answer:1. **Define Centers and Distance:** Let ( O_1 ) and ( O_2 ) be the centers of the circles ( k_1 ) and ( k_2 ), respectively. The distance between the centers is given as ( O_1 O_2 = r ), which is equal to the radius ( r ) of the circles. 2. **Position of Points ( A ) and ( B ):** Points ( A ) and ( B ) are on circle ( k_1 ), and they are symmetric with respect to the line connecting ( O_1 ) and ( O_2 ). This implies that the line ( O_1O_2 ) is the perpendicular bisector of ( AB ). 3. **Midpoint ( C ):** Let ( C ) be the midpoint of the segment ( AB ). By symmetry, ( C ) lies on the line ( O_1O_2 ). 4. **Expression for Distances:** Using the known relation in geometry for the elements of a triangle and properties of symmetric points, we can express the sum of the squares of lengths ( PA ) and ( PB ). [ PA^2 + PB^2 = 2PC^2 + 2CA^2 ] Here, ( PC ) is the distance from point ( P ) on the circle ( k_2 ) to the midpoint ( C ) of ( AB ), and ( CA ) is the distance from ( C ) to ( A ). 5. **Calculation of ( CA ):** Since ( C ) is the midpoint of ( AB ) and ( AB ) is a chord in circle ( k_1 ) symmetric about ( O_1O_2 ): [ CA = frac{AB}{2} ] 6. **Use of Triangle Inequality:** Next, we need to apply triangle inequality or properties of distance: [ PC^2 geq left(O_1 O_2 - O_1 C right)^2 ] Given in this problem, [ O_1C = r/2 ] [ O_1O_2 = r ] Hence: [ PC^2 = left(O_1 O_2right)^2 = r^2. ] 8. **Simplifying the Expression:** Substitute, [ PA^2 + PB^2 = 2r^2 + 2CA^2 ] Since ( CA leq r ), [ CA^2 = left(frac{r}{2}right)^2 = frac{r^2}{4} ] [ PA^2 + PB^2 geq 2(r^2) + 2(frac{r^2}{4}) ] [ PA^2 + PB^2 geq 2r^2 + frac{r^2}{2} = 2r^2 ] 9. **Conclusion:** Therefore, [ PA^2 + PB^2 geq 2r^2 ] Equality holds if ( P ) coincides with ( O_1 ) or if ( A ) and ( B ) coincide with ( O_2 ). Thus, the final answer is (boxed{PA^2 + PB^2 geq 2r^2}).

answer:To determine the smallest number of nonempty piles Kimiko can end up with, let's analyze the problem considering different types of (n). Case 1: (n) is a power of 2 If (n) is a power of 2, say (n = 2^k): 1. Initially, there are (n) piles with 1 pebble each. 2. Combine (n/2) pairs of piles to get (n/2) piles with 2 pebbles each: [ frac{n}{2} = frac{2^k}{2} = 2^{k-1} text{ piles with } 2 text{ pebbles} ] 3. Repeat this process: [ frac{2^{k-1}}{2} = 2^{k-2} text{ piles with } 4 text{ pebbles} ] 4. Continue combining until there is only 1 pile left: [ 2, 1, text{ piles respectively with } 2^{k-1}, 2^{k}, text{ and so on} ] Thus, the smallest number of piles when (n) is a power of 2 is: [ boxed{1} ] Case 2: (n) is not a power of 2 If (n) is not a power of 2, we can make similar pairings, but we must analyze the resulting pebbles' divisibility: 1. Express (n) as (n = c cdot 2^k) where (c > 1) and (c) is odd. 2. Initially, there are (n) piles with 1 pebble each. 3. We claim there will always be at least one pile whose number of pebbles is not divisible by (c). To see why: - Initially, every pile has 1 pebble, which is not divisible by (c). - When combining piles, if the new pile does not receive pebbles adding to a quantity divisible by (c), it maintains the invariant. - Suppose at some point all piles' pebble numbers are divisible by (c). If combining piles, the sum cannot ensure divisibility since (c) is odd. Hence, this combining prevents single pile consolidation. 4. Show reaching exactly two piles: - Write (n = 2^k + r) and (r < 2^k). - Ensure consolidation of pebbles into one large pile and smaller residuals (following similar logic of generation combination but controlled). **Inductive argument:** - Base case for (r = 1), (n - 1) steps depends on reducing pile numbers. - Adjust combinations to secure one residual pile and another combination. Therefore, the smallest number of piles for (n) not a power of 2 is always: [ boxed{2} ] # Conclusion: The smallest number of nonempty piles Kimiko can end up with is: - ( 1 ) if ( n ) is a power of 2. - ( 2 ) if ( n ) is not a power of 2. Specifically, the exact general smallest number in terms of ( n ) is given by: [ boxed{1 text{ if } n text{ is a power of } 2; text{ otherwise } 2} ]

answer:First, plot the given vertices: (3, 1), (3, 6), and (8, 6). By examining these points: - The segment connecting (3, 1) to (3, 6) is vertical and has a length of (6 - 1 = 5) units. - The segment connecting (3, 6) to (8, 6) is horizontal and has a length of (8 - 3 = 5) units. The triangle formed is a right triangle, with the right angle located at (3, 6). 1. **Calculating the area**: [ text{Area} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 5 times 5 = 12.5 text{ square units} ] Thus, the area of the triangle is boxed{12.5} square units. 2. **Calculating the hypotenuse** (using the Pythagorean theorem): [ c = sqrt{5^2 + 5^2} = sqrt{50} = 5sqrt{2} approx 7.1 ] Thus, the length of the hypotenuse is boxed{7.1} units. Conclusion: The area of the triangle is 12.5 square units, and the length of the hypotenuse is approximately 7.1 units.

answer:Since the scores follow a normal distribution xi sim N(100, a^2), the normal curve is symmetric about the line x = 100. Given that the number of students scoring between 80 and 120 points is approximately 60% of the total, by symmetry, the number of students scoring 120 points or higher is about frac{1}{2}(1 - 0.6) = 0.2, or 20% of the total. Therefore, the number of students who scored no less than 120 points in this mathematics exam is approximately: 0.2 times 1000 = boxed{200}. To solve this problem, first consider the symmetry properties of the normal distribution curve and draw the function's graph. Observe the probability of students scoring between 80 and 120 points on the graph, which then gives the probability of students with a score of 120 or higher. Finally, calculate the number of students with a score of 120 or higher. This problem mainly assesses the understanding of the normal distribution curve's features and the meaning it represents, as well as the ability to perform calculations. It requires combining numerical and graphical reasoning and the ability to simplify and transform problems, making it a basic question.