🧠:Okay, so I need to find the value of ( P_{2010}left(-frac{1}{2}right) ), where ( P_k(x) ) is the polynomial that satisfies ( P_k(n) = 1^k + 2^k + cdots + n^k ) for all positive integers ( n ). Hmm, this seems related to Faulhaber's formula, right? Faulhaber's formula expresses sums of powers of integers in terms of Bernoulli numbers. But I'm not entirely sure how to apply that here, especially since we need to evaluate the polynomial at ( x = -frac{1}{2} ), which is not a positive integer. Let me think through this step by step.First, I recall that these polynomials ( P_k(x) ) are known as the Faulhaber polynomials. For example, when ( k = 1 ), we have ( P_1(x) = frac{1}{2}x(x+1) ), which gives the sum of the first ( n ) integers. For higher powers, the polynomials get more complex, involving Bernoulli numbers. The general formula is something like:[P_k(x) = frac{1}{k+1} sum_{j=0}^k binom{k+1}{j} B_j x^{k+1 - j}]where ( B_j ) are Bernoulli numbers, with ( B_0 = 1 ), ( B_1 = -frac{1}{2} ), ( B_2 = frac{1}{6} ), and so on. But I need to confirm if this is the correct formula. Let me check for ( k = 1 ):Using the formula,[P_1(x) = frac{1}{2} left[ binom{2}{0} B_0 x^2 + binom{2}{1} B_1 x^1 right] = frac{1}{2} left[ 1 cdot 1 cdot x^2 + 2 cdot left(-frac{1}{2}right) x right] = frac{1}{2} left( x^2 - x right ) = frac{1}{2}x(x - 1)]Wait, that's not matching the given ( P_1(x) = frac{1}{2}x(x+1) ). Hmm, so maybe I messed up the indices or the formula is slightly different? Let me check again. Maybe the formula starts at ( j = 0 ), but perhaps there's an adjustment for the Bernoulli numbers.Wait, actually, the correct Faulhaber's formula is:[sum_{i=1}^n i^k = frac{1}{k+1} sum_{j=0}^k binom{k+1}{j} B_j n^{k+1 - j}]So the polynomial ( P_k(x) ) is indeed ( frac{1}{k+1} sum_{j=0}^k binom{k+1}{j} B_j x^{k+1 - j} ). But when we plug ( k = 1 ), we get:[P_1(x) = frac{1}{2} left[ B_0 x^2 + binom{2}{1} B_1 x^1 right] = frac{1}{2} left[ 1 cdot x^2 + 2 cdot left(-frac{1}{2}right) x right] = frac{1}{2}(x^2 - x)]But the known formula is ( frac{1}{2}x(x + 1) ). There's a discrepancy here. That suggests that perhaps the formula uses a different convention for Bernoulli numbers or there's an offset. Wait, maybe some sources define Bernoulli numbers with ( B_1 = +frac{1}{2} ) instead of ( -frac{1}{2} ). Let me check.No, actually, the Bernoulli numbers typically have ( B_1 = -frac{1}{2} ). So why is there a discrepancy? Let me compute the sum ( 1 + 2 + cdots + n ) using the formula:[frac{1}{2} left( B_0 n^2 + 2 B_1 n right) = frac{1}{2}(n^2 - n) = frac{n(n - 1)}{2}]But the actual sum is ( frac{n(n + 1)}{2} ). So clearly, something's wrong here. Maybe the formula is different? Or perhaps the indexing?Wait, I think the standard formula for Faulhaber's formula is actually:[sum_{i=1}^n i^k = frac{1}{k+1} sum_{j=0}^k binom{k+1}{j} B_j n^{k+1 - j}]But when ( j = 0 ), the term is ( binom{k+1}{0} B_0 n^{k+1} ), which is ( 1 cdot 1 cdot n^{k+1} ). Then the next term is ( binom{k+1}{1} B_1 n^{k} ), which is ( (k+1)(-1/2) n^k ). Then for ( j geq 2 ), the terms involve higher Bernoulli numbers.Wait, so when ( k = 1 ):[sum_{i=1}^n i = frac{1}{2} left[ B_0 n^2 + 2 B_1 n right] + text{higher terms?}]But since ( k = 1 ), the sum is up to ( j = k = 1 ). Wait, no, the upper limit is ( j = k ). So when ( k = 1 ), the sum over ( j ) is from 0 to 1. So:[frac{1}{2} left[ binom{2}{0} B_0 n^2 + binom{2}{1} B_1 n^{1} right] = frac{1}{2} left[ 1 cdot 1 cdot n^2 + 2 cdot (-1/2) cdot n right] = frac{1}{2}(n^2 - n) = frac{n(n - 1)}{2}]But that contradicts the known formula ( frac{n(n + 1)}{2} ). Therefore, there must be an error in my understanding of the formula. Let me check a reference.Ah, here's the thing: The formula is actually:[sum_{i=1}^n i^k = frac{1}{k+1} sum_{j=0}^k binom{k+1}{j} B_j n^{k+1 - j}]But when ( k = 1 ), this gives:[sum_{i=1}^n i = frac{1}{2} left( B_0 n^2 + 2 B_1 n right ) = frac{1}{2}n^2 + frac{1}{2}(-1) n = frac{n^2 - n}{2} = frac{n(n - 1)}{2}]Wait, but that's not correct. The sum ( 1 + 2 + cdots + n ) is ( frac{n(n + 1)}{2} ). Therefore, there must be a mistake in the formula. Perhaps the formula actually starts the summation at ( i=0 ) instead of ( i=1 ), but no, the standard formula is for ( i=1 ) to ( n ).Alternatively, maybe some sources use a different convention where the Bernoulli numbers are adjusted. Wait, maybe the formula includes ( B_1 = +1/2 ), but that contradicts the standard definition where ( B_1 = -1/2 ).Alternatively, perhaps there's an alternate form where the polynomial is expressed differently. Wait, let's consider that maybe the formula is shifted. For instance, in some presentations, the Bernoulli numbers are defined such that the sum can be written as:[sum_{i=1}^n i^k = frac{1}{k+1} sum_{j=0}^k (-1)^j binom{k+1}{j} B_j n^{k+1 - j}]But with the inclusion of ( (-1)^j ), that might fix the sign. Let's try that for ( k = 1 ):[frac{1}{2} [ (-1)^0 binom{2}{0} B_0 n^2 + (-1)^1 binom{2}{1} B_1 n^1 ] = frac{1}{2}[1 cdot 1 cdot n^2 - 2 cdot (-1/2) n] = frac{1}{2}(n^2 + n) = frac{n(n + 1)}{2}]Ah! That works. So perhaps there is a sign factor involved depending on the convention. Therefore, the correct formula might actually be:[sum_{i=1}^n i^k = frac{1}{k+1} sum_{j=0}^k (-1)^j binom{k+1}{j} B_j n^{k+1 - j}]But I need to verify this. Let me check another source. Hmm, according to the Wikipedia entry on Faulhaber's formula, the formula is indeed:[sum_{i=1}^n i^k = frac{1}{k+1} sum_{j=0}^k binom{k+1}{j} B_j n^{k+1 - j}]without the ( (-1)^j ) factor. But in that case, the calculation for ( k=1 ) gives the wrong result. This suggests that either the formula is presented differently or there's an alternative convention for Bernoulli numbers.Wait a minute, perhaps the difference is due to the fact that some definitions of Bernoulli numbers use ( B_1 = +1/2 ) instead of ( -1/2 ). Let me check that. If we take ( B_1 = +1/2 ), then the formula for ( k=1 ):[frac{1}{2} [ B_0 n^2 + 2 B_1 n ] = frac{1}{2}(n^2 + n) = frac{n(n + 1)}{2}]Which is correct. But in standard definitions, ( B_1 = -1/2 ). So perhaps in some contexts, especially in Faulhaber's formula, they use a different sign convention for ( B_1 ). This is confusing. Let me check the Bernoulli numbers again.Standard Bernoulli numbers: ( B_0 = 1 ), ( B_1 = -1/2 ), ( B_2 = 1/6 ), ( B_3 = 0 ), ( B_4 = -1/30 ), etc. So, if we use these, then the formula as presented in Wikipedia must be correct, but how does it handle the case of ( k=1 )? It seems like there's a contradiction.Wait, perhaps the formula is actually:[sum_{i=0}^n i^k = frac{1}{k+1} sum_{j=0}^k binom{k+1}{j} B_j n^{k+1 - j}]But if we sum from ( i=0 ), then for ( k=1 ):[0 + 1 + 2 + cdots + n = frac{1}{2} [ B_0 (n+1)^2 + 2 B_1 (n+1) ]]Wait, that's getting more complicated. Alternatively, maybe the formula is for ( sum_{i=1}^{n-1} i^k ). Let me check.Wait, no, this is getting too confusing. Let me try to find another approach. Alternatively, perhaps the polynomial ( P_k(x) ) can be related to Bernoulli polynomials. Bernoulli polynomials satisfy certain properties, and they are connected to sums of powers.Alternatively, perhaps there's a generating function approach. The generating function for the sums ( 1^k + 2^k + cdots + n^k ) can be written in terms of exponential generating functions involving Bernoulli numbers.Alternatively, perhaps there's a relation between ( P_k(x) ) and Bernoulli polynomials. The Bernoulli polynomials ( B_k(x) ) satisfy the relation:[B_{k+1}(x + 1) - B_{k+1}(x) = (k + 1) x^k]Therefore, summing both sides from ( x = 0 ) to ( x = n ):[sum_{x=0}^n (B_{k+1}(x + 1) - B_{k+1}(x)) = (k + 1) sum_{x=0}^n x^k]The left-hand side telescopes to ( B_{k+1}(n + 1) - B_{k+1}(0) ). Therefore,[(k + 1) sum_{x=0}^n x^k = B_{k+1}(n + 1) - B_{k+1}(0)]Therefore,[sum_{x=0}^n x^k = frac{1}{k + 1} left( B_{k+1}(n + 1) - B_{k+1}(0) right )]But since ( sum_{x=1}^n x^k = sum_{x=0}^n x^k - 0^k = frac{1}{k + 1} left( B_{k+1}(n + 1) - B_{k+1}(0) right ) )Therefore, the polynomial ( P_k(n) = frac{1}{k + 1} left( B_{k+1}(n + 1) - B_{k+1}(0) right ) )Hence, ( P_k(x) = frac{1}{k + 1} left( B_{k+1}(x + 1) - B_{k+1}(0) right ) )Therefore, since we need ( P_{2010}(-1/2) ), this would be:[P_{2010}left(-frac{1}{2}right) = frac{1}{2011} left( B_{2011}left(-frac{1}{2} + 1right) - B_{2011}(0) right ) = frac{1}{2011} left( B_{2011}left(frac{1}{2}right) - B_{2011}(0) right )]So now, I need to evaluate ( B_{2011}left(frac{1}{2}right) - B_{2011}(0) ). I recall that Bernoulli polynomials at ( frac{1}{2} ) have some symmetry properties. In particular, there is a relation involving Bernoulli polynomials evaluated at ( 1 - x ):[B_n(1 - x) = (-1)^n B_n(x)]For example, ( B_n(1/2) ) can be related to ( B_n(1/2) ) when ( n ) is even or odd. Wait, substituting ( x = 1/2 ):[B_n(1 - 1/2) = B_n(1/2) = (-1)^n B_n(1/2)]Therefore, for odd ( n ), this would imply ( B_n(1/2) = -B_n(1/2) ), which implies ( B_n(1/2) = 0 ). For even ( n ), this gives ( B_n(1/2) = B_n(1/2) ), which is trivially true.Therefore, for odd ( n geq 1 ), ( B_n(1/2) = 0 ). Since 2011 is odd, ( B_{2011}(1/2) = 0 ).Therefore, ( B_{2011}(1/2) - B_{2011}(0) = 0 - B_{2011}(0) = -B_{2011}(0) ). But ( B_{n}(0) ) is equal to the Bernoulli number ( B_n ), except when ( n = 1 ), where ( B_1(0) = B_1 = -1/2 ). However, for ( n geq 1 ), ( B_n(0) = B_n ), the nth Bernoulli number.But Bernoulli numbers ( B_n ) are zero for odd ( n geq 3 ). Since 2011 is an odd integer greater than 1, ( B_{2011} = 0 ).Therefore, ( B_{2011}(0) = B_{2011} = 0 ).Hence, ( B_{2011}(1/2) - B_{2011}(0) = 0 - 0 = 0 ).Therefore, ( P_{2010}(-1/2) = frac{1}{2011} times 0 = 0 ).Wait, but let me verify each step carefully to make sure I didn't skip any critical points.First, we expressed ( P_k(x) ) in terms of Bernoulli polynomials:( P_k(x) = frac{1}{k + 1} (B_{k + 1}(x + 1) - B_{k + 1}(0)) )Then, substituting ( x = -1/2 ):( P_{2010}(-1/2) = frac{1}{2011} (B_{2011}(-1/2 + 1) - B_{2011}(0)) = frac{1}{2011} (B_{2011}(1/2) - B_{2011}(0)) )Then, using the reflection formula ( B_n(1 - x) = (-1)^n B_n(x) ). For ( x = 1/2 ), this gives ( B_n(1/2) = (-1)^n B_n(1/2) ). Therefore, for odd ( n ), ( B_n(1/2) = -B_n(1/2) ), leading to ( B_n(1/2) = 0 ). Since 2011 is odd, ( B_{2011}(1/2) = 0 ).Next, ( B_{2011}(0) = B_{2011} ), the 2011th Bernoulli number. For odd ( n geq 3 ), ( B_n = 0 ). Since 2011 is an odd integer greater than 1, ( B_{2011} = 0 ).Therefore, the difference ( B_{2011}(1/2) - B_{2011}(0) = 0 - 0 = 0 ), leading to ( P_{2010}(-1/2) = 0 ).This seems to check out. Let me see if there's another way to verify this result. Perhaps using generating functions or another property.Alternatively, consider that the polynomial ( P_k(x) ) interpolates the sum ( 1^k + 2^k + cdots + n^k ) at positive integers ( n ). However, evaluating the polynomial at non-integer points, especially negative ones, might relate to some analytic continuation or polynomial extrapolation. But since polynomials are determined by their coefficients, once we have the polynomial, plugging in any value is valid.Another approach could involve the use of the Umbral calculus or generating functions. Let me think about generating functions. The generating function for the sums ( P_k(n) ) can be written as:[sum_{k=0}^{infty} P_k(n) frac{t^k}{k!} = sum_{k=0}^infty left( sum_{m=1}^n m^k right ) frac{t^k}{k!} = sum_{m=1}^n sum_{k=0}^infty frac{(m t)^k}{k!} = sum_{m=1}^n e^{m t} = e^t frac{e^{n t} - 1}{e^t - 1}]Which simplifies to:[frac{e^{(n + 1) t} - e^t}{e^t - 1} = frac{e^{(n + 1) t} - 1}{e^t - 1} - frac{e^t - 1}{e^t - 1} = left( sum_{k=0}^infty frac{B_k (n + 1)}{k!} t^k right ) - 1]Wait, perhaps connecting this to Bernoulli numbers. Alternatively, perhaps through generating functions, we can relate ( P_k(x) ) to Bernoulli polynomials.But given that we already have an answer through the Bernoulli polynomial approach, and the steps check out, perhaps the answer is indeed zero. Let me see if there's any other perspective.Alternatively, consider the general property of polynomials. Since ( P_k(n) = 1^k + 2^k + cdots + n^k ), then ( P_k(x) ) is a polynomial of degree ( k + 1 ). For example, ( P_1(x) ) is quadratic. Then, evaluating ( P_k(-frac{1}{2}) ) might relate to some extension.But according to the previous reasoning, using the relation with Bernoulli polynomials and their properties, the answer is zero. However, let me test this with a lower k where I can compute manually.Take ( k = 1 ). Then ( P_1(x) = frac{1}{2}x(x + 1) ). Then ( P_1(-1/2) = frac{1}{2}(-1/2)(-1/2 + 1) = frac{1}{2}(-1/2)(1/2) = frac{1}{2}(-1/4) = -1/8 ). Wait, but according to the previous reasoning, for ( k = 1 ), ( P_k(-1/2) ) would be:Using ( P_k(x) = frac{1}{k + 1}(B_{k + 1}(x + 1) - B_{k + 1}(0)) )For ( k = 1 ), ( P_1(x) = frac{1}{2}(B_2(x + 1) - B_2(0)) )Bernoulli polynomial ( B_2(x) = x^2 - x + frac{1}{6} ). Therefore,( B_2(x + 1) = (x + 1)^2 - (x + 1) + frac{1}{6} = x^2 + 2x + 1 - x - 1 + frac{1}{6} = x^2 + x + frac{1}{6} )Then,( P_1(x) = frac{1}{2}(x^2 + x + frac{1}{6} - B_2(0)) ). Since ( B_2(0) = frac{1}{6} ),So,( P_1(x) = frac{1}{2}(x^2 + x + frac{1}{6} - frac{1}{6}) = frac{1}{2}(x^2 + x) = frac{1}{2}x(x + 1) ), which matches.Now, evaluating ( P_1(-1/2) ):( frac{1}{2}(-1/2)(-1/2 + 1) = frac{1}{2}(-1/2)(1/2) = -1/8 ). However, according to our previous general formula:For ( k = 1 ), ( P_1(-1/2) = frac{1}{2}(B_2(1/2) - B_2(0)) )Compute ( B_2(1/2) = (1/2)^2 - (1/2) + frac{1}{6} = 1/4 - 1/2 + 1/6 = (3/12 - 6/12 + 2/12) = (-1/12) ). Wait, that's not correct. Wait, Bernoulli polynomial ( B_2(x) = x^2 - x + frac{1}{6} ). Therefore,( B_2(1/2) = (1/2)^2 - (1/2) + 1/6 = 1/4 - 1/2 + 1/6 = (3/12 - 6/12 + 2/12) = (-1/12) )But ( B_2(0) = 0^2 - 0 + 1/6 = 1/6 ). Therefore,( P_1(-1/2) = frac{1}{2}(B_2(1/2) - B_2(0)) = frac{1}{2}(-1/12 - 1/6) = frac{1}{2}(-1/12 - 2/12) = frac{1}{2}(-3/12) = frac{1}{2}(-1/4) = -1/8 ), which matches the direct calculation.But according to the previous logic for ( k = 2010 ), which is even, leading to ( k + 1 = 2011 ), which is odd. Then, ( B_{2011}(1/2) = 0 ), and ( B_{2011}(0) = 0 ), so the difference is zero, hence ( P_{2010}(-1/2) = 0 ). But for ( k = 1 ), even though ( k + 1 = 2 ) is even, the difference ( B_2(1/2) - B_2(0) = -1/12 - 1/6 = -1/4 neq 0 ). So how does the previous logic apply here?Wait, in the case ( k = 1 ), ( k + 1 = 2 ), which is even. Then, according to the reflection formula:( B_n(1 - x) = (-1)^n B_n(x) ). For ( n = 2 ):( B_2(1 - x) = (-1)^2 B_2(x) = B_2(x) )Therefore, ( B_2(1/2) = B_2(1 - 1/2) = B_2(1/2) ), which is a tautology. So for even ( n ), ( B_n(1/2) ) isn't necessarily zero. In our previous example, ( B_2(1/2) = -1/12 ), which is non-zero.Therefore, the previous logic applies only when ( k + 1 ) is odd, i.e., when ( k ) is even. Wait, for ( k = 2010 ), ( k + 1 = 2011 ), which is odd, so ( B_{2011}(1/2) = 0 ). For ( k = 1 ), ( k + 1 = 2 ), even, so ( B_2(1/2) neq 0 ). Therefore, the previous conclusion that ( P_{2010}(-1/2) = 0 ) holds because for odd ( n = 2011 ), ( B_{2011}(1/2) = 0 ), and ( B_{2011}(0) = 0 ), leading to their difference being zero. But for even ( n ), this doesn't hold, as in the case of ( k = 1 ).Therefore, the key point is that when ( k ) is even, ( k + 1 ) is odd, so ( B_{k+1}(1/2) = 0 ), and since ( B_{k+1}(0) = B_{k+1} ), which is also zero for odd ( k+1 geq 3 ), the difference is zero, leading to ( P_{k}(-1/2) = 0 ).But wait, ( k = 2010 ) is even, so ( k + 1 = 2011 ) is odd. Therefore, ( B_{2011}(1/2) = 0 ), and ( B_{2011}(0) = B_{2011} = 0 ) because it's an odd Bernoulli number greater than 1. Thus, the difference is zero, leading to ( P_{2010}(-1/2) = 0 ).This seems consistent. Let me check another case where ( k ) is even. Let's take ( k = 2 ). Then, ( P_2(n) = 1^2 + 2^2 + cdots + n^2 = frac{n(n + 1)(2n + 1)}{6} ). Let's compute ( P_2(-1/2) ).First, directly:( P_2(-1/2) = frac{(-1/2)(-1/2 + 1)(2(-1/2) + 1)}{6} = frac{(-1/2)(1/2)(-1 + 1)}{6} = frac{(-1/2)(1/2)(0)}{6} = 0 ).So indeed, ( P_2(-1/2) = 0 ). Let's verify through the Bernoulli polynomial approach.For ( k = 2 ), ( P_2(x) = frac{1}{3}(B_3(x + 1) - B_3(0)) ).Bernoulli polynomial ( B_3(x) = x^3 - frac{3}{2}x^2 + frac{1}{2}x ).Therefore, ( B_3(x + 1) = (x + 1)^3 - frac{3}{2}(x + 1)^2 + frac{1}{2}(x + 1) ).Expanding:( (x^3 + 3x^2 + 3x + 1) - frac{3}{2}(x^2 + 2x + 1) + frac{1}{2}x + frac{1}{2} )= ( x^3 + 3x^2 + 3x + 1 - frac{3}{2}x^2 - 3x - frac{3}{2} + frac{1}{2}x + frac{1}{2} )Combine like terms:- ( x^3 )- ( 3x^2 - frac{3}{2}x^2 = frac{3}{2}x^2 )- ( 3x - 3x + frac{1}{2}x = frac{1}{2}x )- ( 1 - frac{3}{2} + frac{1}{2} = 0 )Thus, ( B_3(x + 1) = x^3 + frac{3}{2}x^2 + frac{1}{2}x )Therefore, ( P_2(x) = frac{1}{3}(x^3 + frac{3}{2}x^2 + frac{1}{2}x - B_3(0)) ). Since ( B_3(0) = 0^3 - frac{3}{2}0^2 + frac{1}{2}0 = 0 ).Therefore, ( P_2(x) = frac{1}{3}(x^3 + frac{3}{2}x^2 + frac{1}{2}x) = frac{1}{3}x^3 + frac{1}{2}x^2 + frac{1}{6}x ). Factoring:( frac{x}{6}(2x^2 + 3x + 1) = frac{x(2x + 1)(x + 1)}{6} ), which matches the standard formula ( frac{n(n + 1)(2n + 1)}{6} ).Now, evaluating ( P_2(-1/2) ):( frac{1}{3}(-1/2)^3 + frac{1}{2}(-1/2)^2 + frac{1}{6}(-1/2) = frac{1}{3}(-1/8) + frac{1}{2}(1/4) + frac{1}{6}(-1/2) )= ( -1/24 + 1/8 - 1/12 )Convert to 24 denominator:= ( -1/24 + 3/24 - 2/24 = 0/24 = 0 )Which matches the direct calculation. According to the Bernoulli polynomial approach, ( P_2(-1/2) = frac{1}{3}(B_3(1/2) - B_3(0)) )Compute ( B_3(1/2) ):Using ( B_3(x) = x^3 - frac{3}{2}x^2 + frac{1}{2}x )( B_3(1/2) = (1/2)^3 - (3/2)(1/2)^2 + (1/2)(1/2) = 1/8 - 3/2 * 1/4 + 1/4 = 1/8 - 3/8 + 2/8 = 0 )( B_3(0) = 0 )Therefore, ( P_2(-1/2) = frac{1}{3}(0 - 0) = 0 ), which matches.This supports the general result that for even ( k ), ( P_k(-1/2) = 0 ). Similarly, for ( k = 0 ), though ( k = 0 ) is a trivial case where ( P_0(n) = n ), and ( P_0(-1/2) = -1/2 ), but since ( k = 0 ) is even, but ( k + 1 = 1 ), which is odd, but ( B_1(1/2) = 0 ), but ( B_1(0) = -1/2 ). So:( P_0(-1/2) = frac{1}{1}(B_1(1/2) - B_1(0)) = 0 - (-1/2) = 1/2 ), but ( P_0(n) = n ), so ( P_0(-1/2) = -1/2 ). Contradiction. Wait, but ( k = 0 ), the sum ( 1^0 + 2^0 + ... + n^0 = n ), so ( P_0(x) = x ). Therefore, ( P_0(-1/2) = -1/2 ). However, according to the formula:( P_0(x) = frac{1}{1}(B_1(x + 1) - B_1(0)) )( B_1(x) = x - 1/2 ), so ( B_1(x + 1) = x + 1 - 1/2 = x + 1/2 )Then,( P_0(x) = (x + 1/2 - B_1(0)) ), but ( B_1(0) = -1/2 )Thus,( P_0(x) = x + 1/2 - (-1/2) = x + 1 )But this contradicts ( P_0(x) = x ). Therefore, there must be a miscalculation here.Wait, maybe the formula ( P_k(x) = frac{1}{k+1}(B_{k+1}(x + 1) - B_{k+1}(0)) ) is valid for ( k geq 1 ). For ( k = 0 ), the sum ( sum_{i=1}^n i^0 = n ), so ( P_0(x) = x ). But according to the formula:( P_0(x) = frac{1}{1}(B_1(x + 1) - B_1(0)) = B_1(x + 1) - B_1(0) )Since ( B_1(x) = x - 1/2 ), ( B_1(x + 1) = (x + 1) - 1/2 = x + 1/2 )Thus,( P_0(x) = (x + 1/2) - (-1/2) = x + 1 ). But ( P_0(x) = x ). So there's a discrepancy here. Therefore, the formula might be adjusted for ( k = 0 ). Alternatively, perhaps the original formula applies for ( k geq 1 ), and ( k = 0 ) is a special case. Therefore, we should be cautious about edge cases.However, in our problem, ( k = 2010 ), which is a large even number, so ( k + 1 = 2011 ) is odd, and the previous reasoning holds, giving ( P_{2010}(-1/2) = 0 ). This is supported by the ( k = 2 ) case, which worked, and the ( k = 1 ) case where it didn't apply (since ( k + 1 = 2 ) is even), but for ( k = 2 ), it worked.Therefore, consolidating all this, the value of ( P_{2010}(-1/2) ) is zero.Final Answerboxed{0}

✅:To find the value of ( P_{2010}left(-frac{1}{2}right) ), we use the relationship between the polynomial ( P_k(x) ) and Bernoulli polynomials. The polynomial ( P_k(x) ) can be expressed as:[P_k(x) = frac{1}{k+1} left( B_{k+1}(x + 1) - B_{k+1}(0) right)]where ( B_n(x) ) are Bernoulli polynomials. For ( k = 2010 ), we need to evaluate ( P_{2010}left(-frac{1}{2}right) ):[P_{2010}left(-frac{1}{2}right) = frac{1}{2011} left( B_{2011}left(-frac{1}{2} + 1right) - B_{2011}(0) right)]Simplifying the argument of the Bernoulli polynomial:[B_{2011}left(-frac{1}{2} + 1right) = B_{2011}left(frac{1}{2}right)]Using the reflection property of Bernoulli polynomials for odd ( n ):[B_n(1 - x) = (-1)^n B_n(x)]For ( n = 2011 ) (which is odd) and ( x = frac{1}{2} ):[B_{2011}left(1 - frac{1}{2}right) = B_{2011}left(frac{1}{2}right) = -B_{2011}left(frac{1}{2}right)]This implies:[B_{2011}left(frac{1}{2}right) = 0]Additionally, Bernoulli numbers ( B_n ) for odd ( n geq 3 ) are zero. Thus, ( B_{2011}(0) = B_{2011} = 0 ).Therefore:[B_{2011}left(frac{1}{2}right) - B_{2011}(0) = 0 - 0 = 0]Hence:[P_{2010}left(-frac{1}{2}right) = frac{1}{2011} times 0 = 0]The final answer is:[boxed{0}]

🔑:1. Given Identity: Start with the given identity for the sum of first ( n ) positive integers: [ 1 + 2 + cdots + n = frac{1}{2} n(n+1) ] This can be represented by a polynomial ( P_1(x) = frac{1}{2} x(x+1) ).2. General Polynomial ( P_k(x) ): For each positive integer ( k ), there is a unique polynomial ( P_k(x) ) such that: [ P_k(n) = 1^k + 2^k + 3^k + cdots + n^k quad text{for each } n = 1, 2, ldots ]3. Define ( f(x) ): Define ( f(x) = P_k(x) - P_k(x-1) ). This means: [ f(n) = n^k quad text{for all integers } n geq 2 ] Note that ( f ) is a polynomial, so we must have: [ f(x) = x^k ]4. Evaluate ( f(x) ) at Specific Points: In particular, for integers ( n geq 2 ), evaluate differences of ( P_k ) at consecutive integers: [ begin{aligned} P_k(-n+1) - P_k(-n) &= f(-n+1) = (n-1)^k, P_k(-n+2) - P_k(-n+1) &= f(-n+2) = (n-2)^k, vdots P_k(0) - P_k(-1) &= f(0) = 0^k = 0, P_k(1) - P_k(0) &= f(1) = 1^k = 1. end{aligned} ]5. Summing these Equations: Summing all these equalities, we get: [ P_k(1) - P_k(-n) = 1^k + 0^k + 1^k + 2^k + cdots + (n-1)^k ] This simplifies to: [ P_k(1) - P_k(-n) = 1^k + 0 + 1^k + 2^k + cdots + (n-1)^k = (2 cdot 1^k) + 2^k + cdots + (n-1)^k ]6. Define ( g(x) ): Define ( g(x) = P_k(-x) + P_k(x-1) ). Then: [ g(n) = 0 quad text{for all integers } n geq 2 ] Since ( g ) is a polynomial and it equals ( 0 ) for an infinite number of integers, we must have: [ g(x) = 0 ]7. Evaluating at ( -frac{1}{2} ): In particular, evaluate at ( x = -frac{1}{2} ): [ P_kleft(-frac{1}{2}right) + P_kleft(-frac{1}{2}-1right) = 0 ] Simplifying, we get: [ P_kleft(-frac{1}{2}right) + P_kleft(-frac{3}{2}right) = 0 ] This implies: [ P_kleft(-frac{1}{2}right) + P_kleft(-frac{1}{2}right) = 0 implies 2P_kleft(-frac{1}{2}right) = 0 ] Thus: [ P_kleft(-frac{1}{2}right) = 0 ] Conclusion:[boxed{0}]

🧠:Okay, let's try to tackle this problem step by step. The equation given is:sqrt{x+3-4 sqrt{x-1}} + sqrt{x+8-6 sqrt{x-1}} = 1We need to find all integral values of ( x ) that satisfy this equation and then sum them up. Hmm, radicals can be tricky, especially with variables inside them. Let me think about how to approach this.First, I notice that both square roots have expressions involving ( sqrt{x - 1} ). Maybe if I let ( t = sqrt{x - 1} ), that substitution could simplify things. Let me try that.Let ( t = sqrt{x - 1} ). Then, ( t geq 0 ) because square roots are non-negative. Also, squaring both sides gives ( t^2 = x - 1 ), so ( x = t^2 + 1 ). That's useful because we can substitute ( x ) in terms of ( t ) into the original equation. Let me do that.Substituting ( x = t^2 + 1 ) into each part inside the radicals:First radical: ( x + 3 - 4 sqrt{x - 1} = (t^2 + 1) + 3 - 4t = t^2 + 4 - 4t )Second radical: ( x + 8 - 6 sqrt{x - 1} = (t^2 + 1) + 8 - 6t = t^2 + 9 - 6t )So the original equation becomes:sqrt{t^2 + 4 - 4t} + sqrt{t^2 + 9 - 6t} = 1Hmm, let's see if those quadratics under the radicals can be simplified. Maybe they are perfect squares?Looking at ( t^2 - 4t + 4 ). Wait, that's ( (t - 2)^2 ). Yes, because ( (t - 2)^2 = t^2 -4t +4 ). Similarly, ( t^2 -6t +9 ) is ( (t - 3)^2 ). Perfect!So substituting back, the equation simplifies to:sqrt{(t - 2)^2} + sqrt{(t - 3)^2} = 1But the square root of a square is the absolute value. So this becomes:|t - 2| + |t - 3| = 1Ah! Now the equation is in terms of absolute values. This seems more manageable. So now, we need to solve ( |t - 2| + |t - 3| = 1 ), where ( t = sqrt{x - 1} ) and ( t geq 0 ). Also, since ( x ) must be an integer, ( x - 1 ) must be a perfect square (because ( t ) is the square root of ( x - 1 ), and ( t ) has to be a non-negative real number, but since ( x ) is integral, ( x -1 ) must be a perfect square for ( t ) to be rational or integer. Wait, actually, ( t ) doesn't have to be an integer, but ( x ) must be an integer, so ( x = t^2 + 1 ). Therefore, ( t^2 ) must be an integer minus 1. Wait, no. If ( x ) is an integer, then ( t^2 = x -1 ) must also be an integer. Therefore, ( t^2 ) is an integer, which means that ( t ) must be a non-negative real number such that ( t^2 ) is integer. So ( t ) can be the square root of an integer, but since ( x ) is integral, ( x -1 ) is a non-negative integer (since we can't take the square root of a negative number), so ( t ) is the square root of a non-negative integer, hence ( t ) is either an integer or irrational. But if ( t ) is irrational, then ( x = t^2 +1 ) would still be an integer if ( t^2 ) is an integer. Therefore, ( t ) can be sqrt(k) where k is a non-negative integer. However, since we are looking for integral x, the values of t can be sqrt(k) where k is a non-negative integer. But maybe we can consider t as a real number and then check for which x (integers) the equation holds. However, perhaps first solving for t in the equation |t - 2| + |t - 3| =1, then express x in terms of t, and then check which x are integers.So let's first solve |t -2| + |t -3| =1. Let's consider different cases based on the value of t.The absolute value expressions change their behavior at t=2 and t=3. So, the critical points are at t=2 and t=3. Therefore, we can split the problem into three intervals:1. t < 22. 2 ≤ t ≤ 33. t > 3Let's analyze each interval.Case 1: t < 2In this case, |t -2| = 2 - t and |t -3| = 3 - t. Therefore, the equation becomes:(2 - t) + (3 - t) = 1 ⇒ 5 - 2t = 1 ⇒ -2t = -4 ⇒ t = 2But in this case, we assumed t < 2, so t = 2 is not in this interval. Therefore, no solution here.Case 2: 2 ≤ t ≤ 3Here, |t -2| = t -2 and |t -3| = 3 - t. The equation becomes:(t -2) + (3 - t) = 1 ⇒ (t -2 + 3 - t) = 1 ⇒ 1 =1Wait, that simplifies to 1 =1, which is always true. So in this interval, any t between 2 and 3 (inclusive) satisfies the equation.Case 3: t >3Here, |t -2| = t -2 and |t -3| = t -3. So the equation becomes:(t -2) + (t -3) =1 ⇒ 2t -5 =1 ⇒ 2t=6 ⇒ t=3But we assumed t >3 in this case, so t=3 is not in the interval. Therefore, no solution here.So the solution set for t is the closed interval [2,3]. Therefore, any t in [2,3] satisfies the equation.But remember, t = sqrt(x -1). Therefore, t must satisfy t = sqrt(x -1) where t ∈ [2,3], which implies that x -1 = t^2, so x = t^2 +1. Since t is between 2 and 3, inclusive, t^2 is between 4 and 9, so x is between 5 and 10 inclusive. But x must be an integer. So x can be 5,6,7,8,9,10. However, we need to check which of these integers x satisfy the original equation because even though t is in [2,3], when we substitute back, we need to ensure that the radicals are defined and the equation holds. Wait, but if t is in [2,3], then x = t^2 +1, so x ranges from 5 to 10. But perhaps not all integers in 5 to 10 will satisfy the original equation because even though we have solved the absolute value equation, we have to verify if substituting back into the original radicals gives valid results.Wait, but let me think. When we made the substitution ( t = sqrt{x -1} ), then x must be at least 1, which is already considered. Then, the expressions inside the radicals in the original equation:First radical: ( x +3 -4 sqrt{x -1} ). Let's check if this is non-negative for x in 5 to 10. Let's compute for x=5: 5+3 -4*sqrt(4)=8 -4*2=8-8=0. For x=5, the first radical is sqrt(0)=0. The second radical: x+8 -6*sqrt(x -1). For x=5: 5+8 -6*2=13-12=1. sqrt(1)=1. So 0 +1=1, which satisfies the equation. Similarly, for x=10: sqrt(10+3 -4*sqrt(9)) + sqrt(10+8 -6*sqrt(9))= sqrt(13 -12) + sqrt(18 -18)= sqrt(1) + sqrt(0)=1+0=1. That works. For x=6: sqrt(6+3 -4*sqrt(5)) + sqrt(6+8 -6*sqrt(5)). Wait, sqrt(5) is irrational. Let me compute numerically. sqrt(5)≈2.236. So first radical: 9 -4*2.236≈9-8.944≈0.056. sqrt(0.056)≈0.237. Second radical:14 -6*2.236≈14-13.416≈0.584. sqrt(0.584)≈0.764. Sum≈0.237+0.764≈1.001≈1. Close enough? Hmm, but since x=6 is an integer, and we need to check if the original equation holds exactly. Wait, but sqrt(x -1) for x=6 is sqrt(5), so substituting back, the expressions inside the radicals would be:First radical: x +3 -4 sqrt(x -1) =6 +3 -4*sqrt(5)=9 -4*sqrt(5). Similarly, second radical:6 +8 -6*sqrt(5)=14 -6*sqrt(5).But then, sqrt(9 -4*sqrt(5)) + sqrt(14 -6*sqrt(5)) = sqrt( (sqrt(5) - 2)^2 ) + sqrt( (3 - sqrt(5))^2 ) = |sqrt(5) -2| + |3 - sqrt(5)|. Since sqrt(5)≈2.236, which is greater than 2 and less than 3. Therefore, sqrt(5) -2 ≈0.236, which is positive, so |sqrt(5) -2|=sqrt(5)-2. Similarly, 3 - sqrt(5)≈0.764, positive, so |3 - sqrt(5)|=3 - sqrt(5). Therefore, adding them: (sqrt(5)-2) + (3 - sqrt(5)) =1. So yes, exactly 1. Therefore, even though the intermediate steps involve irrational numbers, the equation holds exactly for x=6. Similarly for other x in 5 to10. So, in fact, all integers x from 5 to10 inclusive satisfy the equation.Wait, but let's check x=7. Then sqrt(7 -1)=sqrt(6). Then:First radical:7 +3 -4*sqrt(6)=10 -4*sqrt(6). Second radical:7 +8 -6*sqrt(6)=15 -6*sqrt(6). So sqrt(10 -4*sqrt(6)) + sqrt(15 -6*sqrt(6)).Again, check if these can be written as squares. Let's see:Suppose 10 -4*sqrt(6) can be written as (a - b*sqrt(6))^2. Then expanding:a^2 +6b^2 -2ab*sqrt(6) =10 -4*sqrt(6). Therefore:Equating the rational and irrational parts:a^2 +6b^2=102ab=4 ⇒ ab=2So solving for a and b. Let's see, possible integers a and b. Let's try a=2, b=1: then ab=2*1=2. Then a² +6b²=4 +6=10. Perfect. Therefore, 10 -4*sqrt(6)= (2 - sqrt(6))². But wait, (2 - sqrt(6))²=4 -4*sqrt(6) +6=10 -4*sqrt(6). Yes. Therefore, sqrt(10 -4*sqrt(6))=|2 - sqrt(6)|. Since sqrt(6)≈2.449>2, this is sqrt(6)-2. Similarly, 15 -6*sqrt(6)= (sqrt(6))² -6*sqrt(6) +9= (sqrt(6)-3)^2. So sqrt(15 -6*sqrt(6))=|sqrt(6)-3|=3 -sqrt(6) since sqrt(6)<3. Therefore, adding them: (sqrt(6)-2)+(3 -sqrt(6))=1. Exactly. So x=7 works.Similarly, for x=8:sqrt(8-1)=sqrt(7)First radical:8+3 -4*sqrt(7)=11 -4*sqrt(7). Second radical:8+8 -6*sqrt(7)=16 -6*sqrt(7)Check if these are perfect squares. Let's attempt:Suppose 11 -4*sqrt(7)= (a -b sqrt(7))^2. Then:a² +7b² -2ab sqrt(7)=11 -4 sqrt(7)Thus:a² +7b²=112ab=4 ⇒ ab=2Looking for integers a,b. Possible a=2, b=1: a² +7b²=4+7=11. Perfect. Therefore, 11 -4 sqrt(7)= (2 - sqrt(7))². Therefore sqrt(11 -4 sqrt(7))=|2 - sqrt(7)|=sqrt(7)-2 (since sqrt(7)≈2.645>2)Second radical:16 -6 sqrt(7). Let's check if this is a square. Let's suppose (c - d sqrt(7))²=16 -6 sqrt(7). Then:c² +7d² -2cd sqrt(7)=16 -6 sqrt(7)Thus:c² +7d²=162cd=6 ⇒ cd=3Possible c=3, d=1: c² +7d²=9+7=16. Perfect. Therefore, 16 -6 sqrt(7)= (3 - sqrt(7))². sqrt(16 -6 sqrt(7))=|3 - sqrt(7)|=3 - sqrt(7). Therefore, adding the two radicals: (sqrt(7)-2)+(3 - sqrt(7))=1. So x=8 works.Similarly, x=9:sqrt(9-1)=sqrt(8)=2*sqrt(2)First radical:9 +3 -4*sqrt(8)=12 -4*2*sqrt(2)=12 -8 sqrt(2). Second radical:9 +8 -6*sqrt(8)=17 -12 sqrt(2)Wait, but 12 -8 sqrt(2) might not be a perfect square. Wait, let's check:Assume 12 -8 sqrt(2) = (a - b sqrt(2))^2.Expanding: a² + 2b² - 2ab sqrt(2) =12 -8 sqrt(2)Thus:a² +2b²=122ab=8 ⇒ ab=4Looking for integers a,b. Possible a=4, b=1: a² +2b²=16+2=18≠12. a=2, b=2: a² +2b²=4 +8=12. ab=4. Yes, a=2, b=2. So:(2 - 2 sqrt(2))²=4 -8 sqrt(2) +8=12 -8 sqrt(2). Yes! Therefore sqrt(12 -8 sqrt(2))=|2 - 2 sqrt(2)|=2 sqrt(2)-2 (since 2 sqrt(2)≈2.828>2)Second radical:17 -12 sqrt(2). Let's check if this is a square.Assume (c - d sqrt(2))²=17 -12 sqrt(2). Expanding:c² +2d² -2cd sqrt(2)=17 -12 sqrt(2)Thus:c² +2d²=172cd=12 ⇒ cd=6Possible integer solutions: c=3, d=2: c² +2d²=9 +8=17. Yes. Therefore, (3 - 2 sqrt(2))²=9 -12 sqrt(2) +8=17 -12 sqrt(2). So sqrt(17 -12 sqrt(2))=|3 -2 sqrt(2)|=3 -2 sqrt(2) (since 2 sqrt(2)≈2.828<3). Therefore, adding the two radicals: (2 sqrt(2)-2)+(3 -2 sqrt(2))=1. Exactly. So x=9 works.Similarly, x=10:sqrt(10-1)=sqrt(9)=3First radical:10 +3 -4*3=13-12=1. sqrt(1)=1Second radical:10 +8 -6*3=18-18=0. sqrt(0)=0So 1 +0=1. Therefore x=10 works.Therefore, all integers x from 5 to10 inclusive satisfy the equation. So the integral values are 5,6,7,8,9,10.Now, we need to find the sum of these values: 5+6+7+8+9+10.Calculating: 5+6=11, 11+7=18, 18+8=26, 26+9=35, 35+10=45.So the sum is 45.Wait, but let me double-check if all these x are valid. For example, x=5: as before, radicals become sqrt(0)+sqrt(1)=0+1=1. Good. x=6,7,8,9,10 all checked out in the earlier steps. So yes, all these x satisfy the equation. Therefore, the sum is 45.Final AnswerThe sum of all integral values of ( x ) is boxed{45}.

✅:To solve the equation sqrt{x+3-4 sqrt{x-1}} + sqrt{x+8-6 sqrt{x-1}} = 1,we start by substituting ( t = sqrt{x - 1} ). This gives ( x = t^2 + 1 ). Substituting ( x = t^2 + 1 ) into the equation, we simplify the expressions inside the radicals:- The first radical becomes ( sqrt{(t - 2)^2} = |t - 2| ).- The second radical becomes ( sqrt{(t - 3)^2} = |t - 3| ).Thus, the equation simplifies to:|t - 2| + |t - 3| = 1.We solve this absolute value equation by considering different intervals for ( t ):1. For ( t < 2 ): - The equation becomes ( (2 - t) + (3 - t) = 1 ), which simplifies to ( 5 - 2t = 1 ). Solving gives ( t = 2 ), which is not in this interval.2. For ( 2 leq t leq 3 ): - The equation becomes ( (t - 2) + (3 - t) = 1 ), which simplifies to ( 1 = 1 ). This is always true, so all ( t ) in this interval are solutions.3. For ( t > 3 ): - The equation becomes ( (t - 2) + (t - 3) = 1 ), which simplifies to ( 2t - 5 = 1 ). Solving gives ( t = 3 ), which is not in this interval.Thus, ( t ) must be in the interval ([2, 3]). Converting back to ( x ), we have ( x = t^2 + 1 ). Since ( t in [2, 3] ), ( t^2 ) ranges from 4 to 9, so ( x ) ranges from 5 to 10. The integral values of ( x ) in this range are 5, 6, 7, 8, 9, and 10.We verify that each of these values satisfies the original equation. Summing these integral values:5 + 6 + 7 + 8 + 9 + 10 = 45.The sum of all integral values of ( x ) is (boxed{45}).

🔑:1. First, we start with the original equation: [ sqrt{x+3-4sqrt{x-1}} + sqrt{x+8-6sqrt{x-1}} = 1 ]2. Let us introduce a substitution to simplify the problem. Let: [ y = sqrt{x-1} ] Thus, x = y^2 + 1.3. Substitute ( y ) back into the equation: [ sqrt{(y^2 + 1) + 3 - 4y} + sqrt{(y^2 + 1) + 8 - 6y} = 1 ] Simplify inside the square roots: [ sqrt{y^2 - 4y + 4} + sqrt{y^2 - 6y + 9} = 1 ] Notice that the expressions inside the square roots can be written as perfect squares: [ sqrt{(y-2)^2} + sqrt{(y-3)^2} = 1 ] Hence, it can be further simplified to: [ |y-2| + |y-3| = 1 ]4. Consider different cases based on the value of (y): Case 1: ( y geq 3 ): [ (y-2) + (y-3) = 1 ] Simplify: [ y-2 + y-3 = 1 ] [ 2y - 5 = 1 ] [ 2y = 6 ] [ y = 3 ] Substituting back for (y): [ y = 3 implies x = y^2 + 1 = 3^2 + 1 = 10 ] Case 2: ( y leq 2 ): [ (2-y) + (3-y) = 1 ] Simplify: [ 5 - 2y = 1 ] [ 2y = 4 ] [ y = 2 ] Substituting back for (y): [ y = 2 implies x = y^2 + 1 = 2^2 + 1 = 5 ] Case 3: ( 2 < y < 3 ): Here sqrt{x-1} lies between 2 and 3. So, (2-y) equals (y-2) which equals 1. This case is generally true for all values between: [ 2 < y < 3 implies 5 < x < 10 ] For all these ( x ), the initial equation holds. Therefore, the integer values for ( x ) in this range are 6, 7, 8, and 9.5. Combining all feasible integer solutions: [ text{The values are } 5, 6, 7, 8, 9, 10. ] Thus, the sum of all integral values of ( x ) is: [ 5 + 6 + 7 + 8 + 9 + 10 = 45 ]Conclusion: [ boxed{45} ]

🧠:Okay, let me try to figure out how to prove that AD + DC is less than or equal to AB + BC in triangle ABC with a point D inside. Hmm, so first, I need to visualize the problem. Let me draw triangle ABC. Let me label the vertices A, B, and C. Now, point D is somewhere inside the triangle. So, AD and DC are two line segments from A to D and from D to C. The problem states that the sum of these two segments is less than or equal to AB + BC. Wait, AB and BC are sides of the triangle, right? So AB is one side, BC is another side. So we need to show that the path from A to D to C is shorter than or equal to the path from A to B to C. But since D is inside the triangle, maybe there's a way to relate these lengths using some triangle inequalities or properties.Let me recall the triangle inequality theorem. It says that the sum of the lengths of any two sides of a triangle is greater than the length of the remaining side. So, in triangle ABC, AB + BC > AC. But here, we have a point D inside the triangle, so maybe I can apply the triangle inequality to different triangles involving D.If I look at triangle ADC, then AD + DC > AC by the triangle inequality. But we need AD + DC compared to AB + BC. Since AC is less than AB + BC (from the triangle inequality in ABC), then AD + DC > AC < AB + BC. But this doesn't directly help because we need to show AD + DC ≤ AB + BC, not comparing both to AC.Hmm, maybe another approach. Since D is inside the triangle ABC, perhaps we can extend some lines or construct auxiliary lines to create triangles where we can apply inequalities. For example, maybe reflecting point D over a side or something?Wait, another thought: In a triangle, the shortest path from a point to another point is a straight line. Since D is inside ABC, maybe the path from A to D to C can be compared to the path from A to B to C by constructing a path that goes through D but relates to the sides AB and BC.Alternatively, maybe using the concept of convexity. A triangle is a convex set, so any point inside can be expressed as a convex combination of the vertices. Not sure if that helps here.Wait, maybe using the triangle inequality in different triangles. Let's consider triangles ABD and BDC.In triangle ABD, we have AD ≤ AB + BD. Similarly, in triangle BDC, DC ≤ BD + BC. If I add these two inequalities: AD + DC ≤ AB + BD + BD + BC = AB + BC + 2BD. But BD is a positive length since D is inside the triangle, so this gives AD + DC ≤ AB + BC + 2BD, which is not helpful because it's a larger upper bound. So this approach might not work.Alternatively, maybe considering the path from A to C via D and comparing it to the path via B. If we can show that the path via D is shorter or equal to the path via B, but I'm not sure. Wait, the problem states that AD + DC ≤ AB + BC. Since AB + BC is the length of the path from A to B to C, but D is inside, maybe there's a way to relate these.Another idea: Reflect point C over AB or something. Sometimes reflecting points can create symmetrical situations where inequalities can be applied. Let me try that. If I reflect point C over AB, but not sure if that's the right reflection. Alternatively, maybe reflect D over AB?Wait, here's a technique I remember for problems involving points inside triangles and distances: using the concept of the shortest path. If D is inside triangle ABC, then the path from A to D to C must be compared to the perimeter path from A to B to C. But how?Alternatively, think about the Law of Cosines. Maybe express AD and DC in terms of other sides and angles. But that might get complicated without knowing specific angles.Wait, another approach: Consider the point D and connect it to B. Then, in triangle ABD, AD ≤ AB + BD, and in triangle BDC, DC ≤ BD + BC. If I add these, as before, I get AD + DC ≤ AB + 2BD + BC. Since BD is positive, this gives AD + DC < AB + BC + 2BD. But we need to get rid of the 2BD term. Maybe this approach isn't helpful.Alternatively, perhaps consider using the triangle inequality on a different set of triangles. For example, if we connect A to D and D to C, can we relate AD + DC to AB + BC through another triangle?Wait, maybe using the concept of the ellipse. The set of points D such that AD + DC is constant is an ellipse with foci at A and C. The maximum value of AD + DC inside the triangle would be when D is on the side AC, but that would just be AC. However, the problem states AD + DC ≤ AB + BC, which is a different value. So maybe the maximum occurs at a different point.Alternatively, since AB + BC is the length of two sides of the triangle, perhaps the path from A to B to C is being considered. If D is inside the triangle, maybe AD + DC can't exceed AB + BC because of the triangle's properties.Wait, here's an idea: Let's consider the path from A to D to C. If we can show that this path is shorter than going around the triangle via B, then we are done. But how?Alternatively, consider that the shortest path from A to C is the straight line AC, which is less than AB + BC. But AD + DC is greater than AC (by triangle inequality). But this contradicts the problem statement. Wait, no, the problem says AD + DC ≤ AB + BC, not AC. Since AC < AB + BC, then AD + DC > AC < AB + BC. But that doesn't necessarily give AD + DC ≤ AB + BC. So this approach isn't directly helpful.Wait, maybe there's an inequality that relates the sum of two sides of a triangle to the sum of two other sides. For example, in triangle ABC, AB + BC > AC, but how does that relate to AD + DC?Alternatively, maybe use the concept of Ptolemy's inequality, since we're dealing with a quadrilateral. If D is inside triangle ABC, then ABCD is a quadrilateral (though not necessarily convex). Ptolemy's inequality states that for any quadrilateral, the product of the diagonals is less than or equal to the sum of the products of the opposite sides. But I'm not sure if that applies here.Wait, Ptolemy's inequality for a convex quadrilateral: AC·BD ≤ AB·CD + AD·BC. Not sure if that's helpful here. Maybe not.Alternatively, consider the triangle areas. If D is inside ABC, then the areas of ABD, BDC, and ADC add up to the area of ABC. But I don't see how area relates to the lengths.Wait, another thought: Use the concept of the shortest path. If D is inside ABC, then the minimal path from A to C through D must be at least AC, but we need to compare it to AB + BC. Not sure.Wait, perhaps using vector coordinates. Let me assign coordinates to the triangle. Let me set point B at the origin (0,0), point A at (a,0), and point C at (0,c), making triangle ABC a right triangle for simplicity. Then, point D is inside this triangle, so its coordinates are (x,y) where x > 0, y > 0, and x/a + y/c < 1 (if it's a right triangle). Then, AD is the distance from A (a,0) to D (x,y): sqrt((x - a)^2 + y^2). DC is the distance from D (x,y) to C (0,c): sqrt(x^2 + (y - c)^2). AB is the distance from A (a,0) to B (0,0): a. BC is the distance from B (0,0) to C (0,c): c. So we need to prove that sqrt((x - a)^2 + y^2) + sqrt(x^2 + (y - c)^2) ≤ a + c.But this seems complicated. Maybe using the triangle inequality in coordinate geometry? Let's see. If I consider the path from A to D to C, the total distance is AD + DC. Compare that to going from A to B to C, which is AB + BC = a + c.Alternatively, since moving from A to B to C is a path that goes around the triangle, and D is inside, perhaps the path through D is shorter. Wait, but the problem states that AD + DC is less than or equal to AB + BC, so maybe it's not necessarily shorter, but it's bounded by AB + BC.Wait, in this coordinate system, if D approaches B, then AD approaches AB, and DC approaches BC, so AD + DC approaches AB + BC. So the maximum of AD + DC occurs when D is at B. If D is anywhere else inside the triangle, AD + DC would be less than AB + BC. Is that true?Wait, let me test with an example. Let me take a specific triangle. Let's say ABC is a right triangle with A at (1,0), B at (0,0), and C at (0,1). Then AB = 1, BC = 1, so AB + BC = 2. Let me pick D at the centroid, which is (1/3, 1/3). Then AD is sqrt((1/3 - 1)^2 + (1/3 - 0)^2) = sqrt((-2/3)^2 + (1/3)^2) = sqrt(4/9 + 1/9) = sqrt(5/9) ≈ 0.745. DC is sqrt((1/3 - 0)^2 + (1/3 - 1)^2) = sqrt((1/3)^2 + (-2/3)^2) = sqrt(1/9 + 4/9) = sqrt(5/9) ≈ 0.745. So AD + DC ≈ 1.49, which is less than 2. If D is at (0.5,0.5), then AD is sqrt((0.5 - 1)^2 + (0.5)^2) = sqrt(0.25 + 0.25) = sqrt(0.5) ≈ 0.707. DC is sqrt((0.5)^2 + (0.5 - 1)^2) = same as AD, so total ≈ 1.414, still less than 2. If D approaches B, say D is at (0.1,0.1), then AD is sqrt((0.1 - 1)^2 + 0.1^2) = sqrt(0.81 + 0.01) = sqrt(0.82) ≈ 0.906. DC is sqrt(0.1^2 + (0.1 - 1)^2) = same as AD, so total ≈ 1.812, which is still less than 2. When D approaches B, like D = (0.01,0.01), AD ≈ sqrt(0.98^2 + 0.01^2) ≈ 0.98, DC similarly ≈ 0.98, so total ≈ 1.96, which is close to 2 but still less. So the maximum seems to approach 2 as D approaches B, but never actually reaches it unless D is exactly at B. Since D is inside the triangle, not on the boundary, but wait, the problem says "inside the triangle", but some sources consider boundary points as part of the triangle. If D is allowed to be on the boundary, then when D is at B, AD + DC = AB + BC. So perhaps the maximum is achieved when D coincides with B, hence AD + DC ≤ AB + BC.Therefore, maybe the inequality holds because the maximum value of AD + DC occurs when D is at B, giving equality, and for any other point inside, it's less. So how to formalize this?Perhaps considering reflecting point B over line AC or something else. Wait, here's a method I recall for such inequalities: reflecting a point to turn the problem into a straight line.Let me try reflecting point C over AB to get a point C'. Then, the distance from A to C' is equal to AC, but maybe this can help. Alternatively, reflecting point A over BC.Wait, here's an idea. Let's reflect point B across the line AC to get point B'. Then, AB + BC = AB + BC. Hmm, not sure. Alternatively, if we reflect point C over AB, then the path from A to B to C becomes a straight line from A to C'. Wait, let me think.If we reflect point C over AB to get C', then the length AC' is equal to AB + BC. Because reflecting C over AB, the path A to B to C is equivalent to the straight line from A to C'. Therefore, any path from A to C via a point on AB (like B) corresponds to a straight line to C'. Therefore, if we can show that AD + DC ≤ AC', which is AB + BC, then we have the result.But how does D relate to this reflection? Since D is an arbitrary point inside the triangle, not necessarily on AB. Wait, but if we reflect C over AB to get C', then for any point D inside ABC, the path AD + DC is less than or equal to the path from A to C', which is AB + BC. Wait, maybe this is the way.In reflection problems, the shortest path from A to C via a point on AB is achieved by reflecting C to C' and taking the straight line. But here, D is not restricted to AB. However, since D is inside the triangle ABC, maybe we can use the reflection to compare.Alternatively, consider that AD + DC is maximized when D is on the boundary of the triangle. If that's the case, then the maximum would be at vertex B, giving AD + DC = AB + BC. Therefore, for any interior point, AD + DC is less than this maximum.But how to prove that the maximum occurs at B?Alternatively, parameterize the position of D in the triangle and show that the maximum of AD + DC occurs at B. But this would require calculus and partial derivatives, which might be complicated.Alternatively, use the concept of convex functions. The function f(D) = AD + DC is a convex function, and the maximum over a convex set (the triangle) occurs at an extreme point (vertex). But the vertices are A, B, C. Let's check the values at the vertices. At D = A, AD + DC = 0 + AC. At D = C, AD + DC = AD + 0. At D = B, AD + DC = AB + BC. Since AC < AB + BC (by triangle inequality), then the maximum among the vertices is AB + BC. Therefore, if the maximum of AD + DC over the triangle occurs at B, then for any D inside the triangle, AD + DC ≤ AB + BC.But does the function f(D) = AD + DC attain its maximum at a vertex? In general, for convex functions over convex polygons, the maximum is attained at a vertex. But is f(D) convex?Wait, the sum of two convex functions is convex. The distance from a fixed point (A) is a convex function, and the distance to another fixed point (C) is also convex. Therefore, f(D) = AD + DC is convex. Therefore, over the convex set (triangle ABC), the maximum is attained at one of the vertices. Therefore, since at B, f(D) = AB + BC, which is the maximum among the vertices, then indeed for any D in the triangle, AD + DC ≤ AB + BC.Therefore, this would be the proof: Since f(D) = AD + DC is a convex function on the convex set triangle ABC, its maximum occurs at one of the vertices. Evaluating at each vertex, the maximum is AB + BC when D = B. Therefore, for any point D inside the triangle, AD + DC ≤ AB + BC.But wait, in our earlier example with D approaching B, AD + DC approaches AB + BC, but when D is exactly at B, which is a vertex, then it's equal. So the inequality holds with equality when D is at B, and strictly less otherwise.Therefore, this would be a valid proof using convexity. However, maybe the problem expects a more elementary proof without invoking convex functions.Alternatively, here's another approach using triangle inequality and constructing auxiliary lines.Let me consider triangle ABD and triangle CBD.Wait, connect BD. Then, in triangle ABD, we have AD ≤ AB + BD. In triangle CBD, DC ≤ BD + BC. Adding these two inequalities:AD + DC ≤ AB + BD + BD + BC = AB + BC + 2BD.But this gives AD + DC ≤ AB + BC + 2BD, which is not helpful since BD is positive. So we need a different approach.Wait, another idea: Use the fact that in triangle ABC, for any point D inside, the sum of distances from D to two vertices is less than the sum of two sides. Wait, but this is similar to the original problem.Alternatively, consider extending AD to meet BC at some point E. Then, in triangle AEC, we can apply the triangle inequality.Let me try this. Let's extend AD to meet BC at E. Then, in triangle AEC, AE + EC > AC. But not sure how this relates to AD + DC.Alternatively, in triangle EDC, if E is on BC, then DC ≤ DE + EC. But maybe not helpful.Alternatively, since E is on BC, then AE = AD + DE. So AE + EC = AD + DE + EC. But EC is part of BC. So maybe AE + EC = AD + DC ≤ something.Wait, perhaps using the triangle inequality on path A to E to C. But AE + EC ≥ AC. Not helpful.Alternatively, since E is on BC, then BE + EC = BC. So, AE + EC = AE + (BC - BE). But not sure.Wait, here's a different approach inspired by reflection. Reflect point A over the side BC to get point A'. Then, the distance from A' to C is equal to AC. But perhaps this can be used to relate AD + DC.Alternatively, reflect point D over BC to get D', but I'm not sure.Wait, going back to the reflection idea. If we reflect point C over AB to get C', then AC' = AB + BC. Then, for any point D on AB, AD + DC = AD + DC' (since DC = DC' because of reflection). Therefore, the minimal path from A to C via AB is AC', but since D is not restricted to AB, but inside the triangle, perhaps AD + DC ≤ AC' = AB + BC.But how does this work if D is not on AB? Wait, but if D is inside the triangle, then maybe AD + DC ≤ AC' by some geometric argument.Alternatively, consider that in the plane, the shortest path from A to C via a point D is the straight line AC. But if we constrain D to be inside the triangle, the maximum AD + DC might be achieved when D is at B.Alternatively, think of AD + DC as a path from A to C through D. The maximum possible length of this path within the triangle would be when D is as far as possible from both A and C, which might be at B.But how to formalize this?Wait, using the triangle inequality again. For any point D inside ABC, consider the path AB + BC. If we can show that AD + DC is less than or equal to AB + BC by relating the segments through point B.Alternatively, using the concept of adding inequalities. Let's look at triangle ABD: AD ≤ AB + BD. In triangle BCD: DC ≤ BD + BC. If we add these, as before, we get AD + DC ≤ AB + 2BD + BC. But we need AD + DC ≤ AB + BC. So unless BD is zero, which would mean D = B, this inequality is not tight. Therefore, this approach doesn't give the required result.Wait, another approach: Use the fact that in a triangle, the sum of two sides minus the third side is positive. Maybe consider the difference between AB + BC and AD + DC.Let’s compute AB + BC - (AD + DC). We need to show that this is ≥ 0.AB + BC - AD - DC = (AB - AD) + (BC - DC). Hmm, not sure.Alternatively, consider vectors. Let me place point B at the origin. Let vector BA = a and vector BC = c. Then, point D can be expressed as a convex combination: d = λa + μc, where λ, μ ≥ 0 and λ + μ ≤ 1. Then, AD = ||d - a|| and DC = ||c - d||. Then, need to show that ||d - a|| + ||c - d|| ≤ ||a|| + ||c||.But ||a|| is AB, and ||c|| is BC. Wait, this looks like the triangle inequality for vectors. The left side is the sum of distances from d to a and c to d, which is the same as the distance from a to c via d. The right side is the sum of the distances from a to origin (B) and from origin to c, which is AB + BC.But unless d is on the path from a to origin to c, which is not necessarily the case, how does this relate? Wait, perhaps using the triangle inequality in the vector space:||d - a|| + ||c - d|| ≥ ||c - a||, which is the usual triangle inequality. But we need the opposite inequality, which is not generally true. So this approach might not work.Alternatively, since D is inside the triangle, perhaps d can be expressed as d = ta + sc, with t + s ≤ 1 and t, s ≥ 0. Then, compute AD + DC in terms of t and s, and show it's ≤ AB + BC. Let's try.Let’s set B at the origin, A at vector a, C at vector c. Then, D is ta + sc, where t, s ≥ 0 and t + s ≤ 1.AD = ||d - a|| = ||(ta + sc) - a|| = ||(t - 1)a + sc||.DC = ||c - d|| = ||c - (ta + sc)|| = || - ta + (1 - s)c||.We need to show that ||(t - 1)a + sc|| + || - ta + (1 - s)c|| ≤ ||a|| + ||c||.This seems complicated. Maybe in specific cases, like when a and c are orthogonal, but in general, it's not straightforward.Wait, if a and c are orthogonal, let's assume a = (a, 0) and c = (0, c). Then, D = (ta, sc) where t, s ≥ 0 and t + s ≤ 1.AD = sqrt((ta - a)^2 + (sc - 0)^2) = sqrt((a(1 - t))^2 + (sc)^2).DC = sqrt((ta - 0)^2 + (sc - c)^2) = sqrt((ta)^2 + (c(1 - s))^2).AB = a, BC = c.We need to show that sqrt(a²(1 - t)² + s²c²) + sqrt(t²a² + c²(1 - s)²) ≤ a + c.This still looks complex, but maybe using the Minkowski inequality. The Minkowski inequality states that for vectors u and v, ||u + v|| ≤ ||u|| + ||v||. But here, we have two separate terms. Maybe consider squaring both sides.Let’s square both sides:[sqrt(a²(1 - t)² + s²c²) + sqrt(t²a² + c²(1 - s)²)]² ≤ (a + c)².Expanding the left side:a²(1 - t)² + s²c² + t²a² + c²(1 - s)² + 2*sqrt([a²(1 - t)² + s²c²][t²a² + c²(1 - s)²]).Simplify the terms:a²[(1 - t)² + t²] + c²[s² + (1 - s)²] + 2*sqrt([a²(1 - t)² + s²c²][t²a² + c²(1 - s)²]).Calculate (1 - t)² + t² = 1 - 2t + 2t².Similarly, s² + (1 - s)² = 1 - 2s + 2s².So the left side becomes:a²(1 - 2t + 2t²) + c²(1 - 2s + 2s²) + 2*sqrt([a²(1 - t)² + s²c²][t²a² + c²(1 - s)²]).The right side is (a + c)^2 = a² + 2ac + c².Therefore, subtracting left side from right side:a² + 2ac + c² - [a²(1 - 2t + 2t²) + c²(1 - 2s + 2s²) + 2*sqrt(...)] ≥ 0.Simplify the terms:a²[1 - (1 - 2t + 2t²)] + c²[1 - (1 - 2s + 2s²)] + 2ac - 2*sqrt(...).Which simplifies to:a²(2t - 2t²) + c²(2s - 2s²) + 2ac - 2*sqrt(...).Factor out 2:2[a²(t - t²) + c²(s - s²) + ac] - 2*sqrt(...).Divide both sides by 2:a²(t - t²) + c²(s - s²) + ac - sqrt([a²(1 - t)² + s²c²][t²a² + c²(1 - s)²]) ≥ 0.This is getting too complicated. Maybe there's a better way.Alternatively, since we have t + s ≤ 1, maybe substitute s = 1 - t - k, where k ≥ 0. But this might not help.Alternatively, consider that since D is inside the triangle, t and s are between 0 and 1. Let's consider the maximum of AD + DC.Take partial derivatives with respect to t and s and set to zero to find critical points. But this is calculus and might not be suitable for an elementary proof.Alternatively, use the fact that AD + DC achieves its maximum when D is at B, which is when t = 0 and s = 0 (if we placed B at the origin). Wait, in our coordinate system, D at B is (0,0), so t = 0, s = 0. Then AD = AB = a, DC = BC = c, so AD + DC = a + c. For other points inside, AD + DC is less. Hence, the maximum occurs at B, proving the inequality.But how to justify that without calculus? Maybe by noting that moving D away from B along any direction would either increase AD or DC, but not both. Wait, not necessarily. For example, moving D from B towards A would increase DC but decrease AD. However, the net effect might be a decrease in the sum.Alternatively, since B is the common vertex of AB and BC, and any movement from B would require moving along a direction that is not colinear with both AB and BC, hence causing at least one of the distances AD or DC to increase less than the other decreases, but this is hand-wavy.Alternatively, use the triangle inequality in another form. For any point D inside ABC, consider the path A to B to D to C. Then, AB + BD + DC ≥ AD + DC (by triangle inequality AB + BD ≥ AD). But this again isn't helpful.Wait, another approach: Consider triangle ABC and point D inside. Apply the triangle inequality to quadrilateral ABDC. In a convex quadrilateral, the sum of the lengths of any three sides must exceed the fourth side. But AB + BC + CD ≥ AD? Not sure.Alternatively, think of the problem in terms of the polygon inequality. For a polygon, the sum of all sides except one is greater than the remaining side. But this is for convex polygons.Alternatively, use the concept of mass point or barycentric coordinates. Express D as a combination of A, B, C. But I don't see the connection.Wait, returning to the reflection idea. Suppose we reflect point D over side AB to get D'. Then, AD = AD' and BD = BD'. Maybe this can form a triangle where AD + DC is part of a larger structure.Alternatively, reflect point D over AB to D', then AD = AD' and BD = BD'. Then, in triangle D'DC, we might relate D'C to D'D + DC. But not sure.Alternatively, consider that reflecting D over AB to D', then the path AD + DC becomes AD' + DC, which could be compared to AC'. But I'm not making progress here.Wait, I recall a problem where to prove that the sum of distances from a point inside a triangle to two vertices is less than the sum of two sides. The proof involved extending the segment from the point to one side and applying the triangle inequality.Let me try that. Extend AD beyond D to meet BC at point E. Then, in triangle AEC, we have AC ≤ AE + EC. But AE = AD + DE, so AC ≤ AD + DE + EC. Therefore, AC ≤ AD + (DE + EC). But DE + EC = DC + something? Not sure.Wait, if E is on BC, then EC is part of BC. Maybe express BC as BE + EC. If we can relate DE to BE or something.Alternatively, in triangle DEC, DE + EC ≥ DC. So from above, AC ≤ AD + DE + EC ≥ AD + DC. Therefore, AC ≤ AD + DC, which is the usual triangle inequality. But we need AD + DC ≤ AB + BC.This isn't helpful.Wait, another idea: Use the theorem that in a triangle, the length of any line segment from a vertex to a point on the opposite side is less than the sum of the other two sides. But I need to check.Alternatively, consider that in triangle ABC, for any point D on BC, AD ≤ AB + BD ≤ AB + BC. But D is inside the triangle, not necessarily on BC.Wait, if D is inside the triangle, we can project it onto BC, say point E, then use some inequalities involving E.Alternatively, use the fact that AD ≤ AB + BD and DC ≤ BC - BD (if D is projected onto BC). But this is speculative.Wait, if we project D onto BC at point E, then by the Pythagorean theorem, AD² = AE² + DE² and DC² = EC² + DE². But this only holds if the projection is orthogonal, which complicates things.Alternatively, think in terms of coordinates again. Let’s take triangle ABC with coordinates: A(0,0), B(b,0), C(0,c). Point D(x,y) inside the triangle, so x ≥ 0, y ≥ 0, and x/b + y/c ≤ 1.Then, AD + DC = sqrt(x² + y²) + sqrt((x - b)^2 + (y - c)^2). Wait, no. If A is at (0,0), C is at (0,c), B is at (b,0). Then, point D(x,y) inside the triangle. Then AD is sqrt(x² + y²), DC is sqrt((x - 0)^2 + (y - c)^2) = sqrt(x² + (y - c)^2). Wait, no, C is at (0,c), so DC is sqrt((x - 0)^2 + (y - c)^2) = sqrt(x² + (y - c)^2). AB is the distance from A(0,0) to B(b,0) which is b. BC is the distance from B(b,0) to C(0,c) which is sqrt(b² + c²). So we need to prove that sqrt(x² + y²) + sqrt(x² + (y - c)^2) ≤ b + sqrt(b² + c²). This seems specific to coordinate system, but maybe not helpful.Alternatively, maybe using the Minkowski inequality for integrals or something, but this is too advanced.Wait, perhaps the key lies in considering that when D is at B, AD + DC equals AB + BC, and for any other position, AD + DC is less. To show this, we can argue that moving D from B to any other position inside the triangle causes at least one of AD or DC to decrease.But is this true? If we move D from B towards A, AD decreases but DC increases. Similarly, moving D from B towards C, DC decreases but AD increases. Moving D away from B into the interior, both AD and DC increase, but their sum might not necessarily.Wait, let's test this. In the coordinate system with A(0,0), B(1,0), C(0,1), D at (0.5,0.5), AD = sqrt(0.25 + 0.25) ≈ 0.707, DC = same, sum ≈ 1.414 < 2. If D is at (0.5,0), AD = 0.5, DC = sqrt(0.25 + 1) ≈ 1.118, sum ≈ 1.618 < 2. If D is at (0.5,0.25), AD = sqrt(0.25 + 0.0625) ≈ 0.559, DC = sqrt(0.25 + 0.5625) ≈ 0.901, sum ≈ 1.46 < 2. If D approaches B, sum approaches 2. So it seems that the maximum is achieved at B. So, intuitively, if you move D away from B, even if one distance increases, the other decreases more, so the sum decreases. But how to prove this?Maybe consider the function f(D) = AD + DC and show that it has its maximum at B. To do this, we can use the concept of maxima in a closed and bounded set (triangle is compact), so f(D) attains its maximum. Then, we can check that the maximum occurs at B by showing that the gradient (derivative) points towards B from any point inside.But this requires calculus. Alternatively, use variational principle: assume there is a point D inside where AD + DC > AB + BC, then reach a contradiction.Suppose there exists a point D inside triangle ABC such that AD + DC > AB + BC. Then, consider the path from A to D to C. This path would be longer than AB + BC. But AB + BC is the length of the path from A to B to C. However, in the triangle, the straight line from A to C is the shortest path, which is less than AB + BC. But AD + DC is greater than AC. So AC < AD + DC ≤ AB + BC (if the conjecture is true). But how does the assumption AD + DC > AB + BC contradict something?If AD + DC > AB + BC, then consider triangle ABD: AD ≤ AB + BD. And triangle CBD: DC ≤ BD + BC. Adding, AD + DC ≤ AB + 2BD + BC. So if AD + DC > AB + BC, then AB + BC < AD + DC ≤ AB + BC + 2BD, which implies BD > 0, which is true since D is not B. But this doesn't lead to a contradiction, so this approach is not helpful.Another angle: Use the Law of Cosines in triangle ABD and CBD.In triangle ABD: AD² = AB² + BD² - 2·AB·BD·cosθ, where θ is the angle at B.In triangle CBD: DC² = BC² + BD² - 2·BC·BD·cosφ, where φ is the angle at B.But unless we relate θ and φ, this might not help.Alternatively, since angles θ and φ are related (they are adjacent angles at B summing to angle ABC), but this complicates things.Wait, maybe using Ptolemy's theorem if ABCD is cyclic, but D is inside the triangle, so ABCD is not cyclic.Alternatively, consider the following argument: Since D is inside triangle ABC, the angles at D in triangles ADB and CDB are such that angle ADB + angle CDB > 180 degrees (since D is inside). Maybe use this to relate the sides.Alternatively, think of AD + DC as a function and use the principle that in a convex polygon, the maximum distance sum occurs at a vertex.But to avoid higher-level math, perhaps a geometric construction.Here's a possible elementary proof:Extend AD to meet BC at point E. Then, in triangle ABE, we have AB + BE > AE (triangle inequality). Also, in triangle EDC, EC + DC > ED. Adding these inequalities:AB + BE + EC + DC > AE + ED.But BE + EC = BC, so AB + BC + DC > AE + ED.But AE + ED = AD + ED (since AE = AD + DE). Wait, no: If we extend AD to E on BC, then AE = AD + DE. So, AE + ED = AD + DE + ED = AD + 2ED. Not sure.Alternatively, AE = AD + DE. Then, the inequality becomes AB + BC + DC > AD + DE + ED = AD + 2ED. Not helpful.Another approach: In triangle ABD, AD ≤ AB + BD. In triangle BDC, DC ≤ BD + BC. Adding:AD + DC ≤ AB + 2BD + BC.If we can show that 2BD ≤ AB + BC - AD - DC, but this seems circular.Wait, perhaps subtract the two inequalities:From AD ≤ AB + BD and DC ≤ BC + BD,AD + DC ≤ AB + BC + 2BD.But we need AD + DC ≤ AB + BC, so this would require 2BD ≤ 0, which implies BD = 0, i.e., D = B. Thus, equality holds only when D = B. For other points, the inequality AD + DC < AB + BC + 2BD holds, but since BD > 0, this doesn't help us directly.Hmm. It seems that all approaches are hitting a wall unless we use convexity or reflection. Since time is limited, I think the reflection method is the most promising.Reflect point C over AB to get point C'. Then, AC' = AB + BC (since reflection makes AB + BC = AC'). Now, for any point D on AB, AD + DC = AD + DC' (because DC = DC' due to reflection). Therefore, the minimal path from A to C via AB is a straight line to C'. But for a point D inside the triangle, not on AB, the path AD + DC would be less than or equal to AC' because reflecting DC over AB to DC' forms a straight line path from A to C', which is longer than any indirect path through D inside the triangle.Wait, more precisely, if we reflect DC over AB to DC', then AD + DC = AD + DC' is equal to the path from A to C' via D. The shortest path from A to C' is the straight line AC', so AD + DC' ≥ AC'. But AC' = AB + BC. Wait, this suggests AD + DC' ≥ AB + BC, but we need AD + DC ≤ AB + BC. This seems contradictory.Wait, maybe I mixed up the reflection. If we reflect C over AB to C', then for any point D, DC = DC'. So AD + DC = AD + DC' ≥ AC' by triangle inequality. But AC' = AB + BC. Wait, so this suggests AD + DC ≥ AB + BC, which is the opposite of what we need. So this approach can't be right.Therefore, reflecting C over AB gives a lower bound, not an upper bound. Hence, this reflection trick isn't helpful here.Alternatively, reflect B over AC to B', and then relate AD + DC to AB' or something. Not sure.Given that I'm struggling with an elementary proof, maybe the correct approach is to use convexity as follows:The function f(D) = AD + DC is convex because it is the sum of two convex functions (distance functions). The maximum of a convex function over a convex polygon occurs at a vertex. The vertices of triangle ABC are A, B, and C. Evaluating f at each vertex:- At D = A: f(A) = AA + AC = 0 + AC = AC.- At D = B: f(B) = AB + BC.- At D = C: f(C) = AC + CC = AC + 0 = AC.Therefore, the maximum value of f(D) over the triangle is AB + BC, which occurs at D = B. Since D is inside the triangle (including the boundary), AD + DC ≤ AB + BC.This proof is concise and uses convexity, which might be beyond basic geometry, but it's valid. However, if the problem requires an elementary proof without convexity, another method is needed.Alternatively, here's an elementary proof using the triangle inequality twice and the fact that D is inside ABC:Consider triangle ABD: AD ≤ AB + BD.Consider triangle BDC: DC ≤ BD + BC.Add these two inequalities:AD + DC ≤ AB + 2BD + BC.Now, since D is inside ABC, BD < BC (if D is inside, then BD is less than BC, but actually BD can be up to BC if D is on BC). Wait, no. If D is inside the triangle, BD can be as large as the length of the longest median, which is less than BC in a non-degenerate triangle.Wait, this is not necessarily true. For example, in triangle ABC, if D is close to B, BD can be very small, but if D is on the median from B to AC, BD can be up to the length of the median, which is less than BC in a non-equilateral triangle.Therefore, BD < BC is not always true, so we can't say 2BD ≤ BC, for instance. Therefore, the previous approach doesn't give us AD + DC ≤ AB + BC.Another idea: Use the fact that in triangle ABC, the distance from any interior point D to a vertex is less than the sum of the other two sides. Wait, but that's not generally true. For example, in a very acute triangle, a point near the center can have distances to vertices larger than some sides.But the problem isn't claiming that AD or DC is less than a side, but their sum is less than AB + BC.Wait, considering the triangle inequality in another way: For any point D inside ABC, AB + BC is the sum of two sides, and AD + DC is a path from A to C via D. Since the minimal path from A to C is AC, and the maximal path would be around the perimeter, which is AB + BC, perhaps AD + DC is bounded by AB + BC.But how to formalize this?Perhaps consider that the path AD + DC must go around the obstacle of the triangle, so the longest path inside the triangle is the perimeter path AB + BC.Alternatively, use the following argument from "알기쉬운 기하학" (Easy Geometry) textbook: For a point D inside triangle ABC, extend AD to meet BC at E. Then, AD + DC < AD + DE + EC = AE + EC. But AE + EC < AB + BE + EC = AB + BC. Hence, AD + DC < AB + BC.Wait, let me check this step by step.Extend AD to meet BC at E. Then, in triangle CDE, we have DC < DE + EC (triangle inequality). Therefore, DC < DE + EC. Adding AD to both sides: AD + DC < AD + DE + EC. But AD + DE = AE, so this becomes AD + DC < AE + EC.Now, consider path AE + EC. Since E is on BC, AE can be compared to AB + BE. In triangle ABE, AE < AB + BE. Hence, AE + EC < (AB + BE) + EC = AB + (BE + EC) = AB + BC.Therefore, combining the inequalities:AD + DC < AE + EC < AB + BC.Hence, AD + DC < AB + BC.This seems correct! Therefore, the strict inequality holds for any D inside the triangle (not on the boundary). If D is on the boundary, say at B, then AD + DC = AB + BC. So, depending on whether the problem considers boundary points as "inside", the inequality can be non-strict.Therefore, the final proof is:Extend AD to meet BC at E. By the triangle inequality in triangle CDE, DC < DE + EC. Adding AD to both sides gives AD + DC < AD + DE + EC = AE + EC. In triangle ABE, AE < AB + BE. Therefore, AE + EC < AB + BE + EC = AB + BC. Combining these inequalities, AD + DC < AB + BC. If D is on the boundary, specifically at B, then AD + DC = AB + BC. Hence, for any point D inside or on the triangle, AD + DC ≤ AB + BC.This completes the proof.

✅:To prove that ( AD + DC leq AB + BC ) for a point ( D ) inside triangle ( ABC ), follow these steps:1. Extend ( AD ) to meet ( BC ) at ( E ): Since ( D ) is inside the triangle, extending ( AD ) will intersect ( BC ) at some point ( E ).2. Apply the triangle inequality in ( triangle CDE ): [ DC < DE + EC ] Adding ( AD ) to both sides: [ AD + DC < AD + DE + EC ] Notice that ( AD + DE = AE ), so: [ AD + DC < AE + EC ]3. Apply the triangle inequality in ( triangle ABE ): [ AE < AB + BE ] Adding ( EC ) to both sides: [ AE + EC < AB + BE + EC ] Since ( BE + EC = BC ), this simplifies to: [ AE + EC < AB + BC ]4. Combine the inequalities: From steps 2 and 3: [ AD + DC < AE + EC < AB + BC ] Therefore: [ AD + DC < AB + BC ] If ( D ) coincides with ( B ) (on the boundary), ( AD + DC = AB + BC ). Hence, for any ( D ) inside or on the triangle: [ AD + DC leq AB + BC ]Conclusion: For any point ( D ) inside triangle ( ABC ), the inequality ( AD + DC leq AB + BC ) holds, with equality if and only if ( D ) is at vertex ( B ).

🔑:Given: A triangle ABC with an interior point D.To Prove: AD + DC leq AB + BC.1. Extend the segment AD: - Draw the line segment AD and extend it past point D until it intersects the side BC at point E.2. Comparison using the triangle inequality: - Consider the path A rightarrow E rightarrow C. By the triangle inequality applied to triangle ABE, we get: [ AB + BE geq AE ]3. Introducing intermediary value: - Take AE + EC as an intermediary comparison value. Notice that segment EC is common to both paths B rightarrow E rightarrow C and A rightarrow E rightarrow C. Hence, the total length of the path A rightarrow B rightarrow C can be compared to A rightarrow E rightarrow C.4. Using triangle inequality again: - By the triangle inequality applied to AED, we know: [ AD + DE geq AE ] - Similarly, by the triangle inequality applied to DEC, we observe: [ DE + EC geq DC ]5. Combining the inequalities: - Add up the inequalities AD + DE and DE + EC: [ AD + DE + DE + EC geq AE + DC ] - Simplifying, we get: [ AD + 2DE + EC geq AE + DC ]6. Considering triangle inequalities resulting from segments: - Note DE + EC = DC implies further simplification: [ AD + DC leq AE ]7. Compare two paths: - The path A rightarrow E rightarrow C must be shorter than or equal to the path A rightarrow D rightarrow C because segment EC is common in both paths. Simplifying and integrating all paths: [ AD + DC leq AB + BC ] Conclusion: [boxed{AD + DC leq AB + BC}]

🧠:Okay, so I need to prove that in triangle ABC, when we draw tangents to the incircle that are parallel to each side, forming three smaller triangles AED, BGF, and CIH, the sum of their inradii (r1, r2, r3) equals the inradius (r) of the original triangle ABC. Hmm, let's start by visualizing the problem.First, let me sketch triangle ABC with its incircle O. The incircle touches each side at certain points—let me recall, the points of tangency divide the sides into segments equal to the semiperimeter minus the opposite side. But maybe that's not immediately important here.Now, drawing tangents to the incircle that are parallel to each side. For example, a tangent parallel to side BC would be somewhere inside the triangle, touching the incircle, and forming a smaller triangle with vertex A. Similarly, tangents parallel to AB and AC would form smaller triangles at vertices B and C, respectively.So each smaller triangle is similar to the original triangle because their sides are parallel. That seems right since if a line is drawn parallel to a side of a triangle, the resulting smaller triangle is similar to the original one.Given similarity, the ratio of their inradii should be the same as the ratio of their corresponding sides. But since we need to relate the sum of the inradii to the original inradius, maybe we need to find the scaling factors for each smaller triangle and then express r1, r2, r3 in terms of r, summing them up.But how do we find the scaling factors? Let's think about the distance between the original side and the tangent line. Since the tangent is parallel to the side, the distance between them is twice the inradius? Wait, no. Wait, the inradius is the distance from the incenter to any side. If we draw a tangent line parallel to a side, the distance between the original side and the tangent line would be twice the inradius? Hmm, maybe not. Let's consider.The incircle is tangent to each side. If we draw another tangent line parallel to a side, the distance between the two parallel lines (the original side and the tangent) would be 2r, since the radius is r, and the two tangent lines would be on opposite sides of the incircle. Wait, but in this case, the tangent line is still inside the triangle, so maybe it's only a distance of 2r apart? Wait, no. If the incircle has radius r, then the distance between two parallel tangent lines would be 2r if they are on opposite sides of the circle. But in our case, the original side is a tangent to the incircle, and the new tangent line is on the same side of the incircle? Wait, no. The original side is one tangent, and the new tangent line is another tangent parallel to it but inside the triangle. So the distance between the original side and the new tangent line should be twice the radius? Wait, maybe not.Wait, the distance between two parallel tangent lines of a circle is 2r. But here, the original side is a tangent, and the new tangent is inside the triangle. So the distance between the original side and the new tangent line is 2r. But in the triangle, how does that relate to the position of the smaller triangle?Wait, maybe not. Let me think again. The incircle is tangent to side BC at a point. If we draw a line parallel to BC that is tangent to the incircle, this line is above BC (towards vertex A). The distance between BC and this tangent line is the distance between two parallel tangent lines of the incircle. Since the incircle is tangent to BC, and the other tangent line is parallel, the distance between them is 2r. Wait, that seems too much. Because if the radius is r, then moving from BC upwards by 2r would go beyond the incenter, which is at distance r from BC. Hmm, maybe this is not correct.Wait, the distance between two parallel tangent lines of a circle is equal to twice the diameter if they are on opposite sides. Wait, no. Let's recall that for a circle, the distance between two parallel tangent lines is twice the radius. Because the radius is perpendicular to each tangent, so the distance between the two lines is twice the radius. Yes, that's right. So if you have two parallel tangent lines to a circle, the distance between them is 2r. For example, if you have a horizontal circle, the two horizontal tangent lines are at y = r and y = -r, so the distance between them is 2r.But in our case, the original side BC is a tangent to the incircle, and the other tangent line parallel to BC is inside the triangle. So the distance between BC and this new tangent line is 2r? But if the inradius is r, then the distance from the incenter to BC is r. If the new tangent line is on the other side of the incircle, then the distance between BC and the new tangent line would be 2r. But in the triangle, the incenter is already at distance r from BC. If we draw a line parallel to BC that's tangent to the incircle on the opposite side (towards vertex A), then the distance from BC to that line is 2r. However, in a triangle, the altitude from A to BC is h_a. So if 2r is less than h_a, then such a line exists. Otherwise, not. But since the incircle is tangent to all sides, the inradius is related to the area and semiperimeter, so maybe 2r is less than the altitude.But perhaps this is overcomplicating. Let's approach this step by step.First, let's consider triangle ABC with inradius r. The incircle touches BC at point D, AC at point E, and AB at point F (standard notation). Wait, but the problem mentions points E, F on AB; G, H on BC; and I, D on AC. Maybe the notation is different here, but perhaps I should not get bogged down by notation and instead focus on the concepts.Suppose we draw a tangent to the incircle parallel to BC. This tangent will meet AB and AC at some points, forming a smaller triangle AED (as per the problem statement). Similarly for the other sides.Since AED is similar to ABC, with AED ~ ABC, the ratio of their sides is some k1, and the ratio of their inradii is also k1. So r1 = k1 * r. Similarly, r2 = k2 * r and r3 = k3 * r. Then, if we can show that k1 + k2 + k3 = 1, then r1 + r2 + r3 = r(k1 + k2 + k3) = r. So maybe this is the approach.But how do we find k1, k2, k3?Alternatively, perhaps we can relate the distances between the original sides and the tangent lines. If the distance between BC and the tangent line parallel to BC is d, then the ratio k1 is (h_a - d)/h_a, where h_a is the altitude from A to BC. But the inradius r is related to the area and semiperimeter: r = area / s, where s = (a + b + c)/2.Alternatively, maybe coordinate geometry would help. Let's set up coordinates.Let me place triangle ABC with side BC on the x-axis, B at (0,0), C at (c,0), and A somewhere in the plane. The inradius is r, so the incenter is at (d, r), where d is some distance along the x-axis. Wait, but the incenter coordinates can be determined based on the sides.Alternatively, maybe it's easier to use barycentric coordinates or another system, but perhaps Cartesian coordinates would be clearer.Alternatively, use homothety. The smaller triangle AED is a homothetic image of ABC. Since AED is parallel to BC and tangent to the incircle, the homothety center is at A, and the ratio is such that the incircle of AED is the image of the incircle of ABC. Wait, maybe not exactly, but homothety could relate the two.Wait, if we have a line parallel to BC tangent to the incircle, then the homothety that maps BC to this tangent line would center at A, because the lines are parallel. The ratio of homothety would be the ratio of distances from A. Let’s denote the original altitude from A to BC as h_a. The distance from A to the tangent line would be h_a - 2r, since the distance between BC and the tangent is 2r (as established earlier). Therefore, the ratio of homothety k1 would be (h_a - 2r)/h_a. Therefore, the inradius of AED would be k1 * r = r * (h_a - 2r)/h_a. Wait, but this might not be correct. Let me verify.Wait, the homothety centered at A that sends BC to the tangent line parallel to BC would scale all lengths by the ratio of the distances from A. So if the original distance from A to BC is h_a, and the distance from A to the tangent line is h_a - d, where d is the distance between BC and the tangent line. Earlier, we thought d = 2r. If that's the case, then the ratio k1 = (h_a - 2r)/h_a. Therefore, the inradius of AED would be k1 * r. But then, if we do this for all three sides, the sum of the inradii would be r * [ (h_a - 2r)/h_a + (h_b - 2r)/h_b + (h_c - 2r)/h_c ]. But this seems complicated, and unless h_a, h_b, h_c relate in a way that this sum simplifies to 1, which isn't obvious.Alternatively, maybe my assumption that d = 2r is wrong. Let's recast.Consider the incircle with center I and radius r. The distance from I to BC is r. The tangent line parallel to BC is another tangent to the circle. The distance between BC and this tangent line is the distance between two parallel lines: BC and the tangent. Since both are tangent to the circle, the distance between them is 2r. Because the center I lies between them, and the distance from I to each line is r, so total distance between the lines is 2r.Therefore, the altitude from A to BC is h_a. The distance from A to the tangent line is h_a - 2r. Therefore, the ratio of similarity k1 is (h_a - 2r)/h_a. Therefore, the inradius of triangle AED is r1 = k1 * r = r * (h_a - 2r)/h_a.Similarly, for the other two smaller triangles, we would have r2 = r * (h_b - 2r)/h_b and r3 = r * (h_c - 2r)/h_c.Thus, the sum r1 + r2 + r3 = r [ (h_a - 2r)/h_a + (h_b - 2r)/h_b + (h_c - 2r)/h_c ) ].Simplify this expression: = r [ 3 - 2r (1/h_a + 1/h_b + 1/h_c ) ]But we need this sum to be equal to r. Therefore, 3 - 2r (1/h_a + 1/h_b + 1/h_c ) = 1 => 2 - 2r (1/h_a + 1/h_b + 1/h_c ) = 0 => 1 = r (1/h_a + 1/h_b + 1/h_c )But is this true? Let's check if 1 = r (1/h_a + 1/h_b + 1/h_c )Recall that h_a = 2Δ / a, h_b = 2Δ / b, h_c = 2Δ / c, where Δ is the area of ABC, and a, b, c are the lengths of the sides.Therefore, 1/h_a + 1/h_b + 1/h_c = (a + b + c)/(2Δ). But r = Δ / s, where s = (a + b + c)/2. Therefore, r (1/h_a + 1/h_b + 1/h_c ) = (Δ / s) * ( (a + b + c)/(2Δ) ) = (Δ / s) * (2s)/(2Δ) ) = (Δ / s) * (s / Δ ) = 1. Yes! Therefore, indeed, 1 = r (1/h_a + 1/h_b + 1/h_c ). Therefore, the sum r1 + r2 + r3 = r [ 3 - 2 * 1 ] = r [ 1 ] = r. Therefore, proved.Wait, that seems to work out. Let me recap to ensure I didn't make a leap.1. The distance between each original side and the tangent line parallel to it is 2r.2. Therefore, the altitude of the smaller triangle is the original altitude minus 2r.3. The ratio of similarity for each smaller triangle is (h - 2r)/h, where h is the original altitude.4. Therefore, their inradii are r * (h - 2r)/h.5. Summing these gives r times [sum over (h - 2r)/h] for each altitude.6. Expanding the sum: sum [1 - 2r/h] = 3 - 2r sum(1/h)7. We showed that sum(1/h) = (a + b + c)/(2Δ) = 1/r, since r = Δ/s and s = (a + b + c)/2.8. Therefore, 3 - 2r*(1/r) = 3 - 2 = 1, so the total sum is r*1 = r.Therefore, the conclusion follows.But let me verify step 7 again. Let's compute sum(1/h_a + 1/h_b + 1/h_c). Since h_a = 2Δ/a, h_b = 2Δ/b, h_c = 2Δ/c. Therefore, 1/h_a + 1/h_b + 1/h_c = (a + b + c)/(2Δ). And since r = Δ / s, where s = (a + b + c)/2, so 1/r = s / Δ = (a + b + c)/(2Δ). Therefore, indeed sum(1/h) = 1/r. Therefore, 2r * sum(1/h) = 2r * (1/r) = 2. Hence, 3 - 2 = 1. Therefore, the calculation checks out.Therefore, the sum of the inradii r1 + r2 + r3 equals r. So the proof is complete.But let me think if there's another way to approach this, maybe using areas or other properties.Alternatively, since each smaller triangle is similar to ABC, their inradii are proportional to their corresponding sides. The distance between the original side and the tangent line is 2r, as established. So the height of the smaller triangle AED is h_a - 2r. Since the original height is h_a, the ratio is (h_a - 2r)/h_a, so the inradius is scaled by that factor. Similarly for the others. Then, adding them up as before.Alternatively, using coordinate geometry: place the incenter at (0, r) for simplicity, with BC on the x-axis from (-s, 0) to (t, 0), but this might complicate. Alternatively, take specific coordinates.Let me try coordinate geometry with triangle ABC being an equilateral triangle for simplicity. Wait, but the problem is general, not specific to equilateral triangles. But testing with an equilateral triangle might help verify the formula.Suppose ABC is equilateral with side length a, so all altitudes are h = (√3/2)a. The inradius r = h / 3 = (√3/6)a. Drawing a tangent line parallel to BC, the distance between BC and the tangent is 2r = (√3/3)a. Therefore, the height of the smaller triangle AED is h - 2r = (√3/2)a - (√3/3)a = (√3/6)a, which is equal to r. Therefore, the ratio k1 is (√3/6)a / (√3/2)a = 1/3. Therefore, the inradius of AED is k1 * r = (1/3)r. Similarly, the other two smaller triangles would also have inradii (1/3)r each. Therefore, the sum is (1/3 + 1/3 + 1/3)r = r, which matches the result. So in the equilateral case, it works.But does this hold for any triangle? The previous general proof suggests yes. So the key idea is that the distance between a side and the parallel tangent is 2r, leading to the similarity ratio, and the sum of the scaled inradii adds up due to the relation between r and the altitudes.Alternatively, think about the areas. The inradius is related to the area and semiperimeter. Maybe we can express the areas of the smaller triangles in terms of r and sum them up. But since the smaller triangles are similar, their areas would be k1² * area of ABC, and their inradii would be k1 * r. However, summing the areas might not directly help unless we can relate the sum of the areas to something else.But in the previous approach, using the altitudes and the properties of the inradius, we found a relationship that holds for any triangle. Therefore, the conclusion seems solid.Another way to think about it: The three smaller triangles each "consume" a portion of the original inradius, and their inradii sum up to the original because the distances between the sides and the tangents account for twice the inradius, which when subtracted from the altitudes and scaled appropriately, leads to the sum equaling the original inradius.Overall, the critical steps are:1. Recognizing the distance between a side and the parallel tangent to the incircle is 2r.2. Relating this distance to the similarity ratio of the smaller triangle.3. Expressing each smaller inradius as a scaled version of r.4. Summing them and using the relationship between r and the altitudes to simplify the sum to r.This seems to cover the necessary reasoning. I don't see any gaps or errors in the logic, and the equilateral triangle case checks out. Therefore, the proof is valid.Final Answerboxed{r = r_1 + r_2 + r_3}

✅:To prove that ( r = r_1 + r_2 + r_3 ) where ( r ) is the inradius of ( triangle ABC ) and ( r_1, r_2, r_3 ) are the inradii of the smaller triangles ( triangle AED, triangle BGF, triangle CIH ) formed by tangents to the incircle parallel to the sides of ( triangle ABC ):1. Distance Between Parallel Tangents: The distance between a side of ( triangle ABC ) and the tangent line parallel to it is ( 2r ). This is because the inradius ( r ) is the distance from the incenter to any side, and the distance between two parallel tangent lines of a circle is twice the radius.2. Similarity of Triangles: Each smaller triangle (e.g., ( triangle AED )) is similar to ( triangle ABC ) because their sides are parallel. The ratio of similarity ( k ) for each smaller triangle is determined by the ratio of the distances from the vertex to the original side and the tangent line.3. Similarity Ratio: For each smaller triangle, the altitude from the vertex (e.g., ( A )) is reduced by ( 2r ). If the original altitude is ( h_a ), the new altitude is ( h_a - 2r ). The similarity ratio ( k_1 ) is ( frac{h_a - 2r}{h_a} ). The inradius of the smaller triangle ( triangle AED ) is then ( r_1 = k_1 cdot r = r cdot frac{h_a - 2r}{h_a} ).4. Summing Inradii: Similarly, for the other two smaller triangles ( triangle BGF ) and ( triangle CIH ), the inradii are ( r_2 = r cdot frac{h_b - 2r}{h_b} ) and ( r_3 = r cdot frac{h_c - 2r}{h_c} ). Summing these inradii gives: [ r_1 + r_2 + r_3 = r left( frac{h_a - 2r}{h_a} + frac{h_b - 2r}{h_b} + frac{h_c - 2r}{h_c} right) ]5. Simplifying the Sum: Simplifying the expression inside the parentheses: [ frac{h_a - 2r}{h_a} + frac{h_b - 2r}{h_b} + frac{h_c - 2r}{h_c} = 3 - 2r left( frac{1}{h_a} + frac{1}{h_b} + frac{1}{h_c} right) ]6. Relating to Inradius: Using the relationships ( h_a = frac{2Delta}{a} ), ( h_b = frac{2Delta}{b} ), ( h_c = frac{2Delta}{c} ) (where ( Delta ) is the area of ( triangle ABC )) and ( r = frac{Delta}{s} ) (where ( s ) is the semiperimeter), we find: [ frac{1}{h_a} + frac{1}{h_b} + frac{1}{h_c} = frac{a + b + c}{2Delta} = frac{2s}{2Delta} = frac{s}{Delta} = frac{1}{r} ]7. Final Calculation: Substituting this back: [ 3 - 2r left( frac{1}{r} right) = 3 - 2 = 1 ] Therefore, the sum of the inradii is: [ r_1 + r_2 + r_3 = r cdot 1 = r ]Thus, we conclude that ( r = r_1 + r_2 + r_3 ).[boxed{r = r_1 + r_2 + r_3}]

🔑:1. Consider the inscribed circle odot O_1 of triangle AED. 2. Let odot O_1 intersect ED at point P. Then, we have: [ ED = EP + PD = r_1 left(cot frac{1}{2} angle AED + cot frac{1}{2} angle ADE right). ]3. Clearly, O is the excenter of triangle AED. By the properties of excenters, OE and OD are the external angle bisectors of triangle AED.4. Because angle O_1 EO = 90^circ, it follows that: [ ED = r left(cot angle DEO + cot angle EDOright) = r left(tan frac{1}{2} angle AED + tan frac{1}{2} angle ADE right). ]5. Therefore, we can write: [ frac{r_1}{r} = frac{tan frac{1}{2} angle AED + tan frac{1}{2} angle ADE}{cot frac{1}{2} angle AED + cot frac{1}{2} angle ADE} = tan frac{1}{2} angle AED cdot tan frac{1}{2} angle ADE. ]6. Because ED parallel BC, we have the angles: [ angle AED = angle B quad text{and} quad angle ADE = angle C. ]7. Hence, [ frac{r_1}{r} = tan frac{B}{2} cdot tan frac{C}{2}. ]8. Similarly, for triangle BGF and triangle CIH, we get: [ frac{r_2}{r} = tan frac{C}{2} cdot tan frac{A}{2} quad text{and} quad frac{r_3}{r} = tan frac{A}{2} cdot tan frac{B}{2}. ]9. Note that in triangle ABC: [ tan frac{B}{2} cdot tan frac{C}{2} + tan frac{C}{2} cdot tan frac{A}{2} + tan frac{A}{2} cdot tan frac{B}{2} = 1. ]10. Combining these results, we conclude: [ r = r_1 + r_2 + r_3. ](boxed{r = r_1 + r_2 + r_3})

🧠:Okay, let's tackle this geometry problem step by step. So, we have a triangle where the base is 4 cm longer than its corresponding height. Then, both the base and the height are increased by 12 cm, resulting in a new triangle whose area is five times the original area. We need to find the original base and height.First, let me parse the problem again to make sure I understand all the details. The original triangle has a base that's 4 cm longer than its height. Let's denote the height as h cm. Then, the base would be h + 4 cm. The area of a triangle is (base * height)/2, so the original area would be (h*(h + 4))/2.Now, when both the base and the height are increased by 12 cm, the new base becomes (h + 4) + 12 = h + 16 cm, and the new height becomes h + 12 cm. The area of the new triangle is then ((h + 16)*(h + 12))/2. According to the problem, this new area is five times the original area. So, the equation we can set up is:((h + 16)(h + 12))/2 = 5 * (h(h + 4))/2Hmm, since both sides have a denominator of 2, we can multiply both sides by 2 to eliminate the denominators. That simplifies the equation to:(h + 16)(h + 12) = 5h(h + 4)Alright, let's expand both sides. Starting with the left side:(h + 16)(h + 12) = h^2 + 12h + 16h + 192 = h^2 + 28h + 192And the right side:5h(h + 4) = 5h^2 + 20hSo now the equation is:h^2 + 28h + 192 = 5h^2 + 20hLet's move all terms to one side to form a quadratic equation. Subtract the left side from both sides:0 = 5h^2 + 20h - h^2 - 28h - 192Simplify:0 = 4h^2 - 8h - 192Hmm, maybe we can divide all terms by 4 to simplify further:0 = h^2 - 2h - 48Now, we have a quadratic equation: h^2 - 2h - 48 = 0To solve this, we can use the quadratic formula. The quadratic is in the form ax^2 + bx + c = 0, so here a=1, b=-2, c=-48.The quadratic formula is h = [-b ± sqrt(b^2 - 4ac)]/(2a)Plugging in the values:h = [2 ± sqrt((-2)^2 - 4*1*(-48))]/(2*1)h = [2 ± sqrt(4 + 192)]/2h = [2 ± sqrt(196)]/2sqrt(196) is 14, so:h = [2 ± 14]/2This gives two solutions:h = (2 + 14)/2 = 16/2 = 8h = (2 - 14)/2 = (-12)/2 = -6But since height can't be negative, we discard h = -6. Therefore, the height is 8 cm.Then, the base is h + 4 = 8 + 4 = 12 cm.Wait, let me check if this makes sense. Original area is (12*8)/2 = 48 cm². Then, increasing both by 12 cm, the new base is 24 cm and new height is 20 cm. New area is (24*20)/2 = 240 cm². Is 240 equal to five times 48? 5*48 is 240. Yes, that checks out.But let me go through the steps again to make sure I didn't make a mistake in the algebra.Original equation:((h + 16)(h + 12))/2 = 5*(h*(h + 4))/2Multiply both sides by 2:(h + 16)(h + 12) = 5h(h + 4)Expanding left side:h^2 + 12h + 16h + 192 = h^2 + 28h + 192Right side:5h^2 + 20hSo:h^2 + 28h + 192 = 5h^2 + 20hSubtract left side from right side:0 = 4h^2 - 8h - 192Divide by 4:h^2 - 2h - 48 = 0Quadratic formula gives h = [2 ± sqrt(4 + 192)]/2 = [2 ± 14]/2, so 8 and -6. Correct.Yes, seems correct. So the original height is 8 cm, base is 12 cm. And the verification with areas checks out. So that's the answer.But wait, let me think if there's another way to approach this. Maybe using variables for base and height instead of just height. Let me try that.Let the original height be h cm, then the base is h + 4 cm. Original area: (h*(h + 4))/2.After increasing both by 12 cm, new height is h + 12, new base is (h + 4) + 12 = h + 16. New area: ( (h + 16)*(h + 12) ) /2. This is equal to 5 times original area: 5*(h*(h + 4))/2.So, same equation as before. So whether I use h as height or h as base, the equation comes out the same. Wait, actually, in this case, we let h be the height, base is h + 4. Alternatively, if we let b be the base, then height is b - 4. Let's try that.Let base be b cm, so height is b - 4 cm. Original area: (b*(b - 4))/2.After increasing both by 12 cm: new base = b + 12, new height = (b - 4) + 12 = b + 8. New area: ( (b + 12)*(b + 8) ) /2 = 5*(b*(b - 4))/2.Multiply both sides by 2:(b + 12)(b + 8) = 5b(b - 4)Expand left side:b^2 + 8b + 12b + 96 = b^2 + 20b + 96Right side:5b^2 - 20bSet equation:b^2 + 20b + 96 = 5b^2 - 20bBring all terms to one side:0 = 5b^2 - 20b - b^2 - 20b - 96Simplify:0 = 4b^2 - 40b - 96Divide by 4:0 = b^2 - 10b - 24Then, quadratic equation: b^2 -10b -24=0Use quadratic formula: b = [10 ± sqrt(100 + 96)]/2 = [10 ± sqrt(196)]/2 = [10 ±14]/2So, two solutions:b = (10 +14)/2 =24/2=12b = (10 -14)/2= (-4)/2=-2Again, base can't be negative, so b=12 cm. Then height is 12 -4=8 cm. Same answer. So whether we set up with base or height as variable, we get the same result. That cross-check confirms the answer is correct.Another way to verify is plugging back into the area expressions. Original base 12, height 8, area 48. New base 24, height 20, area 240. 240 is indeed 5*48. So that's solid.Is there any other angle to check? Maybe considering units. All measurements are in cm, which is consistent. The operations performed (quadratic equation) are appropriate. Negative solution discarded as physical lengths can't be negative. So all steps make sense.Therefore, confident that the original base is 12 cm and height is 8 cm.Final AnswerThe original triangle has a base of boxed{12} cm and a height of boxed{8} cm.

✅:Let the height of the original triangle be ( h ) cm. Then, the base is ( h + 4 ) cm. The area of the original triangle is (frac{1}{2} times h times (h + 4)).When both the base and the height are increased by 12 cm, the new base becomes ( (h + 4) + 12 = h + 16 ) cm and the new height becomes ( h + 12 ) cm. The area of the new triangle is (frac{1}{2} times (h + 16) times (h + 12)). This new area is five times the original area, leading to the equation:[frac{1}{2} times (h + 16) times (h + 12) = 5 times left( frac{1}{2} times h times (h + 4) right)]Multiplying both sides by 2 to clear the denominators:[(h + 16)(h + 12) = 5h(h + 4)]Expanding both sides:[h^2 + 28h + 192 = 5h^2 + 20h]Rearranging all terms to one side:[0 = 4h^2 - 8h - 192]Dividing by 4:[0 = h^2 - 2h - 48]Solving the quadratic equation ( h^2 - 2h - 48 = 0 ) using the quadratic formula ( h = frac{-b pm sqrt{b^2 - 4ac}}{2a} ):[h = frac{2 pm sqrt{4 + 192}}{2} = frac{2 pm 14}{2}]This gives two solutions:[h = frac{16}{2} = 8 quad text{and} quad h = frac{-12}{2} = -6]Since height cannot be negative, ( h = 8 ) cm. The base is ( h + 4 = 8 + 4 = 12 ) cm.Verification: Original area is (frac{1}{2} times 12 times 8 = 48 ) cm². New area after increasing both dimensions by 12 cm is (frac{1}{2} times 24 times 20 = 240 ) cm², which is indeed five times the original area.Thus, the original triangle has a base of (boxed{12}) cm and a height of (boxed{8}) cm.

🔑:1. Let (a) be the base of the triangle, and the height associated with this base be (m). The problem states that the base is 4 cm longer than the height: [ a = m + 4 ]2. We are given that when both the base and the height are increased by 12 cm, the area of the new triangle is five times the area of the original triangle. The area of a triangle is given by: [ text{Area} = frac{1}{2} times text{base} times text{height} ]3. Let’s denote the base of the new triangle as (a + 12) and its height as (m + 12). According to the problem: [ frac{1}{2} times (a + 12) times (m + 12) = 5 times left( frac{1}{2} times a times m right) ]4. Simplify both sides by multiplying through by 2: [ (a + 12)(m + 12) = 5am ]5. Plug in (m = a - 4) from step 1: [ (a + 12)((a - 4) + 12) = 5a(a - 4) ]6. Simplify inside the parentheses: [ (a + 12)(a + 8) = 5a(a - 4) ]7. Expand both sides: [ a^2 + 8a + 12a + 96 = 5a^2 - 20a ] [ a^2 + 20a + 96 = 5a^2 - 20a ]8. Bring all terms to one side to form a quadratic equation: [ a^2 + 20a + 96 - 5a^2 + 20a = 0 ] [ -4a^2 + 40a + 96 = 0 ] [ 4a^2 - 40a - 96 = 0 ] [ a^2 - 10a - 24 = 0 ]9. Solve the quadratic equation using the quadratic formula (a = frac{-b pm sqrt{b^2 - 4ac}}{2a}): [ a = frac{10 pm sqrt{100 + 96}}{2} ] [ a = frac{10 pm sqrt{196}}{2} ] [ a = frac{10 pm 14}{2} ]10. Determine the possible solutions for (a): [ a = frac{10 + 14}{2} = 12 quad text{or} quad a = frac{10 - 14}{2} = -2 ] Since (a) represents the length of a base, we discard the negative value. Hence: [ a = 12 , text{cm} ]11. Substitute (a = 12 , text{cm}) back into (m = a - 4): [ m = 12 - 4 = 8 , text{cm} ]Conclusion:[boxed{12 , text{cm} , text{(base)}, , 8 , text{cm} , text{(height)}}]

🧠:Okay, let's tackle this problem. We need to find the maximum possible value of ( x - y + z ) given the equation ( 2x^2 + y^2 + z^2 = 2x - 4y + 2xz - 5 ). Hmm, this looks like a quadratic equation with three variables. I remember that for such optimization problems, completing the square might be a good approach. Let me try that.First, let's rewrite the given equation to group like terms together. Let's move all terms to the left side:( 2x^2 + y^2 + z^2 - 2x + 4y - 2xz + 5 = 0 ).Hmm, okay. Let's see if we can reorganize the terms to make completing squares possible. Let's group the x terms, y terms, z terms, and cross terms. Looking at the equation:- The x terms are ( 2x^2 - 2x - 2xz ).- The y terms are ( y^2 + 4y ).- The z terms are ( z^2 ).- The constant term is +5.Starting with the y terms: ( y^2 + 4y ). To complete the square here, take half of 4, which is 2, square it to get 4. So, ( y^2 + 4y = (y + 2)^2 - 4 ).Okay, that's done. Now let's handle the x and z terms. Let's look at ( 2x^2 - 2x - 2xz ). Factor out a 2 from the x² term: ( 2(x^2 - x - xz) ). Hmm, not sure if that's helpful. Alternatively, maybe factor x out of some terms. Let's see:( 2x^2 - 2x - 2xz = 2x^2 - 2x(1 + z) ). Hmm, maybe complete the square in terms of x here. Let's factor out 2:( 2(x^2 - x(1 + z)) ). To complete the square inside the parentheses, take half of (1 + z), which is ( (1 + z)/2 ), square it: ( (1 + z)^2 / 4 ). So,( x^2 - x(1 + z) = left(x - frac{1 + z}{2}right)^2 - frac{(1 + z)^2}{4} ).Therefore, multiplying by 2:( 2left(x - frac{1 + z}{2}right)^2 - frac{(1 + z)^2}{2} ).So substituting back into the original equation:The entire left side becomes:2(x - (1 + z)/2)^2 - (1 + z)^2 / 2 + (y + 2)^2 - 4 + z^2 + 5 = 0.Simplify the constants: -4 + 5 = 1. So:2(x - (1 + z)/2)^2 - (1 + z)^2 / 2 + (y + 2)^2 + z^2 + 1 = 0.Now, let's handle the terms involving z. Let's expand -(1 + z)^2 / 2 + z^2:First, expand (1 + z)^2 = 1 + 2z + z². So:- (1 + 2z + z²)/2 + z² = -1/2 - z - z²/2 + z² = (-1/2) - z + (z²/2).So substituting back:2(x - (1 + z)/2)^2 + (y + 2)^2 + (-1/2 - z + z²/2) + 1 = 0.Combine constants: -1/2 + 1 = 1/2. So:2(x - (1 + z)/2)^2 + (y + 2)^2 + z²/2 - z + 1/2 = 0.Let me write that out:2left(x - frac{1 + z}{2}right)^2 + (y + 2)^2 + frac{z^2}{2} - z + frac{1}{2} = 0.Now, let's look at the terms with z: ( frac{z^2}{2} - z + frac{1}{2} ). Let's complete the square here as well.Factor out 1/2: ( frac{1}{2}(z^2 - 2z + 1) ). Wait, z² - 2z + 1 is (z - 1)^2. So,( frac{1}{2}(z - 1)^2 ).So substituting back, the entire equation becomes:2left(x - frac{1 + z}{2}right)^2 + (y + 2)^2 + frac{1}{2}(z - 1)^2 + frac{1}{2} = 0.Wait, no. Wait, the term was ( frac{1}{2}(z - 1)^2 ), so combining with the previous constants:Wait, before, we had:2(...) + (y + 2)^2 + frac{1}{2}(z - 1)^2 + 1/2 = 0.Wait, but when we completed the square for z, we had:Original z terms: ( frac{z^2}{2} - z + frac{1}{2} ).Which became ( frac{1}{2}(z - 1)^2 ). Let me check:( frac{1}{2}(z - 1)^2 = frac{1}{2}(z² - 2z + 1) = frac{z²}{2} - z + frac{1}{2} ). Yes, correct. So that term is just ( frac{1}{2}(z - 1)^2 ).Therefore, the entire equation is:2left(x - frac{1 + z}{2}right)^2 + (y + 2)^2 + frac{1}{2}(z - 1)^2 + frac{1}{2} = 0.Wait, but where did the +1/2 at the end come from? Let me check again.Wait, after simplifying constants earlier, we had:After moving all terms to the left side and completing squares step by step, we ended up with:2(...) + (y + 2)^2 + frac{1}{2}(z - 1)^2 + 1/2 = 0.Wait, but if all the other terms are squares multiplied by positive coefficients, then each of those squares is non-negative. So 2(...) ≥ 0, (y + 2)^2 ≥ 0, (1/2)(z - 1)^2 ≥ 0. Then the sum of these non-negative terms plus 1/2 equals zero. But 1/2 is positive, so the sum can't be zero. That can't be. There's a mistake here. Let me retrace.Original equation:2x² + y² + z² = 2x -4y +2xz -5.Moving all terms to the left:2x² -2x + y² +4y + z² -2xz +5 =0.Wait, original equation: 2x² + y² + z² -2x +4y -2xz +5=0.Then, grouping terms:For x and z: 2x² -2x -2xz.For y: y² +4y.For z: z².Constant: +5.Completing the square for y: y² +4y = (y +2)^2 -4.So substituting back:2x² -2x -2xz + (y +2)^2 -4 + z² +5=0.Simplify constants: -4 +5=1.So: 2x² -2x -2xz + (y +2)^2 + z² +1=0.Now, the x and z terms: 2x² -2x -2xz.Let me factor out 2 from the x² term:2(x² -x -xz) =2x² -2x -2xz.Wait, perhaps factor x from x² -x -xz:x² -x -xz =x(x -1 -z).Hmm, not helpful. Alternatively, perhaps treat x and z as variables and complete the square for them.Alternatively, let me consider x and z together. Let me write the quadratic terms in x and z:2x² -2xz + z².Wait, 2x² -2xz + z². Let me write this as a quadratic form. Maybe diagonalize it or find a substitution.Alternatively, treat this as 2x² -2xz + z². Let's see if this can be expressed as a square.Let me write 2x² -2xz + z² = (sqrt(2)x)^2 -2xz + z². Hmm, not a perfect square. Alternatively, perhaps complete the square for x.Let me factor out the coefficient of x² which is 2:2(x² - xz) + z². Then, inside the parentheses: x² -xz.Complete the square here:x² -xz =x² -xz + (z²/4) - (z²/4) = (x - z/2)^2 - z²/4.Therefore, 2(x - z/2)^2 - 2*(z²/4) + z² =2(x - z/2)^2 - z²/2 + z² =2(x - z/2)^2 + z²/2.Therefore, 2x² -2xz + z² =2(x - z/2)^2 + z²/2.So substituting back into the equation:2(x - z/2)^2 + z²/2 -2x + (y +2)^2 +1=0.Wait, hold on. Let's double-check:Original x and z terms after moving to left: 2x² -2x -2xz + z². We rewrote 2x² -2xz + z² as 2(x - z/2)^2 + z²/2.Therefore, 2(x - z/2)^2 + z²/2 -2x + (y +2)^2 +1=0.So now, the equation is:2(x - z/2)^2 + z²/2 -2x + (y +2)^2 +1=0.Now, let's handle the term -2x. Maybe express x in terms of (x - z/2). Let me let u = x - z/2. Then x = u + z/2. Substitute into the equation:2u² + z²/2 -2(u + z/2) + (y +2)^2 +1=0.Expanding -2(u + z/2):-2u - z.Therefore, the equation becomes:2u² -2u + z²/2 - z + (y +2)^2 +1=0.Now, let's complete the square for u and z.For the u terms: 2u² -2u. Factor out 2:2(u² -u). Complete the square inside:u² -u = u² -u + 1/4 -1/4 = (u - 1/2)^2 -1/4.Therefore, 2(u -1/2)^2 -1/2.For the z terms: z²/2 - z. Factor out 1/2:1/2(z² - 2z). Complete the square:z² -2z = (z -1)^2 -1.Therefore, 1/2(z -1)^2 -1/2.Substitute back into the equation:2(u -1/2)^2 -1/2 + 1/2(z -1)^2 -1/2 + (y +2)^2 +1=0.Combine constants: -1/2 -1/2 +1 = -1 +1 =0.Therefore, the equation becomes:2(u -1/2)^2 + 1/2(z -1)^2 + (y +2)^2 =0.Since u = x - z/2, substitute back:2(x - z/2 -1/2)^2 + 1/2(z -1)^2 + (y +2)^2 =0.Now, all the terms on the left are squares multiplied by non-negative coefficients. The sum of squares equals zero only if each square is zero.Therefore, each term must be zero:1. ( 2(x - frac{z}{2} - frac{1}{2})^2 = 0 ) implies ( x - frac{z}{2} - frac{1}{2} = 0 ).2. ( frac{1}{2}(z -1)^2 = 0 ) implies ( z -1 =0 ).3. ( (y +2)^2 =0 ) implies ( y +2 =0 ).So solving these equations:From equation 2: z =1.From equation 3: y = -2.Substitute z=1 into equation 1:x - (1)/2 -1/2 =0 → x -1 =0 → x=1.Therefore, the only solution is x=1, y=-2, z=1.Therefore, the given equation has only one solution at (1, -2, 1). Therefore, the value of x - y + z at this point is 1 - (-2) +1 =1 +2 +1=4.But wait, the problem asks for the maximum possible value of x - y + z. If there's only one solution, then the maximum is 4. But maybe I made a mistake in completing the squares? Let me check again.Wait, when we completed the squares, we ended up with the equation:2(x - z/2 -1/2)^2 + (1/2)(z -1)^2 + (y +2)^2 =0.Which indeed implies that each squared term must be zero. Therefore, only one solution exists. Hence, x - y + z can only be 4. Therefore, the maximum possible value is 4. But let me confirm this because sometimes in these quadratic forms, there might be a way to parametrize variables to allow other solutions. But given the equation reduces to sum of squares equal to zero, which only occurs when each square is zero. So there are no other solutions. Therefore, the only possible value of x - y + z is 4, so it's both the maximum and the minimum.But the problem says "find the maximum possible value", implying that there might be multiple values. However, according to this, there's only one solution. Maybe I made a mistake in the algebra steps.Let me check the steps again.Original equation: 2x² + y² + z² = 2x -4y +2xz -5.Moving all terms to the left:2x² -2x + y² +4y + z² -2xz +5 =0.Completing the square for y: (y +2)^2 -4.So substituting:2x² -2x -2xz + z² + (y +2)^2 -4 +5=0.Simplifies to:2x² -2x -2xz + z² + (y +2)^2 +1=0.Then, grouping x and z terms: 2x² -2xz -2x + z².We rewrote 2x² -2xz + z² as 2(x - z/2)^2 + z²/2, which seems correct.Then substituted back, leading to:2(x - z/2)^2 + z²/2 -2x + (y +2)^2 +1=0.Then set u =x - z/2, so x= u + z/2.Substituted into equation: 2u² + z²/2 -2(u + z/2) + (y +2)^2 +1=0.Expanding: 2u² -2u - z + z²/2 + (y +2)^2 +1=0.Then completed squares for u and z:For u: 2u² -2u =2(u² -u) =2[(u -0.5)^2 -0.25] =2(u -0.5)^2 -0.5.For z: z²/2 - z = (1/2)(z² -2z) = (1/2)[(z -1)^2 -1] = (1/2)(z -1)^2 -0.5.So substituting back:2(u -0.5)^2 -0.5 +0.5(z -1)^2 -0.5 + (y +2)^2 +1=0.Simplifying constants: -0.5 -0.5 +1= -1 +1=0.Therefore:2(u -0.5)^2 +0.5(z -1)^2 + (y +2)^2=0.Which requires each term to be zero. Hence, u=0.5, z=1, y=-2.Since u =x - z/2, then x - z/2 =0.5. With z=1, x -0.5=0.5 →x=1.So x=1, y=-2, z=1. Thus, x - y + z=1 -(-2)+1=4. So indeed, the only solution gives x - y + z=4.Therefore, since this is the only possible value, the maximum is 4. Hence, the answer is 4.But wait, let me check once more. Suppose there is a mistake in the completion of squares. Let me verify by plugging x=1, y=-2, z=1 into the original equation.Left side: 2*(1)^2 + (-2)^2 + (1)^2 =2*1 +4 +1=2+4+1=7.Right side: 2*1 -4*(-2) +2*1*1 -5=2 +8 +2 -5=7.Yes, both sides equal 7. So the solution is valid.Therefore, the given equation has only one solution (1, -2, 1), and thus the expression x - y + z can only take the value 4. Therefore, the maximum possible value is 4.But the problem says "find the maximum possible value", which suggests that there might be multiple solutions with varying values of x - y + z, but according to the analysis, there's only one solution. Therefore, maybe the initial approach is wrong.Wait, perhaps the equation represents a single point, hence the maximum is 4. Alternatively, maybe there's a different way to approach this problem, such as using Lagrange multipliers.Let me try that method to confirm.Let’s set up the Lagrangian. Let f(x, y, z) =x - y + z, which we want to maximize subject to the constraint g(x, y, z)=2x² + y² + z² -2x +4y -2xz +5=0.Form the Lagrangian: L =x - y + z - λ(2x² + y² + z² -2x +4y -2xz +5).Take partial derivatives and set to zero.Partial derivative with respect to x:1 - λ(4x -2 -2z)=0.Partial derivative with respect to y:-1 - λ(2y +4)=0.Partial derivative with respect to z:1 - λ(2z -2x)=0.Partial derivative with respect to λ:- (2x² + y² + z² -2x +4y -2xz +5)=0.So we have four equations:1. 1 - λ(4x -2 -2z)=0 → 1 = λ(4x -2 -2z).2. -1 - λ(2y +4)=0 → -1 = λ(2y +4).3. 1 - λ(2z -2x)=0 →1 = λ(2z -2x).4. 2x² + y² + z² -2x +4y -2xz +5=0.Let’s solve these equations.From equation 2: -1 = λ(2y +4) → λ = -1/(2y +4).From equation 1: 1 = λ(4x -2 -2z). Substitute λ:1 = (-1)/(2y +4) * (4x -2 -2z) → (4x -2 -2z) = - (2y +4).Similarly, from equation 3:1 = λ(2z -2x). Substitute λ:1 = (-1)/(2y +4)*(2z -2x) → (2z -2x) = - (2y +4).So now, we have two equations:From equation 1 substitution: 4x -2 -2z = -2y -4.From equation 3 substitution: 2z -2x = -2y -4.Let’s simplify both equations.First equation: 4x -2 -2z +2y +4=0 →4x +2y -2z +2=0 → divide by 2: 2x + y -z +1=0.Second equation: 2z -2x +2y +4=0 → divide by 2: z -x + y +2=0.So now, we have:Equation A: 2x + y - z +1=0.Equation B: -x + y + z +2=0.Let’s solve these two equations for variables. Let me write them as:A: 2x + y - z = -1.B: -x + y + z = -2.Let’s subtract equation B from equation A:(2x + y - z) - (-x + y + z) = -1 - (-2).Simplify left side:2x + y - z +x -y -z=3x -2z.Right side:1.So 3x -2z =1 → equation C:3x -2z=1.Now, let's add equations A and B:(2x + y - z) + (-x + y + z)= -1 + (-2).Simplify left side: x + 2y = -3 → equation D:x +2y = -3.So now, from equation C:3x -2z=1, equation D:x +2y = -3.But we need another equation to relate variables. Let's use equation B: -x + y + z = -2.Let me express z from equation B: z =x - y -2.Substitute into equation C:3x -2(x - y -2)=1 →3x -2x +2y +4=1 →x +2y +4=1→x +2y= -3. Which is equation D. So no new information. So we need to express variables in terms of others.From equation D: x = -3 -2y.From equation B: z= x - y -2= (-3 -2y) - y -2= -3 -3y -2= -5 -3y.So x= -3 -2y, z= -5 -3y.Now, let's substitute these into equation 2: λ= -1/(2y +4).Also, let's recall equation 3 substitution led to equation C and B, which are already used.Now, substitute x, z in terms of y into the constraint equation 4:2x² + y² + z² -2x +4y -2xz +5=0.Let’s compute each term.x= -3 -2y, z= -5 -3y.Compute x²: (-3 -2y)^2=9 +12y +4y².z²: (-5 -3y)^2=25 +30y +9y².xz: (-3 -2y)(-5 -3y)=15 +9y +10y +6y²=15 +19y +6y².Now, substitute into the equation:2*(9 +12y +4y²) + y² + (25 +30y +9y²) -2*(-3 -2y) +4y -2*(15 +19y +6y²) +5=0.Calculate term by term:2*(9 +12y +4y²) =18 +24y +8y².y² remains.25 +30y +9y² remains.-2*(-3 -2y)=6 +4y.4y remains.-2*(15 +19y +6y²)= -30 -38y -12y².+5 remains.Now, add all terms together:18 +24y +8y² + y² +25 +30y +9y² +6 +4y +4y -30 -38y -12y² +5.Combine like terms:y² terms:8y² +1y² +9y² -12y²=6y².y terms:24y +30y +4y +4y -38y=24+30=54; 54+4=58; 58+4=62; 62-38=24y.Constants:18 +25 +6 -30 +5=18+25=43; 43+6=49; 49-30=19; 19+5=24.So overall equation becomes:6y² +24y +24=0.Divide by 6:y² +4y +4=0 → (y +2)^2=0 → y= -2.Therefore, y= -2. Then x= -3 -2*(-2)= -3 +4=1.z= -5 -3*(-2)= -5 +6=1.Thus, we get x=1, y=-2, z=1, which matches the previous solution. Therefore, using Lagrange multipliers also leads to the same unique solution. Hence, the only possible value of x - y + z is 4. Therefore, the maximum value is 4.But why does the problem ask for the maximum possible value? Maybe there was a typo, or maybe I misread the original equation. Let me check again.Original problem: If ( x, y ), and ( z ) are real numbers such that ( 2 x^{2} + y^{2} + z^{2} = 2 x - 4 y + 2 xz - 5 ), find the maximum possible value of ( x - y + z ).No, the equation seems correctly stated. So based on the analysis, the equation constrains the variables to a single point, hence the expression can only take one value, which is 4, so that's both the maximum and minimum. Therefore, the answer is 4.Alternatively, maybe there is an error in assuming the equation reduces to a single point. Let me visualize the equation.The original equation is a quadratic equation in three variables. Such an equation can represent a variety of quadric surfaces, but given that the sum of squares equals a negative number (when rearranged initially), but after completing squares, it reduces to a single point.Therefore, the equation represents a single point in 3D space, so the variables can't vary, hence x - y + z is fixed. Therefore, the maximum is 4.Therefore, the answer is boxed{4}.Final Answerboxed{4}

✅:Given the equation ( 2x^2 + y^2 + z^2 = 2x - 4y + 2xz - 5 ), we need to find the maximum possible value of ( x - y + z ).First, we rewrite the equation by moving all terms to the left side:[2x^2 + y^2 + z^2 - 2x + 4y - 2xz + 5 = 0]We complete the square for the ( y )-terms:[y^2 + 4y = (y + 2)^2 - 4]Substituting this back into the equation, we get:[2x^2 - 2x - 2xz + z^2 + (y + 2)^2 - 4 + 5 = 0]Simplifying the constants:[2x^2 - 2x - 2xz + z^2 + (y + 2)^2 + 1 = 0]Next, we handle the ( x ) and ( z )-terms. We rewrite ( 2x^2 - 2xz + z^2 ) as:[2left(x - frac{z}{2}right)^2 + frac{z^2}{2}]Substituting this back into the equation, we get:[2left(x - frac{z}{2}right)^2 + frac{z^2}{2} - 2x + (y + 2)^2 + 1 = 0]Let ( u = x - frac{z}{2} ). Then ( x = u + frac{z}{2} ), and substituting this into the equation, we get:[2u^2 + frac{z^2}{2} - 2left(u + frac{z}{2}right) + (y + 2)^2 + 1 = 0]Expanding and simplifying:[2u^2 - 2u - z + frac{z^2}{2} + (y + 2)^2 + 1 = 0]Completing the square for ( u ) and ( z ):[2left(u - frac{1}{2}right)^2 - frac{1}{2} + frac{1}{2}(z - 1)^2 - frac{1}{2} + (y + 2)^2 + 1 = 0]Simplifying constants:[2left(u - frac{1}{2}right)^2 + frac{1}{2}(z - 1)^2 + (y + 2)^2 = 0]Each term must be zero:1. ( 2left(u - frac{1}{2}right)^2 = 0 ) implies ( u = frac{1}{2} )2. ( frac{1}{2}(z - 1)^2 = 0 ) implies ( z = 1 )3. ( (y + 2)^2 = 0 ) implies ( y = -2 )Substituting back, ( x = u + frac{z}{2} = frac{1}{2} + frac{1}{2} = 1 ).Thus, the solution is ( x = 1 ), ( y = -2 ), ( z = 1 ). Evaluating ( x - y + z ):[1 - (-2) + 1 = 1 + 2 + 1 = 4]Using Lagrange multipliers confirms this solution is unique. Therefore, the maximum possible value of ( x - y + z ) is (boxed{4}).

🔑:1. Consider the given equation: [ 2x^2 + y^2 + z^2 = 2x - 4y + 2xz - 5 ]2. To simplify, begin by moving all terms to one side of the equation: [ 2x^2 + y^2 + z^2 - 2x + 4y - 2xz + 5 = 0 ]3. Group the quadratic terms. Notice that they can potentially form perfect squares: [ 2x^2 - 2x + y^2 + 4y + z^2 - 2xz + 5 = 0 ]4. Handle the (x)-terms by completing the square: [ 2x^2 - 2x = 2(x^2 - x) = 2left(x^2 - x + frac{1}{4} - frac{1}{4}right) = 2left(left(x - frac{1}{2}right)^2 - frac{1}{4}right) ] So this simplifies to: [ 2left(x - frac{1}{2}right)^2 - frac{1}{2} ]5. Similarly, complete the square for the (y)-terms: [ y^2 + 4y = left(y^2 + 4y + 4 - 4right) = (y + 2)^2 - 4 ]6. Let's handle the (z)-terms and cross term (2xz): [ z^2 - 2xz = (z^2 - 2xz + x^2 - x^2) = (z - x)^2 - x^2 ]7. Substitute these simplified forms back into the equation: [ 2left(x - frac{1}{2}right)^2 - frac{1}{2} + (y + 2)^2 - 4 + (z - x)^2 - x^2 + 5 = 0 ]8. Combine constants: [ 2left(x - frac{1}{2}right)^2 + (y + 2)^2 + (z - x)^2 - frac{1}{2} - 4 - x^2 + 5 = 0 ] This simplifies to: [ 2left(x - frac{1}{2}right)^2 + (y + 2)^2 + (z - x)^2 - frac{1}{2} - 4 + 5 - x^2 = 0 ] Combining the constants further: [ 2left(x - frac{1}{2}right)^2 + (y + 2)^2 + (z - x)^2 - frac{1}{2} - 4 + 5 = 0 ] [ 2left(x - frac{1}{2}right)^2 + (y + 2)^2 + (z - x)^2 + frac{1}{2} - x^2 = 0 ] [ 2left(x - frac{1}{2}right)^2 + (y + 2)^2 + (z - x)^2 + frac{1}{2} = 0 ]9. With coefficients correct, let's regroup and simplify terms aimed at ((x, frac{1}{2}), ((-2, y)), ((z, x))): [ left(x-frac{1}{2}right)^2 +(y+2)^2+(z-x)^2 = 0 implies x-0.25 = 0 implies and x=zimplies ] 10. Equalities fufill ,( y +2 =0), implies -(x=z -1, -y = 2 thus solve Equation & derive.11. As they deduce :(boxed{x-y+z} = 0 -4)= 4 as maximum achievableConclusion:[boxed{4}.]