answer:1. **Observations:** - We observe that ( L_{max} ) is strictly greater than ( L_{min} ), as the lattice points create a triangle with rational area. However, the area of an equilateral triangle is irrational. - Both ( L_{max} ) and ( L_{min} ) are square roots of some positive integers. This follows due to the fact they can be expressed as ( sqrt{x^2 + y^2} ), where ( x, y ) are their respective lengths of projections on the coordinate axis. - By examining small triangles, we can get a lower bound ( k ge 4 ), which is achieved in the triangle ( A(0,0), B(0,2), C(2,1) ). It is easy to verify this. 2. **Verification for ( k = 4 ):** - Rewrite the inequality as: [ L_{max}^2 - L_{max} L_{min} ge frac{1}{sqrt{k}} ] - Denote ( L_{max} = sqrt{a} ) and ( L_{min} = sqrt{a - b} ), where ( b < a ) and ( a, b in mathbb{N} ). - The inequality becomes: [ a - sqrt{a(a - b)} ge frac{1}{2} ] - We need to show that: [ a - sqrt{a(a - b)} ge a - sqrt{a(a - 1)} ] - Simplify the inequality: [ a - sqrt{a(a - 1)} ge frac{1}{2} ] - This can be rewritten as: [ left( a - frac{1}{2} right)^2 ge a(a - 1) ] - Expanding and simplifying: [ a^2 - a + frac{1}{4} ge a^2 - a ] - This inequality is obviously true since: [ frac{1}{4} ge 0 ] Thus, the smallest positive constant ( k ) such that the given inequality holds for any three lattice points is ( k = 4 ). The final answer is ( boxed{4} ).

answer:1. **Define Variables**: Let (x) be the number of stones in the third pile. 2. **Express Stones in Terms of (x)**: - According to the problem, the number of stones in the fifth pile is six times that in the third pile: [ text{Fifth pile} = 6x ] - The number of stones in the second pile is twice the sum of stones in the third and fifth piles: [ text{Second pile} = 2(x + 6x) = 14x ] - The number of stones in the first pile is three times less than those in the fifth pile and 10 less than those in the fourth pile: [ text{First pile} = frac{6x}{3} = 2x ] - Additionally, we know that the fourth pile has ten more stones than the first pile: [ text{Fourth pile} = 2x + 10 ] - The fourth pile is also half of the second pile: [ text{Fourth pile} = frac{14x}{2} = 7x ] 3. **Equate Expressions for the Fourth Pile**: Given the above two expressions for the fourth pile: [ 2x + 10 = 7x ] 4. **Solve for (x)**: Subtract (2x) from both sides: [ 10 = 5x ] Divide by 5: [ x = 2 ] 5. **Determine the Number of Stones in Each Pile**: - Third pile: (x = 2) - Fifth pile: (6x = 12) - Second pile: (14x = 28) - First pile: (2x = 4) - Fourth pile: (7x = 14) 6. **Calculate Total Number of Stones**: [ text{Total} = 2 + 12 + 28 + 4 + 14 = 60 ] # Conclusion: [ boxed{60} ]

answer:1. **First, determine if we can transform the initial card ((5, 19)) into ((1, 50)).** We note the operations of the machines: - Machine 1: ( (a, b) rightarrow (a+1, b+1) ) - Machine 2: ( (a, b) rightarrow left(frac{a}{2}, frac{b}{2}right) ) (only when (a) and (b) are both even) - Machine 3: ( (a, b), (b, c) rightarrow (a, c) ) 2. **Transform the initial card ( (5, 19) ) into ( (1, 8) ):** [ begin{aligned} & (5,19) rightarrow (6,20) & text{(Machine 1)} & (6,20) rightarrow (3,10) & text{(Machine 2)} & cdots & text{(Various operations)} & rightarrow (10,17) rightarrow (3,17) & text{(Machine 1)} & (3,17) rightarrow (4,18) & (4,18) rightarrow (2,9) & text{(Machine 2)} & cdots & text{(Various operations)} & rightarrow (9,16) rightarrow (2,16) & (2,16) rightarrow (1,8) & text{(Machine 2)} end{aligned} ] 3. **Once ( (1, 8) ) is obtained:** [ begin{aligned} & (1, 8) stackrel{text{M1}}{rightarrow} (2, 9) stackrel{text{M1}}{rightarrow} cdots & rightarrow (8, 15) rightarrow (15, 22) & rightarrow (22, 29) rightarrow (29, 36) & rightarrow (36, 43) rightarrow (43, 50) & rightarrow (50) end{aligned} ] Finally, use Machine 3 to connect the cards and get ((1, 50)): [ begin{aligned} (1, 8) rightarrow (8, 15) rightarrow cdots rightarrow (43, 50) & (1, 50) end{aligned} ] Conclusion: It is possible to obtain the card ( (1, 50) ). 4. **Now, consider obtaining card ( (1, 100) ):** **Note**: The difference between the numbers on the card remains a multiple of 7. [ |b - a| quad text{is always a multiple of 7.} ] Here, ( 50 = 1 + 7 cdot 7 ), so operations preserve this factor. ( 99 neq 7 cdot k ) for any integer ( k ). Therefore, it is impossible to obtain card ( (1, 100) ) because 99 is not a multiple of 7. 5. **General solution for ( (a, b) rightarrow (1, n) ):** Let ( d ) be the greatest odd factor of ( b - a ). [ begin{aligned} text{If } n = 1 + k cdot d, & (k in N), text{ then } (a, b) rightarrow (1, n). text{For example, } (a = 5, b = 19): & d = 14. text{Smallest }d=2. end{aligned} ] So for a general card ( (a, b) ): [ text{It is possible if ( n = 1 + k cdot d. )} ] **Proof**: - Necessity is evident as operations maintain the same ( d ). - Sufficiency: Given ( (a, b) ) with ( a < b ) and odd/even combination satisfying (d). Indications: - Converging to ( 1, 1+d ). - Possible if odd/even permutation converges. Therefore, from card ( (a, b) ): [ begin{aligned} & (a, b) rightarrow (1, 1+d) & quad text{If possible.} & (a, b) rightarrow (1, n) end{aligned} ] Finally, in achieving the given card transformation and ( 1, 100 ) solution: [ boxed{text{It's not possible to get } (1, 100). text{Valid only for multiples.}} ]

answer:First, factor the equation (x+y)(x-y) = 77. We seek positive integer solutions for x and y, so x+y > x-y > 0. The factor pairs of 77 are (77, 1) and (11, 7). **Case 1: x+y = 77 and x-y = 1.** Solving these: [ x = frac{77 + 1}{2} = 39, ] [ y = frac{77 - 1}{2} = 38. ] **Case 2: x+y = 11 and x-y = 7.** Solving these: [ x = frac{11 + 7}{2} = 9, ] [ y = frac{11 - 7}{2} = 2. ] Thus, the pairs (x, y) that satisfy the equation are (39, 38) and (9, 2). Conclusion: There are boxed{2} pairs of positive integers (x, y) that solve the equation x^2 - y^2 = 77.