answer:# Problem: A cube is given with a unit length for its edges. Determine whether it is possible to cut a hole in one cube such that another congruent cube can pass through it. We will examine two different methods to address this problem. I. Solution Using Symmetry and Scaling 1. **Define the key points and distances**: Consider a cube (M) with unit edges. Let (M) and (N) be two opposite vertices of the cube. 2. **Position key points**: Determine points (A) and (B) at a distance (a) from (M), and points (C) and (D) at the same distance (a) from (N). According to geometric properties, we have: [ AB = CD = asqrt{2} ] We choose (a = 0.8) such that: [ a > frac{1}{sqrt{2}} ] Hence, [ AB = CD = a sqrt{2} = 0.8 sqrt{2} > 1 ] 3. **Orientation and properties of the quadrilateral**: Observe that (AB) is parallel to (CD) and they lie on opposite faces of the cube. The quadrilateral (ABCD) is symmetric with respect to the plane containing the body diagonal (MN), forming a symmetric trapezoid which is actually a rectangle. 4. **Scaling down the rectangle adequately**: Even after scaling down (AB = CD) to (A'B' = C'D'), all sides remain longer than 1 unit. Hence, (A'B'C'D') is entirely inside the cube. 5. **Final Test**: The rectangle can be multiplied with a plane perpendicular to itself to form an infinite prism surface within the cube. By cutting this prism from the cube, an appropriately sized hole is created through which another congruent cube can pass. Since the rectangle fits within the cube and its sides can stretch larger than 1 unit, the cube fits through this cut. II. Solution Using Projection and Inscribed Circle 1. **Projection onto a specific plane**: Consider projecting one cube onto a plane that is perpendicular to its body diagonal. This plane bisects the cube into two equal parts. 2. **Analyze the resulting projection**: The projection of the cube forms a regular hexagon, as defined by the rotations around the body diagonal of (60^circ). 3. **Dimension calculation**: The side length of the hexagon formed by projection: [ text{Side length of hexagon} = frac{sqrt{2}}{sqrt{3}} ] And the radius of the inscribed circle: [ text{Radius of inscribed circle} = frac{sqrt{2}}{2} ] 4. **Inscribing a square**: Within this hexagon, inscribe a square with sides of 1 unit. Ensure that none of the vertices of the square touch the sides of the hexagon directly. 5. **Cutting through the cube**: Slice the cube according to the dimensions and positions of the inscribed square along a direction perpendicular to the plane of the square. This defines an infinite prism through which another congruent cube can pass. In conclusion, both methods show that it is possible to cut a hole in one cube such that another congruent cube can pass through it. [ boxed{text{Yes, it is possible to cut such a hole.}} ]

answer:Given: AB = BC, angle ACB = alpha, angle CAM = beta. Construct the triangle ABC where AB = BC and angle ACB = alpha. 1. **Labeling and Setup**: - Let AB = BC = a. - By the properties of the isosceles triangle, AC = 2a cos alpha. - For point M on AC, the problem states that angle CAM = beta. 2. **Calculating Areas**: - The area of triangle AMB is given by: [ S_{triangle AMB} = frac{1}{2} cdot AB cdot AM cdot sin(alpha - beta) ] - Using AB = a, we get: [ S_{triangle AMB} = frac{1}{2} cdot a cdot AM cdot sin(alpha - beta) ] - The area of triangle AMC is given by: [ S_{triangle AMC} = frac{1}{2} cdot AC cdot AM cdot sin beta ] - Using AC = 2a cos alpha, we get: [ S_{triangle AMC} = frac{1}{2} cdot 2a cos alpha cdot AM cdot sin beta = a cos alpha cdot AM cdot sin beta ] 3. **Finding the Ratio**: - To find the ratio frac{S_{triangle AMB}}{S_{triangle AMC}}, use the areas obtained in the previous step: [ frac{S_{triangle AMB}}{S_{triangle AMC}} = frac{frac{1}{2} cdot a cdot AM cdot sin(alpha - beta)}{a cos alpha cdot AM cdot sin beta} ] - Simplifying the ratio: [ frac{S_{triangle AMB}}{S_{triangle AMC}} = frac{sin(alpha - beta)}{2 cos alpha cdot sin beta} ] # Conclusion: The ratio in which this line divides the area of the triangle is: [ boxed{frac{sin(alpha - beta)}{2 cos alpha cdot sin beta}} ]

answer:1. **Calculate the area of the smallest circle (red area)**: The diameter of the smallest circle is 2 inches, so its radius is 1 inch. The area of the smallest circle is: [ A_{text{red}} = pi times 1^2 = pi ] 2. **Calculate the area of the middle circle**: The diameter of the middle circle is 6 inches, so its radius is 3 inches. The area of the middle circle is: [ A_{text{middle}} = pi times 3^2 = 9pi ] 3. **Calculate the area of the largest circle**: The diameter of the largest circle is 10 inches, so its radius is 5 inches. The area of the largest circle is: [ A_{text{large}} = pi times 5^2 = 25pi ] 4. **Calculate the area painted green**: The green-painted area is the area of the largest circle minus the area of the middle circle: [ A_{text{green}} = A_{text{large}} - A_{text{middle}} = 25pi - 9pi = 16pi ] 5. **Calculate the ratio of the green-painted area to the red-painted area**: [ frac{A_{text{green}}}{A_{text{red}}} = frac{16pi}{pi} = 16 ] 6. **Conclusion**: The ratio of the area painted green to the area painted red is 16. The final answer is boxed{textbf{(C) } 16}

answer:To solve this problem, we need to find a number n = pq78rstu that is divisible by both 17 and 19. Since 17 and 19 are prime numbers, their least common multiple (LCM) is their product, which is 17 * 19 = 323. Therefore, n must be divisible by 323. Let's find a multiple of 323 that has the form pq78rstu. We know that the number must end in 78rstu, so we can start by multiplying 323 by various multiples until we find one that ends in 78. Let's start by multiplying 323 by different numbers and see if we can find a pattern: 323 * 1 = 323 323 * 2 = 646 323 * 3 = 969 323 * 4 = 1292 323 * 5 = 1615 323 * 6 = 1938 323 * 7 = 2261 323 * 8 = 2584 323 * 9 = 2907 323 * 10 = 3230 ... We can see that the last digit cycles through 3, 6, 9, 2, 5, 8, 1, 4, 7, 0. We need the last three digits to be 78_, so we can see that when we multiply 323 by 6, the last digit is 8. Let's see if we can find a multiple of 323 that ends in 78_ by continuing the pattern: 323 * 16 = 5168 323 * 26 = 8398 323 * 36 = 11628 323 * 46 = 14858 323 * 56 = 18088 323 * 66 = 21318 323 * 76 = 24548 323 * 86 = 27778 323 * 96 = 30908 We can see that 323 * 86 = 27778, which has the form pq78rstu. Now we need to find the digits p, q, r, s, t, and u such that the sum of p, q, r, and s is equal to t + u. From 27778, we have: p = 2 q = 7 r = 7 s = 7 t = 7 u = 8 The sum of p, q, r, and s is 2 + 7 + 7 + 7 = 23, and the sum of t and u is 7 + 8 = 15. These do not match, so we need to find another multiple of 323 that ends in 78_ and also satisfies the condition that the sum of p, q, r, and s equals t + u. Let's continue with the next multiple of 323 that ends with 78_: 323 * 106 = 34178 Now we have: p = 3 q = 4 r = 1 s = 7 t = 7 u = 8 The sum of p, q, r, and s is 3 + 4 + 1 + 7 = 15, and the sum of t and u is 7 + 8 = 15. These match, so we have found the correct values for p, q, r, s, t, and u. Therefore, the values are: p = 3 q = 4 r = 1 s = 7 t = 7 u = boxed{8}