal vector (1, -√3, 0):1*(x - 0) - √3*(y - 0) + 0*(z - 2) = 0 → x - √3 y = 0. So plane PAC is x - √3 y = 0.Wait, but let's verify with point A(0,0,0): x - √3 y = 0 - 0 = 0, which satisfies. Point C(√3,1,0): √3 - √3*1 = 0, which also satisfies. Point P(0,0,2): 0 - √3*0 = 0, which satisfies. So yes, the equation is x - √3 y = 0.Now, NE is perpendicular to plane PAC. Since the normal vector of plane PAC is (1, -√3, 0), then vector NE must be parallel to this normal vector. Therefore, vector NE is a scalar multiple of (1, -√3, 0).Point E is (0, 0.5, 1). Point N is (x, 0, z) on face PAB. So vector NE is E - N = (0 - x, 0.5 - 0, 1 - z) = (-x, 0.5, 1 - z).This vector NE must be parallel to the normal vector (1, -√3, 0). So there exists a scalar k such that:(-x, 0.5, 1 - z) = k*(1, -√3, 0)Which gives the system of equations:1. -x = k*1 → k = -x2. 0.5 = k*(-√3)3. 1 - z = k*0 → 1 - z = 0 → z = 1From equation 3: z = 1.From equation 2: 0.5 = k*(-√3) → k = -0.5 / √3 = -1/(2√3) = -√3 / 6 (rationalized).From equation 1: k = -x → -x = -√3/6 → x = √3/6.Therefore, coordinates of N are (x, 0, z) = (√3/6, 0, 1).Now, we need to verify that N is indeed on face PAB. Since PAB is the triangle with vertices P(0,0,2), A(0,0,0), and B(√3,0,0). The coordinates of N are (√3/6, 0, 1). Since y=0, it's on the plane of PAB. Let's check if it's inside the triangle.Parametrizing the face PAB, we can use barycentric coordinates or check if the point is within the edges. Let's see: since N is at x=√3/6 (~0.288) and z=1. The face PAB in the x-z plane goes from P(0,0,2) to A(0,0,0) to B(√3,0,0). The point (√3/6,0,1) lies along the line connecting P to B?Wait, the line from P(0,0,2) to B(√3,0,0) can be parametrized as x = √3*t, y=0, z=2 - 2t, where t ∈ [0,1]. Let's see if (√3/6,0,1) is on this line.Set x = √3*t = √3/6 → t = 1/6.Then z = 2 - 2*(1/6) = 2 - 1/3 = 5/3 ≈ 1.666, which is not 1. So N is not on PB.Alternatively, the line from A(0,0,0) to B(√3,0,0) is along x-axis, z=0. But N has z=1, so not on AB.Alternatively, the line from P(0,0,2) to A(0,0,0) is along z-axis, x=0. N has x=√3/6, so not on PA.Therefore, N is inside the face PAB, not on any of the edges. To confirm, perhaps parameterize the face.Alternatively, express N as a combination of points. Let me see, in terms of PA and PB.But since the coordinates are (√3/6, 0,1), and since in the face PAB, all points have y=0, and x and z coordinates must satisfy the triangle's constraints. Since from P(0,0,2) to B(√3,0,0), any point on the line PB is (√3t, 0, 2-2t). Similarly, from A(0,0,0) to B(√3,0,0), it's (√3s,0,0) where s ∈ [0,1]. The face PAB can be parametrized by points that are combinations of these.Alternatively, since N is inside the face, perhaps express it as a linear combination. But maybe it's sufficient that N's coordinates (√3/6,0,1) satisfy the face PAB's existence. Since in x-z plane, between x=0 to √3 and z=0 to 2, and within the triangle. Since x=√3/6 ≈ 0.288, z=1. If we imagine the triangle, this point is somewhere inside.Now, we need to confirm that NE is perpendicular to plane PAC. Vector NE is (-√3/6, 0.5, 0). Wait, let me compute NE again.Point N is (√3/6,0,1). Point E is (0,0.5,1). So vector NE is E - N = (0 - √3/6, 0.5 - 0, 1 - 1) = (-√3/6, 0.5, 0). The normal vector of plane PAC is (1, -√3, 0). So check if NE is a scalar multiple of (1, -√3, 0).Let's see: if (-√3/6, 0.5, 0) = k*(1, -√3, 0). Then:-√3/6 = k*1 → k = -√3/60.5 = k*(-√3) → 0.5 = (-√3/6)*(-√3) = (√3*√3)/6 = 3/6 = 0.5. Correct.Third component 0 = 0. So yes, NE is parallel to the normal vector, hence NE is perpendicular to plane PAC. Therefore, point N is correctly found as (√3/6,0,1).Now, the problem asks for the distances from N to AB and AP.First, distance from N to AB. AB is the edge from A(0,0,0) to B(√3,0,0). Since AB is along the x-axis at y=0, z=0. The distance from point N(√3/6,0,1) to line AB.But wait, AB is a line in 3D space. Since AB lies on the x-axis from (0,0,0) to (√3,0,0). The distance from N to AB can be calculated as the perpendicular distance from N to AB.But since AB is along the x-axis, the distance from a point (x,y,z) to AB is sqrt(y^2 + z^2). Because the closest point on AB is (x,0,0), so the difference vector is (0, y, z), hence the distance is sqrt(y² + z²). For point N(√3/6, 0,1), y=0, z=1, so distance is sqrt(0 + 1²) = 1. Wait, that seems too easy.Wait, but N is (√3/6,0,1). The line AB is in the x-axis. The distance from N to AB is the distance from (√3/6,0,1) to the x-axis. The formula for the distance from a point to the x-axis is sqrt(y² + z²). Since y=0 and z=1, distance is 1. So yes, distance from N to AB is 1.Alternatively, using the formula for distance from point to line in 3D:The distance from point P(x0,y0,z0) to line through points A(x1,y1,z1) and B(x2,y2,z2) is |(AP × AB)| / |AB|.Here, line AB is from A(0,0,0) to B(√3,0,0). Vector AB is (√3,0,0). Point N is (√3/6,0,1). Vector AN is (√3/6,0,1). Cross product AN × AB:|i j k||√3/6 0 1||√3 0 0|= i*(0*0 - 1*0) - j*(√3/6*0 - 1*√3) + k*(√3/6*0 - 0*√3)= i*0 - j*(-√3) + k*0 = √3 jMagnitude of cross product is √(0² + (√3)² + 0²) = √3.Magnitude of AB is √( (√3)^2 + 0 + 0 ) = √3.Therefore, distance is √3 / √3 = 1. So yes, distance from N to AB is 1.Next, distance from N to AP. AP is the edge from A(0,0,0) to P(0,0,2). AP is along the z-axis. The distance from N(√3/6,0,1) to line AP.Similarly, the formula for distance from point to line AP. Since AP is along the z-axis from (0,0,0) to (0,0,2). The distance from point (x,y,z) to the z-axis is sqrt(x² + y²). For N(√3/6,0,1), x=√3/6, y=0. So distance is sqrt( (√3/6)^2 + 0 ) = √3/6.Alternatively, using the cross product method. Line AP is from A(0,0,0) to P(0,0,2). Vector AP is (0,0,2). Point N(√3/6,0,1). Vector AN is (√3/6,0,1). Cross product AN × AP:|i j k||√3/6 0 1||0 0 2|= i*(0*2 - 1*0) - j*(√3/6*2 - 1*0) + k*(√3/6*0 - 0*0)= i*0 - j*( (2√3)/6 ) + k*0 = - (√3/3) jMagnitude of cross product is √(0 + ( (√3/3)^2 ) + 0 ) = √(1/3) = 1/√3.Magnitude of AP is 2. So distance is (1/√3)/2 = 1/(2√3) = √3/6. So same result. Therefore, distance from N to AP is √3/6.Therefore, the distances are 1 and √3/6.Let me recap:1. For the first part, cosine of the angle between AC and PB is 3√7/14.2. Point N is (√3/6, 0, 1), distances to AB and AP are 1 and √3/6.I need to confirm once more that N is indeed on face PAB. Since face PAB is the triangle with vertices at (0,0,2), (0,0,0), and (√3,0,0). The point (√3/6,0,1) has x=√3/6 (~0.288), z=1. Since x is between 0 and √3 (~1.732), and z is between 0 and 2, and y=0, it's within the bounds. To check if it's inside the triangle, we can use barycentric coordinates or check the parametric equations.Parametrizing the face PAB as a combination of PA and PB edges. If we consider parameters u and v such that N = u*A + v*B + (1 - u - v)*P, but since N has coordinates (√3/6,0,1), let's solve for u and v.But maybe it's better to check if the point is inside the triangle by verifying that it can be expressed as a convex combination. Let's consider the line from P to B: parametric equation is P + t*(B - P) = (0,0,2) + t*(√3,0,-2), which gives (√3 t, 0, 2 - 2t). If we set this equal to N(√3/6,0,1), then:√3 t = √3/6 → t = 1/62 - 2t = 1 → 2 - 2*(1/6) = 2 - 1/3 = 5/3 ≈ 1.666, which is not 1. So N is not on PB.Alternatively, check if N is on the plane PAB. Since face PAB is the triangle, all points on it satisfy y=0 and can be expressed as a combination of the three vertices. Since N has y=0 and coordinates (√3/6,0,1), it's inside the triangle because x and z are between the extreme values. For example, when moving from P(0,0,2) down to A(0,0,0) and across to B(√3,0,0), the point (√3/6,0,1) is somewhere in the middle.Alternatively, project N onto the x-z plane. The triangle PAB in x-z plane has vertices at (0,2), (0,0), and (√3,0). The point (√3/6,1) is inside this triangle. To confirm, in the x-z plane, the triangle is formed by the points (0,2), (0,0), and (√3,0). The line from (0,2) to (√3,0) is z = (-2/√3)x + 2. For x=√3/6, z = (-2/√3)(√3/6) + 2 = (-2/6) + 2 = (-1/3) + 2 = 5/3 ≈ 1.666. But N has z=1, which is below this line. Hence, N is actually outside the triangle? Wait, that's a problem.Wait, if in the x-z plane, the triangle PAB has vertices at P(0,2), A(0,0), B(√3,0). The point N is (√3/6,1). Plotting this, the line from P(0,2) to B(√3,0) is z = (-2/√3)x + 2. At x=√3/6 ≈ 0.288, z= (-2/√3)*(√3/6) + 2 = (-2/6) + 2 = (-1/3) + 2 = 5/3 ≈ 1.666. But N is at z=1, which is below that line. Therefore, in the x-z plane, N is below the line PB, which means it's outside the triangle PAB. Wait, that can't be. Did I make a mistake?Wait, this is critical. If N is supposed to be on face PAB, but according to the coordinates, in the x-z plane projection, it's outside the triangle. That suggests an error in calculation.Wait, maybe the parametrization is different. Let me think again. Wait, in 3D space, face PAB is a triangle with points P(0,0,2), A(0,0,0), B(√3,0,0). So in 3D, the face is not planar in x-z plane? Wait, no. Since all three points have y=0, the face PAB is indeed in the y=0 plane, so the x-z plane. So the face is the triangle with vertices at (0,2), (0,0), and (√3,0) in x-z coordinates.The point N is (√3/6,1) in x-z. The line from P(0,2) to B(√3,0) is z = - (2/√3)x + 2. At x=√3/6, z= - (2/√3)(√3/6) + 2 = - (2/6) + 2 = -1/3 + 2 = 5/3 ≈ 1.666. So the point N(√3/6,1) is below this line, hence outside the triangle. Therefore, this is a problem. That means the earlier assumption that N is inside the face PAB is incorrect. There's a mistake here.Wait, how is that possible? We derived N as (√3/6,0,1), but this point is outside the face PAB. That means there's an error in the calculation. Let's go back.When we found point N as (√3/6,0,1), we assumed that NE is perpendicular to plane PAC. But if this point is not on face PAB, then our solution is invalid.Wait, where did the mistake happen? Let me check the cross product and the equations again.We have vector NE = E - N = (0 - x, 0.5 - 0, 1 - z). We set this equal to k*(1, -√3, 0). So:- x = k0.5 = -√3 k1 - z = 0 → z = 1From 0.5 = -√3 k → k = -0.5 / √3 = -1/(2√3) = -√3/6Then from -x = k → x = -k = √3/6So N is (√3/6, 0, 1). But as we saw, this point is outside the triangle PAB in the x-z plane. That means there's a contradiction. Therefore, the mistake must be in the assumption that such a point N exists on face PAB with NE perpendicular to plane PAC. But the problem states that such a point exists, so our error is elsewhere.Wait, perhaps I made a mistake in the normal vector of plane PAC. Let me recalculate the normal vector.Plane PAC has points P(0,0,2), A(0,0,0), and C(√3,1,0).Vectors PA = A - P = (0,0,-2)Vector PC = C - P = (√3,1,-2)Cross product PA × PC:|i j k||0 0 -2||√3 1 -2|= i*(0*(-2) - (-2)*1) - j*(0*(-2) - (-2)*√3) + k*(0*1 - 0*√3)= i*(0 + 2) - j*(0 + 2√3) + k*(0 - 0)= 2i - 2√3 j + 0kYes, that's correct. So normal vector is (2, -2√3, 0) or (1, -√3, 0). So plane equation x - √3 y = 0, which is correct.Then vector NE must be parallel to (1, -√3, 0). So vector NE = (-x, 0.5, 1 - z) = k*(1, -√3, 0). As before, leading to x=√3/6, z=1. But this point is outside face PAB. Therefore, there's a contradiction.Wait, but maybe the face PAB is not the entire y=0 plane, but just the triangle. So even though N is in y=0, it has to be within the triangle. But according to our calculation, N is outside. That suggests an error in the approach.Alternative approach: maybe using projections or considering parametric equations for the face PAB.Let me parameterize the face PAB. Since it's a triangle with vertices P(0,0,2), A(0,0,0), and B(√3,0,0). Any point on PAB can be expressed as a combination of two parameters. Let’s use parameters s and t such that N = A + s*(B - A) + t*(P - A), where s ≥ 0, t ≥ 0, and s + t ≤ 1.But B - A = (√3,0,0), P - A = (0,0,2). So coordinates of N would be (√3 s, 0, 2 t). With s ≥ 0, t ≥ 0, s + t ≤ 1.So N is (√3 s, 0, 2 t). Now, we need to find s and t such that NE is perpendicular to plane PAC. As before, vector NE = E - N = (0 - √3 s, 0.5 - 0, 1 - 2 t) = (-√3 s, 0.5, 1 - 2 t). This must be parallel to the normal vector (1, -√3, 0). So:(-√3 s, 0.5, 1 - 2 t) = k*(1, -√3, 0)Which gives:-√3 s = k0.5 = -√3 k1 - 2 t = 0 → t = 0.5From 0.5 = -√3 k → k = -0.5 / √3 = -1/(2√3)From -√3 s = k → s = -k / √3 = -(-1/(2√3)) / √3 = 1/(2*3) = 1/6Thus, s = 1/6, t = 0.5. Check s + t = 1/6 + 1/2 = 1/6 + 3/6 = 4/6 = 2/3 ≤ 1. So valid.Therefore, N is (√3*(1/6), 0, 2*(0.5)) = (√3/6, 0, 1). Same coordinates as before. But according to the parametrization, s=1/6, t=0.5, which sum to 2/3, so it's inside the triangle. Wait, but earlier when projecting onto x-z plane, it seemed outside. What's the issue?Ah, the mistake was in the projection. The parametrization s and t with N = (√3 s, 0, 2 t) where s + t ≤ 1. Here, s=1/6, t=1/2, so s + t = 1/6 + 1/2 = 2/3 ≤ 1. Therefore, it is inside the face PAB. The earlier confusion was due to interpreting the triangle in x-z plane as having vertices at (0,2), (0,0), (√3,0), which is correct, but the line from (0,2) to (√3,0) is z = - (2/√3)x + 2. However, when s + t ≤ 1, the coordinates are (√3 s, 0, 2 t). So for each s and t such that s + t ≤1, x=√3 s, z=2 t. Therefore, substituting t = 1 - s - u (but perhaps a better way). Let me express t in terms of s.If we set s + t = 1 - u, where u ≥ 0. Alternatively, since the parametrization allows any s and t such that s + t ≤1, the region in x-z plane is all points (√3 s, 2 t) where s and t are non-negative and s + t ≤1. The boundary s + t =1 corresponds to the line from B(√3,0,0) to P(0,0,2). For example, when s=1, t=0: (√3,0,0). When s=0, t=1: (0,0,2). When s=0.5, t=0.5: (√3/2, 0, 1). So the line in x-z plane is x=√3 s, z=2(1 - s). So s varies from 0 to1, t=1 -s. Therefore, the line from P(0,2) to B(√3,0) in x-z plane is parameterized by s from 0 to1: x=√3 s, z=2(1 - s). So, for example, when s=1/6, x=√3/6, z=2*(1 -1/6)=2*(5/6)=5/3≈1.666, but our point N is at z=1, which is below this line.But according to the parametrization N = (√3 s, 0, 2 t) with s=1/6, t=0.5. Here, s + t =1/6 +1/2=2/3 ≤1, so it's inside the triangle. How is this possible?Wait, perhaps the confusion arises from interpreting the parametrization. The face PAB is a triangle in 3D space, but when projected onto the x-z plane, it's a different triangle. Wait, no. In this case, the face PAB is in the y=0 plane, so the x-z coordinates are as given. The triangle has vertices at (0,2), (0,0), and (√3,0). The parametrization using s and t where N=(√3 s, 0, 2 t) with s + t ≤1 covers all points within the triangle. Therefore, the point (√3/6,1) in x-z is inside the triangle. How?Wait, because when s=1/6 and t=1/2, s + t=2/3 ≤1, so it is inside. But when projecting onto the x-z plane, the line from P(0,2) to B(√3,0) is the boundary where s + t=1. So points inside have s + t <1. The point N has s + t=2/3 <1, so it's inside the triangle. Therefore, my previous conclusion that it's outside was incorrect. The line in x-z plane for s + t=1 is x=√3 s, z=2(1 - s). For s=1/6, that would be x=√3/6, z=2*(1 -1/6)=5/3. So the point N(√3/6,1) is below this line because 1 <5/3≈1.666. However, according to the parametrization, the point is inside the triangle. This seems contradictory.Wait, I think the confusion is between the 3D face and its projection. In the 3D face PAB, the point N is inside because it's a convex combination with s + t=2/3. However, when projected onto the x-z plane, the triangle appears as the triangle with vertices at (0,2), (0,0), (√3,0). The parametrization using s and t covers this triangle. So the coordinates (√3 s, 2 t) with s + t ≤1. But this seems to suggest that z=2t and x=√3 s, so when s and t vary, x and z are related by x=√3 s, z=2 t, and s + t ≤1. Therefore, substituting t ≤1 -s, we have z=2 t ≤2(1 -s) =2 -2s. But x=√3 s ⇒ s=x/√3. So z ≤2 -2*(x/√3). Therefore, the upper boundary in x-z plane is z=2 - (2/√3)x. The point N is (√3/6,1). Let's check if 1 ≤2 - (2/√3)(√3/6)=2 - (2/6)=2 -1/3=5/3≈1.666. Yes, 1 ≤5/3, so the point is below the upper boundary, hence inside the triangle. Therefore, my previous mistake was thinking it's outside, but actually it's inside.Therefore, the coordinates of N are correct and it's inside the face PAB.To further confirm, let's plot the triangle in x-z plane:- Point P: (0,2)- Point A: (0,0)- Point B: (√3,0)The triangle is formed by connecting these three points. The upper edge from P to B is the line z= - (2/√3)x +2. The lower edges are from A to B (z=0) and from A to P (x=0, z from 0 to2).The point N is at (√3/6,1). This lies above z=0 and below the upper edge z= - (2/√3)x +2, which at x=√3/6 is z=5/3≈1.666. Since 1 <5/3, the point is inside the triangle.Therefore, N is indeed on face PAB. My earlier confusion was due to miscalculating the position, but with the correct parametrization, it's confirmed to be inside.Thus, the answers for part 2 are:- Distance from N to AB: 1- Distance from N to AP: √3/6So summarizing:1. The cosine of the angle between AC and PB is 3√7/14.2. The distances from N to AB and AP are 1 and √3/6 respectively.Final Answer1. The cosine of the angle is boxed{dfrac{3sqrt{7}}{14}}.2. The distances from point ( N ) to ( AB ) and ( AP ) are boxed{1} and boxed{dfrac{sqrt{3}}{6}} respectively.

✅:To solve the given problem about the pyramid ( P-ABCD ):1. Finding the cosine of the angle between lines ( AC ) and ( PB ): - Set up a coordinate system with ( A ) at the origin ((0, 0, 0)). - Coordinates of points: ( A(0, 0, 0) ), ( B(sqrt{3}, 0, 0) ), ( C(sqrt{3}, 1, 0) ), ( D(0, 1, 0) ), ( P(0, 0, 2) ). - Vectors ( overrightarrow{AC} = (sqrt{3}, 1, 0) ) and ( overrightarrow{PB} = (sqrt{3}, 0, -2) ). - Calculate the dot product: ( overrightarrow{AC} cdot overrightarrow{PB} = 3 ). - Magnitudes: ( |overrightarrow{AC}| = 2 ), ( |overrightarrow{PB}| = sqrt{7} ). - Cosine of the angle: ( cos theta = frac{3}{2sqrt{7}} = frac{3sqrt{7}}{14} ).2. Finding point ( N ) in face ( PAB ) such that ( NE perp ) plane ( PAC ): - Coordinates of ( E ), the midpoint of ( PD ): ( (0, 0.5, 1) ). - Normal vector to plane ( PAC ) is ( (1, -sqrt{3}, 0) ). - Vector ( overrightarrow{NE} ) must be parallel to the normal vector. - Solving for ( N ) gives coordinates ( (sqrt{3}/6, 0, 1) ). - Distances from ( N ) to ( AB ) and ( AP ): - Distance to ( AB ) (x-axis line): ( sqrt{0^2 + 1^2} = 1 ). - Distance to ( AP ) (z-axis line): ( sqrt{(sqrt{3}/6)^2 + 0^2} = sqrt{3}/6 ).Final answers:1. The cosine of the angle between ( AC ) and ( PB ) is (boxed{dfrac{3sqrt{7}}{14}}).2. The distances from point ( N ) to ( AB ) and ( AP ) are (boxed{1}) and (boxed{dfrac{sqrt{3}}{6}}) respectively.

🔑:# Part (1):1. We use point (A) as the origin ((0, 0, 0)) to establish a right-angle coordinate system in space.2. The coordinates of the vertices (A, B, C, D, P, E) are: [ begin{aligned} & A(0, 0, 0), & B(sqrt{3}, 0, 0), & C(sqrt{3}, 1, 0), & D(0, 1, 0), & P(0, 0, 2), & Eleft(0, frac{1}{2}, 1right). end{aligned} ]3. We calculate the vectors (overrightarrow{AC}) and (overrightarrow{PB}): [ overrightarrow{AC} = (sqrt{3}, 1, 0), quad overrightarrow{PB} = (sqrt{3}, 0, -2). ] 4. Let the angle between (overrightarrow{AC}) and (overrightarrow{PB}) be (theta). We use the dot product to find (cos theta): [ cos theta = frac{overrightarrow{AC} cdot overrightarrow{PB}}{|overrightarrow{AC}| cdot |overrightarrow{PB}|}. ]5. Compute the dot product: [ overrightarrow{AC} cdot overrightarrow{PB} = (sqrt{3}) cdot (sqrt{3}) + (1) cdot (0) + (0) cdot (-2) = 3. ]6. Compute the magnitudes of the vectors: [ |overrightarrow{AC}| = sqrt{(sqrt{3})^2 + 1^2 + 0^2} = sqrt{4} = 2, ] [ |overrightarrow{PB}| = sqrt{(sqrt{3})^2 + 0^2 + (-2)^2} = sqrt{3 + 4} = sqrt{7}. ]7. Substitute into the formula for (cos theta): [ cos theta = frac{3}{2 sqrt{7}} = frac{3 sqrt{7}}{14}. ]8. Thus, the value of the cosine of the angle between (AC) and (PB) is: [ boxed{frac{3 sqrt{7}}{14}}. ]# Part (2):1. Since point (N) is in the side plane (PAB), assume the coordinates of (N) are ((x, 0, z)). We have the vector: [ overrightarrow{NE} = left(-x, frac{1}{2}, 1-zright). ]2. Because (overrightarrow{NE} perp) plane (PAC): [ left{ begin{array}{l} overrightarrow{NE} cdot overrightarrow{AP} = 0, overrightarrow{NE} cdot overrightarrow{AC} = 0. end{array} right. ]3. Substitute the coordinates to get: [ overrightarrow{AP} = (0, 0, 2), quad overrightarrow{AC} = (sqrt{3}, 1, 0). ]4. Expressing ( overrightarrow{NE} cdot overrightarrow{AP} = 0 ): [ left(-x, frac{1}{2}, 1-zright) cdot (0, 0, 2) = 0 Rightarrow 2(1 - z) = 0 Rightarrow z = 1. ]5. Expressing (overrightarrow{NE} cdot overrightarrow{AC} = 0): [ left(-x, frac{1}{2}, 1 - zright) cdot (sqrt{3}, 1, 0) = 0 Rightarrow -sqrt{3}x + frac{1}{2} = 0 Rightarrow x = frac{sqrt{3}}{6}. ]6. Thus, the coordinates of (N) are: [ left(frac{sqrt{3}}{6}, 0, 1right). ]7. The distance from point (N) to line (AB): [ |N_{AB}| = 1. ]8. The distance from point (N) to line (AP): [ |N_{AP}| = frac{sqrt{3}}{6}. ]9. Therefore, the distances are: [ boxed{1, frac{sqrt{3}}{6}}. ]

🧠:Okay, so I need to prove that the dihedral angle φ of the polyhedral angle formed by tangent planes at the vertices of a regular n-sided polygon inscribed in a great circle of a sphere satisfies φ ≥ π(1 - 2/n). Hmm, let me start by understanding the problem step by step.First, let me visualize the setup. A regular n-gon is inscribed in a great circle of a sphere. A great circle is any circle on the sphere's surface with the same center and radius as the sphere itself. So, the polygon is lying on this great circle, and each vertex is a point on the sphere. At each vertex, there's a tangent plane to the sphere. The intersection of these tangent planes forms a polyhedral angle, and φ is the dihedral angle between two adjacent tangent planes.Wait, the dihedral angle is the angle between two planes. In this case, the two tangent planes at adjacent vertices. So, if I take two adjacent vertices of the n-gon, each has a tangent plane, and the dihedral angle between them is φ. The problem states that this angle φ is at least π(1 - 2/n). So, I need to show that φ cannot be smaller than this value.Let me recall some properties. The tangent plane at a point on a sphere is perpendicular to the radius at that point. So, for each vertex of the polygon, the tangent plane is perpendicular to the radius vector from the sphere's center to that vertex.Since the polygon is regular and inscribed in a great circle, all its vertices are equidistant from each other along the great circle. The central angle between two adjacent vertices is 2π/n radians. So, the angle between the radii to two adjacent vertices is 2π/n.Now, the dihedral angle between two tangent planes. The dihedral angle between two planes can be found using the angle between their normal vectors. The normal vector to each tangent plane is the radius vector at the point of tangency. Therefore, the dihedral angle φ between two adjacent tangent planes is equal to the angle between the normals of the planes, which are the radii of the sphere at the two adjacent vertices. Wait, but the angle between the normals is the central angle between the two vertices, which is 2π/n. But the dihedral angle between two planes is equal to the angle between their normals? Wait, no. Wait, the dihedral angle is actually the angle between the two planes, which is supplementary to the angle between their normals if the normals are pointing towards each other. Hmm, maybe I need to think carefully here.Let me recall: The dihedral angle between two planes is the angle between two lines, each lying on one plane and perpendicular to the line of intersection of the two planes. Alternatively, if we have two planes with normals n₁ and n₂, the dihedral angle φ can be found using the formula:cos φ = (n₁ ⋅ n₂) / (|n₁| |n₂|)But in this case, since both normals are radii of the sphere, their magnitudes are equal to the sphere's radius, which we can assume to be 1 for simplicity. So, cos φ = n₁ ⋅ n₂. But the dot product of two unit vectors is the cosine of the angle between them. So, if θ is the angle between n₁ and n₂, then cos φ = cos θ. Wait, does that mean φ = θ? But that can't be right, because the dihedral angle between two planes is not the same as the angle between their normals. Wait, actually, no. Wait, if the angle between the normals is θ, then the dihedral angle φ is either θ or π - θ, depending on orientation. Hmm.Wait, perhaps I need to correct this. The dihedral angle is defined as the angle you would measure between the two planes when looking along their line of intersection. The angle between the normals is supplementary to the dihedral angle. Let me verify this with an example. If two planes intersect along a line, and their normals make an angle θ, then the dihedral angle φ between the planes is π - θ. Because if the normals are close together (small θ), the planes themselves are almost opposite, forming a large dihedral angle. Conversely, if the normals are almost opposite (θ close to π), the dihedral angle is small.Wait, actually, I think it's the other way around. Let me think. Suppose two planes have normals that form an angle θ. If θ is acute, then the dihedral angle between the planes is acute as well. Wait, no. Let me take two coordinate planes as an example. The xy-plane and xz-plane have normals along the z-axis and y-axis, respectively. The angle between the normals is 90 degrees, and the dihedral angle between the planes is also 90 degrees. So, in this case, the dihedral angle equals the angle between the normals. Hmm.Another example: Take two planes that are almost parallel. Then their normals are almost aligned, so the angle θ between normals is small. The dihedral angle between the planes is also small. So, in that case, the dihedral angle φ is equal to θ. Wait, so perhaps in general, the dihedral angle is equal to the angle between the normals? But that contradicts my earlier thought. Hmm.Wait, maybe the confusion arises from how the dihedral angle is measured. The dihedral angle can be considered as the angle between the two half-planes when looking along the line of intersection. Alternatively, the angle between the normals is measured in the space outside the planes. Let me check with the formula. If we have two planes with normals n₁ and n₂, then the dihedral angle φ satisfies cos φ = (n₁ ⋅ n₂) / (|n₁||n₂|). Wait, no, that's not correct. Wait, actually, the dihedral angle can be computed using the normals, but the formula is a bit different. Let me recall that the dihedral angle φ is related to the angle θ between the normals by the formula cos φ = ±(n₁ ⋅ n₂)/(|n₁||n₂|). Wait, but depending on the orientation, the dihedral angle could be θ or π - θ.Alternatively, maybe the dihedral angle is equal to π - θ, where θ is the angle between the normals. Let me check with the coordinate planes. For the xy-plane and xz-plane, the normals are along z and y, angle between normals is 90 degrees, dihedral angle is 90 degrees. So, in this case, φ = θ. If two planes have normals with angle θ between them, then φ = θ. So, maybe the dihedral angle is equal to the angle between the normals. That seems to hold in this case. Wait, but if two planes are oriented such that their normals point towards each other, then maybe the angle between the normals is θ, but the dihedral angle is π - θ? Hmm, perhaps.Wait, let's take two planes that form a dihedral angle φ. The angle between their normals is θ. Then, the relationship between φ and θ is φ + θ = π. Is that correct? Let me think. If two planes are almost coinciding, the dihedral angle φ is close to 0, and the angle between normals θ is also close to 0. So, in that case, φ = θ. Wait, but if the planes are oriented so that their normals are pointing in opposite directions, then the dihedral angle would be π, and the angle between normals is π. Wait, so maybe actually the dihedral angle is equal to the angle between the normals. Hmm.Wait, perhaps the confusion arises because dihedral angles can be measured in two different ways: the angle between the planes on one side or the other. The dihedral angle is the smaller angle between the two planes, so it's always between 0 and π. The angle between the normals is also between 0 and π. However, depending on the orientation, the dihedral angle could be equal to the angle between the normals or π minus that angle. Wait, but how to determine which one it is?Alternatively, maybe the dihedral angle is equal to the angle between the normals if we take the acute angle, or π minus the angle between normals if it's obtuse. But in our case, the normals are pointing outward from the sphere, right? Each tangent plane is at a point on the sphere, so the normal vector (radius) points outward. So, if two adjacent vertices are on the great circle, their normals are separated by an angle of 2π/n. So, the angle between normals is 2π/n, so then the dihedral angle would be either 2π/n or π - 2π/n. But which one?Wait, let's take an example. Let n=4, so we have a square inscribed in a great circle. The angle between normals (radii) at adjacent vertices is 90 degrees (π/2 radians). Then, the dihedral angle between the tangent planes. Let's think about the tangent planes at two adjacent vertices of a square on a sphere. Each tangent plane is perpendicular to the radius. The dihedral angle between them would be the angle you have to turn from one tangent plane to the other around their line of intersection. The line of intersection of two tangent planes at adjacent vertices would be the tangent line to the sphere where the two planes meet. Wait, actually, the line of intersection of two tangent planes at two points on a sphere is the common tangent line to the two points. Hmm, but in 3D, two planes intersect along a line. For two tangent planes at adjacent vertices of the square, the line of intersection is the line where both planes are tangent to the sphere. Since the two points are adjacent on the great circle, the distance between them is minimal. The dihedral angle between the planes would be determined by the angle between their normals.Wait, if the normals are separated by 90 degrees (for n=4), then according to the earlier example with coordinate planes, the dihedral angle would be 90 degrees as well. But according to the formula we are supposed to prove, when n=4, φ ≥ π(1 - 2/4) = π(1 - 1/2) = π/2. So, φ ≥ π/2. If in reality φ = π/2, then equality holds. So, that seems correct.Wait, but if the angle between normals is θ, then dihedral angle φ is equal to θ? For the coordinate planes, yes. So, perhaps in general, φ = θ, the angle between the normals. Therefore, in the case of a regular n-gon, the angle between normals is 2π/n, so φ = 2π/n. But wait, that would mean φ = 2π/n, but the problem states that φ ≥ π(1 - 2/n). For n=4, 2π/4 = π/2, and π(1 - 2/4) = π/2, so equality holds. For n=3, 2π/3 ≈ 2.094, and π(1 - 2/3) = π/3 ≈ 1.047. But 2π/3 is greater than π/3, so φ would be 2π/3 which is greater than π/3. So, the inequality holds. Wait, but in this case, if φ is equal to the angle between normals, which is 2π/n, then 2π/n compared to π(1 - 2/n). Let's compute 2π/n ≥ π(1 - 2/n). Dividing both sides by π: 2/n ≥ 1 - 2/n → 2/n + 2/n ≥ 1 → 4/n ≥ 1 → n ≤4. But for n=5, 4/5=0.8 <1, so this would not hold. Therefore, this suggests that φ is not equal to the angle between the normals. Therefore, my previous conclusion is incorrect.Wait, that's a contradiction. So, perhaps my initial assumption that φ is equal to the angle between normals is wrong. Therefore, I need to re-examine this.Let me think again. The dihedral angle between two planes is the angle between the planes when viewed along their line of intersection. To compute this angle, we can use the normals. The formula for the dihedral angle φ is:cos φ = ( n₁ ⋅ n₂ ) / ( |n₁| |n₂| )But wait, actually, no. The angle between two planes is given by the angle between their normals, but it could be acute or obtuse. However, the dihedral angle is the angle between the two planes measured in a specific way. Let me recall that the dihedral angle can be calculated as the angle whose cosine is equal to the dot product of the normals divided by the product of their magnitudes, but actually, that gives the angle between the normals. So, if the normals make an angle θ, then the dihedral angle is π - θ. Wait, but in the coordinate plane example, if normals are at 90 degrees, dihedral angle is 90 degrees. So, π - θ would be π - π/2 = π/2, which matches. Wait, but in that case, θ is the angle between normals, and dihedral angle is π - θ. But then, in the case where two planes are almost parallel, the angle between normals θ is small, so the dihedral angle would be π - θ, which would be close to π. But that contradicts the intuition that the dihedral angle between almost parallel planes is small. So, perhaps the formula is actually that the dihedral angle is equal to θ when θ is acute, and π - θ when θ is obtuse? Wait, no. Wait, maybe the dihedral angle is the acute angle between the planes, so if θ is acute, then dihedral angle is θ, and if θ is obtuse, then dihedral angle is π - θ. But in the coordinate plane example, θ is 90 degrees, so dihedral angle is 90 degrees. If θ is acute, then dihedral angle is θ; if θ is obtuse, dihedral angle is π - θ. Wait, but 90 degrees is neither acute nor obtuse. Hmm. Maybe this is not the correct approach.Alternatively, let's use the formula for dihedral angle in terms of normals. Let me look it up mentally. The dihedral angle between two planes can be calculated using the normals n₁ and n₂ as follows:cos φ = ( n₁ ⋅ n₂ ) / ( |n₁| |n₂| )But wait, this formula gives the cosine of the angle between the normals, which is θ, not φ. So, if φ is the dihedral angle, then cos θ = n₁ ⋅ n₂ / (|n₁| |n₂|). So, θ = angle between normals, and φ = dihedral angle. Then, how are θ and φ related? If we consider that the dihedral angle is the angle between the two planes, and θ is the angle between the normals, then θ + φ = π. Because the normals are pointing outward from their respective planes, so if the dihedral angle is φ, the angle between normals is π - φ. Wait, that makes sense. Let me verify with the coordinate planes. The dihedral angle between the xy and xz planes is π/2, and the angle between their normals (along z and y) is also π/2. So, θ = φ. Wait, that contradicts θ + φ = π. Hmm, so maybe this relationship isn't universal.Alternatively, perhaps the dihedral angle is equal to the angle between the normals if the normals are pointing towards each other, or something else. This is getting confusing. Maybe I need a different approach.Alternatively, consider two tangent planes at points A and B on the sphere. The line of intersection of these planes is the common tangent line to the sphere at points A and B. Wait, but in 3D, two tangent planes at two distinct points on a sphere will intersect along a line that is the set of points equidistant or something? Wait, perhaps not. Let me think. If I have two points A and B on the sphere, each with their own tangent plane. The tangent plane at A is all points r such that (r - A) ⋅ A = 0, assuming the sphere is centered at the origin. Similarly, the tangent plane at B is (r - B) ⋅ B = 0. So, the line of intersection of these two planes is the set of points r that satisfy both (r ⋅ A) = |A|² and (r ⋅ B) = |B|². Since A and B are on the sphere, |A| = |B| = R, the radius. Let's assume R=1 for simplicity. Then, the equations become r ⋅ A = 1 and r ⋅ B = 1. The line of intersection is the set of points r that satisfy both equations. The direction vector of this line is A × B, since it's perpendicular to both A and B.Now, to find the dihedral angle between the two planes, we can use the normals. The normal vectors to the planes are A and B, as the tangent planes at A and B have normals A and B, respectively. The dihedral angle φ between the two planes is the angle between the planes, which can be calculated using the normals. The formula for the dihedral angle is:cos φ = ( A ⋅ B ) / ( |A| |B| )But wait, if A and B are unit vectors, then A ⋅ B = cos θ, where θ is the angle between A and B. So, cos φ = cos θ, which implies φ = θ. But in the coordinate plane example, this holds. However, in our case, the angle between A and B is 2π/n, since the polygon is regular and inscribed in a great circle. Therefore, φ = 2π/n. But according to the inequality we need to prove, φ ≥ π(1 - 2/n). Let's check for n=3: φ = 2π/3 ≈ 2.094, and π(1 - 2/3) ≈ π/3 ≈ 1.047. So, 2π/3 > π/3, so inequality holds. For n=4: φ= π/2, and π(1 - 2/4)= π/2, equality. For n=5: φ=2π/5≈1.257, π(1 - 2/5)=3π/5≈1.885. Wait, 2π/5≈1.257 is less than 3π/5≈1.885. So, this contradicts the inequality. Therefore, my conclusion that φ=2π/n must be wrong. Therefore, there's a mistake in my reasoning.Wait, so if for n=5, the dihedral angle φ is supposed to be ≥ 3π/5≈1.885, but according to the angle between normals it's 2π/5≈1.257, which is smaller, this contradicts the inequality. Hence, my assumption that φ equals the angle between normals is incorrect. Therefore, I need to revisit my approach.Perhaps the dihedral angle is not directly the angle between the normals. Maybe it's a different angle. Let's consider the actual geometry here. The polyhedral angle is formed by the tangent planes at all n vertices. So, at each vertex of the polygon, there's a tangent plane, and the intersection of these planes creates a pyramid-like structure with n faces meeting at a vertex. Wait, but the dihedral angle φ is the angle between two adjacent tangent planes. So, each pair of adjacent tangent planes intersects along a line, and the dihedral angle between them is φ. The problem states that this dihedral angle φ must be at least π(1 - 2/n).Wait, perhaps I need to model this in 3D coordinates. Let's suppose the sphere is centered at the origin with radius 1. The regular n-gon is inscribed in a great circle, say the equator. Then, the vertices are equally spaced around the equator, each separated by an angle of 2π/n. Let's take two adjacent vertices, A and B, located at (1, 0, 0) and (cos(2π/n), sin(2π/n), 0). The tangent planes at these points are given by x = 1 and (cos(2π/n)x + sin(2π/n)y) = 1. Wait, because the tangent plane at A is x=1, since the normal vector at A is (1,0,0), and similarly for B, the normal vector is (cos(2π/n), sin(2π/n), 0), so the tangent plane is cos(2π/n)x + sin(2π/n)y = 1.Now, to find the dihedral angle between these two planes. The dihedral angle can be found by taking the angle between the normals of the planes. The normal vector to the first plane is (1, 0, 0), and the normal vector to the second plane is (cos(2π/n), sin(2π/n), 0). The angle θ between these normals is 2π/n. However, as we saw earlier, if we use the formula for dihedral angle, which is cos φ = ( n₁ ⋅ n₂ ) / ( |n₁| |n₂| ), then cos φ = cos(2π/n), so φ = 2π/n. But this leads to the contradiction when n=5. So, there must be something wrong here.Wait, but the problem states that φ ≥ π(1 - 2/n). For n=3, φ=2π/3≈2.094, and π(1 - 2/3)=π/3≈1.047. So, 2π/3 > π/3, so inequality holds. For n=4, φ=π/2, and π(1 - 2/4)=π/2, equality. For n=5, φ=2π/5≈1.257, while π(1 - 2/5)=3π/5≈1.885. But 2π/5 < 3π/5, which contradicts the inequality. Therefore, my approach must be flawed. Therefore, there's a misunderstanding in how the dihedral angle is defined here.Wait, perhaps the polyhedral angle mentioned in the problem is not the dihedral angle between two adjacent tangent planes, but rather the angle at the edge where multiple tangent planes meet? Wait, the problem says "the dihedral angle of the resulting polyhedral angle by φ". A polyhedral angle is a vertex where multiple faces meet, and the dihedral angles are the angles between adjacent faces. So, if we have a polyhedral angle formed by n tangent planes, each at a vertex of the n-gon, then each dihedral angle φ is between two adjacent tangent planes. So, in that case, φ is the dihedral angle between each pair of adjacent tangent planes. However, according to the previous calculation, φ=2π/n, which for n≥3 is less than π(1 - 2/n) when n>4. Wait, let's check:For n=5, 2π/5 ≈1.2566, π(1 - 2/5)=3π/5≈1.884, so 1.2566 <1.884, which contradicts the inequality. Therefore, either the problem statement is incorrect, or my understanding is wrong.Alternatively, perhaps the dihedral angle is not the angle between the tangent planes but something else. Wait, maybe I need to consider the spherical polygon formed by the tangent planes. Wait, each tangent plane cuts the sphere along a circle. The intersection of multiple tangent planes would form a spherical polygon on the sphere. Wait, but tangent planes touch the sphere at a single point each. Hmm. Alternatively, the polyhedral angle is the angle formed by the edges where the tangent planes intersect. But since all tangent planes are tangent to the sphere, their intersections are lines outside the sphere.Alternatively, maybe the polyhedral angle is the angle between the tangent planes as seen from inside the polyhedron formed by the tangent planes. Wait, but the tangent planes all lie outside the sphere, so the intersection of the half-spaces defined by the tangent planes would form a convex polyhedron containing the sphere. The dihedral angles of this polyhedron would be the angles between adjacent faces (tangent planes). So, in this case, the dihedral angle φ is the angle between two adjacent tangent planes as part of the polyhedron. However, according to the previous calculation, this angle is 2π/n, but the problem states that φ must be at least π(1 - 2/n). So, for n=5, 2π/5≈1.2566 is less than 3π/5≈1.884, which violates the inequality. Therefore, there must be a different approach.Wait, maybe the dihedral angle is not the angle between the tangent planes but the angle between the planes after considering the orientation. Wait, but how?Alternatively, perhaps I need to parameterize the problem differently. Let's consider the tangent planes at two adjacent vertices. Let's take two points A and B on the sphere, separated by an arc length of 2π/n. The tangent planes at A and B are each perpendicular to the radii OA and OB, where O is the center of the sphere. The dihedral angle between these two planes can be found by considering the angle between their normals, which are OA and OB. As previously, the angle between OA and OB is 2π/n. However, depending on the orientation, the dihedral angle could be this angle or π minus this angle.Wait, but the dihedral angle is the angle you would measure when moving from one plane to the other along their line of intersection. Let me try to compute this angle using another method. Let's consider a point on the line of intersection of the two tangent planes. Since both planes are tangent to the sphere at A and B, the line of intersection is the common external tangent to the sphere at points A and B. Wait, but in 3D, two tangent planes at two points on a sphere will intersect along a line that's outside the sphere. The dihedral angle between the planes is the angle between their respective normals, but maybe projected onto some plane.Alternatively, let's use vector algebra to compute the dihedral angle. Suppose we have two planes with normals n₁ and n₂. The dihedral angle φ between them can be found using the formula:cos φ = ( (n₁ × n₂) ⋅ t ) / ( |n₁ × n₂| |t| )where t is a vector along the line of intersection of the two planes. Wait, no, that doesn't seem right. Wait, the dihedral angle can also be found using the angle between the two planes around their line of intersection. To compute this, we can take two vectors, one in each plane, both perpendicular to the line of intersection, and then find the angle between these vectors.Let me try this approach. Let t be the direction vector of the line of intersection of the two planes. Then, in each plane, we can find a vector perpendicular to t. The angle between these two vectors is the dihedral angle φ.So, for the tangent planes at A and B, their normals are n₁ = A and n₂ = B. The line of intersection of the two planes has direction vector t = n₁ × n₂ = A × B. To find vectors in each plane perpendicular to t, we can take v₁ = n₁ × t and v₂ = n₂ × t.Then, the angle between v₁ and v₂ is the dihedral angle φ. Let's compute this.First, t = A × B. Then, v₁ = A × (A × B) = A(A ⋅ B) - B(A ⋅ A) = A(cos θ) - B(1), where θ is the angle between A and B (which is 2π/n). Similarly, v₂ = B × (A × B) = B(A ⋅ B) - A(B ⋅ B) = B(cos θ) - A(1).Now, compute the angle between v₁ and v₂. The cosine of this angle is (v₁ ⋅ v₂) / (|v₁| |v₂|).First, compute v₁ ⋅ v₂:= [A cos θ - B] ⋅ [B cos θ - A]= A ⋅ B cos² θ - A ⋅ A cos θ - B ⋅ B cos θ + B ⋅ A= cos θ * cos² θ - 1 * cos θ - 1 * cos θ + cos θ= cos³ θ - 2 cos θ + cos θ= cos³ θ - cos θNext, compute |v₁| and |v₂|. Let's compute |v₁|:|v₁| = sqrt( [A cos θ - B] ⋅ [A cos θ - B] )= sqrt( A ⋅ A cos² θ - 2 A ⋅ B cos θ + B ⋅ B )= sqrt( cos² θ - 2 cos θ * cos θ + 1 )= sqrt( cos² θ - 2 cos² θ + 1 )= sqrt(1 - cos² θ )= sin θSimilarly, |v₂| = sin θTherefore, cos φ = (cos³ θ - cos θ) / (sin θ * sin θ )= cos θ (cos² θ - 1) / sin² θ= cos θ (-sin² θ) / sin² θ= -cos θTherefore, cos φ = -cos θSo, φ = arccos(-cos θ) = π - θTherefore, the dihedral angle φ is π - θ, where θ is the angle between the normals A and B, which is 2π/n. Therefore, φ = π - 2π/n = π(1 - 2/n). Wait, so this gives φ = π(1 - 2/n), which is exactly the lower bound given in the problem. Therefore, the dihedral angle is exactly equal to π(1 - 2/n), which would imply that φ ≥ π(1 - 2/n) is always true with equality. But the problem states to prove that φ ≥ π(1 - 2/n), which would be an inequality, not an equality. So, there's a contradiction here. Unless my calculation is correct, but there's a missing consideration.Wait, but in my calculation, I assumed that the polygon is inscribed in a great circle. If the polygon is inscribed in an arbitrary great circle, but if the sphere is arbitrary, but a great circle is fixed. However, perhaps if the polygon is not regular, or if the great circle is not the equator, but another great circle, does the dihedral angle change? Wait, but the problem states a regular n-sided polygon is inscribed in an arbitrary great circle. However, all regular polygons inscribed in a great circle are congruent up to rotation, so the dihedral angle should be the same regardless of the orientation. Therefore, according to this calculation, φ = π(1 - 2/n), which would make the inequality φ ≥ π(1 - 2/n) an equality. Therefore, the problem's inequality is actually an equality, which seems to contradict the problem statement. So, there's a confusion here.Wait, maybe the error is in assuming that the polygon is regular. Wait, the problem states a regular n-sided polygon inscribed in a great circle. Therefore, the regular polygon is uniform, and the calculation above shows that the dihedral angle is exactly π(1 - 2/n). Therefore, the inequality φ ≥ π(1 - 2/n) holds with equality. But the problem says to prove that φ is at least this value, implying that it could be larger. So, perhaps when the polygon is not regular, the dihedral angle could be larger or smaller? But the problem specifies a regular polygon, so perhaps the inequality is actually an equality. Therefore, maybe the problem has a typo, or perhaps I'm missing something.Wait, let me check the problem statement again: "A regular n-sided polygon is inscribed in an arbitrary great circle of a sphere. At each vertex of this polygon, a tangent plane to the sphere is laid. Denote the dihedral angle of the resulting polyhedral angle by φ. Prove that φ ≥ π(1−2/n)."Wait, perhaps "arbitrary great circle" implies that the sphere is arbitrary, but the result should hold for any sphere. However, in my calculation, the sphere's radius cancels out because we used unit vectors. Therefore, the dihedral angle is independent of the sphere's radius and only depends on n. Therefore, φ = π(1 - 2/n) regardless of the sphere's size. So, the inequality φ ≥ π(1 - 2/n) holds with equality, so the minimum value of φ is π(1 - 2/n), and it cannot be smaller. Therefore, the problem is to show that this dihedral angle cannot be smaller than π(1 - 2/n), which my calculation shows it's exactly equal to that value. Therefore, the inequality holds as an equality. Hence, the problem is likely asking to prove that φ is at least this value, and since it's exactly equal, the inequality holds.But why would the problem phrase it as an inequality if it's an equality? Maybe there's a generalization where the polygon is not regular, but in this problem, it is regular. Wait, the problem specifically states a regular polygon. Therefore, in that case, the dihedral angle is exactly π(1 - 2/n). Therefore, the inequality φ ≥ π(1 - 2/n) holds with equality. Perhaps in a non-regular polygon inscribed in a great circle, the dihedral angles could vary, and the regular polygon gives the minimal case? Wait, but if the polygon is regular, the dihedral angles are all equal. If the polygon is irregular, maybe some dihedral angles become larger and others smaller? But the problem specifies a regular polygon, so perhaps in that case, it's exactly equal. Therefore, maybe the problem intended to state an equality, but phrased it as an inequality. Alternatively, perhaps my calculation is missing a factor.Wait, let's go through the calculation again. We found that cos φ = -cos θ, where θ is the angle between the normals, which is 2π/n. Therefore, φ = π - 2π/n = π(1 - 2/n). Therefore, this is exact. Therefore, the dihedral angle φ is exactly π(1 - 2/n), so the inequality is actually an equality. Hence, the problem statement might have a mistake, or perhaps there is a different interpretation.Wait, maybe the problem is not about the dihedral angle between two adjacent tangent planes, but the dihedral angle at each edge of the polyhedral angle formed by all tangent planes. However, in a regular polyhedron formed by tangent planes at the vertices of a regular polygon, all dihedral angles are equal. Therefore, the calculation still holds. Therefore, φ = π(1 - 2/n), so the inequality holds as equality.Alternatively, maybe the problem is considering a different kind of dihedral angle. Wait, perhaps the polyhedral angle mentioned is not between two adjacent tangent planes, but the angle inside the polyhedron formed by all tangent planes. Wait, but the polyhedron formed by the tangent planes is an infinite region where all the tangent planes bound the sphere. The intersection of all the half-spaces defined by the tangent planes (i.e., the regions containing the sphere) would form a convex polyhedron. The dihedral angles between adjacent faces (tangent planes) of this polyhedron are the angles we're calculating. In this case, for the regular polygon, those dihedral angles are all equal to π(1 - 2/n). Therefore, the inequality holds as equality.Alternatively, perhaps the problem is considering a different sphere, but the calculation shows it's independent of the sphere's radius. Therefore, regardless of the sphere's size, the dihedral angle is π(1 - 2/n). Hence, the inequality is actually an equality. Therefore, the problem statement might have a typo, or maybe there's a different configuration.Alternatively, perhaps the polygon is not required to be convex, but since it's regular, it is convex. Hmm.Given that my detailed calculation leads to φ = π(1 - 2/n), which matches the lower bound in the inequality, I think that the problem might have intended to state an equality, but due to a misstatement, it's phrased as an inequality. However, given the problem asks to prove φ ≥ π(1 - 2/n), and in the regular case, it's equal, perhaps the inequality holds in more general cases, and the regular case is the minimal one. Wait, but the problem specifies a regular polygon. So, in that case, if it's always equal, then the inequality is trivial. Therefore, there must be a different interpretation.Wait, another possibility: maybe the polygon is inscribed in an arbitrary great circle, but the sphere is allowed to have any radius. However, as shown earlier, the dihedral angle is independent of the sphere's radius because it's determined by the angular separation between the vertices. Therefore, even for different radii, φ remains the same. Therefore, the inequality still holds as equality.Alternatively, maybe the problem considers a non-regular polygon, but the problem states it's a regular polygon. Therefore, I'm confused.Alternatively, perhaps the polyhedral angle is not the dihedral angle between two adjacent planes but another angle in the structure. Wait, a polyhedral angle has multiple dihedral angles, but since the polygon is regular and the configuration is symmetric, all dihedral angles are equal. Therefore, my calculation should hold.Wait, to confirm, let's take n=3. A regular triangle inscribed in a great circle. The angle between normals is 2π/3, so the dihedral angle should be π - 2π/3 = π/3. But according to the problem's inequality, φ ≥ π(1 - 2/3) = π/3. So, equality holds. For a square, n=4, dihedral angle φ= π - 2π/4 = π - π/2 = π/2, and the inequality is φ ≥ π(1 - 2/4) = π/2, equality again. For n=5, φ= π - 2π/5= 3π/5, which is exactly the lower bound given by the inequality. Therefore, in all cases, φ equals π(1 - 2/n), hence the inequality holds with equality. Therefore, the problem's inequality is actually an equality, and perhaps the problem statement has a typo. However, given that the user wants a proof of the inequality, I need to reconcile this.Wait, maybe in some other configurations, the dihedral angle could be larger. For example, if the polygon is not inscribed in a great circle. But the problem states it is inscribed in a great circle. Alternatively, if the tangent planes are not laid at the vertices but somewhere else. No, the problem specifies tangent planes at the vertices. Therefore, in this specific case, the dihedral angle is exactly π(1 - 2/n), which is the lower bound. Therefore, the inequality holds.Hence, the answer is that φ = π(1 - 2/n), and thus φ is always equal to this value, which satisfies the inequality φ ≥ π(1 - 2/n). Therefore, the inequality holds with equality.But then, the problem says "prove that φ ≥ ...", which would be straightforward if φ is exactly equal. But maybe the original problem had a different configuration, like the polygon being on a non-great circle, leading to φ being larger? Or perhaps considering the polygon on a great circle but allowing for non-regular polygons? But the problem specifies a regular polygon.Alternatively, perhaps the problem is in higher dimensions, but I think it's in 3D.Alternatively, maybe my calculation is incorrect. Let me verify with n=3. For an equilateral triangle on a great circle of a sphere. The angle between the normals is 120 degrees. The dihedral angle between the two tangent planes is π - 120 degrees = 60 degrees, which is π/3 radians. According to the formula, π(1 - 2/3)=π/3, which matches. For n=4, square, dihedral angle π - π/2=π/2, which is π(1 - 2/4)=π/2. For n=5, dihedral angle π - 2π/5=3π/5, which is π(1 - 2/5)=3π/5. So, equality holds. Therefore, in all cases, φ=π(1 - 2/n). Therefore, the inequality is redundant, but technically correct since φ cannot be less than that value, it's exactly equal. Therefore, the proof is to show that φ=π(1 - 2/n), which automatically satisfies φ≥π(1 - 2/n).Therefore, the solution is to compute the dihedral angle between two adjacent tangent planes as π - θ, where θ is the central angle between two adjacent vertices (2π/n), leading to φ=π - 2π/n=π(1 - 2/n), hence the inequality holds with equality.But the problem says "prove that φ ≥ ...". So, maybe the problem is more general, allowing the polygon to be non-regular, but inscribed in a great circle. Then, the regular polygon would minimize or maximize the dihedral angle. However, since the problem specifies a regular polygon, in that case, φ is exactly π(1 - 2/n). If the problem had said "inscribed in a circle" (not necessarily great), then the dihedral angle could vary. However, given it's a great circle, and regular, the result is fixed.Given the problem specifies a regular polygon inscribed in a great circle, the dihedral angle is uniquely determined, and the inequality holds as equality. Therefore, the proof is as follows.Final Answerboxed{varphi geq pileft(1-frac{2}{n}right)}

✅:To prove that the dihedral angle (varphi) of the polyhedral angle formed by tangent planes at the vertices of a regular (n)-sided polygon inscribed in a great circle of a sphere satisfies (varphi geq pileft(1 - frac{2}{n}right)), we start by considering the geometry of the problem.1. Setup and Properties: - A regular (n)-sided polygon is inscribed in a great circle of a sphere. - Each vertex of the polygon is a point on the sphere, and the tangent plane at each vertex is perpendicular to the radius at that point. - The central angle between two adjacent vertices of the polygon is (frac{2pi}{n}).2. Dihedral Angle Calculation: - The dihedral angle between two tangent planes at adjacent vertices is determined by the angle between the normals to these planes. - The normals to the tangent planes at adjacent vertices are the radii of the sphere at those vertices, separated by the central angle (frac{2pi}{n}). - The dihedral angle (varphi) between the tangent planes is supplementary to the angle between the normals. This is because the dihedral angle is the angle between the planes when viewed along their line of intersection, which is supplementary to the angle between the normals.3. Key Formula: - The angle between the normals (radii) is (frac{2pi}{n}). - The dihedral angle (varphi) is therefore (pi - frac{2pi}{n}).4. Simplifying the Result: - Simplifying (pi - frac{2pi}{n}) gives (pileft(1 - frac{2}{n}right)).5. Conclusion: - Since the dihedral angle (varphi) is exactly (pileft(1 - frac{2}{n}right)) for a regular (n)-sided polygon inscribed in a great circle, it follows that (varphi geq pileft(1 - frac{2}{n}right)) holds with equality.Thus, the dihedral angle (varphi) satisfies the inequality (varphi geq pileft(1 - frac{2}{n}right)) with equality when the polygon is regular and inscribed in a great circle.[boxed{varphi geq pileft(1-frac{2}{n}right)}]

🔑:1. Define the geometry and notation: Let the radius of the sphere be ( R ). Consider a circle on the sphere of radius ( r ). We inscribe a regular ( n )-sided polygon inside this circle. At each vertex ( P_i ) of this polygon, we draw a tangent plane to the sphere.2. Understand the geometric relationship: The radii connecting the center ( O ) of the sphere to these vertices ( P_i ) are perpendicular to the tangent planes at those points. The angle between the sphere's radii to two adjacent vertices ( P_i ) and ( P_{i+1} ) is ( omega ).3. Relate the angles: The angle ( omega ) between the radii at adjacent vertices is the supplement of the angle ( varphi ) between the two tangent planes at these vertices. Thus, [ varphi = pi - omega. ]4. Calculate the central angles for the polygon: Within the circle of radius ( r ) on the sphere, the central angle corresponding to each side of the ( n )-sided polygon is [ alpha = frac{2pi}{n}. ] 5. Compare central angles in different circles: A chord of the same length in a circle of larger radius subtends a smaller central angle than in a circle of smaller radius. Since ( R > r ) for the sphere and the smaller circle within it, we have: [ omega < alpha = frac{2pi}{n}. ]6. Combine the inequalities: Using the inequality for ( omega ) and the relationship between ( varphi ) and ( omega ), we get: [ varphi = pi - omega > pi - frac{2pi}{n}. ] 7. Handle the extreme case: When ( r = R ), the smaller circle coincides with the sphere itself, and thus, the central angle for each side of the polygon matches exactly: [ omega = frac{2pi}{n}. ] Consequently, in this case: [ varphi = pi - frac{2pi}{n}. ] 8. Concluding the inequality: From the step 4 and step 7, we can generalize: [ varphi geq pi left( 1 - frac{2}{n} right). ] Conclusion:[boxed{varphi geq pi left(1 - frac{2}{n}right)}]

🧠:Okay, so I need to solve this math problem where appending a zero to the right of a given number, subtracting that result from 143, and the outcome is three times the original number. Hmm, let me break this down step by step.First, let's parse the problem again to make sure I understand it. The given number is some number, let's call it N. If we append a zero to the right of N, that's like multiplying N by 10, right? For example, if N was 5, appending a zero would make it 50. So, in mathematical terms, appending a zero is equivalent to 10*N.Then, the problem says we subtract this new number (10*N) from 143. So the expression for that would be 143 - 10*N. The result of this subtraction is equal to three times the original number, which is 3*N. So putting it all together, the equation should be:143 - 10*N = 3*NNow, I need to solve for N. Let me write that equation down again:143 - 10N = 3NTo solve for N, I can get all the terms with N on one side and the constants on the other. Let's add 10N to both sides to eliminate the -10N on the left. That would give:143 = 13NThen, divide both sides by 13 to solve for N:N = 143 / 13Now, let me compute that division. 13 times 10 is 130, so 143 minus 130 is 13. So 13*11 is 143. Wait, no, 13*10=130, plus 13 is 143. So 13*11=143. Therefore, N is 11.But let me check this again to make sure. If N is 11, appending a zero gives 110. Subtracting that from 143 gives 143 - 110 = 33. Three times the original number is 3*11 = 33. Yep, that works. So 33 equals 33, which is correct.Wait, just to make sure I didn't make any mistakes in setting up the equation. The problem states that appending a zero to the right of the number and subtracting that from 143 gives three times the original number. So it's 143 minus (10*N) equals 3*N. Yes, that's exactly what I did. So the setup seems correct.Alternatively, maybe I can approach it differently to confirm. Let's let the original number be N. Appending a zero gives 10N. Then, subtracting that from 143: 143 - 10N = 3N. Solving for N, same steps: 143 = 13N => N = 11.Yes, same result. So I think that's correct. Maybe another way to check is plugging N=11 back into the original problem. Original number is 11. Append a zero: 110. Subtract from 143: 143 - 110 = 33. Three times 11 is 33. Correct. So the answer must be 11.I don't see any mistakes here. The key was translating the words into the correct equation. Sometimes people might misinterpret "appended to the right" as adding a zero digit, which is the same as multiplying by 10, so that part was important. If someone thought appending a zero was adding 0 to the number, they'd get a different result, but that's not the case here. So yes, 11 is the correct answer.

✅:The given number is determined through the following steps:1. Let the original number be ( N ).2. Appending a zero to ( N ) transforms it into ( 10N ).3. Subtracting this value from 143 gives the equation: [ 143 - 10N = 3N ]4. Solving for ( N ): [ 143 = 13N quad Rightarrow quad N = frac{143}{13} = 11 ]5. Verification: Appending a zero to 11 gives 110. Subtracting from 143 yields ( 143 - 110 = 33 ), which equals ( 3 times 11 ).Answer: The given number is (boxed{11}).

🔑:1. Let the thought number be ( x ).2. According to the problem, attaching a zero to ( x ) from the right gives us ( 10x ). 3. The problem states that if we subtract this new number ( 10x ) from 143, the result is three times the original number ( 3x ). Therefore, we write the equation: [ 143 - 10x = 3x ]4. Next, we solve this equation for ( x ): - Combine like terms by adding ( 10x ) to both sides: [ 143 - 10x + 10x = 3x + 10x ] Simplifying this, we get: [ 143 = 13x ] - Isolate ( x ) by dividing both sides by 13: [ x = frac{143}{13} ] - Perform the division: [ x = 11 ]5. Verify the solution by substituting ( x = 11 ) back into the condition given in the problem: - Calculate ( 10x ): [ 10 cdot 11 = 110 ] - Check the difference ( 143 - 110 ): [ 143 - 110 = 33 ] - Verify that this equals three times the original number: [ 3 cdot 11 = 33 ] The calculations confirm the correctness.Conclusion: The thought number is:[boxed{11}]