🧠:Okay, so I have this problem here: Let α be an element in the finite field 𝔽_{q^n}. I need to show that if the trace of α from 𝔽_{q^n} down to 𝔽_q is zero, then there exists some element β in 𝔽_{q^n} such that α = β^q - β. Hmm, alright. Let me start by recalling what the trace function is.The trace function, Tr_{𝔽_{q^n}/𝔽_q}(α), is defined as the sum of the conjugates of α with respect to the field extension. Specifically, it's α + α^q + α^{q^2} + ... + α^{q^{n-1}}. So, if the trace of α is zero, that means this sum equals zero in 𝔽_q. The problem is asking to show that if α has trace zero, then it can be expressed as β^q - β for some β in the same field. This seems related to additive characters or maybe additive versions of Hilbert's Theorem 90. Wait, Hilbert's Theorem 90 is about multiplicative characters, right? It states that an element of norm 1 can be written as σ(β)/β, where σ is a generator of the Galois group. But here, we're dealing with additive stuff. Maybe there's an additive version of this theorem?Let me check. Yes, actually, the additive version of Hilbert's Theorem 90 states that if you have a cyclic extension L/K with Galois group generated by σ, then any element of L with trace zero can be written as σ(β) - β for some β in L. That sounds exactly like what we need here! Since our extension 𝔽_{q^n}/𝔽_q is cyclic of degree n, with Galois group generated by the Frobenius automorphism σ: x ↦ x^q. So, applying the additive Hilbert 90, if Tr_{𝔽_{q^n}/𝔽_q}(α) = 0, then there exists β such that α = σ(β) - β = β^q - β. But wait, maybe the user expects a more elementary proof, not directly citing Hilbert 90? Since finite fields are pretty concrete, maybe we can construct β explicitly or use some linear algebra approach?Let me try to think of it in terms of linear equations. The trace is a linear functional from 𝔽_{q^n} to 𝔽_q. The kernel of this functional is the set of elements with trace zero, which has dimension n-1 over 𝔽_q. On the other hand, the image of the map β ↦ β^q - β is a subgroup of the additive group of 𝔽_{q^n}, and we need to show that this image is exactly the kernel of the trace.First, let's check that the image of β ↦ β^q - β is contained in the kernel of the trace. Suppose α = β^q - β. Then, compute Tr(α):Tr(α) = Tr(β^q - β) = Tr(β^q) - Tr(β). Since the trace is invariant under the Frobenius automorphism (because Tr(β^q) = β^q + (β^q)^q + ... + (β^q)^{q^{n-1}}} = β^q + β^{q^2} + ... + β^{q^n} = β + β^q + ... + β^{q^{n-1}}} because β^{q^n} = β). Therefore, Tr(β^q) = Tr(β), so Tr(α) = Tr(β) - Tr(β) = 0. Hence, the image is indeed in the kernel of the trace.Now, we need to show that the image has the same dimension as the kernel, which is n-1. The dimension of the image is equal to the dimension of 𝔽_{q^n} minus the dimension of the kernel of the map β ↦ β^q - β. So, if the kernel of this map is just the elements β such that β^q = β, which is 𝔽_q, since those are the elements fixed by the Frobenius automorphism. Therefore, the kernel is 𝔽_q, which has dimension 1 over 𝔽_q. Therefore, the image has dimension n - 1, which matches the dimension of the kernel of the trace. Since the image is a subspace of the kernel and they have the same dimension, they must be equal. Therefore, every element with trace zero is in the image, i.e., can be written as β^q - β.Alternatively, maybe we can use additive polynomials. The equation x^q - x = α is an additive polynomial. The roots of x^q - x - α form an affine space over 𝔽_q. The number of solutions is either 0 or q. So, if the equation x^q - x = α has a solution, then it has q solutions. The key is to show that if Tr(α) = 0, then such a solution exists.Alternatively, maybe consider the mapping f: 𝔽_{q^n} → 𝔽_{q^n} defined by f(β) = β^q - β. This is an additive homomorphism. The kernel is {β | β^q = β} = 𝔽_q. So, by the first isomorphism theorem, the image of f is isomorphic to 𝔽_{q^n}/𝔽_q, which has dimension n-1 over 𝔽_q. But the trace map is a surjective linear functional, so its kernel is also dimension n-1. Therefore, image of f is exactly the kernel of the trace. Therefore, the two subspaces coincide, so any α with Tr(α)=0 is in the image of f. So there exists β such that α = β^q - β.Alternatively, if I want to be more explicit, maybe use a basis for 𝔽_{q^n} over 𝔽_q and write the trace as a linear combination. But that might get messy. The previous argument using linear algebra and dimensions seems sufficient.Wait, but let me verify that the image of f is indeed contained in the kernel of the trace, which we did earlier, and that their dimensions match. Since the image has dimension n - 1 and the kernel of the trace also has dimension n - 1, over 𝔽_q, so they must be the same. Therefore, the result follows.So, putting it all together: since the map β ↦ β^q - β has kernel 𝔽_q, its image has dimension n - 1 over 𝔽_q. The trace map is a linear functional with kernel of dimension n - 1. Since the image is contained in the kernel and they have the same dimension, they must coincide. Therefore, any α with trace zero is in the image, so such a β exists.Another way: Maybe use additive characters and orthogonality. The additive characters of 𝔽_{q^n} are homomorphisms from (𝔽_{q^n}, +) to the multiplicative group of complex numbers. The orthogonality relations might relate the trace to these characters, but this might be overcomplicating.Alternatively, consider solving the equation β^q - β = α. Let's see, if we set up a system of equations. Let’s suppose that β is an element of 𝔽_{q^n}. Then, raising both sides to the q-th power, we get β^{q^2} - β^q = α^q. Subtracting the original equation from this, we get β^{q^2} - 2β^q + β = α^q - α. Continuing this process, recursively applying Frobenius and subtracting previous equations... Hmm, this might lead to a telescoping series. If we keep doing this up to the n-th power, since β^{q^n} = β, we might end up with an equation involving Tr(α). Let me try:Starting with β^q - β = α.Applying Frobenius: β^{q^2} - β^q = α^q.Subtract the previous equation: (β^{q^2} - β^q) - (β^q - β) = α^q - α ⇒ β^{q^2} - 2β^q + β = α^q - α.Do it again: apply Frobenius to the second equation: β^{q^3} - β^{q^2} = α^{q^2}.Subtract the previous result: (β^{q^3} - β^{q^2}) - (β^{q^2} - 2β^q + β) = α^{q^2} - (α^q - α) ⇒ β^{q^3} - 2β^{q^2} + 2β^q - β = α^{q^2} - α^q + α.Continuing this, after n steps, since β^{q^n} = β, the left-hand side would telescope. Let's try with n=2 first. If n=2, then after two steps:First equation: β^q - β = α.Second equation: β^{q^2} - β^q = α^q.But since β^{q^2} = β, substitute into the second equation: β - β^q = α^q.Now, adding the first equation: (β^q - β) + (β - β^q) = α + α^q ⇒ 0 = Tr(α). So in this case, if Tr(α) = 0, then the two equations are compatible. So, when n=2, we can set up the system:β^q - β = α,β - β^q = α^q.Adding both equations gives 0 = α + α^q = Tr(α), which holds. Therefore, in this case, we can solve for β. For example, take the first equation β^q = β + α. Then, substitute into the second equation: β - (β + α) = α^q ⇒ -α = α^q ⇒ α + α^q = 0. Which is exactly Tr(α) = 0 since n=2. Therefore, in this case, if Tr(α)=0, then the system has a solution. So for n=2, we can solve it directly.Similarly, for general n, if we write the system of equations obtained by applying Frobenius repeatedly, the system will be consistent if and only if Tr(α) = 0. Therefore, there exists a solution β if and only if Tr(α) = 0. This is similar to solving a linear system over a field. Since the system is consistent when Tr(α)=0, then β exists.Alternatively, maybe use induction on n. Suppose the result holds for extensions of degree less than n. Let’s see. If n=1, then 𝔽_q, and the trace is just the identity map. So if Tr(α)=0, then α=0, and 0=β^q - β. Since β^q = β in 𝔽_q, so any β in 𝔽_q satisfies this. So, holds trivially. For n=2, as above. Suppose true for n-1. Hmm, not sure how induction would work here. Maybe not the best approach.Alternatively, using additive polynomials. The polynomial L(x) = x^q - x is a linearized polynomial, meaning it's additive: L(β + γ) = L(β) + L(γ). The kernel of L is 𝔽_q, as we saw. The image is therefore a subgroup of the additive group of 𝔽_{q^n} of index q. The size of the image is q^{n-1}. The kernel of the trace map is also a subgroup of size q^{n-1}. Since the image of L is contained in the kernel of the trace, and they have the same size, they must be equal. Therefore, the result follows.Alternatively, think in terms of duality. The trace map is a non-degenerate bilinear form on 𝔽_{q^n} over 𝔽_q. So, the dual space of 𝔽_{q^n} over 𝔽_q is 1-dimensional, generated by the trace function. Therefore, the annihilator of the image of L (the set of linear functionals that vanish on the image) is... Hmm, maybe overcomplicating.But in any case, all these different approaches converge to the conclusion that the image of β ↦ β^q - β is exactly the kernel of the trace. Therefore, any α with Tr(α)=0 can be written as β^q - β for some β.Let me check with an example. Let’s take q=2, n=2, so 𝔽_4. Let’s pick an element α with trace 0. The trace from 𝔽_4 to 𝔽_2 is Tr(α) = α + α^2. Let’s list the elements of 𝔽_4: 0, 1, a, a+1, where a^2 = a + 1.Compute Tr(0) = 0 + 0 = 0.Tr(1) = 1 + 1 = 0.Tr(a) = a + a^2 = a + (a + 1) = 1.Tr(a+1) = (a + 1) + (a + 1)^2 = (a +1) + (a^2 + 1) = (a +1) + ((a +1) +1) = a +1 + a + 2 = 2a + 3 = 1 (since coefficients are mod 2).So elements with trace 0 are 0 and 1. Now, can these be written as β^2 - β?Compute for β in 𝔽_4:If β=0: 0^2 -0=0.β=1:1^2 -1=0.β=a: a^2 -a = (a+1) -a =1.β=a+1: (a+1)^2 - (a+1) = (a^2 +1) -a -1 = (a+1 +1) -a -1 = a +2 -a -1 =1 -1=0 (mod 2).Wait, so β=a gives α=1, β=a+1 gives α=0. So the image of the map β ↦ β^2 - β in 𝔽_4 is {0,1}, which are exactly the elements with trace 0. So that works. For example, α=1 has Tr(1)=0 and can be written as a^2 - a. Similarly, α=0 is trivial. So in this case, it works.Another example: q=3, n=2. 𝔽_9. Let’s construct it as 𝔽_3[x]/(x^2 +1), so elements are a + bα where α^2 = -1. The trace is Tr(β) = β + β^3. Let’s pick an element with trace 0. For example, take β = α. Tr(α) = α + α^3 = α + (α^3). Since α^2 = -1, α^3 = α*(-1) = -α. So Tr(α) = α - α =0. So α is an element with trace 0. Now, can we find γ such that γ^3 - γ = α? Let's suppose γ = a + bα. Then γ^3 = (a + bα)^3 = a^3 + 3a^2 b α + 3 a b^2 α^2 + b^3 α^3. But in characteristic 3, this simplifies to a^3 + 0 + 0 + b^3 α^3. Since α^3 = -α, so γ^3 = a^3 - b^3 α. Then γ^3 - γ = (a^3 - a) + (-b^3 α - b α). We need this equal to α. Therefore:(a^3 - a) + (-b^3 - b)α = α.Thus, equate coefficients:a^3 - a = 0,(-b^3 - b) =1.In 𝔽_3, a^3 = a for all a ∈ 𝔽_3, so a^3 - a =0 holds for any a ∈ 𝔽_3. So the first equation is always satisfied. The second equation: -b^3 - b =1 ⇒ b^3 + b = -1 = 2 in 𝔽_3. So we need to solve b^3 + b = 2 in 𝔽_3. Let's test b=0: 0 +0=0≠2. b=1:1 +1=2. Yes! So b=1. Therefore, γ = a + α where a ∈ 𝔽_3. So for example, γ=0 + α=α: γ^3 - γ = α^3 - α = -α - α = -2α = α (since -2=1 in 𝔽_3). Wait, that gives γ^3 - γ = α, which is exactly what we wanted. So yes, β=α is a solution. So this works.Therefore, in these examples, the statement holds. So that's reassuring.In summary, the key idea is recognizing that this is an instance of the additive Hilbert's Theorem 90, or using linear algebra to show that the image of the map β ↦ β^q - β is exactly the kernel of the trace map, by comparing dimensions and showing inclusion. Since both are (n-1)-dimensional 𝔽_q-subspaces of 𝔽_{q^n}, they must coincide, hence the result follows.Final Answerboxed{alpha = beta^q - beta}

✅:To show that if (operatorname{tr}_{mathbb{F}_{q^{n}} / mathbb{F}_{q}}(alpha)=0), then there exists (beta in mathbb{F}_{q^{n}}) such that (alpha=beta^{q}-beta), we can use the additive version of Hilbert's Theorem 90 and linear algebra arguments.1. Trace Function Definition: The trace function (operatorname{tr}_{mathbb{F}_{q^{n}} / mathbb{F}_{q}}(alpha)) is defined as the sum of the conjugates of (alpha): [ operatorname{tr}_{mathbb{F}_{q^{n}} / mathbb{F}_{q}}(alpha) = alpha + alpha^q + alpha^{q^2} + cdots + alpha^{q^{n-1}} ] Given (operatorname{tr}_{mathbb{F}_{q^{n}} / mathbb{F}_{q}}(alpha) = 0), this sum equals zero.2. Additive Hilbert's Theorem 90: The additive version of Hilbert's Theorem 90 states that for a cyclic extension (L/K) with Galois group generated by (sigma), any element (alpha in L) with trace zero can be written as (sigma(beta) - beta) for some (beta in L). For the extension (mathbb{F}_{q^n}/mathbb{F}_q), the Frobenius automorphism (sigma(beta) = beta^q) generates the Galois group. Thus, if (operatorname{tr}_{mathbb{F}_{q^{n}} / mathbb{F}_{q}}(alpha) = 0), then (alpha = beta^q - beta) for some (beta in mathbb{F}_{q^n}).3. Linear Algebra Approach: - The trace map is a linear functional with kernel of dimension (n-1) over (mathbb{F}_q). - The map (beta mapsto beta^q - beta) is an additive homomorphism with kernel (mathbb{F}_q), leading to an image of dimension (n-1) over (mathbb{F}_q). - Since the image of this map is contained in the kernel of the trace and both have the same dimension, they must coincide. Therefore, any (alpha) with trace zero is in the image of this map.4. Verification with Examples: - For (mathbb{F}_4) and (mathbb{F}_9), explicit calculations show that elements with trace zero can indeed be expressed as (beta^q - beta).Thus, combining these arguments, we conclude that if (operatorname{tr}_{mathbb{F}_{q^{n}} / mathbb{F}_{q}}(alpha) = 0), there exists (beta in mathbb{F}_{q^{n}}) such that (alpha = beta^q - beta).[boxed{alpha = beta^q - beta}]

🔑:1. Definition of Sets A and B: - Let A subset mathbb{F}_{q^n} be the set of elements with trace zero. - Let B subset mathbb{F}_{q^n} be the set of elements that can be expressed as beta^q - beta for some beta in mathbb{F}_{q^n}. - Goal: Show that A subseteq B.2. Inclusion B subseteq A: - If beta in mathbb{F}_{q^n}, then operatorname{tr}_{mathbb{F}_{q^n}/mathbb{F}_q}(beta^q - beta) = 0. Therefore, every element of B has trace zero, i.e., B subseteq A.3. Cardinalities: - We need to show that operatorname{Card} A leq operatorname{Card} B. - We know that operatorname{Card} A = q^{n-1}.4. Defining a Function: - Define a function f: mathbb{F}_{q^n} rightarrow B by f(beta) = beta^q - beta. - Every element alpha in B can have at most q preimages under f since it solves a polynomial equation of degree q at most: (f(beta) = beta^q - beta = alpha).5. Counting Elements: - Since f is a polynomial of degree q, the image of f (i.e., B) has at least frac{q^n}{q} = q^{n-1} elements. - So, operatorname{Card} B geq q^{n-1}.6. Conclusion: - Given that operatorname{Card} A = operatorname{Card} B = q^{n-1} and B subseteq A, it follows that A = B. - Therefore, if operatorname{tr}_{mathbb{F}_{q^n}/mathbb{F}_q}(alpha) = 0, then there exists beta in mathbb{F}_{q^n} such that alpha = beta^q - beta.Thus, we have shown A subseteq B and operatorname{Card} A = operatorname{Card} B, concluding that A = B.[boxed{Q.E.D.}]

🧠:Okay, so I need to find the minimum value of the expression |MA|² + |MB|² + |MC|² where M is an arbitrary point in the plane, and A, B, C are the vertices of a triangle with side lengths a, b, c. Hmm, this seems like a geometry optimization problem. Let me think about how to approach this.First, I remember that in coordinate geometry, the sum of squared distances from a point to several points can be minimized by finding the centroid or something similar. Wait, isn't there a formula related to the centroid minimizing the sum of squared distances? Let me recall. Yes, I think the centroid (or the center of mass) minimizes the sum of squared distances from a set of points. So maybe the answer is that the minimum occurs at the centroid of triangle ABC, and then we can compute the value there.But let me verify this. Suppose I place the triangle in a coordinate system to make the calculations easier. Let me assign coordinates to points A, B, C. Let's say point A is at (x₁, y₁), B at (x₂, y₂), and C at (x₃, y₃). Then, for a point M with coordinates (x, y), the expression becomes:|MA|² + |MB|² + |MC|² = [(x - x₁)² + (y - y₁)²] + [(x - x₂)² + (y - y₂)²] + [(x - x₃)² + (y - y₃)²]If I expand this, it becomes:3x² - 2x(x₁ + x₂ + x₃) + (x₁² + x₂² + x₃²) + 3y² - 2y(y₁ + y₂ + y₃) + (y₁² + y₂² + y₃²)Which can be rewritten as:3x² - 2x(x₁ + x₂ + x₃) + 3y² - 2y(y₁ + y₂ + y₃) + (x₁² + x₂² + x₃² + y₁² + y₂² + y₃²)This is a quadratic in x and y. To find the minimum, we can take partial derivatives with respect to x and y and set them to zero.Partial derivative with respect to x:6x - 2(x₁ + x₂ + x₃) = 0 ⇒ x = (x₁ + x₂ + x₃)/3Similarly, partial derivative with respect to y:6y - 2(y₁ + y₂ + y₃) = 0 ⇒ y = (y₁ + y₂ + y₃)/3So the critical point is at the centroid of the triangle, which has coordinates ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3). Therefore, the minimum occurs at the centroid G.Now, substituting this back into the expression to find the minimum value. Let me compute |GA|² + |GB|² + |GC|².First, note that the centroid G has coordinates ( (x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3 )So, |GA|² = [( (x₁ + x₂ + x₃)/3 - x₁ )² + ( (y₁ + y₂ + y₃)/3 - y₁ )² ]Simplify:= [ ( (-2x₁ + x₂ + x₃)/3 )² + ( (-2y₁ + y₂ + y₃)/3 )² ]Similarly for |GB|² and |GC|²:|GB|² = [ ( (x₁ - 2x₂ + x₃)/3 )² + ( (y₁ - 2y₂ + y₃)/3 )² ]|GC|² = [ ( (x₁ + x₂ - 2x₃)/3 )² + ( (y₁ + y₂ - 2y₃)/3 )² ]Adding these together:Sum = (1/9)[ ( (-2x₁ + x₂ + x₃ )² + (x₁ - 2x₂ + x₃ )² + (x₁ + x₂ - 2x₃ )² ) + ( (-2y₁ + y₂ + y₃ )² + (y₁ - 2y₂ + y₃ )² + (y₁ + y₂ - 2y₃ )² ) ]Let me compute the x terms first. Let's denote each term:Term1 = (-2x₁ + x₂ + x₃ )²Term2 = (x₁ - 2x₂ + x₃ )²Term3 = (x₁ + x₂ - 2x₃ )²Similarly for the y terms.Expanding Term1:= ( -2x₁ + x₂ + x₃ )² = [ (-2x₁) + (x₂ + x₃) ]² = 4x₁² + (x₂ + x₃)² - 4x₁(x₂ + x₃ )Similarly, expanding Term2:= (x₁ - 2x₂ + x₃ )² = [ (x₁ + x₃) - 2x₂ ]² = (x₁ + x₃)² + 4x₂² - 4x₂(x₁ + x₃ )Term3:= (x₁ + x₂ - 2x₃ )² = [ (x₁ + x₂) - 2x₃ ]² = (x₁ + x₂)² + 4x₃² - 4x₃(x₁ + x₂ )Adding Term1 + Term2 + Term3:= 4x₁² + (x₂ + x₃)² - 4x₁(x₂ + x₃) + (x₁ + x₃)² + 4x₂² - 4x₂(x₁ + x₃) + (x₁ + x₂)² + 4x₃² - 4x₃(x₁ + x₂ )Let me compute each part:First, the quadratic terms:4x₁² + 4x₂² + 4x₃²Then the squared terms:(x₂ + x₃)² + (x₁ + x₃)² + (x₁ + x₂)²Expanding these:(x₂² + 2x₂x₃ + x₃²) + (x₁² + 2x₁x₃ + x₃²) + (x₁² + 2x₁x₂ + x₂² )Adding up:x₂² + 2x₂x₃ + x₃² + x₁² + 2x₁x₃ + x₃² + x₁² + 2x₁x₂ + x₂²= 2x₁² + 2x₂² + 2x₃² + 2x₁x₂ + 2x₁x₃ + 2x₂x₃So total quadratic terms:4x₁² + 4x₂² + 4x₃² + 2x₁² + 2x₂² + 2x₃² + 2x₁x₂ + 2x₁x₃ + 2x₂x₃Wait, no. Wait, the initial terms were:Term1 + Term2 + Term3 = [4x₁² + (x₂ + x₃)² - 4x₁(x₂ + x₃)] + [similar terms]So after expanding, we have:Quadratic terms: 4x₁² + 4x₂² + 4x₃² (from the first parts) plus the expanded squared terms.Wait, no, let's do it step by step:Term1: 4x₁² + (x₂² + 2x₂x₃ + x₃²) -4x₁x₂ -4x₁x₃Term2: (x₁² + 2x₁x₃ + x₃²) + 4x₂² -4x₁x₂ -4x₂x₃Term3: (x₁² + 2x₁x₂ + x₂²) + 4x₃² -4x₁x₃ -4x₂x₃Now, summing Term1 + Term2 + Term3:For x₁² terms: 4x₁² + x₁² + x₁² = 6x₁²x₂² terms: (x₂² + 4x₂²) + x₂² = 6x₂²x₃² terms: (x₃² + x₃²) + 4x₃² = 6x₃²Cross terms:From Term1: 2x₂x₃ -4x₁x₂ -4x₁x₃From Term2: 2x₁x₃ -4x₁x₂ -4x₂x₃From Term3: 2x₁x₂ -4x₁x₃ -4x₂x₃Adding cross terms:2x₂x₃ + 2x₁x₃ + 2x₁x₂ -4x₁x₂ -4x₁x₃ -4x₂x₃ -4x₁x₂ -4x₁x₃ -4x₂x₃Let me compute term by term:Positive terms: 2x₂x₃ + 2x₁x₃ + 2x₁x₂Negative terms: -4x₁x₂ -4x₁x₃ -4x₂x₃ -4x₁x₂ -4x₁x₃ -4x₂x₃So total positive: 2(x₁x₂ + x₁x₃ + x₂x₃)Total negative: -8x₁x₂ -8x₁x₃ -8x₂x₃So overall cross terms: 2(x₁x₂ + x₁x₃ + x₂x₃) -8(x₁x₂ + x₁x₃ + x₂x₃) = -6(x₁x₂ + x₁x₃ + x₂x₃)Therefore, the total x terms sum:6x₁² + 6x₂² + 6x₃² -6(x₁x₂ + x₁x₃ + x₂x₃)Similarly, the y terms will be the same:6y₁² + 6y₂² + 6y₃² -6(y₁y₂ + y₁y₃ + y₂y₃)Therefore, the total sum is:(1/9)[6(x₁² + x₂² + x₃²) -6(x₁x₂ + x₁x₃ + x₂x₃) +6(y₁² + y₂² + y₃²) -6(y₁y₂ + y₁y₃ + y₂y₃)]Factor out the 6:= (6/9)[ (x₁² + x₂² + x₃² - x₁x₂ - x₁x₃ - x₂x₃ ) + (y₁² + y₂² + y₃² - y₁y₂ - y₁y₃ - y₂y₃ ) ]Simplify 6/9 to 2/3:= (2/3)[ (x₁² + x₂² + x₃² - x₁x₂ - x₁x₃ - x₂x₃ ) + (y₁² + y₂² + y₃² - y₁y₂ - y₁y₃ - y₂y₃ ) ]Now, notice that x₁² + x₂² + x₃² - x₁x₂ - x₁x₃ - x₂x₃ can be written as (1/2)[(x₁ - x₂)^2 + (x₁ - x₃)^2 + (x₂ - x₃)^2]Similarly for the y terms. Let's verify:For three variables, (x₁ - x₂)^2 + (x₁ - x₃)^2 + (x₂ - x₃)^2 = 2(x₁² + x₂² + x₃² - x₁x₂ - x₁x₃ - x₂x₃ )Yes, expanding:(x₁ - x₂)^2 = x₁² - 2x₁x₂ + x₂²(x₁ - x₃)^2 = x₁² - 2x₁x₃ + x₃²(x₂ - x₃)^2 = x₂² - 2x₂x₃ + x₃²Sum: 2x₁² + 2x₂² + 2x₃² - 2x₁x₂ - 2x₁x₃ - 2x₂x₃ = 2(x₁² + x₂² + x₃² - x₁x₂ - x₁x₃ - x₂x₃ )So, x₁² + x₂² + x₃² - x₁x₂ - x₁x₃ - x₂x₃ = 1/2 [ (x₁ - x₂)^2 + (x₁ - x₃)^2 + (x₂ - x₃)^2 ]Therefore, substituting back into the expression:(2/3)[ (1/2)[(x₁ - x₂)^2 + (x₁ - x₃)^2 + (x₂ - x₃)^2 ] + (1/2)[(y₁ - y₂)^2 + (y₁ - y₃)^2 + (y₂ - y₃)^2 ] ]Factor out the 1/2:= (2/3)(1/2)[ (sum of squared differences in x) + (sum of squared differences in y) ]= (1/3)[ (sum of (x_i - x_j)^2 for i < j) + (sum of (y_i - y_j)^2 for i < j) ]But note that (x_i - x_j)^2 + (y_i - y_j)^2 is the squared distance between points i and j. So, the sum over i < j of |A_iA_j|², which for triangle ABC is |AB|² + |AC|² + |BC|².Therefore, the expression simplifies to (1/3)( |AB|² + |AC|² + |BC|² )Since the side lengths are given as a, b, c, then |AB|² + |AC|² + |BC|² = a² + b² + c². Therefore, the minimum value is (1/3)(a² + b² + c² )Wait, so putting it all together, the minimum value of |MA|² + |MB|² + |MC|² is (a² + b² + c²)/3, achieved when M is the centroid of triangle ABC.But let me check if this makes sense. For example, if the triangle is equilateral with side length s, then the centroid is also the center, and the sum of squared distances should be (3s²)/3 = s². Let's see for an equilateral triangle. Suppose each side is s, and coordinates can be set as A(0,0), B(s,0), C(s/2, (s√3)/2 ). Then centroid is at ( (0 + s + s/2)/3, (0 + 0 + (s√3)/2 )/3 ) = ( (3s/2)/3, (s√3/2)/3 ) = (s/2, s√3/6 )Compute |MA|² + |MB|² + |MC|² at centroid:|GA|² = (s/2 - 0)^2 + (s√3/6 - 0)^2 = (s²/4) + (s²*3/36) = s²/4 + s²/12 = (3s² + s²)/12 = 4s²/12 = s²/3Similarly, |GB|²: (s/2 - s)^2 + (s√3/6 - 0)^2 = (-s/2)^2 + (s√3/6)^2 = s²/4 + s²/12 = same as above, s²/3|GC|²: (s/2 - s/2)^2 + (s√3/6 - s√3/2)^2 = 0 + (-s√3/3)^2 = (s²*3)/9 = s²/3So total sum is 3*(s²/3) = s², which is indeed (a² + b² + c²)/3 since a = b = c = s, so (s² + s² + s²)/3 = s². That checks out.Another test case: right triangle with legs 3 and 4, hypotenuse 5. Then the minimum value should be (9 + 16 + 25)/3 = 50/3 ≈ 16.666...Let me compute it using coordinates. Let A(0,0), B(3,0), C(0,4). Centroid G is at ( (0 + 3 + 0)/3, (0 + 0 + 4)/3 ) = (1, 4/3 )Compute |GA|²: (1 - 0)^2 + (4/3 - 0)^2 = 1 + 16/9 = 25/9 ≈ 2.777...|GB|²: (1 - 3)^2 + (4/3 - 0)^2 = 4 + 16/9 = 52/9 ≈ 5.777...|GC|²: (1 - 0)^2 + (4/3 - 4)^2 = 1 + (-8/3)^2 = 1 + 64/9 = 73/9 ≈ 8.111...Total sum: 25/9 + 52/9 + 73/9 = 150/9 = 50/3 ≈ 16.666..., which matches (9 + 16 + 25)/3. Perfect.Therefore, the conclusion seems solid. The minimum is achieved at the centroid, and the value is (a² + b² + c²)/3.But let me think again if there's another way to approach this without coordinates. Maybe using vectors?Suppose we consider vectors with origin at some point O. Let vectors OA, OB, OC be a, b, c respectively. Then, for any point M with position vector m, the expression is:|MA|² + |MB|² + |MC|² = |m - a|² + |m - b|² + |m - c|²Expanding each term:= (|m|² - 2m·a + |a|²) + (|m|² - 2m·b + |b|²) + (|m|² - 2m·c + |c|²)Combine like terms:= 3|m|² - 2m·(a + b + c) + (|a|² + |b|² + |c|²)To minimize this quadratic expression in m, take the derivative with respect to m and set to zero. The gradient is 6m - 2(a + b + c). Setting to zero gives m = ( a + b + c ) / 3, which is the centroid G. Substituting back:The minimum value is 3|G|² - 2G·(a + b + c) + (|a|² + |b|² + |c|²)But since G = ( a + b + c ) / 3, |G|² = ( |a + b + c|² ) / 9So plugging in:= 3*( |a + b + c|² / 9 ) - 2*( ( a + b + c ) / 3 )·( a + b + c ) + (|a|² + |b|² + |c|² )Simplify each term:First term: |a + b + c|² / 3Second term: -2/3 |a + b + c|²Third term: |a|² + |b|² + |c|²Combine them:= (1/3)|a + b + c|² - (2/3)|a + b + c|² + (|a|² + |b|² + |c|² )= (-1/3)|a + b + c|² + (|a|² + |b|² + |c|² )Now, expand |a + b + c|²:= |a|² + |b|² + |c|² + 2a·b + 2a·c + 2b·cSubstitute back:= (-1/3)( |a|² + |b|² + |c|² + 2a·b + 2a·c + 2b·c ) + |a|² + |b|² + |c|²= (-1/3)( sum |v|² + 2sum v·w ) + sum |v|²= ( -sum |v|²/3 - 2sum v·w /3 ) + sum |v|²= (2/3 sum |v|² ) - (2/3 sum v·w )= (2/3)( sum |v|² - sum v·w )But note that sum |v|² - sum v·w can be rewritten as (1/2)sum |v - w|² over all pairs. Wait, similar to before.Indeed, sum_{i < j} |v_i - v_j|² = (1/2)[sum_{i ≠ j} |v_i - v_j|² ] = (1/2)[sum_{i,j} |v_i - v_j|² - sum_{i=j} |v_i - v_j|² ]But since when i = j, |v_i - v_j|² = 0, so it's (1/2)sum_{i,j} |v_i - v_j|²But expanding |v_i - v_j|² = |v_i|² + |v_j|² - 2v_i·v_jTherefore, sum_{i < j} |v_i - v_j|² = (1/2)sum_{i ≠ j} ( |v_i|² + |v_j|² - 2v_i·v_j )= (1/2)[ (n-1)sum |v_i|² + (n-1)sum |v_j|² - 2sum_{i ≠ j} v_i·v_j ) ] where n = 3.But since i and j are dummy indices, sum_{i ≠ j} |v_i|² = 2(n-1) sum |v_i|². Wait, no:Wait, for each i, sum over j ≠ i: sum_{j ≠ i} |v_i|² = (n-1)|v_i|². So sum_{i ≠ j} |v_i|² = sum_{i=1}^n (n-1)|v_i|² = (n-1)sum |v_i|². Similarly for |v_j|².But in our case, n = 3, so sum_{i < j} |v_i - v_j|² = (1/2)[ 2(3 - 1)sum |v_i|² - 2sum_{i < j} v_i·v_j ) ]Wait, let's compute step by step:sum_{i ≠ j} ( |v_i|² + |v_j|² - 2v_i·v_j ) = sum_{i ≠ j} |v_i|² + sum_{i ≠ j} |v_j|² - 2sum_{i ≠ j} v_i·v_jBut sum_{i ≠ j} |v_i|² = sum_{i=1}^3 sum_{j ≠ i} |v_i|² = sum_{i=1}^3 2|v_i|² = 2(|a|² + |b|² + |c|² )Similarly, sum_{i ≠ j} |v_j|² = same thing, so 2(|a|² + |b|² + |c|² )Sum_{i ≠ j} v_i·v_j = 2sum_{i < j} v_i·v_jTherefore,sum_{i < j} |v_i - v_j|² = (1/2)[ 2(|a|² + |b|² + |c|² ) + 2(|a|² + |b|² + |c|² ) - 4sum_{i < j} v_i·v_j ) ]Wait, no:Wait, sum_{i ≠ j} ( |v_i|² + |v_j|² - 2v_i·v_j ) = sum_{i ≠ j} |v_i|² + sum_{i ≠ j} |v_j|² - 2sum_{i ≠ j} v_i·v_j )Which is 2sum |v_i|² + 2sum |v_j|² - 2*2sum_{i < j} v_i·v_jWait, actually sum_{i ≠ j} |v_i|² = 2sum |v_i|² because for each pair (i,j) where i ≠ j, you count |v_i|² once. There are 3 points, so for each i, there are 2 j's, so total sum is 2(|a|² + |b|² + |c|² )Similarly sum_{i ≠ j} |v_j|² is the same, so total sum for the first two terms is 4(|a|² + |b|² + |c|² )Then, sum_{i ≠ j} v_i·v_j = 2sum_{i < j} v_i·v_jTherefore, putting it together:sum_{i < j} |v_i - v_j|² = (1/2)[4(|a|² + |b|² + |c|² ) - 4sum_{i < j} v_i·v_j ) ]= 2(|a|² + |b|² + |c|² ) - 2sum_{i < j} v_i·v_jBut in our previous expression from the vector approach, we had:sum |v|² - sum v·w = (|a|² + |b|² + |c|² ) - ( a·b + a·c + b·c )But sum_{i < j} |v_i - v_j|² = 2( |a|² + |b|² + |c|² ) - 2( a·b + a·c + b·c )Therefore, sum |v|² - sum v·w = (1/2) sum_{i < j} |v_i - v_j|²So substituting back into the expression for the minimum value:(2/3)( sum |v|² - sum v·w ) = (2/3)( (1/2) sum_{i < j} |v_i - v_j|² ) = (1/3) sum_{i < j} |v_i - v_j|²But sum_{i < j} |v_i - v_j|² is |AB|² + |AC|² + |BC|² = a² + b² + c²Therefore, the minimum value is (1/3)(a² + b² + c²), same as before. So this confirms the result using vector approach as well.Therefore, regardless of coordinate system or vector approach, the minimum value is (a² + b² + c²)/3, achieved at the centroid.Is there another possible point that could give a lower sum? For example, maybe the circumcenter or orthocenter? But in general, those points don't minimize the sum of squared distances. The centroid is specifically the point that minimizes the sum of squared distances because it's the mean of the coordinates, analogous to the mean in statistics minimizing the sum of squared deviations.Additionally, in physics, the centroid corresponds to the center of mass if the triangle is made of a uniform sheet. The moment of inertia around the centroid is minimized, which relates to the sum of squared distances.Therefore, considering all these approaches and verifications, I'm confident the minimum value is indeed (a² + b² + c²)/3.Final AnswerThe minimum value is boxed{dfrac{a^2 + b^2 + c^2}{3}}.

✅:To find the minimum value of the expression (|MA|^2 + |MB|^2 + |MC|^2) where (M) is an arbitrary point in the plane and (A, B, C) are the vertices of a triangle with side lengths (a, b, c), we start by considering the centroid (G) of the triangle. The centroid is known to minimize the sum of squared distances from a point to a set of points.1. Coordinate Geometry Approach: - Place the triangle in a coordinate system with vertices (A(x_1, y_1)), (B(x_2, y_2)), and (C(x_3, y_3)). - The expression (|MA|^2 + |MB|^2 + |MC|^2) is expanded and simplified to a quadratic form in (x) and (y). - By taking partial derivatives and setting them to zero, we find the critical point at the centroid (Gleft(frac{x_1 + x_2 + x_3}{3}, frac{y_1 + y_2 + y_3}{3}right)).2. Vector Approach: - Using vectors, we express the sum of squared distances from a point (M) with position vector (mathbf{m}) to the vertices (A, B, C). - The expression is expanded and simplified, leading to the conclusion that the minimum occurs at (mathbf{m} = frac{mathbf{a} + mathbf{b} + mathbf{c}}{3}), which is the centroid.3. Verification: - The minimum value is calculated by substituting the centroid back into the expression. This involves expanding and simplifying the squared distances from the centroid to each vertex. - The result is shown to be (frac{1}{3}(a^2 + b^2 + c^2)), which is confirmed using both coordinate and vector methods.4. Physical Interpretation: - The centroid corresponds to the center of mass, minimizing the moment of inertia, which is analogous to minimizing the sum of squared distances.After thorough analysis and verification, the minimum value of the expression (|MA|^2 + |MB|^2 + |MC|^2) is found to be:[boxed{dfrac{a^2 + b^2 + c^2}{3}}]

🔑:Given:- ( a, b, c ) are the lengths of the sides of triangle ( triangle ABC )- ( M ) is an arbitrary point in the planeWe need to find the minimum value of the expression:[|MA|^2 + |MB|^2 + |MC|^2]To solve this, we will utilize the concept of the centroid (center of mass) of the triangle and some properties of coordinate geometry.1. Centroid of the Triangle (triangle ABC): The centroid ( G ) of triangle ( triangle ABC ) with vertices ( A(x_1, y_1) ), ( B(x_2, y_2) ), and ( C(x_3, y_3) ) is given by: [ G left( frac{x_1 + x_2 + x_3}{3}, frac{y_1 + y_2 + y_3}{3} right) ]2. Distance Squared from (M) to Vertices: Consider an arbitrary point ( M(x, y) ). The squared distances from (M) to (A), (B), and (C) respectively are: [ |MA|^2 = (x - x_1)^2 + (y - y_1)^2 ] [ |MB|^2 = (x - x_2)^2 + (y - y_2)^2 ] [ |MC|^2 = (x - x_3)^2 + (y - y_3)^2 ]3. Sum of Distances Squared: Now, we need to find the sum: [ S = |MA|^2 + |MB|^2 + |MC|^2 ] [ S = [(x - x_1)^2 + (y - y_1)^2] + [(x - x_2)^2 + (y - y_2)^2] + [(x - x_3)^2 + (y - y_3)^2] ]4. Expanding the Expression: Expanding each term results in: [ S = (x^2 - 2xx_1 + x_1^2) + (y^2 - 2yy_1 + y_1^2) + (x^2 - 2xx_2 + x_2^2) + (y^2 - 2yy_2 + y_2^2) + (x^2 - 2xx_3 + x_3^2) + (y^2 - 2yy_3 + y_3^2) ] Grouping similar terms, we get: [ S = 3x^2 + 3y^2 - 2x(x_1 + x_2 + x_3) - 2y(y_1 + y_2 + y_3) + (x_1^2 + x_2^2 + x_3^2) + (y_1^2 + y_2^2 + y_3^2) ]5. Setting ( M ) to the Centroid ( G ): To minimize ( S ), we place ( M ) at the centroid ( G ): [ x = frac{x_1 + x_2 + x_3}{3}, quad y = frac{y_1 + y_2 + y_3}{3} ] Substituting these values simplifies the expression due to symmetry, resulting in cancellations of the linear terms: [ S_{text{min}} = left[ 3left(frac{x_1 + x_2 + x_3}{3}right)^2 + 3left(frac{y_1 + y_2 + y_3}{3}right)^2 - 2left(frac{x_1 + x_2 + x_3}{3} cdot frac{x_1 + x_2 + x_3}{3}right) - 2left(frac{y_1 + y_2 + y_3}{3} cdot frac{y_1 + y_2 + y_3}{3}right) right] ] Further simplifying and condensing: [ S_{text{min}} = frac{1}{3}[ (x_1^2 + y_1^2) + (x_2^2 + y_2^2) + (x_3^2 + y_3^2) ] ]Conclusion:[boxed{frac{a^2 + b^2 + c^2}{3}}]

🧠:Okay, let's see. The problem is about vectors in geometry. We need to express the sum of vectors AB and AD in terms of vectors a and b, where AC is given as vector a and BD is vector b. There are some relationships provided between these vectors. Hmm. Let me parse the problem again step by step.First, the problem states that vector AC is the sum of vectors BA and BD. So, mathematically, that would be:[overrightarrow{AC} = overrightarrow{BA} + overrightarrow{BD}]And also, vector BD is the sum of vector AD and (2/3) of vector AB. So:[overrightarrow{BD} = overrightarrow{AD} + frac{2}{3} overrightarrow{AB}]We are given that AC = a and BD = b. Our goal is to express AB + AD in terms of a and b.Alright, let's write down what we need to find: (overrightarrow{AB} + overrightarrow{AD}). Let's denote this as the target expression.First, let's note that vectors can be expressed in terms of each other, so maybe we can set up some equations and solve for the unknowns. Let's list all the given information and see how they can be connected.Given:1. (overrightarrow{AC} = overrightarrow{BA} + overrightarrow{BD}) (which is equal to vector a)2. (overrightarrow{BD} = overrightarrow{AD} + frac{2}{3} overrightarrow{AB}) (which is equal to vector b)3. We need to find (overrightarrow{AB} + overrightarrow{AD}) in terms of a and b.Let me assign variables to the unknown vectors to make it clearer. Let's let:Let (overrightarrow{AB} = vec{x}) and (overrightarrow{AD} = vec{y}). Then our target expression is (vec{x} + vec{y}).Now, let's rewrite the given equations in terms of these variables.First equation:(overrightarrow{AC} = overrightarrow{BA} + overrightarrow{BD})But (overrightarrow{BA}) is the opposite of (overrightarrow{AB}), so (overrightarrow{BA} = -vec{x}).And (overrightarrow{BD}) is given as vector b, so substituting in:(vec{a} = -vec{x} + vec{b})Wait, hold on. Wait, the first equation is:[overrightarrow{AC} = overrightarrow{BA} + overrightarrow{BD}]But (overrightarrow{AC}) is given as (vec{a}), (overrightarrow{BA}) is (-overrightarrow{AB} = -vec{x}), and (overrightarrow{BD}) is (vec{b}). So:[vec{a} = -vec{x} + vec{b}]So that gives us one equation: (vec{a} = -vec{x} + vec{b}). Let's note that as equation (1).Second equation:(overrightarrow{BD} = overrightarrow{AD} + frac{2}{3} overrightarrow{AB})But (overrightarrow{BD} = vec{b}), (overrightarrow{AD} = vec{y}), and (overrightarrow{AB} = vec{x}). So:[vec{b} = vec{y} + frac{2}{3}vec{x}]That's equation (2): (vec{b} = vec{y} + frac{2}{3}vec{x}).So now, we have two equations:1. (vec{a} = -vec{x} + vec{b}) (equation 1)2. (vec{b} = vec{y} + frac{2}{3}vec{x}) (equation 2)Our target is to find (vec{x} + vec{y}). Let's see how to express this in terms of a and b.First, from equation 1, let's solve for (vec{x}):From equation 1:[vec{a} = -vec{x} + vec{b}][Rightarrow vec{x} = vec{b} - vec{a}]So, (vec{x} = vec{b} - vec{a}). Got that.Now, let's substitute (vec{x}) into equation 2 to find (vec{y}):Equation 2:[vec{b} = vec{y} + frac{2}{3}vec{x}]Substitute (vec{x} = vec{b} - vec{a}):[vec{b} = vec{y} + frac{2}{3}(vec{b} - vec{a})][Rightarrow vec{y} = vec{b} - frac{2}{3}(vec{b} - vec{a})][= vec{b} - frac{2}{3}vec{b} + frac{2}{3}vec{a}][= left(1 - frac{2}{3}right)vec{b} + frac{2}{3}vec{a}][= frac{1}{3}vec{b} + frac{2}{3}vec{a}]So, (vec{y} = frac{2}{3}vec{a} + frac{1}{3}vec{b}).Now, our target expression is (vec{x} + vec{y}):Let's compute that:[vec{x} + vec{y} = (vec{b} - vec{a}) + left(frac{2}{3}vec{a} + frac{1}{3}vec{b}right)][= vec{b} - vec{a} + frac{2}{3}vec{a} + frac{1}{3}vec{b}][= (vec{b} + frac{1}{3}vec{b}) + (-vec{a} + frac{2}{3}vec{a})][= left(1 + frac{1}{3}right)vec{b} + left(-1 + frac{2}{3}right)vec{a}][= frac{4}{3}vec{b} - frac{1}{3}vec{a}]So, (vec{x} + vec{y} = -frac{1}{3}vec{a} + frac{4}{3}vec{b}), which can be written as:[overrightarrow{AB} + overrightarrow{AD} = frac{4}{3}vec{b} - frac{1}{3}vec{a}]Alternatively, factoring out 1/3:[= frac{1}{3}(4vec{b} - vec{a})]But the problem says to express in terms of (vec{a}) and (vec{b}). So this seems like the answer.Wait, let me check the calculations again to make sure I didn't make a mistake.Starting from equation 1: (vec{a} = -vec{x} + vec{b}) → (vec{x} = vec{b} - vec{a}). That's straightforward.Equation 2: (vec{b} = vec{y} + frac{2}{3}vec{x}). Substitute (vec{x}):(vec{b} = vec{y} + frac{2}{3}(vec{b} - vec{a}))Then solving for (vec{y}):(vec{y} = vec{b} - frac{2}{3}vec{b} + frac{2}{3}vec{a})Which is (vec{y} = frac{1}{3}vec{b} + frac{2}{3}vec{a}). That looks correct.Then, (vec{x} + vec{y} = (vec{b} - vec{a}) + (frac{2}{3}vec{a} + frac{1}{3}vec{b}))Combine like terms:For (vec{b}): (1 + frac{1}{3} = frac{4}{3})For (vec{a}): (-1 + frac{2}{3} = -frac{1}{3})So total: (frac{4}{3}vec{b} - frac{1}{3}vec{a}). Yes, that's correct.Alternatively, the problem might prefer the answer written with (vec{a}) first, but since they just ask to express in terms of (vec{a}) and (vec{b}), either order is acceptable. So, the final answer is (frac{4}{3}vec{b} - frac{1}{3}vec{a}), which can also be written as (-frac{1}{3}vec{a} + frac{4}{3}vec{b}).Hmm. Let me just check if there's another way to approach this problem, maybe using vector geometry diagrams.Alternatively, maybe express all vectors in terms of position vectors. But since all vectors are relative, maybe that's more complicated. Let me think.Suppose we take point A as the origin. Then, position vectors:Let’s denote points A, B, C, D with position vectors (vec{A}), (vec{B}), (vec{C}), (vec{D}). But since A is the origin, (vec{A} = vec{0}).Therefore:(overrightarrow{AB} = vec{B} - vec{A} = vec{B} - vec{0} = vec{B})Similarly, (overrightarrow{AD} = vec{D})(overrightarrow{AC} = vec{C} - vec{A} = vec{C} = vec{a})Given that (overrightarrow{AC} = overrightarrow{BA} + overrightarrow{BD})But (overrightarrow{BA} = vec{A} - vec{B} = -vec{B})(overrightarrow{BD} = vec{D} - vec{B})Therefore, the first equation:(vec{C} = -vec{B} + (vec{D} - vec{B}) = -2vec{B} + vec{D})But (vec{C} = vec{a}), so:[vec{a} = -2vec{B} + vec{D} quad text{(Equation 1)}]Second equation: (overrightarrow{BD} = overrightarrow{AD} + frac{2}{3}overrightarrow{AB})But (overrightarrow{BD} = vec{D} - vec{B}), (overrightarrow{AD} = vec{D} - vec{A} = vec{D}), and (overrightarrow{AB} = vec{B} - vec{A} = vec{B}). So:[vec{D} - vec{B} = vec{D} + frac{2}{3}vec{B}][Rightarrow -vec{B} = frac{2}{3}vec{B}]Wait, that can't be right. Wait, hold on. Let me check that again.Wait, the problem states: (overrightarrow{BD}) is the sum of (overrightarrow{AD}) and (frac{2}{3}overrightarrow{AB}). So:[overrightarrow{BD} = overrightarrow{AD} + frac{2}{3}overrightarrow{AB}]But in position vectors, with A as origin:(overrightarrow{AD} = vec{D})(overrightarrow{AB} = vec{B})Therefore:[overrightarrow{BD} = vec{D} + frac{2}{3}vec{B}]But (overrightarrow{BD}) is also equal to (vec{D} - vec{B}). Therefore:[vec{D} - vec{B} = vec{D} + frac{2}{3}vec{B}][Rightarrow -vec{B} = frac{2}{3}vec{B}][Rightarrow -vec{B} - frac{2}{3}vec{B} = 0][Rightarrow -frac{5}{3}vec{B} = 0][Rightarrow vec{B} = 0]Wait, that would imply that point B is at the origin, but A is already the origin. That can't be. This suggests a contradiction, which probably means I made a mistake in interpreting the vectors.Wait, maybe taking A as origin complicates things because if (overrightarrow{BD}) is expressed in terms of (overrightarrow{AD}) and (overrightarrow{AB}), which are vectors from A. Hmm, perhaps this coordinate approach is leading to inconsistency because of the way the vectors are related.Wait, let's try again. If we take A as origin, then:- (overrightarrow{AB} = vec{B})- (overrightarrow{AD} = vec{D})- (overrightarrow{AC} = vec{C} = vec{a})- (overrightarrow{BD} = vec{D} - vec{B} = vec{b})Given the first equation: (overrightarrow{AC} = overrightarrow{BA} + overrightarrow{BD})(overrightarrow{BA} = -overrightarrow{AB} = -vec{B})(overrightarrow{BD} = vec{D} - vec{B} = vec{b})So, (vec{a} = -vec{B} + (vec{D} - vec{B}) = vec{D} - 2vec{B})Therefore, equation 1: (vec{D} - 2vec{B} = vec{a})Equation 2 comes from the other given:(overrightarrow{BD} = overrightarrow{AD} + frac{2}{3}overrightarrow{AB})Which in position vectors is:(vec{D} - vec{B} = vec{D} + frac{2}{3}vec{B})Wait, this simplifies to:(vec{D} - vec{B} = vec{D} + frac{2}{3}vec{B})Subtract (vec{D}) from both sides:(-vec{B} = frac{2}{3}vec{B})Bring terms together:(-vec{B} - frac{2}{3}vec{B} = 0)(-frac{5}{3}vec{B} = 0)Thus, (vec{B} = 0), which would mean point B is the same as point A (since A is the origin). But that can't be, because then vector AB would be zero, which might not make sense in the problem context. This suggests an inconsistency.Hmm, this is confusing. Where did I go wrong? Let's see. The problem says (overrightarrow{BD}) is the sum of (overrightarrow{AD}) and (frac{2}{3}overrightarrow{AB}). If we take A as the origin, then (overrightarrow{AD}) is vector D and (overrightarrow{AB}) is vector B. So, BD should be AD + (2/3)AB, which is D + (2/3)B. But BD is also D - B. Therefore, D - B = D + (2/3)B. Which leads to B = 0. So, this suggests that either the problem has an inconsistency, or my approach is wrong.But since the problem didn't state that points are distinct or anything, maybe B is indeed at A? But then AB would be a zero vector, which might complicate things. However, in the initial equations, if B is at A, then BA is zero, BD would be AD, and so on. Let me check with that.If B is at A, then BA is zero, so AC = 0 + BD, so AC = BD. Therefore, a = b. But in the problem, a and b are arbitrary vectors? Not necessarily. So this suggests that unless a = b, the problem is inconsistent. But the problem doesn't specify that a = b, so perhaps there is a mistake in my coordinate approach.Alternatively, maybe the mistake is in interpreting the vector equations. Let's go back.Original equations:1. (overrightarrow{AC} = overrightarrow{BA} + overrightarrow{BD})2. (overrightarrow{BD} = overrightarrow{AD} + frac{2}{3}overrightarrow{AB})Maybe expressing these equations without coordinates is better.From equation 1:AC is BA + BD. Let's recall that BA = -AB, so AC = -AB + BD.Given AC = a and BD = b, so:a = -AB + b → AB = b - a.From equation 2:BD = AD + (2/3)AB. BD is b, so:b = AD + (2/3)AB → AD = b - (2/3)AB.But AB is already found as (b - a). Therefore:AD = b - (2/3)(b - a) = b - (2/3)b + (2/3)a = (1/3)b + (2/3)a.Therefore, AB + AD = (b - a) + [(2/3)a + (1/3)b] = b - a + (2/3)a + (1/3)b = (b + (1/3)b) + (-a + (2/3)a) = (4/3)b - (1/3)a.So that's the same result as before. So even though in the coordinate approach, assuming A as origin led to a contradiction (unless a = b), but when approached algebraically without coordinates, the equations are consistent.Therefore, the answer is (-1/3)a + (4/3)b.Therefore, the problem is solved correctly. The coordinate approach led to an inconsistency probably because of over-constraining the position vectors, but treating the vectors purely algebraically without assigning coordinates works.Therefore, the final answer is (boxed{-frac{1}{3}vec{a} + frac{4}{3}vec{b}}) or equivalently (boxed{frac{4}{3}vec{b} - frac{1}{3}vec{a}}). Depending on the preferred order, but both are equivalent. Since the problem doesn't specify an order, either form is acceptable, but it's common to write terms in alphabetical order, so maybe -1/3 a + 4/3 b.Final AnswerThe expression (overrightarrow{A B} + overrightarrow{A D}) in terms of (vec{a}) and (vec{b}) is boxed{frac{4}{3}vec{b} - frac{1}{3}vec{a}}.

✅:Given that (overrightarrow{AC}) is the sum of (overrightarrow{BA}) and (overrightarrow{BD}), and (overrightarrow{BD}) is the sum of (overrightarrow{AD}) and (frac{2}{3}overrightarrow{AB}), we need to express (overrightarrow{AB} + overrightarrow{AD}) in terms of (vec{a}) and (vec{b}), where (overrightarrow{AC} = vec{a}) and (overrightarrow{BD} = vec{b}).First, we convert the given relationships into vector equations:1. (overrightarrow{AC} = overrightarrow{BA} + overrightarrow{BD})2. (overrightarrow{BD} = overrightarrow{AD} + frac{2}{3}overrightarrow{AB})Given (overrightarrow{AC} = vec{a}) and (overrightarrow{BD} = vec{b}), we rewrite these equations:1. (vec{a} = -overrightarrow{AB} + vec{b})2. (vec{b} = overrightarrow{AD} + frac{2}{3}overrightarrow{AB})Let (overrightarrow{AB} = vec{x}) and (overrightarrow{AD} = vec{y}). Our target expression is (vec{x} + vec{y}).From the first equation:[vec{a} = -vec{x} + vec{b} implies vec{x} = vec{b} - vec{a}]From the second equation:[vec{b} = vec{y} + frac{2}{3}vec{x}]Substituting (vec{x} = vec{b} - vec{a}) into this equation:[vec{b} = vec{y} + frac{2}{3}(vec{b} - vec{a}) implies vec{y} = vec{b} - frac{2}{3}(vec{b} - vec{a}) = frac{1}{3}vec{b} + frac{2}{3}vec{a}]Now, we find (vec{x} + vec{y}):[vec{x} + vec{y} = (vec{b} - vec{a}) + left(frac{2}{3}vec{a} + frac{1}{3}vec{b}right)]Combining like terms:[= vec{b} - vec{a} + frac{2}{3}vec{a} + frac{1}{3}vec{b} = left(1 + frac{1}{3}right)vec{b} + left(-1 + frac{2}{3}right)vec{a} = frac{4}{3}vec{b} - frac{1}{3}vec{a}]Thus, the expression (overrightarrow{AB} + overrightarrow{AD}) in terms of (vec{a}) and (vec{b}) is boxed{frac{4}{3}vec{b} - frac{1}{3}vec{a}}.

🔑:1. We start by translating the vector relations into equations. We are given: [ overrightarrow{AC} = overrightarrow{BA} + overrightarrow{BD} ] and [ overrightarrow{BD} = overrightarrow{AD} + frac{2}{3} overrightarrow{AB} ] Given that (overrightarrow{AC} = vec{a}) and (overrightarrow{BD} = vec{b}), we will use these notations to express the vectors in terms of (vec{a}) and (vec{b}).2. First, rewrite ( overrightarrow{BA} ) as ( -overrightarrow{AB} ) because the direction is reversed, then substitute into the first equation: [ vec{a} = -overrightarrow{AB} + vec{b} ] Solving for (overrightarrow{AB}), we get: [ overrightarrow{AB} = vec{b} - vec{a} ]3. Next, use the second equation: [ vec{b} = overrightarrow{AD} + frac{2}{3} (vec{b} - vec{a}) ] To isolate (overrightarrow{AD}), first simplify and solve for (overrightarrow{AD}), let ( overrightarrow{AB} = vec{b} - vec{a} ): [ vec{b} = overrightarrow{AD} + frac{2}{3} vec{b} - frac{2}{3} vec{a} ] [ vec{b} - frac{2}{3} vec{b} = overrightarrow{AD} - frac{2}{3} vec{a} ] [ frac{1}{3} vec{b} = overrightarrow{AD} - frac{2}{3} vec{a} ] [ overrightarrow{AD} = frac{1}{3} vec{b} + frac{2}{3} vec{a} ]4. Now to find the sum ( overrightarrow{AB} + overrightarrow{AD} ): [ overrightarrow{AB} + overrightarrow{AD} = (vec{b} - vec{a}) + left(frac{1}{3} vec{b} + frac{2}{3} vec{a}right) ] Combine the terms: [ overrightarrow{AB} + overrightarrow{AD} = vec{b} - vec{a} + frac{1}{3} vec{b} + frac{2}{3} vec{a} ] Group the (vec{b}) and (vec{a}) terms together: [ overrightarrow{AB} + overrightarrow{AD} = left(vec{b} + frac{1}{3} vec{b}right) + left(-vec{a} + frac{2}{3} vec{a}right) ] Simplify the coefficients: [ overrightarrow{AB} + overrightarrow{AD} = left(frac{4}{3} vec{b}right) + left(-frac{1}{3} vec{a}right) ] [ overrightarrow{AB} + overrightarrow{AD} = frac{4}{3} vec{b} - frac{1}{3} vec{a} ] Conclusion:[boxed{frac{4}{3} vec{b} - frac{1}{3} vec{a}}]

🧠:Okay, so I need to find the ratio BE/ED in quadrilateral ABCD where the sides are AB=5, BC=6, CD=5, DA=4, and angle ABC is 90 degrees. The diagonals AC and BD intersect at point E. Hmm, let's start by visualizing the quadrilateral. Since angle ABC is 90 degrees, triangle ABC is a right-angled triangle. Maybe I can draw coordinate axes here to make things easier.Let me place point B at the origin (0,0). Then, since AB=5 and angle ABC is 90 degrees, AB is along the y-axis and BC is along the x-axis. Wait, no, if angle at B is 90 degrees, then AB and BC are perpendicular. So if B is at (0,0), then A can be at (0,5) because AB=5, and C can be at (6,0) because BC=6. Let me confirm that. AB would be from (0,0) to (0,5), which is length 5. BC is from (0,0) to (6,0), which is length 6. Then angle ABC is 90 degrees, which is correct.Now, we need to find coordinates for D. We know DA=4 and CD=5. So point D is connected to C (6,0) and to A (0,5). Let me denote coordinates for D as (x,y). Then the distance from D to C is 5, so:√[(x - 6)^2 + (y - 0)^2] = 5 → (x - 6)^2 + y^2 = 25.Also, the distance from D to A is 4:√[(x - 0)^2 + (y - 5)^2] = 4 → x^2 + (y - 5)^2 = 16.So we have two equations:1. (x - 6)^2 + y^2 = 252. x^2 + (y - 5)^2 = 16Let me expand both equations:First equation:(x^2 - 12x + 36) + y^2 = 25 → x^2 + y^2 -12x +36 =25 → x^2 + y^2 -12x = -11Second equation:x^2 + (y^2 -10y +25) =16 → x^2 + y^2 -10y +25=16 → x^2 + y^2 -10y = -9Now, subtract the second equation from the first equation:(x^2 + y^2 -12x) - (x^2 + y^2 -10y) = -11 - (-9)Simplify:-12x - (-10y) = -11 +9 → -12x +10y = -2Divide both sides by 2:-6x +5y = -1So we have -6x +5y = -1. Let's solve for one variable in terms of the other. Let's say we solve for y:5y =6x -1 → y = (6x -1)/5Now plug this into one of the original equations. Let's use the second equation:x^2 + (y -5)^2 =16Substitute y:x^2 + [(6x -1)/5 -5]^2 =16Simplify the expression in the brackets:(6x -1)/5 -25/5 = (6x -1 -25)/5 = (6x -26)/5So:x^2 + [(6x -26)/5]^2 =16Compute [(6x -26)/5]^2:(36x^2 - 312x +676)/25Thus, equation becomes:x^2 + (36x^2 -312x +676)/25 =16Multiply both sides by 25 to eliminate denominator:25x^2 +36x^2 -312x +676 =400Combine like terms:(25x^2 +36x^2) + (-312x) + (676 -400) =061x^2 -312x +276=0Divide all terms by common factor if possible. Let's check GCD of 61,312,276. 61 is prime. 312 ÷61 is 5.114, not integer. So no common factor. So quadratic equation is:61x^2 -312x +276=0Use quadratic formula: x = [312 ±√(312^2 -4*61*276)]/(2*61)Compute discriminant:312^2 = (300 +12)^2 = 90000 +7200 +144=973444*61*276 =4*61*276. Compute 61*276:61*200=12200, 61*76=4636, total 12200+4636=16836Multiply by 4: 16836*4=67344Thus discriminant is 97344 -67344=30000√30000=√(10000*3)=100√3≈173.205Thus x=(312 ±100√3)/122Simplify:Divide numerator and denominator by 2:x=(156 ±50√3)/61So x= (156 +50√3)/61 or x=(156 -50√3)/61Now compute y=(6x -1)/5 for each x.First for x=(156 +50√3)/61:y=(6*(156 +50√3)/61 -1)/5= (936 +300√3 -61)/305= (875 +300√3)/305Simplify numerator and denominator:Divide numerator and denominator by 5:= (175 +60√3)/61Similarly for x=(156 -50√3)/61:y=(6*(156 -50√3)/61 -1)/5= (936 -300√3 -61)/305= (875 -300√3)/305= (175 -60√3)/61So we have two possible coordinates for D:D1= ((156 +50√3)/61, (175 +60√3)/61)D2= ((156 -50√3)/61, (175 -60√3)/61)Now, need to figure out which one is correct. Since ABCD is a quadrilateral, and we know the order of the vertices, we need to ensure that the points are connected in order. Let me check which of these points would make the quadrilateral simple (non-intersecting). Since ABC is already a right triangle, D should be positioned such that CD=5 and DA=4. The two solutions correspond to D being on different sides of AC. Hmm. Maybe both are possible, but since the problem doesn't specify, perhaps we need to use additional constraints. Wait, but in a competition problem like this, usually there is a unique solution, so maybe both solutions lead to the same ratio BE/ED. Let me check.Alternatively, perhaps I can use coordinate geometry to find the intersection point E of diagonals AC and BD, then compute BE/ED.First, let's find equations of diagonals AC and BD.Coordinates:A is (0,5), C is (6,0). So diagonal AC is from (0,5) to (6,0). The parametric equation for AC can be written as:x = 0 +6ty =5 -5t, where t ranges from 0 to1.Similarly, diagonal BD connects B(0,0) to D. Let's take D as ((156 +50√3)/61, (175 +60√3)/61) first.So BD is from (0,0) to ((156 +50√3)/61, (175 +60√3)/61). The parametric equations:x = ((156 +50√3)/61)sy = ((175 +60√3)/61)s, where s ranges from 0 to1.The intersection point E is where the two diagonals meet. So we need to solve for t and s such that:6t = ((156 +50√3)/61)s5 -5t = ((175 +60√3)/61)sLet me solve the first equation for t:t = ((156 +50√3)/61)s /6 = ((156 +50√3)/366)sNow substitute into second equation:5 -5*((156 +50√3)/366)s = ((175 +60√3)/61)sMultiply through by 366*61 to eliminate denominators. Wait, this might get messy. Alternatively, let's compute numerically. Maybe approximate √3≈1.732 to check.Compute 156 +50√3 ≈156 +50*1.732≈156+86.6≈242.6. Then divide by 61:≈242.6/61≈3.977. So x-coordinate of D≈3.977. Similarly, 175 +60√3≈175 +103.92≈278.92, divided by 61≈4.572. So D≈(3.977,4.572). Let's check if this makes sense.Alternatively, if we take D2: (156 -50√3)/61≈(156 -86.6)/61≈69.4/61≈1.138, and y≈(175 -60√3)/61≈(175 -103.92)/61≈71.08/61≈1.165. So D2≈(1.138,1.165). But if D is near (1.1,1.1), then DA distance would be from (0,5) to (1.1,1.1): sqrt(1.21 + 15.21)=sqrt(16.42)≈4.05, which is close to 4, and CD from (6,0) to (1.1,1.1): sqrt( (6-1.1)^2 + (0-1.1)^2 )=sqrt(24.01 +1.21)=sqrt(25.22)≈5.02, close to 5. So both D1 and D2 satisfy the distance conditions. Therefore, there are two possible quadrilaterals. However, the problem states "quadrilateral ABCD", implying a convex quadrilateral. Let me check if both D1 and D2 result in convex quadrilaterals.If D is at D1≈(3.977,4.572), then the order A-B-C-D would likely form a convex quadrilateral. Similarly, D2 at≈(1.138,1.165) would create a concave quadrilateral? Because from A(0,5) to B(0,0) to C(6,0) to D(1.1,1.1), the turn at C might be concave. Hmm, but without a diagram, it's hard to tell. However, in competition problems, usually, the ratio BE/ED is the same regardless of the position of D, but let's verify.Wait, if D is in two different positions, the intersection point E might give different ratios. Therefore, maybe the problem implicitly assumes a convex quadrilateral. Since ABC is a right angle, and given the side lengths, perhaps D1 is the correct one. Alternatively, maybe the problem is designed such that the ratio is the same regardless. Let me compute both possibilities.First, let's proceed with D1: ((156 +50√3)/61, (175 +60√3)/61). Let's denote this as (d1x, d1y).Diagonal BD1 is from B(0,0) to D1(d1x,d1y). The parametric equations for BD1 are x = d1x * s, y = d1y * s.Diagonal AC is from A(0,5) to C(6,0), parametric equations x=6t, y=5 -5t.At intersection E, we have:6t = d1x * s5 -5t = d1y * sFrom first equation: s = (6t)/d1xSubstitute into second equation:5 -5t = d1y * (6t)/d1xMultiply both sides by d1x:5d1x -5t d1x =6t d1yBring terms with t to one side:5d1x = t(5d1x +6d1y)Thus,t = (5d1x)/(5d1x +6d1y)Then,s = (6t)/d1x = (6*(5d1x)/(5d1x +6d1y))/d1x = 30/(5d1x +6d1y)Therefore, the coordinates of E are (6t,5 -5t) where t is as above.But we need to find BE/ED. Since BD1 is parametrized by s from 0 to1, BE corresponds to s from 0 to s, and ED corresponds to s from s to1. So BE/ED = s/(1 - s).From s =30/(5d1x +6d1y). So BE/ED= [30/(5d1x +6d1y)]/[1 -30/(5d1x +6d1y)] =30/[ (5d1x +6d1y) -30 ].But let's compute 5d1x +6d1y.Given d1x=(156 +50√3)/61, d1y=(175 +60√3)/61Thus,5d1x +6d1y =5*(156 +50√3)/61 +6*(175 +60√3)/61= [5*156 +5*50√3 +6*175 +6*60√3]/61Compute each term:5*156=7805*50√3=250√36*175=10506*60√3=360√3Total numerator:780+1050 +250√3+360√3=1830 +610√3Thus,5d1x +6d1y=(1830 +610√3)/61= (1830/61) + (610√3)/61=30 +10√3Therefore,BE/ED=30/[ (30 +10√3) -30 ]=30/(10√3)=3/√3=√3But √3 is irrational, but in terms of ratio, but the problem might want a rational number. Wait, maybe I made a mistake here.Wait, wait. Wait, when I calculated 5d1x +6d1y=30 +10√3. Then:BE/ED=30 / [ (5d1x +6d1y) -30 ]=30/(30 +10√3 -30)=30/(10√3)=3/√3=√3≈1.732. But the problem is likely expecting a rational number. So maybe I made a mistake in assuming D1. Let's check D2.For D2: ((156 -50√3)/61, (175 -60√3)/61)Similarly, compute 5d2x +6d2y.d2x=(156 -50√3)/61, d2y=(175 -60√3)/615d2x +6d2y=5*(156 -50√3)/61 +6*(175 -60√3)/61= [5*156 -5*50√3 +6*175 -6*60√3]/61= [780 -250√3 +1050 -360√3]/61= (1830 -610√3)/61=30 -10√3Therefore,BE/ED=30/[ (30 -10√3) -30 ]=30/(-10√3)= -3/√3. But length can't be negative, so the absolute value is 3/√3=√3. Wait, but this suggests BE/ED=√3 for both cases? But how come? Because in one case, E is inside the quadrilateral, and in the other case, maybe E is outside? But since diagonals intersect at E, it should be inside for a convex quadrilateral. However, if D is D2, which might make the quadrilateral concave, but even so, the ratio remains the same magnitude. But √3 is approximately 1.732. But the problem might want an exact value, which is √3. However, in olympiad problems, ratios like this are often rational numbers. Maybe I made a mistake in calculation.Wait, let's think differently. Instead of using coordinates, maybe there's a theorem or property that can give the ratio directly.Since ABCD has a right angle at B, and sides AB=5, BC=6. So ABC is a right-angled triangle with legs 5 and 6. Then, points A(0,5), B(0,0), C(6,0). Then we have point D such that CD=5 and DA=4. Maybe using mass point geometry or coordinate geometry.Alternatively, use vectors.Alternatively, use the concept of Ceva's theorem or Menelaus' theorem.Alternatively, since diagonals intersect at E, the ratio BE/ED can be found by areas or using similar triangles.Wait, another approach: in a quadrilateral, the ratio in which the diagonals divide each other can be found using coordinates. The formula is that if diagonals AC and BD intersect at E, then:BE/ED = [AB * BC * sin(θ1)] / [AD * DC * sin(θ2)]But I'm not sure. Alternatively, use the formula from coordinate geometry where the ratio is determined by the coordinates.Wait, but given that I already have coordinates for A, B, C, D, perhaps using the section formula.If we have two lines intersecting, the coordinates of E can be found by solving the two parametric equations, and then the ratio can be computed.But earlier, when I tried with D1, I got BE/ED=√3, and similarly for D2, absolute value same. However, the problem might require a rational answer. Let me check my calculations again.Wait, let's take the coordinates:For D1: ((156 +50√3)/61, (175 +60√3)/61)For D2: ((156 -50√3)/61, (175 -60√3)/61)Now, for diagonal AC: parametric equations from A(0,5) to C(6,0).For diagonal BD: from B(0,0) to D.Let’s use D1 first. Let’s denote D as (d_x, d_y).The parametric equations:AC: (6t, 5 -5t)BD: (d_x s, d_y s)Intersection when 6t = d_x s and 5 -5t = d_y s.From first equation: s =6t / d_xSubstitute into second equation:5 -5t = d_y*(6t / d_x)Multiply both sides by d_x:5d_x -5t d_x =6t d_yRearrange:5d_x = t(5d_x +6d_y)So t=5d_x / (5d_x +6d_y)Similarly, s=6t / d_x=6*(5d_x)/(5d_x +6d_y)/d_x=30/(5d_x +6d_y)Therefore, BE/ED= s/(1 -s)= [30/(5d_x +6d_y)] / [1 -30/(5d_x +6d_y)] =30/[5d_x +6d_y -30]For D1, 5d_x +6d_y=30 +10√3Thus, BE/ED=30/(10√3)=3/√3=√3Similarly, for D2, 5d_x +6d_y=30 -10√3Thus, BE/ED=30/(-10√3)= -3/√3, but since lengths are positive, take absolute value: 3/√3=√3Therefore, regardless of D being D1 or D2, the ratio BE/ED is √3. But √3 is approximately 1.732, which is irrational. However, the problem might require rationalizing the denominator, expressing it as √3, or perhaps there's a mistake.Alternatively, maybe there's a different approach using similar triangles.Let me think. In quadrilateral ABCD, with ABC right angle. Let's compute the coordinates again.Coordinates:A(0,5), B(0,0), C(6,0). D is to be found.DA=4, CD=5.We found two possible D points. Let's take D1 and see the structure.Alternatively, use vectors. The position vector of E can be expressed in terms of A, B, C, D.But since E is the intersection of diagonals AC and BD, we can express E as a weighted average.In general, the ratio in which E divides AC can be found by solving the two parametric equations. Alternatively, use the formula for the intersection of two lines in coordinate geometry.Given two lines:Line AC: from (0,5) to (6,0). Slope is (0-5)/(6-0)= -5/6. Equation: y= -5/6 x +5Line BD: from (0,0) to (d_x,d_y). Slope is d_y/d_x. Equation: y= (d_y/d_x)xIntersection at E: solve -5/6 x +5 = (d_y/d_x)xMultiply both sides by 6d_x:-5d_x x +30d_x =6d_y xRearrange:x( -5d_x -6d_y )= -30d_xThus, x= (30d_x)/(5d_x +6d_y )Then y= (d_y/d_x)x= (d_y/d_x)*(30d_x)/(5d_x +6d_y)=30d_y/(5d_x +6d_y)Thus, coordinates of E are (30d_x/(5d_x +6d_y),30d_y/(5d_x +6d_y))Now, the ratio BE/ED is the ratio along BD from B to E to E to D.Since BD is from (0,0) to (d_x,d_y), and E is at (30d_x/(5d_x +6d_y),30d_y/(5d_x +6d_y)), then the parameter s such that E divides BD in the ratio BE:ED=s:(1-s). Therefore, the coordinates of E can also be written as (s*d_x, s*d_y). Comparing with earlier expressions:s*d_x=30d_x/(5d_x +6d_y) → s=30/(5d_x +6d_y)Therefore, BE/ED= s/(1-s)= [30/(5d_x +6d_y)]/[1 -30/(5d_x +6d_y)] =30/[5d_x +6d_y -30]But earlier calculation showed that for D1 and D2, 5d_x +6d_y=30±10√3, leading to BE/ED=30/(±10√3)=±3/√3=±√3. Since the ratio is positive, we take the absolute value, so √3.But √3 seems unexpected. Maybe the answer is indeed √3, but in the problem statement, if it's from a competition, maybe it's rational. Wait, perhaps I made a mistake in the coordinate setup.Wait, let's verify with actual numbers. Take D1 coordinates approximately (3.977,4.572). Then BD1 has length sqrt((3.977)^2 + (4.572)^2)≈sqrt(15.82 +20.9)=sqrt(36.72)=6.06. Which is approximately, but not exactly, but the distances DA=4 and CD=5 are satisfied. Then E is the intersection point. Let's compute E's coordinates.Using the formula:x=30d_x/(5d_x +6d_y)For D1, d_x≈3.977, d_y≈4.572Compute denominator:5*3.977 +6*4.572≈19.885 +27.432≈47.317x≈30*3.977 /47.317≈119.31/47.317≈2.522y≈30*4.572 /47.317≈137.16/47.317≈2.898So E is approximately (2.522,2.898). Then BE is the distance from B(0,0) to E(2.522,2.898)=sqrt(2.522² +2.898²)≈sqrt(6.36 +8.40)=sqrt(14.76)≈3.84ED is distance from E to D1≈sqrt( (3.977 -2.522)^2 + (4.572 -2.898)^2 )=sqrt( (1.455)^2 + (1.674)^2 )≈sqrt(2.117 +2.803)=sqrt(4.92)=2.22Thus BE/ED≈3.84/2.22≈1.73, which is approximately √3≈1.732, which matches our earlier result. So the ratio is indeed √3. But let's check with D2.Take D2≈(1.138,1.165). Then BD2 length≈sqrt(1.138² +1.165²)≈sqrt(1.295 +1.357)=sqrt(2.652)≈1.628. But CD2 distance is 5, DA≈4. So diagonals would intersect outside? Wait, but if D is D2, the quadrilateral might be self-intersecting. However, the problem states "quadrilateral ABCD", which usually implies simple (non-intersecting). Therefore, D1 is the correct position.Therefore, the ratio BE/ED is √3. But the problem might require a rationalized form or fraction. But √3 is already simplified. Alternatively, maybe I made a mistake in the coordinate setup.Wait, another approach: use areas. In a quadrilateral, the ratio of the segments of the diagonals can be related to the areas of triangles.Let’s consider triangles ABC and ADC.But ABC is a right triangle with area (5*6)/2=15.But we don't know the area of ADC. However, if we can find the areas of triangles AED and BEC, maybe we can relate the ratios.Alternatively, use the theorem that in a quadrilateral, the ratio of the segments of one diagonal is equal to the ratio of the products of the adjacent sides. Wait, that's in a harmonic quadrilateral or something else. Maybe not.Alternatively, use vector cross products.Let me denote vectors:Let’s place B at origin, so coordinates:B(0,0), A(0,5), C(6,0). D(x,y).Vectors BA= A - B = (0,5)Vectors BC= C - B = (6,0)Vectors BD= D - B = (x,y)Vectors AC= C - A = (6,-5)Vectors AD= D - A = (x,y-5)Vectors CD= D - C = (x-6,y)We know |AD|=4 and |CD|=5.We already solved this earlier.But another idea: use barycentric coordinates or parameterize the diagonals.Alternatively, note that in the ratio BE/ED=√3, which is approximately 1.732, but maybe the problem expects an exact value in terms of fractions with radicals.But let's think again. Maybe using mass point geometry.Mass point assigns masses to vertices such that the masses are inversely proportional to the lengths.But mass point is typically for concurrency of cevians, but here we have two diagonals intersecting.Alternatively, assign masses at A and C such that the masses are proportional to the weights that balance at E.Wait, the mass at A would be proportional to EC, and mass at C proportional to AE. Similarly, masses at B and D proportional to ED and BE.But I'm not sure.Alternatively, use the formula from analytic geometry: if two lines intersect, the ratio can be found by determinants.Alternatively, use the concept that the ratio BE/ED is equal to the ratio of the areas of triangles ABE and ADE.But to compute areas, we need heights, which might not be straightforward.Alternatively, note that triangles ABE and ADE share the same base AE, so their areas are proportional to their heights from B and D to AE. Similarly, heights are related to the distances from B and D to line AE. But without knowing the equation of AE, this might be complicated.Alternatively, since ABC is a right triangle, and we have coordinates, maybe it's best to stick with the coordinate method, which gave BE/ED=√3.But let's verify once again with exact values.For D1:5d_x +6d_y=30 +10√3Thus, BE/ED=30/(10√3)=3/√3=√3Rationalizing the denominator: √3.But often, ratios are expressed as fractions with rational denominators, so √3 is acceptable.Alternatively, the answer might be 1/√3, but that would be if the ratio was inverted. But according to the calculation, it's BE/ED=√3.But let's check with another approach.Consider triangle ABD and CBD.Wait, but not sure.Alternatively, use Menelaus' theorem on triangle ABC with transversal BD.Wait, Menelaus' theorem states that for a triangle, if a line crosses the three sides (or their extensions), the product of the segment ratios is equal to 1.But BD intersects AC at E, so Menelaus' theorem on triangle ABC with transversal BD:(AF/FB) * (BE/EC) * (CD/DA)=1. Wait, not sure.Alternatively, in triangle ABC, point E lies on AC, and line BD passes through E. Menelaus' theorem would state:(AE/EC) * (CB/BB) * (BD/DA)=1. Not sure.Maybe not the right approach.Alternatively, use coordinate geometry with exact values.From previous calculations, BE/ED=√3. So unless there's a miscalculation, this should be the answer.Wait, but the problem is likely expecting a rational number. Is there a mistake in assuming coordinates? Let me check distances again for D1.Coordinates of D1: x=(156 +50√3)/61≈(156 +86.6)/61≈242.6/61≈3.977y=(175 +60√3)/61≈(175 +103.92)/61≈278.92/61≈4.572Distance from D1 to A(0,5):sqrt( (3.977)^2 + (4.572 -5)^2 )≈sqrt(15.82 +0.183)≈sqrt(16.0)≈4.0, which matches DA=4.Distance from D1 to C(6,0):sqrt( (3.977 -6)^2 + (4.572)^2 )≈sqrt( (-2.023)^2 +20.9 )≈sqrt(4.09 +20.9)=sqrt(25)≈5, which matches CD=5.Similarly, for D2, distances are approximately 4 and5.Thus, coordinates are correct.Therefore, the ratio BE/ED=√3.But the problem might require the answer in a different form. Wait, maybe rationalizing the denominator:√3=√3/1, but sometimes written as a fraction. However, √3 is already simplest form.Alternatively, maybe the answer is 3, but that doesn't align with the calculation. Alternatively, perhaps I made a mistake in the sign.Wait, in the case of D2, 5d_x +6d_y=30 -10√3≈30 -17.32≈12.68, and then BE/ED=30/(12.68 -30)=30/(-17.32)=-1.732, but since ratio is positive, take absolute value, so 1.732=√3. Therefore, regardless of D1 or D2, the absolute value is √3.Therefore, the answer should be √3.But let me recall that in some cases, the ratio can be found using similar triangles. If there are similar triangles involving E, maybe we can find the ratio without coordinates.Consider triangles BEA and DEC. If they are similar, then the ratio could be found. But I need to check angles.Alternatively, since ABC is right-angled, and if we can find coordinates, which we did, leading to the ratio √3.Alternatively, maybe use the concept of homothety.But given that the coordinate method, which is systematic, gives BE/ED=√3, and the approximate calculation confirms it, I think this is the correct answer.However, to ensure no mistakes, let's recheck the key steps.Computed coordinates for D correctly with distances 4 and5.Solved the two equations for D's coordinates correctly.Set up parametric equations for diagonals AC and BD correctly.Solved for intersection point E correctly.Computed the ratio BE/ED as s/(1-s)=30/(5d_x +6d_y -30).Substituted 5d_x +6d_y=30±10√3 correctly.Thus, BE/ED=30/(±10√3)=±3/√3=±√3, taking absolute value √3.Therefore, the answer is √3.But let me check with another method for confirmation.Using the formula for the ratio of division by diagonals in a quadrilateral:In a convex quadrilateral, the ratio in which the diagonals divide each other can be found by the formula:BE/ED = (AB * BC) / (AD * DC)Wait, is this a valid formula? Not sure. Let me think.If the quadrilateral is a kite, then diagonals bisect each other at certain ratios. But here, it's not a kite.Alternatively, in general quadrilaterals, there's no direct formula, unless specific conditions are met.Alternatively, use area method.The ratio BE/ED can be found by the ratio of areas of triangles sharing the same base or height.Consider triangles BEA and DEA. They share the base EA. The ratio of their areas is equal to the ratio of their heights from B and D to EA, which is the same as BE/ED.Similarly, triangles BEC and DEC share base EC, ratio of areas is BE/ED.But to find the areas, we might need more information.Alternatively, express areas in terms of coordinates.Coordinates:A(0,5), B(0,0), C(6,0), D(d_x,d_y), E(e_x,e_y).Area of triangle ABE: (1/2)| (0*(0 -e_y) +0*(e_y -5) +e_x*(5 -0) | = (1/2)|5e_x|Similarly, area of triangle ADE: (1/2)|0*(d_y -e_y) +d_x*(e_y -5) +e_x*(5 -d_y)|This seems complicated. Alternatively, use shoelace formula for areas.But perhaps this is more involved than the coordinate method we already did.Given that the coordinate method gives a clear answer and checks out numerically, I will conclude that the ratio BE/ED is √3.But to confirm, let me check if any properties of quadrilaterals with perpendicular adjacent sides and given side lengths result in this ratio.Alternatively, consider that in a quadrilateral with ABC right angle, the ratio BE/ED might be determined by the sides AB, BC, CD, DA.But I don't recall a specific theorem for this.Alternatively, use complex numbers.Place B at the origin in the complex plane. A is 5i, C is6. D is a complex number such that |D -6|=5 and |D -5i|=4.This is the same as the coordinate approach, leading to the same equations.Thus, solving as before, leading to the same ratio.Therefore, after multiple approaches and verification, the ratio BE/ED is √3.But wait, the problem might want the answer as a fraction involving radicals, which is acceptable. Therefore, the final answer is boxed{sqrt{3}}.Final Answerboxed{sqrt{3}}

✅:In quadrilateral (ABCD), we have (AB = 5), (BC = 6), (CD = 5), (DA = 4), and (angle ABC = 90^circ). We need to compute the ratio (frac{BE}{ED}) where diagonals (AC) and (BD) intersect at (E).1. Coordinate Setup: - Place (B) at the origin ((0, 0)). - (A) is at ((0, 5)) and (C) is at ((6, 0)) since (angle ABC = 90^circ).2. Finding Coordinates of (D): - Let (D) have coordinates ((x, y)). - Using the distance formula for (DA = 4) and (CD = 5): [ sqrt{x^2 + (y - 5)^2} = 4 quad text{and} quad sqrt{(x - 6)^2 + y^2} = 5 ] - Solving these equations, we find two possible coordinates for (D): [ D_1 = left( frac{156 + 50sqrt{3}}{61}, frac{175 + 60sqrt{3}}{61} right) quad text{and} quad D_2 = left( frac{156 - 50sqrt{3}}{61}, frac{175 - 60sqrt{3}}{61} right) ]3. Intersection of Diagonals: - Diagonal (AC) is parameterized as (x = 6t), (y = 5 - 5t). - Diagonal (BD) is parameterized as (x = d_x s), (y = d_y s) where (d_x) and (d_y) are coordinates of (D). - Solving for the intersection (E) by equating coordinates from both parameterizations, we find the parameter (s) and the ratio (frac{BE}{ED}).4. Calculating the Ratio: - For both (D_1) and (D_2), the ratio (frac{BE}{ED}) is calculated using the formula derived from the intersection parameters: [ frac{BE}{ED} = frac{30}{5d_x + 6d_y - 30} ] - Substituting the values for (D_1) and (D_2), we find: [ 5d_x + 6d_y = 30 pm 10sqrt{3} ] - This results in (frac{BE}{ED} = sqrt{3}) for both cases.Thus, the ratio (frac{BE}{ED}) is (boxed{sqrt{3}}).

🔑:1. Identify Given Information and Sketch the Quadrilateral: [ begin{aligned} &text{Given quadrilateral ABCD:} &AB = 5, &BC = 6, &CD = 5, &DA = 4, &angle ABC = 90^circ. end{aligned} ] The quadrilateral has a right angle at (B), which implies triangle ( Delta ABC ) is a right triangle.2. Calculate the Length of the Diagonal (AC): From the Pythagorean theorem applied to the right triangle ( Delta ABC ), we have: [ AC = sqrt{AB^2 + BC^2} = sqrt{5^2 + 6^2} = sqrt{25 + 36} = sqrt{61}. ]3. Using Law of Cosines to Find ( angle ADC ): In triangle ( Delta ACD ), to find ( angle ADC ): [ text{Let } cos theta = cos(angle ADC). ] By the law of cosines: [ AD^2 + CD^2 - 2 cdot AD cdot CD cdot cos (theta) = AC^2 ] Substituting the known values: [ 4^2 + 5^2 - 2 cdot 4 cdot 5 cdot cos (theta) = 61, ] which simplifies to: [ 16 + 25 - 40 cos (theta) = 61, ] therefore: [ 41 - 40 cos (theta) = 61, ] thus: [ -40 cos (theta) = 20 implies cos (theta) = -frac{1}{2}. ] This implies: [ theta = 120^circ. ] Thus, ( angle ADC = 120^circ ).4. Determine the Ratio ( frac{BE}{ED} ): The ratio ( frac{BE}{ED} ) can be determined by using the ratio of areas of triangles ( Delta ABC ) and ( Delta ACD ).5. Find the Areas of Triangles ( Delta ABC ) and ( Delta ACD ): [ text{Area of } Delta ABC = frac{1}{2} cdot AB cdot BC = frac{1}{2} cdot 5 cdot 6 = 15. ] [ text{Area of } Delta ACD = frac{1}{2} cdot AD cdot CD cdot sin(angle ADC) = frac{1}{2} cdot 4 cdot 5 cdot frac{sqrt{3}}{2} = frac{1}{2} cdot 4 cdot 5 cdot 0.866 = 10 cdot frac{sqrt{3}}{2} = 10sqrt{3}. ]6. Calculate the Ratio of Segment Lengths: Given that the ratio of ( frac{BE}{ED} ) is the ratio of the areas of the triangles: [ frac{BE}{ED} = frac{text{Area of } Delta ABC}{text{Area of } Delta ACD} = frac{15}{10 sqrt{3}} = frac{15}{10 sqrt{3}} = frac{3}{2sqrt{3}} = frac{3 times sqrt{3}}{2 times 3} = sqrt{3}. ]Conclusion:[boxed{sqrt{3}}]

🧠:Okay, so I need to prove that the sum of three consecutive natural numbers is divisible by three. Let me start by understanding the problem. Natural numbers are like 1, 2, 3, 4, and so on. Consecutive means they follow each other without gaps. So three consecutive numbers could be something like 5, 6, 7 or 10, 11, 12. The claim is that if you add any three such numbers together, the total will always be divisible by three. First, maybe I should try some examples to see if this holds true. Let me pick a few sets of numbers. Let's take 1, 2, 3. Their sum is 6, which divided by 3 is 2. That works. Another set: 4, 5, 6. Sum is 15, divided by 3 is 5. Okay, good. How about 7, 8, 9? Sum is 24, divided by 3 is 8. Hmm, seems to check out. Let me try a larger set, like 10, 11, 12. Sum is 33, divided by 3 is 11. Still works. Maybe even a single-digit and double-digit mix: 2, 3, 4. Sum is 9, which is 3 times 3. Alright, so the examples are supporting the claim. But examples aren't a proof. I need to show this works for any three consecutive numbers.So, how do I generalize this? Let me think about representing the three consecutive numbers algebraically. If the first number is n, then the next two numbers should be n+1 and n+2. That makes sense because consecutive numbers increase by 1 each time. So the three numbers are n, n+1, n+2. Now, let's add them up: n + (n + 1) + (n + 2). Let me compute that sum step by step. Combining the terms: n + n + n is 3n. Then 1 + 2 is 3. So the total sum is 3n + 3. That simplifies to 3(n + 1). Oh, wait! 3 times (n + 1) is clearly a multiple of 3. Therefore, the sum is divisible by 3. But let me make sure I didn't skip any steps here. Let's verify this with another example. Let's take n = 5. Then the numbers are 5, 6, 7. Sum is 18. According to the formula, 3(n + 1) = 3(5 + 1) = 18. Yep, matches. Another example: n = 10. Numbers are 10, 11, 12. Sum is 33. 3(10 + 1) = 33. Correct again. Wait, but what if n is 0? Wait, hold on. Natural numbers usually start from 1, right? So 0 isn't considered a natural number. So the first number n has to be at least 1. But even if someone considers natural numbers starting at 0, then taking n=0 gives 0,1,2. Sum is 3, which is 3*1. Still divisible by 3. So regardless of the starting point, as long as the numbers are consecutive, the formula holds. Is there another way to think about this? Maybe in terms of modular arithmetic. If we have three consecutive numbers, they must cover all residues modulo 3. For example, any number is either 0, 1, or 2 mod 3. If you take three consecutive numbers, they would be n ≡ 0, 1, 2 mod 3 in some order. Adding them together: 0 + 1 + 2 = 3 ≡ 0 mod 3. Therefore, the sum is divisible by 3. But wait, does this hold for any three consecutive numbers? Let's see. Suppose the first number is congruent to 1 mod 3. Then the next two would be 2 and 0 mod 3. Adding them: 1 + 2 + 0 = 3 ≡ 0 mod 3. If the first number is 2 mod 3, then the next two are 0 and 1. Sum is 2 + 0 + 1 = 3 ≡ 0 mod 3. And if the first is 0 mod 3, then sum is 0 + 1 + 2 = 3. So regardless of where you start, the residues add up to 0 mod 3. That's another way to see it. But maybe someone might question if this modular approach is valid. Let's check with numbers. Take 3, 4, 5. 3 is 0 mod 3, 4 is 1, 5 is 2. Sum is 12, which is 0 mod 3. Another set: 7,8,9. 7 is 1 mod 3, 8 is 2, 9 is 0. Sum is 24, which is 0 mod 3. Yep, works. So both algebraic and modular approaches confirm the result. The algebraic approach shows the sum is 3(n + 1), which is obviously a multiple of 3, and the modular approach shows that the residues cycle through 0,1,2, adding up to 0 mod 3. Is there a way this could fail? Suppose someone argues about negative numbers, but the problem states natural numbers, which are positive integers. So negatives aren't considered here. Another angle: The middle number of three consecutive numbers is the average. For example, 4,5,6: average is 5, and 5*3=15. Similarly, 10,11,12: average is 11, 11*3=33. So the average is the middle number, which is an integer, so multiplying by 3 gives a multiple of 3. Therefore, the sum is three times the middle number, hence divisible by three. Wait, that's another way to see it. Let me formalize that. Let the three numbers be n, n+1, n+2. The middle number is n+1. The sum is n + (n +1) + (n +2) = 3n + 3 = 3(n +1). So that's 3 times the middle number. Since the middle number is an integer, the sum is a multiple of 3. Therefore, regardless of the value of n, the sum is three times an integer, hence divisible by three. I think that covers all possible angles. The key insight is representing the numbers algebraically, summing them, and factoring out the 3. Alternatively, using modular arithmetic or considering the average. All methods converge to the same conclusion. So the step-by-step proof would be:Let three consecutive natural numbers be n, n+1, n+2.Sum = n + (n+1) + (n+2) = 3n + 3 = 3(n + 1).Since 3(n +1) is a multiple of 3, the sum is divisible by 3.Therefore, the sum of any three consecutive natural numbers is divisible by three.

✅:To prove that the sum of three consecutive natural numbers is divisible by three, follow these steps:1. Define the Numbers: Let the three consecutive natural numbers be ( n ), ( n+1 ), and ( n+2 ), where ( n ) is a natural number.2. Calculate the Sum: [ n + (n + 1) + (n + 2) ]3. Simplify the Expression: [ 3n + 3 = 3(n + 1) ]4. Factor Out the Common Term: The expression ( 3(n + 1) ) is clearly a multiple of 3.Conclusion: Since the sum simplifies to ( 3(n + 1) ), it is divisible by 3 for any natural number ( n ). Thus, the sum of three consecutive natural numbers is always divisible by three.(boxed{text{The sum of three consecutive natural numbers is divisible by three.}})

🔑:1. Introduce the sequence form: Let's consider three consecutive natural numbers. These can be written in the form ( n ), ( n+1 ), and ( n+2 ) where ( n ) is a natural number.2. Sum the sequence components: We now calculate the sum of these three consecutive natural numbers: [ s = n + (n+1) + (n+2) ]3. Combine like terms: By combining the like terms in the sum, we obtain: [ s = n + n + 1 + n + 2 = 3n + 3 ]4. Factor out the common term: The sum ( s ) can be factored as follows: [ s = 3n + 3 = 3(n + 1) ]5. Highlight divisibility: From the factored form, we see that ( s ) is a product of 3 and another integer ( (n + 1) ). This explicitly shows that the sum ( s ) is divisible by 3. Conclusion: Hence, the sum of three consecutive natural numbers is always divisible by 3.[boxed{s}]

🧠:Okay, let's try to solve this problem step by step. So, the question is about Polycarp and Kolka each creating the smallest 5-digit number with distinct even digits. Polycarp did it correctly, but Kolka made a mistake. However, the difference between Kolka's incorrect number and the correct one is less than 100. We need to find both numbers.First, let's figure out what the correct smallest 5-digit number with distinct even digits is. The even digits are 0, 2, 4, 6, 8. Since it's a 5-digit number, the first digit can't be 0. So, the smallest possible first digit is 2, right? Wait, no. Wait, if we use 2 as the first digit, then the remaining digits would be 0, 4, 6, 8. But we need to arrange them in ascending order to get the smallest number. So, maybe the correct number starts with 2, followed by 0, 4, 6, 8. Wait, but that would be 20468. Let me check.Wait, but maybe starting with a smaller digit. Wait, the digits have to be even and distinct. The available digits are 0,2,4,6,8. The first digit can't be 0, so the smallest possible first digit is 2. Then, the next digits should be the smallest remaining digits. So, after 2, the next smallest is 0, then 4, then 6, then 8. So that would make 20468. Is that the smallest possible?Wait, let me confirm. Let's try different combinations. If the first digit is 2, then the next digits should be 0,4,6,8. So 20468. If we start with 2, then 0, then 4, 6, 8, that's the smallest. If instead of 20468, could we have a smaller number by rearranging? Let's see: 20468 versus 20486? No, 20468 is smaller. Similarly, any other permutation would be larger. So Polycarp's number is 20468.Now, Kolka made a mistake but his number is within 100 less than the correct one. Wait, wait, the problem says the difference between Kolka's number and the correct answer was less than 100. So it's |Kolka's number - Polycarp's number| < 100. But since Kolka's number is a 5-digit number with distinct even digits, but incorrect, maybe he rearranged digits incorrectly.So Polycarp's number is 20468. Let's see what numbers are within 100 of 20468. That would be from 20468 - 99 = 20369 up to 20468 + 99 = 20567. But Kolka's number must also be a 5-digit number with distinct even digits. So Kolka's number is in [20369, 20567], but all digits must be even, distinct, and 5-digit. But wait, Kolka's number must be even digits, so all digits in Kolka's number are from 0,2,4,6,8, and distinct. Also, first digit can't be 0.So the possible range for Kolka's number is between 20369 and 20567, but all digits even. Let's see the lower bound: 20369 is not all even. The next even number in that range would be 20400, but digits must be distinct. Wait, perhaps Kolka's number is higher than 20468, but within 100, so up to 20567. But 20468 + 99 = 20567. Let's check if there's a number in that range made with distinct even digits.Alternatively, maybe Kolka's number is lower than 20468, but since 20468 is the smallest possible, that can't be. Wait, but wait—maybe Kolka thought differently. For example, maybe he didn't arrange the digits correctly. Let's think: if the correct smallest number is 20468, then any other permutation would be larger, right? Because the digits are in ascending order. So if Kolka's number is less than 20468, that would require digits smaller than 20468, but since 20468 is the smallest possible, that's impossible. Therefore, Kolka's number must be greater than 20468 but less than 20468 + 100 = 20568. But Kolka's number is supposed to be incorrect, so it's not the minimal one.Wait, but the problem states that Kolka made a mistake in creating the smallest number, so his number is also a 5-digit number with distinct even digits, but not the minimal one, and the difference between his number and the correct one is less than 100. So Kolka's number is either just a bit larger than 20468, or maybe even smaller but impossible, since 20468 is the smallest. So the difference is Kolka's number minus Polycarp's number is less than 100. So Kolka's number is between 20468 and 20468 + 99 = 20567.Now, we need to find a number in that range that uses distinct even digits. Let's check possible numbers. Let's start from 20468 and look for the next possible numbers.The next number after 20468 would be 20480, but wait, let's check step by step. Let's increment 20468 by 2 each time (since all digits must be even) and check if the digits are distinct and even.20468 + 2 = 20470. But digits are 2,0,4,7,0. Wait, 7 is not even, so invalid. Also, duplicate 0s. So invalid.20468 + 4 = 20472. Now digits are 2,0,4,7,2. 7 is not even, and duplicate 2s. Invalid.20468 + 6 = 20474. Again, 7 not even. Same issue.Wait, maybe adding 10 instead? Let's see. 20468 + 10 = 20478. But 7 and 8. 7 is not even. So invalid.Hmm, maybe the next valid number would be 20480. Let's check 20480. Digits: 2,0,4,8,0. Duplicate 0s. Not allowed.Next, 20482. Duplicate 2s. 20484: duplicate 4s. 20486: digits 2,0,4,8,6. Wait, that's 20486. Are all digits even and distinct? 2,0,4,8,6. Yes, all even and distinct. So 20486 is a possible number. The difference from 20468 is 20486 - 20468 = 18. Which is less than 100. So that's a candidate.But wait, is there a closer number? Let's check between 20468 and 20486.Wait, 20468 + 12 = 20480, which we saw has duplicate 0s. 20468 + 14 = 20482, duplicate 2s. 20468 + 16 = 20484, duplicate 4s. 20468 + 18 = 20486, which is valid. So the next valid number is 20486, which is 18 more than 20468.But maybe Kolka's number is higher than that. Let's check other possibilities.Alternatively, maybe Kolka swapped some digits. For example, maybe he thought the smallest number starts with 0, but that's impossible because it's a 5-digit number. Another mistake could be using a digit more than once. But the problem states that both Polycarp and Kolka created numbers with distinct even digits. So Kolka's mistake is in the arrangement, not in repeating digits or using odd digits.Wait, the problem says Kolka made a mistake in creating the smallest number, but his number is still composed of distinct even digits. So his number is a valid 5-digit number with distinct even digits, just not the smallest possible. And the difference between his number and the correct one is less than 100.So, possible candidates are numbers just above 20468, like 20486, 20648, etc. Let's see.Another approach: Let's list all 5-digit numbers with distinct even digits that are greater than 20468 but less than 20468 + 100 = 20568.Starting from 20468, next possible number is 20486 (difference 18). Then, maybe 20648? Wait, that's much larger. Let's check numbers in the 204xx range.Wait, after 20486, the next number could be 2048 followed by 6, but 20486 is already considered. Then, 20490? But 9 is odd. 204a6 where a is the next digit. Hmm, maybe 20524? Let's see.Wait, perhaps another approach: generate all numbers from 20468 to 20568, step by 2 (since they must end with even digit), check if all digits are even and distinct.Starting at 20468:20468: correct number.Next even number: 20470. Check digits: 2,0,4,7,0. 7 is odd and duplicate 0. Invalid.20472: 7 is odd. Invalid.20474: 7 again. Invalid.20476: 7 again. Invalid.20478: 7 and 8. 7 is odd. Invalid.20480: digits 2,0,4,8,0. Duplicate 0. Invalid.20482: duplicate 2. Invalid.20484: duplicate 4. Invalid.20486: digits 2,0,4,8,6. All even and distinct. Valid. Difference is 18.Next: 20488: duplicate 8. Invalid.20490: 9 is odd. Invalid.20492: 9 is odd. Invalid.20494: 9 is odd. Invalid.20496: 9 is odd. Invalid.20498: 9 is odd. Invalid.20500: duplicate 0s. Invalid.20502: duplicate 0 and 2. Invalid.20504: duplicate 0 and 4. Invalid.20506: duplicate 0 and 5 is odd. Wait, 20506 has 5, which is odd. Invalid.20508: 5 is odd. Invalid.20520: duplicate 0 and 2. Invalid.20522: duplicate 2. Invalid.20524: duplicate 2. Wait, 2,0,5,2,4. 5 is odd. Invalid.Wait, this is getting tedious. Let's see: between 20468 and 20568, the only valid number with distinct even digits is 20486. Let's confirm.Wait, 20486 is valid. What's next? Let's see: after 20486, the next possible number would be 20648, but that's way beyond 20568. So between 20468 and 20568, the only valid number is 20486. Is that the case?Wait, perhaps there are others. Let's check 20524. Wait, 2,0,5,2,4. No, 5 is odd. 20546: 5 is odd. 20564: 5 is odd. So no, all numbers in the 205xx range would have the third digit as 5, which is odd, so invalid. Therefore, the only valid number in that range is 20486.Wait, but wait, what about 20648? Wait, 20648 is 20648, which is way beyond 20568. So no. So the only possible number within 100 is 20486, difference 18.But wait, maybe Kolka made a different mistake. For example, he might have swapped the last two digits. The correct number is 20468. If he wrote 20486 instead, that's the same as 20486, which we already considered. Difference 18.Alternatively, maybe he swapped the third and fourth digits: 20468 → swapping 4 and 6 gives 20648, which is much larger, difference 180, which is more than 100. So that's out.Alternatively, maybe he arranged the digits in a different incorrect order. Let's think: the correct number is 20468. The next smallest number would be 20486. Then 20648, 20684, 20846, etc. But those are all more than 100 away except 20486.So if Kolka's number is 20486, the difference is 18, which is less than 100. Therefore, that could be the answer.But wait, is there any other number in that range? Let's think again. For example, 20468 + 20 = 20488, which has duplicate 8s. 20468 + 22 = 20490, which has 9. 20468 + 24 = 20492, 9 again. 20468 + 26 = 20494, 9 again. 20468 + 28 = 20496, 9. 20468 + 30 = 20498, 9. 20468 + 32 = 20500, duplicates. So up to 20500, no valid numbers. Then 20502, duplicates. Etc.So the only valid number in that range is 20486. Therefore, Kolka's number is 20486, and Polycarp's is 20468. But wait, the problem states that the difference between Kolka's number and the correct answer is less than 100. So if Kolka's number is 20486, which is 18 more than the correct one, that's acceptable.But wait, the problem says that Kolka made a mistake in forming the smallest number. So he tried to make the smallest possible, but made an error. So his number should be larger than the correct one. Therefore, the difference is Kolka's number minus Polycarp's number, which is positive and less than 100.Therefore, the only possible number is 20486.Wait, but is there another possibility where Kolka's number is lower but invalid? But as we discussed earlier, 20468 is the smallest possible, so any number lower would either start with a 0 (invalid) or have smaller digits arranged, but that's not possible because 20468 is already the minimal arrangement.Therefore, the answer must be Polycarp's number 20468 and Kolka's number 20486, with a difference of 18.But wait, let's check another angle. Maybe Kolka omitted a smaller digit. For example, instead of using 0 as the second digit, he used a higher digit. Wait, but how? The correct number uses 0 as the second digit. If he didn't use 0 there, but used another digit, then the number would be larger. For example, if he started with 2, then used 4 instead of 0, making 24068. That's larger than 20468, difference 24068 - 20468 = 3600, which is way more than 100. So that's not possible.Alternatively, maybe he swapped the 0 with another digit later. For example, 20648. But that difference is 20648 - 20468 = 180, which is more than 100. So invalid.Alternatively, maybe he used a different digit in the thousands place. Wait, the correct number is 20468. If Kolka used a different digit in the second position, but still kept the number as small as possible. For example, if he thought that 0 can't be used in the second position (which is not true), so he used 2 in the first position, then 4, resulting in 24068. But that's much larger.Alternatively, maybe he misplaced the 0. For example, he wrote 20046, but that has duplicate 0s. Invalid.Alternatively, maybe he thought that the digits should be in strictly increasing order but made a mistake. Wait, the correct number 20468 is in increasing order except for the 0 after 2. Wait, no, the digits are 2,0,4,6,8, which is not increasing. Wait, but to form the smallest number, you want the leftmost digits as small as possible. So starting with 2, then the next smallest digit is 0, then 4, 6, 8. So the number is 20468. But that's not in digit order, but it's the smallest possible because putting 0 in the second position gives a smaller number than putting 4 there.Wait, for example, 20468 is smaller than 24068, 24068, etc. So the correct approach is to place the smallest available digits from left to right, considering that after the first digit, 0 can be used.So yes, 20468 is correct.If Kolka instead arranged the digits in ascending order, but that would require the number to start with 0, which is impossible. So he must have arranged them differently.Another possibility: Kolka thought that the digits should be arranged in increasing order excluding the first digit. But that's not the case. For example, he might have arranged them as 24680, which is the digits in order, but that's a much larger number. The difference would be 24680 - 20468 = 4212, which is way over 100.Alternatively, maybe he swapped two digits. For example, 20468 vs 20486. As before, difference 18.Alternatively, maybe he placed 8 before 6, making 20486. That seems plausible.Alternatively, maybe he made a mistake in choosing the digits. Wait, but the problem states that both used distinct even digits. So both used all five even digits: 0,2,4,6,8. Because there are exactly five even digits, so each number must use all five. Wait, wait, the problem says "composed of distinct even digits". It doesn't specify that all five even digits must be used, just that the digits used are distinct and even. Wait, but the smallest 5-digit number would require using the smallest possible digits. Since there are five even digits (0,2,4,6,8), and a 5-digit number must have five digits, all distinct and even. Therefore, both Polycarp and Kolka used all five even digits. So the numbers must be permutations of 0,2,4,6,8, with the first digit not 0.Therefore, Kolka's number must be a permutation of 0,2,4,6,8 starting with 2, and the difference from 20468 is less than 100.So let's list all permutations of 0,2,4,6,8 starting with 2, and calculate their difference from 20468.The permutations would be all numbers starting with 2, followed by permutations of the remaining digits (0,4,6,8).The next smallest number after 20468 would be 20486, then 20648, 20684, 20846, 20864, 24068, etc.Let's compute their differences:20468: correct, difference 0.20486: 20486 - 20468 = 18.20648: 20648 - 20468 = 180.20684: 20684 - 20468 = 216.20846: 20846 - 20468 = 378.20864: 20864 - 20468 = 396.24068: 24068 - 20468 = 3600.And so on. So the only permutation within 100 difference is 20486, difference 18.Therefore, Kolka's number is 20486, Polycarp's is 20468.But wait, let's check if there are any other permutations starting with 2 that have a difference less than 100.Next possible permutation after 20468 would be 20486 (as we saw). Let's see if there's another permutation between 20468 and 20468 + 100.Let's consider numbers where the first two digits are 20, then the next digits are permutations of 4,6,8.After 20468, the next would be 20486. Then, perhaps 20648, but that's 180 more. So no.Alternatively, is there a number like 20648? No, too big. What about numbers starting with 20 and then different arrangements:The thousands digit is 0, so the number is 20xxx. The next digits after 20 are 4,6,8 in some order.The smallest is 20468, then 20486, then 20648, 20684, 20846, 20864.So only 20486 is within 100. Therefore, the answer must be Polycarp: 20468, Kolka: 20486.But let's just confirm once more. Are there any other permutations of 0,2,4,6,8 starting with 2 that fall within 20468 + 100?For example, 20468 + 100 = 20568. Any permutations between 20468 and 20568?Let's consider permutations where the first three digits are 204, then the last two digits are 6 and 8. So 20468 and 20486. The next would be 206..., which is beyond 20568.Wait, 206xx would be 20648, which is 20648, which is way above 20568. So no.Therefore, the only possible number is 20486.Thus, the answer is Polycarp's number is 20468, and Kolka's number is 20486.

✅:Step-by-Step Explanation:1. Identify Correct Number (Polycarp's Number): - To form the smallest 5-digit number with distinct even digits (0, 2, 4, 6, 8), start with the smallest non-zero digit (2), followed by the next smallest digits in ascending order: 2, 0, 4, 6, 8. - Polycarp's Number: 204682. Determine Kolka's Possible Numbers: - Kolka's number must be a permutation of the digits 0, 2, 4, 6, 8 (distinct and even) starting with 2. - The difference between Kolka's number and 20468 must be less than 100.3. Check Valid Permutations Near 20468: - The next larger permutation after 20468 is 20486 (swapping the last two digits). - Calculate the difference: (20486 - 20468 = 18), which is less than 100. - Other permutations (e.g., 20648, 20846) result in differences exceeding 100.Final Answer:- Polycarp's number: 20468- Kolka's number: 20486

🔑:To solve the problem, we need to form the smallest 5-digit number using distinct even digits. Here's a step-by-step breakdown of the solution:1. Form a Number with Distinct Even Digits: The even digits available are: 0, 2, 4, 6, and 8. To form the smallest possible 5-digit number, the digits should be arranged in ascending order. However, the first digit of a number in the decimal system cannot be 0, since it would then be considered less than a 5-digit number. - First Digit: Choose the smallest even digit that is not 0. The smallest digit after 0 is 2. - Remaining Digits: Arrange the remaining digits including 0 in ascending order. Thus, the digits 0, 4, 6, 8 should follow 2. The smallest valid number Polikarp could form is: [ 20468 ]2. Find Kolya's Number: According to the problem, Kolya made a mistake but the difference between the number Kolya formed and the correct number Polikarp formed is less than 100. - Kolya's number must also use the digits 0, 2, 4, 6, and 8. - Since changing any of the first three digits significantly would result in a difference greater than 100, Kolya would only have changed the last two digits. Thus, by swapping the 4th and 5th digits, Kolya’s number becomes: [ 20486 ] - Check the Difference: The difference between Polikarp’s and Kolya’s numbers is: [ 20486 - 20468 = 18 ] Since 18 < 100, Kolya’s number satisfies the condition. Conclusion:The numbers formed by Polikarp and Kolya are:[ boxed{20468, 20486} ]