answer:Given: - Triangle ABC with sides AB = a, AC = b, and angle BAC = 120^circ - We need to find the length of the bisector AM of angle angle BAC 1. **Calculate the area of triangle ABC**: [ S_{triangle ABC} = frac{1}{2} cdot AB cdot AC cdot sin angle BAC ] Since angle BAC = 120^circ: [ sin 120^circ = frac{sqrt{3}}{2} ] Therefore: [ S_{triangle ABC} = frac{1}{2} cdot a cdot b cdot frac{sqrt{3}}{2} = frac{1}{2} a b cdot frac{sqrt{3}}{2} = frac{a b sqrt{3}}{4} ] 2. **Express the areas of triangles ABM and ACM**: [ S_{triangle ABM} = frac{1}{2} cdot AB cdot AM cdot sin angle BAM ] Since AM is the bisector of angle BAC, angle BAM = angle CAM = 60^circ: [ sin 60^circ = frac{sqrt{3}}{2} ] Therefore: [ S_{triangle ABM} = frac{1}{2} cdot a cdot AM cdot frac{sqrt{3}}{2} = frac{a cdot AM cdot sqrt{3}}{4} ] Similarly, [ S_{triangle ACM} = frac{1}{2} cdot b cdot AM cdot frac{sqrt{3}}{2} = frac{b cdot AM cdot sqrt{3}}{4} ] 3. **Sum of the areas of triangles ABM and ACM**: The total area S_{triangle ABC} is the sum of the areas of triangles ABM and ACM: [ S_{triangle ABC} = S_{triangle ABM} + S_{triangle ACM} ] Substituting the areas from above: [ frac{a b sqrt{3}}{4} = frac{a cdot AM cdot sqrt{3}}{4} + frac{b cdot AM cdot sqrt{3}}{4} ] 4. **Solve for AM**: Factor out frac{sqrt{3}}{4}: [ frac{a b sqrt{3}}{4} = frac{sqrt{3}}{4} left( a cdot AM + b cdot AM right) ] Simplify: [ a b sqrt{3} = sqrt{3} left( a cdot AM + b cdot AM right) ] Cancel out sqrt{3} from both sides: [ a b = a cdot AM + b cdot AM ] Factor out AM on the right-hand side: [ a b = AM (a + b) ] Solve for AM: [ AM = frac{a b}{a + b} ] # Conclusion: [ boxed{frac{a b}{a + b}} ]

answer:# Problem: 33.13. In a sequence of ( n ) distinct numbers, it is required to simultaneously find the largest and the next largest number in the sequence. a) Prove that this can be done by comparing ( n + m - 2 ) pairs of numbers, where the integer ( m ) is determined by the inequalities ( m-1 < log_2 n leqslant m ). b) Prove that it is generally impossible to do without comparing fewer than ( n + m - 2 ) pairs of numbers. Part (a) To find the largest and the second largest number in a sequence of ( n ) distinct numbers, we divide the numbers approximately in halves, similar to the method described in problem 33.10. According to the solution to that problem, the process stops after ( m ) steps. The groups obtained in the ( ( m-1 ))-th step consist of pairs of numbers and individual numbers. In each such group, we compare the numbers and find the largest among them. The remaining individual numbers are not compared at this step. Then, similarly, in each group obtained at step ( m-2 ), we find the largest number and so on. In the end, we find the largest number in ( n-1 ) comparisons, because with each comparison, one candidate for the largest number is ruled out, and ( n-1 ) numbers are ruled out in total from the ( n ) numbers. Next, let's find the second largest number. It is clear that it was excluded from the competition as a result of a comparison with the largest number. Consequently, the second largest number must be among the numbers compared directly to the largest number. There are no more than ( m ) such numbers. Selecting the largest from these numbers can be done in ( m-1 ) comparisons. Combining these results, it takes ( n-1 ) comparisons to find the largest number and an additional ( m-1 ) comparisons to find the second largest number among the ( m ) candidates. Therefore, the total number of comparisons required is: [ (n-1) + (m-1) = n + m - 2. ] Conclusion [ boxed{The total number of comparisons required is ( n + m - 2 ).} ] Part (b) We will now prove that it is impossible, in general, to get by with fewer than ( n + m - 2 ) pair comparisons to find the largest and the second largest number. Let's denote by ( k_i ) the number of numbers that have lost at least ( i ) comparisons (note that indirect losses, where we deduce ( a < c ) from ( a < b ) and ( b < c ), are not counted here). The number of total losses is equal to the number of comparisons, because after each comparison, exactly one of the ( k_i ) increases by 1. The sum of all ( k_i ), therefore, equals the number of comparisons because, after each comparison, one number ( k_i ) increases by 1. Assume that number ( j ) loses a comparison. Then ( k_{j+1} ) increases by 1, and the other ( k_i ) remain unchanged. First, note that once the algorithm finishes working, it must satisfy ( k_1 geq n-1 ), since all numbers except one must have lost to someone (if two numbers never lost to anyone, we wouldn't know which one is bigger). Therefore, once the second largest number is discovered, the following must also hold: ( k_1 + k_2 geq n + m - 2 ). Consider the worst-case scenario and use the following assumptions about the outcomes of comparisons: 1. In a comparison of two non-leaders, the result is arbitrary. 2. In a comparison of a leader with a non-leader, the leader always wins. 3. In a comparison of two leaders, the winner is the one with more wins (if the number of wins is the same, the result is arbitrary). These outcomes are possible because the leader has no restrictions from above: if we increase it, the results of previous comparisons do not change. We introduce the concept of subordination to a leader. Initially, all numbers are subordinated to themselves only. In comparisons of types (1) and (2), subordination to the leaders does not change. In a comparison of type (3), the loser and all its subordinates are subordinated to the new leader. We now show by induction on ( k ) that if a leader won ( k ) comparisons, it is subordinated to (including itself) no more than ( 2^k ) numbers. For ( k=0 ), the leader is subordinated only to itself. Suppose the leader who won ( k ) comparisons wins against someone next. According to our comparison outcome agreement of type (3), both the winner and the loser are subordinated to no more than ( 2^k ) numbers respectively; so, when uniting these two groups, no more than ( 2^{k+1} ) numbers remain, as required. Thus, the largest number won at least ( m ) comparisons because it will end up being subordinate to all ( n ) numbers. Out of the ( m ) numbers losing to the largest number, only one number, which didn't lose to anyone else, matters. Therefore, ( k_2 geq m-1 ). Combining these results, the lower bound of pair comparisons needed is: [ k_1 + k_2 geq n - 1 + m - 1 = n + m - 2. ] # Conclusion [ boxed{In general, it is impossible to find the largest and the second largest number with fewer than ( n + m - 2 ) comparisons.} ]

answer:Let's denote the value of the third type of coin as "x" rupees. Since there are an equal number of each type of coin, and there are 20 coins of each type, we can set up the following equation to represent the total value of the coins: Total value = (Number of one rupee coins * Value of one rupee coin) + (Number of 50 paise coins * Value of 50 paise coin) + (Number of x rupee coins * Value of x rupee coin) Given that the total value is 35 rupees, we can write: 35 = (20 * 1) + (20 * 0.50) + (20 * x) Now, let's calculate the total value of the one rupee coins and the 50 paise coins: Total value of one rupee coins = 20 * 1 = 20 rupees Total value of 50 paise coins = 20 * 0.50 = 10 rupees Substituting these values into the equation, we get: 35 = 20 + 10 + (20 * x) Now, let's solve for x: 35 = 30 + 20x 35 - 30 = 20x 5 = 20x Divide both sides by 20 to find the value of x: x = 5 / 20 x = 0.25 So, the value of the third type of coin is boxed{0.25} rupees, which is equivalent to 25 paise.

answer:Let's denote the number of shirts Leo dropped off as S and the number of pairs of trousers as T. We are given that Leo was charged 5 per shirt and 9 per pair of trousers, and the total bill was 140. So we can write the following equation: 5S + 9T = 140 We are also told that the attendant insisted Leo had only dropped off 2 shirts, but it was found that 8 shirts were missing. This means that Leo actually dropped off 2 + 8 = 10 shirts. So S = 10. Now we can substitute S = 10 into the equation to find T: 5(10) + 9T = 140 50 + 9T = 140 9T = 140 - 50 9T = 90 T = 90 / 9 T = 10 So, Leo dropped off boxed{10} pairs of trousers.