answer:1. **Consider the Infinite Chessboard Representation:** Imagine the infinite grid as an infinite chessboard, where the cells are alternately colored white and black. We will place 8 pieces of 1 times 1 dominoes (monominoes) in a specific pattern on this infinite chessboard such that no two monominoes share any edges or vertices.  2. **Identifying the "Good" Dominoes:** Assume we still succeeded in covering the remaining part of the grid with 1 times 2 dominoes (dominoes) despite the presence of the monominoes. Let's call those dominoes "good" dominoes if they are placed entirely within the borders of the outlined rectangular area on the chessboard. 3. **Analyzing Coverage:** Note that each 1 times 2 domino covers exactly one white cell and one black cell. Since the monominoes were placed on the grid in such a way that the total number of uncovered cells in the outlined rectangular area is 13, we must consider the feasibility of coverage by "good" dominoes. 4. **Counting Cells and Domino Feasibility:** Within the outlined rectangular area, let’s observe the number of white and black cells. - Initially, without considering the monominoes, the number of white and black cells would be equal because of the alternation in the chessboard patterning. - Having placed the 8 monominoes in a way that does not share any edges or vertices results in having some white and black cells exposed. But in the outlined area, we claim that we still need to fit the remaining unoccupied area, which amounts to 13 cells after placing the specified monominoes. However, since each domino covers exactly two cells, the maximum number of dominoes that could fit into 13 cells is: [ leftlfloor frac{13}{2} rightrfloor = 6 text{ dominoes} ] This is because 6 dominoes would cover 12 cells, leaving an extra cell uncovered. 5. **Concluding Infeasibility:** This makes it impossible to cover the remaining part perfectly using 1 times 2 dominoes, as we are left with 1 cell out of the 13 not covered: [ 13 - 12 = 1 text{ cell uncovered} ] # Conclusion: It is not possible to place a finite number of 1 times 1 monominoes on an infinite grid in such a manner that no two monominoes share an edge or vertex, and then cover the remaining parts without leaving any gaps using 1 times 2 dominoes. [ boxed{ } ]

answer:Given the quadrilateral ABCD with the following angles: [ angle BAC = 30^circ, quad angle ABD = 26^circ, quad angle DBC = 51^circ, quad angle ACD = 13^circ ] We are required to find the value of angle CAD. 1. Let angle ADC = theta. According to the Law of Sines in triangle ACD: [ frac{AC}{CD} = frac{sin theta}{sin (theta + 13^circ)} ] This is based on the property that the ratio of the sides of a triangle is equal to the ratio of the sines of their opposite angles. 2. Applying the Law of Sines in triangle ABC: [ frac{AC}{BC} = frac{sin 77^circ}{sin 30^circ} ] where 77^circ is derived from 180^circ - 30^circ - 73^circ because angle BCA = 73^circ and angle BAC = 30^circ. 3. Also, applying the Law of Sines in triangle BCD: [ frac{BC}{CD} = frac{sin 43^circ}{sin 51^circ} ] where 43^circ is derived from 180^circ - 51^circ - 86^circ because angle BDC = 86^circ and angle ABD = 26^circ. 4. Combining the above ratios, we get: [ frac{AC}{CD} = frac{AC}{BC} cdot frac{BC}{CD} = frac{sin 77^circ}{sin 30^circ} cdot frac{sin 43^circ}{sin 51^circ} ] 5. Equating the two expressions for frac{AC}{CD}, we have: [ frac{sin theta}{sin (theta + 13^circ)} = frac{sin 77^circ}{sin 30^circ} cdot frac{sin 43^circ}{sin 51^circ} ] 6. Simplifying the right-hand side: [ frac{2 sin 77^circ sin 43^circ}{sin 51^circ} = frac{2 cos 13^circ cos 47^circ cos 73^circ}{sin 51^circ cos 73^circ} = frac{cos 39^circ}{2 sin 51^circ cos 73^circ} = frac{1}{2 cos 73^circ} ] 7. Thus, we have: [ frac{sin theta}{sin (theta + 13^circ)} = frac{1}{2 cos 73^circ} ] 8. Simplifying, we get: [ cos 13^circ + cot theta sin 13^circ = 2 cos 73^circ ] 9. Solving for cot theta: [ cot theta = frac{2 cos 73^circ - cos 13^circ}{sin 13^circ} = frac{2 cos (60^circ + 13^circ) - cos 13^circ}{sin 13^circ} ] Simplifying further: [ cot theta = frac{-2 sin 60^circ sin 13^circ}{sin 13^circ} = -sqrt{3} ] Thus, theta = 150^circ. 10. Therefore: [ angle CAD = 180^circ - 150^circ - 13^circ = 17^circ ] Conclusion: [ boxed{17^circ} ]

answer:Given the problem: [ text{The sum of positive numbers } a, b, c text{ and } d text{ does not exceed 4. Find the maximum value of the expression } ] [ sqrt[4]{a^{2}+3ab} + sqrt[4]{b^{2}+3bc} + sqrt[4]{c^{2}+3cd} + sqrt[4]{d^{2}+3da} ] Let's solve this step-by-step using three different methods. First Solution: Using Inequality of Arithmetic and Geometric Means (AM-GM) 1. **Applying AM-GM Inequality:** For any non-negative real numbers (x_1, x_2, x_3, x_4): [ frac{x_1 + x_2 + x_3 + x_4}{4} geq sqrt[4]{x_1 x_2 x_3 x_4} ] Setting (x_1 = 4a), (x_2 = a+3b), (x_3 = 4), (x_4 = 4) in the above inequality: [ sqrt[4]{a^2 + 3ab} = frac{sqrt[4]{4a cdot (a + 3b) cdot 4 cdot 4}}{2sqrt{2}} leq frac{1}{2sqrt{2}} cdot frac{4a + (a + 3b) + 4 + 4}{4} ] 2. **Summing Up the Inequalities:** By summing up four similar inequalities corresponding to each term, we get: [ begin{aligned} & sqrt[4]{a^2 + 3ab} + sqrt[4]{b^2 + 3bc} + sqrt[4]{c^2 + 3cd} + sqrt[4]{d^2 + 3da} & leq frac{1}{2sqrt{2}} left(frac{5a + 3b + 8}{4} + frac{5b + 3c + 8}{4} + frac{5c + 3d + 8}{4} + frac{5d + 3a + 8}{4} right) & = frac{1}{2sqrt{2}} cdot frac{8(a + b + c + d) + 32}{4} leq frac{16}{2sqrt{2}} = 4sqrt{2} end{aligned} ] The equality holds when (a = b = c = d = 1). Second Solution: Using Cauchy-Schwarz Inequality 1. **Applying Cauchy-Schwarz to Fourth Roots and Squares:** For the sequences (u_i = sqrt[4]{a}, sqrt[4]{b}, sqrt[4]{c}, sqrt[4]{d}) and (v_i = sqrt[4]{a + 3b}, sqrt[4]{b + 3c}, sqrt[4]{c + 3d}, sqrt[4]{d + 3a}): [ left(sqrt[4]{a(a + 3b)} + sqrt[4]{b(b + 3c)} + sqrt[4]{c(c + 3d)} + sqrt[4]{d(d + 3a)} right)^2 leq (sqrt{a} + sqrt{b} + sqrt{c} + sqrt{d})(sqrt{a + 3b} + sqrt{b + 3c} + sqrt{c + 3d} + sqrt{d + 3a}) ] 2. **Bounding the Sum of Roots and Cross Terms:** [ (sqrt{a} + sqrt{b} + sqrt{c} + sqrt{d})^2 leq (a + b + c + d)(1 + 1 + 1 + 1) leq 4^2 ] Hence, [ sqrt{a} + sqrt{b} + sqrt{c} + sqrt{d} leq 4 ] 3. **Bounding the Sum of Cross Terms' Roots:** [ begin{aligned} & (sqrt{a + 3b} + sqrt{b + 3c} + sqrt{c + 3d} + sqrt{d + 3a})^2 leq & leq (a + 3b + b + 3c + c + 3d + d + 3a)(1 + 1 + 1 + 1) = 16(a + b + c + d) leq 64 end{aligned} ] Therefore, [ sqrt{a + 3b} + sqrt{b + 3c} + sqrt{c + 3d} + sqrt{d + 3a} leq 8 ] 4. **Combining the Results:** By combining the results, [ sqrt[4]{a^2 + 3ab} + sqrt[4]{b^2 + 3bc} + sqrt[4]{c^2 + 3cd} + sqrt[4]{d^2 + 3da} leq sqrt{4 cdot 8} = 4sqrt{2} ] The equality holds when (a = b = c = d = 1). Conclusion: The maximum value of the given expression is ( boxed{4sqrt{2}} ).

answer:In this problem, we consider four six-sided dice. The process to find the solution is similar: - The minimum sum when rolling four dice, each showing one, is (4 times 1 = 4). - The maximum sum, with each die showing six, is (4 times 6 = 24). - Therefore, the possible sums range from 4 to 24, inclusive. This gives (24 - 4 + 1 = 21) distinct possible sums. By the Pigeonhole Principle, to ensure that the same sum is rolled at least twice, you need to roll the dice one more time than the number of possible distinct sums. Thus, you must roll the dice (21 + 1 = 22) times. [ boxed{22} ]