🧠:<think>First, let's denote ( x ) as the minutes past 2 PM when Kolya started solving the problem. At this time, the minute hand has moved ( 6x ) degrees from the 12 o'clock position, and the hour hand has moved ( 30^circ + 0.5x^circ ) from the 12 o'clock position.After spending exactly one hour on the solution, the minute hand will have completed another full revolution, so it will be at ( 6x ) degrees again. The hour hand, however, will have moved an additional ( 30^circ ), making its position ( 60^circ + 0.5x^circ ).To ensure that the angle between the hour and minute hands remains unchanged, we set up the equation:[6x = frac{(30 + 0.5x) + (60 + 0.5x)}{2}]Simplifying this equation:[6x = frac{90 + x}{2}][12x = 90 + x][11x = 90][x = frac{90}{11} approx 8.1818]This gives us the time when Kolya started solving the problem as ( 2 text{ PM} + 8 frac{2}{11} text{ minutes} ).Considering the other scenario where the angle for the minute hand is outside the immediate hourly range, we set up the equation:[6x - 180 = frac{(30 + 0.5x) + (60 + 0.5x)}{2}]Simplifying this equation:[6x - 180 = frac{90 + x}{2}][12x - 360 = 90 + x][11x = 450][x = frac{450}{11} approx 40.9091]This gives us the other possible time when Kolya started solving the problem as ( 2 text{ PM} + 40 frac{10}{11} text{ minutes} ). Final AnswerThe times when Kolya started solving the problem are:[boxed{2 text{ PM } 8 frac{2}{11} text{ minutes} text{ or } 2 text{ PM } 40 frac{10}{11} text{ minutes}}]</think>

🧠:<think>First, I need to simplify the inner square root expression (sqrt{153 + 24 sqrt{3}}). I'll assume it can be expressed in the form (sqrt{x} + sqrt{y}), where (x) and (y) are positive integers. Squaring both sides gives me:[153 + 24 sqrt{3} = x + y + 2sqrt{xy}]From this equation, I set up the system of equations:[x + y = 153 quad text{and} quad 2sqrt{xy} = 24sqrt{3}]Solving the second equation for (sqrt{xy}) yields (sqrt{xy} = 12sqrt{3}), so (xy = 432).Next, I use Vieta's formulas to find (x) and (y) as the roots of the quadratic equation:[t^2 - 153t + 432 = 0]Factoring this equation, I get:[(t - 144)(t - 9) = 0]Thus, (x = 144) and (y = 9), so:[sqrt{153 + 24 sqrt{3}} = sqrt{144} + sqrt{9} = 12 + 3 = 15]Now, substituting back into the original expression:[sqrt{49 + 15} = sqrt{64} = 8]I express 8 as the sum of two square roots:[8 = sqrt{1} + sqrt{49}]Therefore, the ordered pair of positive integers is ((1, 49)).</think>

🧠:<think>Alright, let's tackle this geometry problem step by step. I'll start by visualizing the problem and then break it down into smaller parts to understand each component.Visualizing the Problem:We have three circles: A, B, and C. Circles A and B each have a radius of 1, and they are tangent to each other at one point. Circle C has a radius of 2 and is tangent to the midpoint of the line segment connecting the centers of circles A and B. We need to find the area inside circle C but outside circles A and B.Understanding the Configuration:1. Circles A and B: Since both circles have a radius of 1 and are tangent to each other, the distance between their centers is equal to the sum of their radii, which is 1 + 1 = 2 units.2. Midpoint of AB: The midpoint M of the line segment AB (connecting the centers of A and B) will be exactly halfway between the two centers. Since the distance between A and B is 2 units, the midpoint M is 1 unit away from each center.3. Circle C: Circle C has a radius of 2 and is tangent to this midpoint M. So, the center of circle C must be 2 units away from M. But how is it positioned in relation to A and B?Determining the Position of Circle C:- Since circle C is tangent to the midpoint M, and M is 1 unit away from both A and B, the center of circle C must lie along the perpendicular bisector of AB. This means it's either above or below the line AB.- Given that circles A and B are tangent, the centers are 2 units apart, and M is the midpoint. If we consider the standard configuration, circle C is likely positioned above or below this line. For simplicity, let's assume it's above.Distance from C to A and B:- The center of circle C is 2 units away from M. Since M is 1 unit away from both A and B, we can use the Pythagorean theorem to find the distance from the center of C to the centers of A and B.- Imagine a right triangle where one leg is the distance from M to A (which is 1 unit), and the other leg is the vertical distance from M to the center of C (which is 2 units). The hypotenuse of this triangle will be the distance between the centers of C and A (or C and B).- Calculating the hypotenuse: √(1² + 2²) = √(1 + 4) = √5 ≈ 2.236 units.Checking for Overlap:- Circle C has a radius of 2, and the distance from its center to the centers of A and B is approximately 2.236 units. Since 2.236 > 2, circle C does not overlap with circles A and B. This means the area inside C but outside A and B is simply the area of C minus the areas of A and B.Calculating Areas:- Area of Circle C: π * (2)² = 4π- Area of Circle A: π * (1)² = π- Area of Circle B: π * (1)² = π- Total Area of A and B: π + π = 2πFinal Area Calculation:The area inside circle C but outside circles A and B is:4π (Area of C) - 2π (Combined Area of A and B) = 2πConclusion:After carefully considering the configuration, calculating distances, and determining the areas, the final area inside circle C but outside circles A and B is 2π.Final Answerboxed{D}</think>

🧠:<think>To determine how many 11-digit palindromes can be formed using the digits 1, 1, 2, 2, 2, 2, 4, 4, 5, 5, 5, we need to consider the structure of a palindrome. An 11-digit palindrome has the form (overline{abcdeffedcba}), where the digits mirror around the middle digit.1. Identify the middle digit: Since an 11-digit palindrome has a middle digit and this digit must appear an odd number of times, the digit 5 (which appears three times) will be placed in the middle (f = 5).2. Determine the remaining digits: After placing one 5 in the middle, we are left with the digits 1, 1, 2, 2, 2, 2, 4, 4, 5, 5. These digits need to be arranged symmetrically around the middle digit.3. Count possible arrangements for half of the palindrome: We need to arrange the first five digits of (abcde) such that the palindrome condition is met. The remaining 5 digits (not counting the middle 5) are 1, 1, 2, 2, 2, 2, 4, 4, 5, 5.4. Calculate permutations: The total number of ways to choose digits for (abcde) is the number of distinct permutations of these digits. We can choose two digits from ({1, 2, 4, 5}) (since we have two of each). - Two same digits: (binom{4}{1}) ways to choose the type, and then permute the remaining digits. - Two different digits: (binom{4}{2}) ways to choose the types, and then permute the remaining digits. Calculating these: - Two same types: (binom{4}{1} times frac{5!}{2!2!1!} = 4 times 30 = 120). - Two different types: (binom{4}{2} times frac{5!}{2!2!1!} = 6 times 30 = 180).5. Total combinations: Adding these together, we get (120 + 180 = 300).Thus, the total number of 11-digit palindromes is (boxed{text{(D) } 300}).</think>