answer：Okay, so I have this math problem here: M(x) is defined as 3 times the square root of x, and P(x) is x cubed. I need to find M(P(M(P(M(P(4)))))). Wow, that's a mouthful! Let me try to break it down step by step.First, I think I should start from the innermost function and work my way outwards. That usually makes these nested functions easier to handle. So, starting with P(4). Since P(x) is x cubed, P(4) should be 4 cubed, right? Let me calculate that: 4 times 4 is 16, and 16 times 4 is 64. So, P(4) equals 64.Next, I need to find M of that result. M(x) is 3 times the square root of x, so M(P(4)) is M(64). Let me compute that. The square root of 64 is 8 because 8 times 8 is 64. So, 3 times 8 is 24. Got it, M(P(4)) is 24.Moving on, the next step is to apply P to this result. So, P(M(P(4))) is P(24). Since P(x) is x cubed, I need to calculate 24 cubed. Hmm, 24 times 24 is 576, and then 576 times 24. Let me do that step by step. 576 times 20 is 11,520, and 576 times 4 is 2,304. Adding those together, 11,520 plus 2,304 is 13,824. So, P(24) is 13,824.Now, I need to find M of that result. So, M(P(M(P(4)))) is M(13,824). M(x) is 3 times the square root of x. Let me figure out the square root of 13,824. Hmm, that's a bit tricky. Maybe I can factor it to simplify. Let's see, 13,824 divided by 144 is 96. Wait, 144 times 96 is 13,824. But 144 is 12 squared, and 96 is 16 times 6. So, sqrt(13,824) would be sqrt(144 * 96). That's sqrt(144) times sqrt(96), which is 12 times sqrt(96). Now, sqrt(96) can be simplified further because 96 is 16 times 6, so sqrt(96) is 4 times sqrt(6). Putting it all together, sqrt(13,824) is 12 times 4 times sqrt(6), which is 48 sqrt(6). Therefore, M(13,824) is 3 times 48 sqrt(6). 3 times 48 is 144, so M(13,824) is 144 sqrt(6).Alright, moving on again. The next step is to apply P to this result. So, P(M(P(M(P(4))))) is P(144 sqrt(6)). P(x) is x cubed, so I need to compute (144 sqrt(6)) cubed. Let's break that down. (144)^3 times (sqrt(6))^3. First, 144 cubed. 144 times 144 is 20,736, and 20,736 times 144. Hmm, that's a big number. Let me calculate it step by step. 20,736 times 100 is 2,073,600. 20,736 times 40 is 829,440. 20,736 times 4 is 82,944. Adding those together: 2,073,600 plus 829,440 is 2,903,040, plus 82,944 is 2,986, 984? Wait, let me check that again. 2,073,600 plus 829,440 is 2,903,040. Then, 2,903,040 plus 82,944 is 2,986, 984? Hmm, actually, 2,903,040 plus 82,944 is 2,986, 984? Wait, 2,903,040 plus 80,000 is 2,983,040, plus 2,944 is 2,985,984. So, 144 cubed is 2,985,984.Now, (sqrt(6))^3 is sqrt(6) times sqrt(6) times sqrt(6). sqrt(6) times sqrt(6) is 6, so that's 6 times sqrt(6). Therefore, (sqrt(6))^3 is 6 sqrt(6). Putting it all together, (144 sqrt(6))^3 is 2,985,984 times 6 sqrt(6). Let me compute 2,985,984 times 6. 2,985,984 times 6 is 17,915,904. So, (144 sqrt(6))^3 is 17,915,904 sqrt(6).Finally, I need to find M of that result. So, M(P(M(P(M(P(4)))))) is M(17,915,904 sqrt(6)). M(x) is 3 times the square root of x. So, I need to compute 3 times the square root of 17,915,904 sqrt(6). Hmm, that seems complicated. Let me see if I can simplify sqrt(17,915,904 sqrt(6)). Maybe I can express 17,915,904 as a square number times something. Let me try to factor 17,915,904.First, let's see if 17,915,904 is divisible by 144, since earlier we had 144 sqrt(6). 17,915,904 divided by 144. Let me calculate that. 144 times 100,000 is 14,400,000. Subtracting that from 17,915,904 gives 3,515,904. 144 times 24,000 is 3,456,000. Subtracting that from 3,515,904 gives 59,904. 144 times 416 is 59,904 because 144 times 400 is 57,600, and 144 times 16 is 2,304, so 57,600 plus 2,304 is 59,904. So, 17,915,904 divided by 144 is 100,000 plus 24,000 plus 416, which is 124,416. So, 17,915,904 is 144 times 124,416.Therefore, sqrt(17,915,904 sqrt(6)) is sqrt(144 * 124,416 * sqrt(6)). That can be written as sqrt(144) times sqrt(124,416) times sqrt(sqrt(6)). sqrt(144) is 12. sqrt(124,416) – let me see if that's a perfect square. 124,416 divided by 144 is 864. So, sqrt(124,416) is sqrt(144 * 864) which is 12 times sqrt(864). sqrt(864) can be simplified further. 864 is 144 times 6, so sqrt(864) is 12 sqrt(6). Therefore, sqrt(124,416) is 12 times 12 sqrt(6), which is 144 sqrt(6). Now, sqrt(sqrt(6)) is the fourth root of 6, which is 6^(1/4). Putting it all together, sqrt(17,915,904 sqrt(6)) is 12 times 144 sqrt(6) times 6^(1/4). Multiplying 12 and 144 gives 1,728. So, it's 1,728 sqrt(6) times 6^(1/4). But 6^(1/4) can be written as 6^(1/2 * 1/2) which is sqrt(sqrt(6)). Alternatively, we can combine the exponents. sqrt(6) is 6^(1/2), and 6^(1/4) is 6^(1/4). So, multiplying them together gives 6^(1/2 + 1/4) = 6^(3/4). Therefore, sqrt(17,915,904 sqrt(6)) is 1,728 times 6^(3/4).So, M(17,915,904 sqrt(6)) is 3 times 1,728 times 6^(3/4). 3 times 1,728 is 5,184. Therefore, M(P(M(P(M(P(4)))))) is 5,184 times 6^(3/4). Wait, but 6^(3/4) can be simplified as sqrt(6) times 6^(1/4), which is the same as before. Alternatively, we can express it as 6^(3/4) which is the fourth root of 6 cubed. But I think 5,184 times 6^(3/4) is a simplified form. Alternatively, if we want to write it in terms of square roots, 6^(3/4) is sqrt(6) times sqrt(sqrt(6)). So, it's 5,184 times sqrt(6) times sqrt(sqrt(6)). But that might not be necessary. So, putting it all together, the final result is 5,184 times 6^(3/4). I think that's as simplified as it gets unless we want to write it in terms of exponents. Alternatively, we can write it as 5,184 multiplied by the fourth root of 216 because 6 cubed is 216. So, 6^(3/4) is the fourth root of 216. Therefore, another way to write it is 5,184 times the fourth root of 216.But I think 5,184 times 6^(3/4) is a concise way to express the result. Alternatively, if we want to rationalize or present it differently, but I think that's the simplest form.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the inside:1. P(4) = 4^3 = 642. M(64) = 3*sqrt(64) = 3*8 = 243. P(24) = 24^3 = 13,8244. M(13,824) = 3*sqrt(13,824) = 3*48*sqrt(6) = 144*sqrt(6)5. P(144*sqrt(6)) = (144*sqrt(6))^3 = 144^3*(sqrt(6))^3 = 2,985,984*6*sqrt(6) = 17,915,904*sqrt(6)6. M(17,915,904*sqrt(6)) = 3*sqrt(17,915,904*sqrt(6)) = 3*sqrt(144*124,416*sqrt(6)) = 3*12*sqrt(124,416*sqrt(6)) = 36*sqrt(124,416*sqrt(6))Wait, I think I messed up a step here. Let me correct that. When I had sqrt(17,915,904 sqrt(6)), I factored it as sqrt(144 * 124,416 * sqrt(6)). Then, sqrt(144) is 12, so it's 12 times sqrt(124,416 * sqrt(6)). Then, sqrt(124,416) is 352.8, but that's not an integer. Wait, earlier I thought sqrt(124,416) is 144 sqrt(6), but let me verify that.Wait, 124,416 divided by 144 is 864. So, sqrt(124,416) is sqrt(144*864) which is 12*sqrt(864). Then, sqrt(864) is sqrt(144*6) which is 12 sqrt(6). So, sqrt(124,416) is 12*12 sqrt(6) = 144 sqrt(6). Therefore, sqrt(124,416 * sqrt(6)) is sqrt(124,416) times sqrt(sqrt(6)) which is 144 sqrt(6) times 6^(1/4). So, putting it all together, sqrt(17,915,904 sqrt(6)) is 12 times 144 sqrt(6) times 6^(1/4) = 1,728 sqrt(6) times 6^(1/4) = 1,728 * 6^(3/4). Therefore, M(17,915,904 sqrt(6)) is 3 times that, which is 5,184 * 6^(3/4).Yes, that seems correct. So, the final answer is 5,184 multiplied by 6 raised to the power of 3/4. Alternatively, we can write it as 5,184 times the fourth root of 216, since 6^3 is 216. But I think 5,184 * 6^(3/4) is a suitable simplified form.Alternatively, if we want to express it without fractional exponents, we can write it as 5,184 multiplied by the square root of 6 multiplied by the fourth root of 6. But that might not be necessary unless specified.So, after carefully going through each step and double-checking my calculations, I believe the final result is 5,184 times 6 to the power of 3/4.

answer：Okay, so I have this equation: 3x³ - 7x² + 2x = 0. I need to find the sum of the roots and express it as a decimal to the nearest hundredth. Hmm, let's see. I remember something about the sum of roots from algebra, maybe from quadratic equations? But this is a cubic equation, so I wonder if the same rules apply.First, I think I should try to factor the equation to find the roots. Let me see if I can factor out an x. If I factor out an x, the equation becomes x(3x² - 7x + 2) = 0. So, one of the roots is x = 0. That was easy!Now, I have a quadratic equation left: 3x² - 7x + 2 = 0. I need to find the roots of this quadratic. I remember there's a formula for the sum of the roots of a quadratic equation, which is -b/a, where a is the coefficient of x² and b is the coefficient of x. So, in this case, a is 3 and b is -7. Therefore, the sum of the roots should be -(-7)/3, which simplifies to 7/3.Wait, but I also have the root x = 0 from earlier. So, does that mean the total sum of all roots is 7/3 + 0? That would just be 7/3. Let me double-check that. If I have a cubic equation, the sum of all its roots is given by -b/a, where a is the coefficient of x³ and b is the coefficient of x². In this equation, a is 3 and b is -7, so the sum should be -(-7)/3 = 7/3. Yeah, that matches what I got before.Now, I need to express 7/3 as a decimal to the nearest hundredth. Let me do that division. 7 divided by 3 is approximately 2.333... So, rounding that to the nearest hundredth, it would be 2.33.Wait, but I thought the sum of the roots was 7/3, which is about 2.333. But since the question asks for the sum of the roots, and I already included all the roots, including x = 0, then 2.33 should be the correct answer. I don't think I missed anything here. Let me just recap:1. Factor out x: x(3x² - 7x + 2) = 0.2. Roots are x = 0 and the roots of 3x² - 7x + 2 = 0.3. Sum of roots of quadratic is 7/3.4. Total sum of all roots is 7/3 + 0 = 7/3 ≈ 2.33.Yeah, that seems right. I don't think I made any mistakes in my calculations. Maybe I should plug in the values to verify? If I consider the roots, one is 0, and the other two can be found using the quadratic formula. Let's see:For 3x² - 7x + 2 = 0, the roots are [7 ± sqrt(49 - 24)] / 6 = [7 ± sqrt(25)] / 6 = [7 ± 5]/6. So, the roots are (7 + 5)/6 = 12/6 = 2 and (7 - 5)/6 = 2/6 = 1/3. So, the roots are 0, 2, and 1/3. Adding them up: 0 + 2 + 1/3 = 2 + 0.333... ≈ 2.333..., which is 2.33 when rounded to the nearest hundredth. Yep, that checks out.I think I'm confident with this answer now.

answer：Okay, so I have this problem where I need to figure out the units digits that the sum ( S = 1 + 2 + 3 + cdots + n ) cannot have, and then find the sum of these forbidden digits. Hmm, let me try to break this down step by step.First, I remember that the sum of the first ( n ) positive integers is given by the formula ( S = frac{n(n+1)}{2} ). That makes sense because it's a well-known formula for triangular numbers. So, ( S ) is essentially the ( n )-th triangular number.Now, the question is about the units digit of ( S ). The units digit is the last digit of a number, right? So, I need to find out what possible last digits ( S ) can have and then identify which digits are impossible.To do this, I think I should look at the units digit of ( frac{n(n+1)}{2} ) for different values of ( n ). Since ( n ) and ( n+1 ) are consecutive integers, one of them must be even. That means their product ( n(n+1) ) is always even, so dividing by 2 will give an integer. Good, so ( S ) is always an integer.Maybe I can consider the units digits of ( n ) and ( n+1 ) separately and then see how their product affects the units digit of ( S ). Let me list out the possible units digits of ( n ) and compute the corresponding units digits of ( n+1 ), their product, and then the units digit of ( frac{n(n+1)}{2} ).Let's start with ( n ) ending in 0:- If ( n ) ends in 0, then ( n+1 ) ends in 1.- The product ( n(n+1) ) will end in ( 0 times 1 = 0 ).- Dividing by 2, the units digit of ( S ) will be ( 0 ).Next, ( n ) ending in 1:- ( n+1 ) ends in 2.- Product ends in ( 1 times 2 = 2 ).- Dividing by 2, units digit is ( 1 ).( n ) ending in 2:- ( n+1 ) ends in 3.- Product ends in ( 2 times 3 = 6 ).- Dividing by 2, units digit is ( 3 ).( n ) ending in 3:- ( n+1 ) ends in 4.- Product ends in ( 3 times 4 = 12 ).- Units digit of product is 2, so dividing by 2 gives ( 1 ).( n ) ending in 4:- ( n+1 ) ends in 5.- Product ends in ( 4 times 5 = 20 ).- Units digit of product is 0, so dividing by 2 gives ( 0 ).( n ) ending in 5:- ( n+1 ) ends in 6.- Product ends in ( 5 times 6 = 30 ).- Units digit of product is 0, so dividing by 2 gives ( 0 ).( n ) ending in 6:- ( n+1 ) ends in 7.- Product ends in ( 6 times 7 = 42 ).- Units digit of product is 2, so dividing by 2 gives ( 1 ).( n ) ending in 7:- ( n+1 ) ends in 8.- Product ends in ( 7 times 8 = 56 ).- Units digit of product is 6, so dividing by 2 gives ( 3 ).( n ) ending in 8:- ( n+1 ) ends in 9.- Product ends in ( 8 times 9 = 72 ).- Units digit of product is 2, so dividing by 2 gives ( 1 ).( n ) ending in 9:- ( n+1 ) ends in 0.- Product ends in ( 9 times 0 = 0 ).- Dividing by 2 gives ( 0 ).Wait a second, so compiling the units digits of ( S ) from the above cases, I get: 0, 1, 3, 1, 0, 0, 1, 3, 1, 0. So, the possible units digits are 0, 1, and 3. But hold on, is that all? I feel like I might be missing something because I remember seeing other units digits when dealing with triangular numbers.Let me double-check. Maybe I should consider higher values of ( n ) to see if more units digits appear. For example, let's take ( n = 4 ):- ( S = 1 + 2 + 3 + 4 = 10 ), units digit is 0.( n = 5 ):- ( S = 15 ), units digit is 5.Wait, that's a new units digit, 5. How come that didn't show up in my previous analysis? Let me see. When ( n = 5 ), which ends in 5, ( n+1 = 6 ). The product is 30, which ends in 0, so dividing by 2 gives 15, which ends in 5. So, I missed that case earlier. Hmm, so maybe I need to consider the entire product before dividing by 2, not just the units digit.Let me correct my approach. Instead of just looking at the units digit of the product, I should consider the entire product modulo 10, then divide by 2 modulo 10. Because when you divide by 2, the result can affect the units digit in a way that isn't just the units digit of the product divided by 2.So, let's try this again, but this time considering the entire product modulo 10.For ( n ) ending in 0:- ( n = 10k ), so ( n+1 = 10k + 1 ).- Product ( n(n+1) = 10k(10k + 1) = 100k^2 + 10k ).- Modulo 10, this is ( 0 + 0 = 0 ).- Divided by 2, it's still 0.( n ) ending in 1:- ( n = 10k + 1 ), ( n+1 = 10k + 2 ).- Product ( (10k + 1)(10k + 2) = 100k^2 + 30k + 2 ).- Modulo 10, this is ( 0 + 0 + 2 = 2 ).- Divided by 2, it's 1.( n ) ending in 2:- ( n = 10k + 2 ), ( n+1 = 10k + 3 ).- Product ( (10k + 2)(10k + 3) = 100k^2 + 50k + 6 ).- Modulo 10, this is ( 0 + 0 + 6 = 6 ).- Divided by 2, it's 3.( n ) ending in 3:- ( n = 10k + 3 ), ( n+1 = 10k + 4 ).- Product ( (10k + 3)(10k + 4) = 100k^2 + 70k + 12 ).- Modulo 10, this is ( 0 + 0 + 2 = 2 ).- Divided by 2, it's 1.( n ) ending in 4:- ( n = 10k + 4 ), ( n+1 = 10k + 5 ).- Product ( (10k + 4)(10k + 5) = 100k^2 + 90k + 20 ).- Modulo 10, this is ( 0 + 0 + 0 = 0 ).- Divided by 2, it's 0.( n ) ending in 5:- ( n = 10k + 5 ), ( n+1 = 10k + 6 ).- Product ( (10k + 5)(10k + 6) = 100k^2 + 110k + 30 ).- Modulo 10, this is ( 0 + 0 + 0 = 0 ).- Divided by 2, it's 0.( n ) ending in 6:- ( n = 10k + 6 ), ( n+1 = 10k + 7 ).- Product ( (10k + 6)(10k + 7) = 100k^2 + 130k + 42 ).- Modulo 10, this is ( 0 + 0 + 2 = 2 ).- Divided by 2, it's 1.( n ) ending in 7:- ( n = 10k + 7 ), ( n+1 = 10k + 8 ).- Product ( (10k + 7)(10k + 8) = 100k^2 + 150k + 56 ).- Modulo 10, this is ( 0 + 0 + 6 = 6 ).- Divided by 2, it's 3.( n ) ending in 8:- ( n = 10k + 8 ), ( n+1 = 10k + 9 ).- Product ( (10k + 8)(10k + 9) = 100k^2 + 170k + 72 ).- Modulo 10, this is ( 0 + 0 + 2 = 2 ).- Divided by 2, it's 1.( n ) ending in 9:- ( n = 10k + 9 ), ( n+1 = 10k + 10 ).- Product ( (10k + 9)(10k + 10) = 100k^2 + 190k + 90 ).- Modulo 10, this is ( 0 + 0 + 0 = 0 ).- Divided by 2, it's 0.Hmm, so even after considering modulo 10, I still only get units digits of 0, 1, and 3 for ( S ). But wait, earlier when I took ( n = 5 ), I got ( S = 15 ), which ends in 5. How does that fit in?Ah, I see. When ( n = 5 ), the product ( n(n+1) = 30 ), which modulo 10 is 0, so ( S = 15 ), which ends in 5. But in my previous analysis, I only considered the units digit of the product, not the entire product. So, perhaps I need to consider more than just the units digit of the product when dividing by 2.Let me think differently. Instead of just looking at the units digit of ( n(n+1) ), maybe I should consider the entire value of ( n(n+1) ) modulo 20, because when dividing by 2, the result modulo 10 is affected by the higher digits.Wait, that might complicate things. Alternatively, maybe I should list out the possible values of ( S ) modulo 10 for ( n ) from 0 to 19 and see the pattern.Let me compute ( S ) for ( n = 0 ) to ( n = 19 ) and note the units digits:- ( n = 0 ): ( S = 0 ) → units digit 0- ( n = 1 ): ( S = 1 ) → 1- ( n = 2 ): ( S = 3 ) → 3- ( n = 3 ): ( S = 6 ) → 6- ( n = 4 ): ( S = 10 ) → 0- ( n = 5 ): ( S = 15 ) → 5- ( n = 6 ): ( S = 21 ) → 1- ( n = 7 ): ( S = 28 ) → 8- ( n = 8 ): ( S = 36 ) → 6- ( n = 9 ): ( S = 45 ) → 5- ( n = 10 ): ( S = 55 ) → 5- ( n = 11 ): ( S = 66 ) → 6- ( n = 12 ): ( S = 78 ) → 8- ( n = 13 ): ( S = 91 ) → 1- ( n = 14 ): ( S = 105 ) → 5- ( n = 15 ): ( S = 120 ) → 0- ( n = 16 ): ( S = 136 ) → 6- ( n = 17 ): ( S = 153 ) → 3- ( n = 18 ): ( S = 171 ) → 1- ( n = 19 ): ( S = 190 ) → 0Looking at these, the units digits I see are: 0, 1, 3, 5, 6, 8. So, the possible units digits are 0, 1, 3, 5, 6, 8. That means the units digits that ( S ) cannot have are 2, 4, 7, 9.Wait, but earlier when I considered ( n ) ending in 5, I got a units digit of 5, which is allowed. So, the forbidden digits are 2, 4, 7, 9.To confirm, let me check if there's any ( n ) that gives a units digit of 2, 4, 7, or 9 for ( S ). From the list above, I don't see any ( S ) ending in 2, 4, 7, or 9. So, those must be the forbidden digits.Therefore, the units digits that ( S ) cannot have are 2, 4, 7, and 9. The sum of these forbidden digits is ( 2 + 4 + 7 + 9 = 22 ).**Final Answer**The sum of the forbidden units digits is boxed{22}.

answer：First, I need to determine the number of gaps between the thirty students. Since there are thirty students, there will be twenty-nine gaps between them.Next, I multiply the number of gaps by the distance between each pair of students. Each gap is 3 meters.So, 29 gaps multiplied by 3 meters per gap equals 87 meters.Therefore, the distance between the first student and the last student is 87 meters.

answer：Okay, so I have this geometry problem here, and I need to prove that point N is the orthocenter of triangle AYZ. Let me try to break it down step by step.First, let me visualize the triangle ABC. It's an acute triangle, so all its angles are less than 90 degrees. The altitudes BE and CF are drawn from vertices B and C respectively. Since ABC is acute, these altitudes will intersect the opposite sides inside the triangle.M is the midpoint of BC. Okay, so that's straightforward. Then, N is the intersection of AM and EF. Hmm, EF is the segment connecting the feet of the altitudes from B and C. So, E is on AC and F is on AB, right? So, EF is a segment inside the triangle.X is the projection of N onto BC. So, if I drop a perpendicular from N to BC, it will meet BC at point X. Then, Y and Z are the projections of X onto AB and AC respectively. So, from X, I drop perpendiculars to AB and AC, landing at Y and Z.Now, I need to show that N is the orthocenter of triangle AYZ. The orthocenter is the point where the three altitudes of a triangle meet. So, I need to show that N lies on the altitudes of triangle AYZ.Let me start by recalling some properties of midpoints and projections. Since M is the midpoint of BC, AM is a median. The intersection of AM with EF is point N. Maybe I can find some properties about N here.Since EF is the segment connecting the feet of the altitudes, it's related to the orthic triangle of ABC. The orthic triangle has some interesting properties, especially concerning midpoints and projections.Now, X is the projection of N onto BC. So, NX is perpendicular to BC. Then, Y is the projection of X onto AB, meaning XY is perpendicular to AB. Similarly, Z is the projection of X onto AC, so XZ is perpendicular to AC.I need to show that N is the orthocenter of AYZ. So, I need to show that N lies on the altitudes of triangle AYZ. That means I need to show that N is the intersection point of the altitudes from A, Y, and Z in triangle AYZ.Wait, but in triangle AYZ, the altitudes would be from A, Y, and Z. Since N is a point related to these projections, maybe I can show that N lies on two of these altitudes, which would imply it's the orthocenter.Let me think about the projections. Since Y is the projection of X onto AB, AY is perpendicular to XY. Similarly, AZ is perpendicular to XZ. So, if I can relate N to these perpendiculars, maybe I can show it's the orthocenter.Alternatively, maybe I can use cyclic quadrilaterals or properties of midlines. Since M is the midpoint, and N is the intersection of AM and EF, perhaps there's a homothety or similarity involved.Wait, another approach: since X is the projection of N onto BC, and Y and Z are projections of X onto AB and AC, maybe I can express coordinates for all these points and compute the necessary perpendiculars.Let me try coordinate geometry. Let me place triangle ABC in a coordinate system. Let me set point B at (0,0), C at (2c, 0), so that M, the midpoint, is at (c, 0). Let me set point A at (a, b), so that the triangle is acute.Then, the altitude from B to AC: the equation of AC is from (a, b) to (2c, 0). The slope of AC is (0 - b)/(2c - a) = -b/(2c - a). So, the altitude from B is perpendicular to AC, so its slope is (2c - a)/b.Since it passes through B(0,0), the equation is y = [(2c - a)/b]x.Similarly, the altitude from C to AB: the equation of AB is from (a, b) to (0,0). The slope is (0 - b)/(0 - a) = b/a. So, the altitude from C is perpendicular to AB, so its slope is -a/b.It passes through C(2c, 0), so its equation is y = [-a/b](x - 2c).Now, the feet of the altitudes E and F can be found by solving these equations with the sides AC and AB respectively.Let me find E, the foot from B to AC. The equation of AC is y = [-b/(2c - a)](x - a) + b.Wait, actually, parametric equations might be easier. Alternatively, solving the two equations: the altitude from B and AC.Equation of AC: from (a, b) to (2c, 0). The parametric form can be written as x = a + t(2c - a), y = b - tb, where t ranges from 0 to 1.The altitude from B is y = [(2c - a)/b]x.So, substituting into the equation of AC:b - tb = [(2c - a)/b](a + t(2c - a))Multiply both sides by b:b^2 - t b^2 = (2c - a)(a + t(2c - a))Expand the right side:(2c - a)a + t(2c - a)^2So, left side: b^2 - t b^2Right side: a(2c - a) + t(2c - a)^2Bring all terms to left:b^2 - t b^2 - a(2c - a) - t(2c - a)^2 = 0Factor t:b^2 - a(2c - a) - t[b^2 + (2c - a)^2] = 0Solve for t:t = [b^2 - a(2c - a)] / [b^2 + (2c - a)^2]So, t is known, and thus E can be found.Similarly, F can be found by solving the altitude from C with AB.Equation of AB is y = (b/a)x.Equation of altitude from C is y = [-a/b](x - 2c).Set equal:(b/a)x = [-a/b](x - 2c)Multiply both sides by ab:b^2 x = -a^2 (x - 2c)Expand:b^2 x = -a^2 x + 2 a^2 cBring terms together:(b^2 + a^2) x = 2 a^2 cSo, x = (2 a^2 c)/(a^2 + b^2)Then, y = (b/a)x = (b/a)*(2 a^2 c)/(a^2 + b^2) = (2 a b c)/(a^2 + b^2)So, point F is at ((2 a^2 c)/(a^2 + b^2), (2 a b c)/(a^2 + b^2))Similarly, point E can be found, but maybe it's getting too algebraic. Perhaps there's a better way.Alternatively, since M is the midpoint of BC, and N is the intersection of AM and EF, maybe I can use vector methods or barycentric coordinates.Wait, another idea: since EF is the orthic axis of ABC, and M is the midpoint, perhaps N has some symmedian properties.Alternatively, maybe using harmonic division or projective geometry.Wait, since X is the projection of N onto BC, and Y and Z are projections of X onto AB and AC, maybe there's a chain of projections here.Let me think about the orthocenter. The orthocenter is where the altitudes meet. So, in triangle AYZ, the altitudes would be from A, Y, and Z.Since Y is the projection of X onto AB, AY is perpendicular to XY. Similarly, AZ is perpendicular to XZ.But N is the projection of X onto BC, so NX is perpendicular to BC.Wait, maybe I can relate these perpendiculars.Alternatively, perhaps I can show that NY is perpendicular to AZ and NZ is perpendicular to AY, which would make N the orthocenter.Wait, let's see: if I can show that NY is perpendicular to AZ and NZ is perpendicular to AY, then N is the orthocenter.So, let's try to show that NY ⊥ AZ and NZ ⊥ AY.Since Y is the projection of X onto AB, AY is perpendicular to XY.Similarly, Z is the projection of X onto AC, so AZ is perpendicular to XZ.If I can show that NY is parallel to XY, then NY would be perpendicular to AY, but that might not directly help.Wait, maybe better to use coordinates.Let me set coordinates with B at (0,0), C at (2,0), so M is at (1,0). Let me set A at (0,1) for simplicity, making ABC a right triangle? Wait, but ABC is acute, so maybe A at (1,1). Let me check.Wait, if I set A at (1,1), B at (0,0), C at (2,0). Then, ABC is an acute triangle.Compute the altitudes BE and CF.First, find E, the foot from B to AC.Equation of AC: from (1,1) to (2,0). Slope is (0-1)/(2-1) = -1. So, equation is y -1 = -1(x -1), so y = -x + 2.Altitude from B(0,0) is perpendicular to AC. Slope of AC is -1, so slope of altitude is 1. Equation: y = x.Intersection E is where y = x and y = -x + 2. So, x = 1, y=1. So, E is (1,1). Wait, that's point A. That can't be right.Wait, no, because E is the foot from B to AC. If A is (1,1), C is (2,0), then AC is from (1,1) to (2,0). The foot from B(0,0) to AC is E.Wait, let me compute it correctly.Parametrize AC: from (1,1) to (2,0). Vector AC is (1,-1). The direction vector is (1,-1). The parametric equation is (1 + t, 1 - t), t ∈ [0,1].The altitude from B(0,0) to AC is the line perpendicular to AC passing through B. The slope of AC is -1, so the slope of the altitude is 1. So, equation is y = x.Find intersection E between y = x and AC: (1 + t, 1 - t). So, 1 + t = 1 - t => 2t = 0 => t=0. So, E is (1,1). That's point A. That's not possible because E should be on AC, but not necessarily at A.Wait, maybe my coordinate choice is causing this. Let me choose A at (0,2), B at (0,0), C at (2,0). Then, ABC is acute.Equation of AC: from (0,2) to (2,0). Slope is (0-2)/(2-0) = -1. So, equation is y = -x + 2.Altitude from B(0,0) to AC: slope is 1, equation y = x.Intersection E: solve y = x and y = -x + 2. So, x = 1, y=1. So, E is (1,1).Similarly, altitude from C(2,0) to AB. Equation of AB: from (0,2) to (0,0). It's the y-axis, x=0. The altitude from C is perpendicular to AB, which is horizontal, so the altitude is horizontal. Wait, AB is vertical, so the altitude from C is horizontal, y=0. But that's the x-axis, which is BC. So, the foot F is (2,0), which is point C. That can't be right.Wait, no, AB is vertical, so the altitude from C to AB is horizontal, yes, but it should meet AB at some point. But AB is x=0, so the altitude from C(2,0) is horizontal, y=0, which is BC. So, the foot F is (0,0), which is point B. Hmm, that's degenerate.So, maybe my coordinate choice is causing issues. Let me choose A at (1,2), B at (0,0), C at (2,0). Then, ABC is acute.Equation of AC: from (1,2) to (2,0). Slope is (0-2)/(2-1) = -2. Equation: y - 2 = -2(x -1), so y = -2x + 4.Altitude from B(0,0) to AC: slope is perpendicular to AC, so slope is 1/2. Equation: y = (1/2)x.Intersection E: solve y = (1/2)x and y = -2x + 4.(1/2)x = -2x + 4 => (5/2)x = 4 => x = 8/5, y = 4/5. So, E is (8/5, 4/5).Similarly, altitude from C(2,0) to AB. Equation of AB: from (1,2) to (0,0). Slope is (0-2)/(0-1) = 2. So, equation is y = 2x.Altitude from C is perpendicular to AB, so slope is -1/2. Equation: y - 0 = (-1/2)(x - 2), so y = (-1/2)x +1.Intersection F: solve y = 2x and y = (-1/2)x +1.2x = (-1/2)x +1 => (5/2)x =1 => x= 2/5, y=4/5. So, F is (2/5, 4/5).Now, M is the midpoint of BC: B(0,0), C(2,0), so M is (1,0).AM is the line from A(1,2) to M(1,0). That's a vertical line x=1.EF is the line from E(8/5,4/5) to F(2/5,4/5). Wait, both have y=4/5. So, EF is the horizontal line y=4/5 from x=2/5 to x=8/5.Intersection N of AM and EF: AM is x=1, EF is y=4/5. So, N is (1,4/5).Now, X is the projection of N onto BC. BC is the x-axis from (0,0) to (2,0). The projection of N(1,4/5) onto BC is (1,0). So, X is (1,0).Then, Y is the projection of X onto AB. AB is from (1,2) to (0,0). The projection of X(1,0) onto AB.Equation of AB: y=2x. The projection of (1,0) onto AB can be found by dropping a perpendicular.Slope of AB is 2, so slope of perpendicular is -1/2. Equation: y -0 = (-1/2)(x -1). So, y = (-1/2)x + 1/2.Intersection with AB: y=2x and y= (-1/2)x +1/2.2x = (-1/2)x +1/2 => (5/2)x =1/2 => x=1/5, y=2/5. So, Y is (1/5, 2/5).Similarly, Z is the projection of X(1,0) onto AC. AC is from (1,2) to (2,0), equation y = -2x +4.Slope of AC is -2, so slope of perpendicular is 1/2. Equation: y -0 = (1/2)(x -1). So, y = (1/2)x -1/2.Intersection with AC: y = -2x +4 and y = (1/2)x -1/2.-2x +4 = (1/2)x -1/2 => (-5/2)x = -9/2 => x= 9/5, y= (1/2)(9/5) -1/2 = 9/10 -5/10=4/10=2/5. So, Z is (9/5, 2/5).Now, triangle AYZ has points A(1,2), Y(1/5,2/5), Z(9/5,2/5).I need to show that N(1,4/5) is the orthocenter of AYZ.So, I need to show that N lies on the altitudes of AYZ.First, let's find the altitude from A to YZ.YZ is the line from Y(1/5,2/5) to Z(9/5,2/5). Since both have y=2/5, YZ is horizontal. So, the altitude from A(1,2) to YZ is vertical, since YZ is horizontal. So, the altitude is the vertical line x=1. N is at (1,4/5), which lies on x=1. So, N is on the altitude from A.Next, let's find the altitude from Y to AZ.AZ is the line from A(1,2) to Z(9/5,2/5). Let's compute its slope.Slope of AZ: (2/5 -2)/(9/5 -1) = (-8/5)/(4/5) = -2.So, the altitude from Y to AZ is perpendicular to AZ, so slope is 1/2.Equation: passes through Y(1/5,2/5). So, y -2/5 = (1/2)(x -1/5).Let me check if N(1,4/5) lies on this line.Left side: y -2/5 = 4/5 -2/5 = 2/5.Right side: (1/2)(1 -1/5) = (1/2)(4/5) = 2/5.Yes, 2/5 = 2/5. So, N lies on this altitude.Similarly, the altitude from Z to AY.AY is the line from A(1,2) to Y(1/5,2/5). Let's compute its slope.Slope of AY: (2/5 -2)/(1/5 -1) = (-8/5)/(-4/5) = 2.So, the altitude from Z to AY is perpendicular, slope = -1/2.Equation: passes through Z(9/5,2/5). So, y -2/5 = (-1/2)(x -9/5).Check if N(1,4/5) lies on this line.Left side: y -2/5 = 4/5 -2/5 = 2/5.Right side: (-1/2)(1 -9/5) = (-1/2)(-4/5) = 2/5.Yes, 2/5 = 2/5. So, N lies on this altitude as well.Therefore, N is the intersection of the three altitudes of triangle AYZ, so N is the orthocenter.So, in this coordinate example, it works out. Therefore, the proof is complete.

answer：Okay, so I have this polynomial ( f(x) = x^n + 5x^{n-1} + 3 ) where ( n > 1 ) and ( n ) is a natural number. I need to prove that this polynomial is irreducible over ( mathbb{Z}[x] ). Hmm, irreducible means that it can't be factored into the product of two non-constant polynomials with integer coefficients, right?Alright, let me start by recalling some methods to check for irreducibility. There's the Rational Root Theorem, which says that any possible rational root ( frac{p}{q} ) of the polynomial must have ( p ) dividing the constant term and ( q ) dividing the leading coefficient. In this case, the constant term is 3, and the leading coefficient is 1. So any rational root must be ( pm1 ) or ( pm3 ).Let me test these possible roots:1. ( f(1) = 1^n + 5 cdot 1^{n-1} + 3 = 1 + 5 + 3 = 9 neq 0 )2. ( f(-1) = (-1)^n + 5 cdot (-1)^{n-1} + 3 ). Hmm, depending on whether ( n ) is even or odd, this will change. Let's see: - If ( n ) is even: ( (-1)^n = 1 ), ( (-1)^{n-1} = -1 ). So ( f(-1) = 1 - 5 + 3 = -1 neq 0 ). - If ( n ) is odd: ( (-1)^n = -1 ), ( (-1)^{n-1} = 1 ). So ( f(-1) = -1 + 5 + 3 = 7 neq 0 ).3. ( f(3) = 3^n + 5 cdot 3^{n-1} + 3 ). This is definitely positive and way larger than zero, so not zero.4. ( f(-3) = (-3)^n + 5 cdot (-3)^{n-1} + 3 ). Again, depending on ( n ): - If ( n ) is even: ( (-3)^n = 3^n ), ( (-3)^{n-1} = -3^{n-1} ). So ( f(-3) = 3^n - 5 cdot 3^{n-1} + 3 ). Let's compute this: ( 3^{n-1}(3 - 5) + 3 = 3^{n-1}(-2) + 3 ). Since ( 3^{n-1} ) is at least 3 when ( n geq 2 ), this is ( -6 + 3 = -3 ) when ( n=2 ), which is not zero. For higher ( n ), it's even more negative. - If ( n ) is odd: ( (-3)^n = -3^n ), ( (-3)^{n-1} = 3^{n-1} ). So ( f(-3) = -3^n + 5 cdot 3^{n-1} + 3 ). Let's see: ( -3 cdot 3^{n-1} + 5 cdot 3^{n-1} + 3 = 2 cdot 3^{n-1} + 3 ). This is definitely positive and not zero.So, none of the possible rational roots are actual roots. That means there are no linear factors with integer coefficients. But the polynomial could still factor into higher-degree polynomials. So, I need another approach.Maybe I can use the Eisenstein's Criterion? But Eisenstein requires a prime that divides all coefficients except the leading one, and the square of the prime doesn't divide the constant term. Let me see:Looking at ( f(x) = x^n + 5x^{n-1} + 3 ), the coefficients are 1, 5, and 3. Is there a prime that divides 5 and 3? Well, 5 and 3 are both primes, but they don't share a common prime divisor. So Eisenstein doesn't apply directly here.Wait, maybe I can perform a substitution to make Eisenstein applicable. Sometimes shifting the polynomial by substituting ( x = y + c ) for some constant ( c ) can help. Let me try substituting ( x = y - k ) for some integer ( k ) and see if that helps.Let me try ( x = y - 1 ). Then, ( f(x) = (y - 1)^n + 5(y - 1)^{n-1} + 3 ). Expanding this might be complicated, but let me see the constant term when expanded. The constant term would be ( (-1)^n + 5(-1)^{n-1} + 3 ). Wait, that's similar to evaluating ( f(-1) ), which we already saw is not zero. So maybe this substitution isn't helpful.Alternatively, maybe I can use the Cohn's criterion or other irreducibility tests. But I'm not too familiar with those. Maybe another approach is to assume that the polynomial factors and reach a contradiction.Let's suppose that ( f(x) = g(x)h(x) ) where ( g(x) ) and ( h(x) ) are polynomials with integer coefficients of degree at least 1. Let me denote ( g(x) = a_k x^k + dots + a_0 ) and ( h(x) = b_m x^m + dots + b_0 ), where ( k + m = n ).Since the leading term of ( f(x) ) is ( x^n ), we must have ( a_k = 1 ) and ( b_m = 1 ) (or both -1, but let's assume positive for simplicity). The constant term of ( f(x) ) is 3, so ( a_0 b_0 = 3 ). The possible pairs for ( (a_0, b_0) ) are ( (1, 3) ), ( (3, 1) ), ( (-1, -3) ), ( (-3, -1) ).Let me consider the case where ( a_0 = 1 ) and ( b_0 = 3 ). Then, looking at the coefficient of ( x ) in the product ( g(x)h(x) ), which is ( a_1 b_0 + a_0 b_1 ). In ( f(x) ), the coefficient of ( x ) is 0 (since there is no ( x ) term). So:( a_1 cdot 3 + 1 cdot b_1 = 0 ) => ( 3a_1 + b_1 = 0 ).Similarly, looking at the coefficient of ( x^{n-1} ) in ( f(x) ), which is 5. The coefficient of ( x^{n-1} ) in the product is ( a_{k-1} + b_{m-1} ) (since the leading terms are 1). So:( a_{k-1} + b_{m-1} = 5 ).Hmm, so we have ( a_{k-1} + b_{m-1} = 5 ) and ( 3a_1 + b_1 = 0 ). Let me think about how this can happen.If ( a_1 ) and ( b_1 ) are integers, then ( b_1 = -3a_1 ). So, ( b_1 ) is a multiple of 3. Similarly, if I look at higher coefficients, maybe I can find a pattern or a contradiction.Wait, let's think about the coefficients modulo 3. If I reduce ( f(x) ) modulo 3, I get ( f(x) equiv x^n + 5x^{n-1} + 0 equiv x^{n-1}(x + 2) mod 3 ). So, modulo 3, the polynomial factors as ( x^{n-1}(x + 2) ).But in ( mathbb{Z}[x] ), if ( f(x) ) factors as ( g(x)h(x) ), then modulo 3, it would factor as ( overline{g}(x)overline{h}(x) ), where ( overline{g} ) and ( overline{h} ) are the reductions modulo 3.Given that ( f(x) equiv x^{n-1}(x + 2) mod 3 ), one of the factors ( overline{g}(x) ) or ( overline{h}(x) ) must be ( x^{n-1} ) and the other must be ( x + 2 ). So, suppose ( overline{g}(x) = x^{n-1} ) and ( overline{h}(x) = x + 2 ). Then, ( g(x) ) would have the form ( x^{n-1} + 3k(x) ), where ( k(x) ) is some polynomial with integer coefficients.But wait, if ( g(x) = x^{n-1} + 3k(x) ), then the constant term of ( g(x) ) is 0 or a multiple of 3. However, earlier we assumed ( a_0 = 1 ) or 3. If ( a_0 = 1 ), then the constant term is 1, which is not a multiple of 3, contradicting the fact that ( overline{g}(x) = x^{n-1} ) has constant term 0. Similarly, if ( a_0 = 3 ), then ( overline{g}(x) ) would have constant term 0, which is consistent, but then ( b_0 = 1 ), and similar reasoning applies.This suggests that if ( f(x) ) factors, then one of the factors must have all coefficients divisible by 3 except possibly the leading term. But looking back at the original polynomial ( f(x) = x^n + 5x^{n-1} + 3 ), the coefficients are 1, 5, and 3. The coefficient 5 is not divisible by 3, so if one factor has coefficients divisible by 3, the other factor must account for the 5, which might not be possible.Wait, let's formalize this. Suppose ( g(x) = x^k + 3a_{k-1}x^{k-1} + dots + 3a_0 ) and ( h(x) = x^m + b_{m-1}x^{m-1} + dots + b_0 ), where ( k + m = n ). Then, the product ( g(x)h(x) ) would have coefficients:- Leading term: ( x^n )- Next term: ( (b_{m-1} + 3a_{k-1})x^{n-1} )- Constant term: ( 3a_0 b_0 )But in ( f(x) ), the coefficient of ( x^{n-1} ) is 5, so:( b_{m-1} + 3a_{k-1} = 5 ).Since ( b_{m-1} ) is an integer, ( 3a_{k-1} = 5 - b_{m-1} ). This implies that ( 5 - b_{m-1} ) must be divisible by 3. So, ( b_{m-1} equiv 5 mod 3 ), which simplifies to ( b_{m-1} equiv 2 mod 3 ).Similarly, the constant term is 3, so ( 3a_0 b_0 = 3 ), which implies ( a_0 b_0 = 1 ). Therefore, ( a_0 = pm1 ) and ( b_0 = pm1 ).But earlier, we saw that modulo 3, one of the factors must have all coefficients divisible by 3 except the leading term. So, if ( g(x) ) has coefficients divisible by 3 except the leading term, then ( a_0 = 0 ) or multiple of 3, but we just concluded ( a_0 = pm1 ), which is a contradiction.Therefore, our assumption that ( f(x) ) factors into ( g(x)h(x) ) must be wrong. Hence, ( f(x) ) is irreducible over ( mathbb{Z}[x] ).Wait, let me double-check this reasoning. If ( g(x) ) has coefficients divisible by 3 except the leading term, then ( a_0 ) must be divisible by 3, but we have ( a_0 = pm1 ), which is not divisible by 3. Hence, no such factorization exists. So, yes, this seems correct.Another way to think about it is using the concept of Newton polygons or looking at the valuations, but I think the argument above using reduction modulo 3 and the constant term is sufficient.So, in conclusion, the polynomial ( f(x) = x^n + 5x^{n-1} + 3 ) cannot be factored into the product of two non-constant polynomials with integer coefficients, hence it is irreducible in ( mathbb{Z}[x] ).