answer:1. **Apply the pigeonhole principle**: Given there are 8 squirrels and each throws a cone at another squirrel independently, the pigeonhole principle tells us that if each squirrel threw at most one cone at each, there must be a squirrel ( A ) who received no more than 1 cone. Otherwise, the total number of cones thrown, ( 8 ), would exceed the number of possible pairs (binom{8}{2} = 28). 2. **Form an initial group**: Place squirrel ( A ) in a separate group. Next, we remove the squirrel whom ( A ) has thrown a cone at (if any), and the one that threw a cone at ( A ). 3. **Update count and reapply principle**: After this removal, we are left with either 6 or 7 squirrels. By reapplying the pigeonhole principle to these remaining squirrels, we find another squirrel ( B ), such that ( B ) received no more than 1 cone. 4. **Form a pair**: Add this squirrel ( B ) to the group that includes ( A ). Again, we remove the squirrel whom ( B ) has thrown a cone at, and the one that threw a cone at ( B ), if any. 5. **Final group formation**: Now, we're left with at least 4 squirrels who did not throw cones at ( A ) and ( B ), nor did ( A ) or ( B ) throw cones at them. 6. **Add the last member to form the group**: Among these remaining squirrels, choose any squirrel ( C ) and add it to the group with ( A ) and ( B ). Given these conditions, none of these squirrels ( A ), ( B ), or ( C ) have thrown cones at each other. # Conclusion: Thus, we have successfully demonstrated that there always exists a group of three squirrels that have not thrown cones at each other. [boxed{}]

answer:Given the problem: 1. In a newly populated apartment building, workers have installed a block consisting of 80 mailboxes. 2. They randomly placed keys corresponding to different apartments inside these locked mailboxes. 3. Our absent-minded scientist, who lives in apartment 37, is trying to find his key using a sushi stick he found and applying an algorithm. The steps to show the algorithm is successful (part a) and calculate the expected number of residents who need to use the stick (part b) are as follows: Part (a) **Proof of Algorithm Effectiveness:** 1. **Numbering and Initialization:** - Number the mailboxes and keys the same way as the apartments, from 1 to 80. - Assume the scientist starts by opening mailbox 37. 2. **Extraction Process:** - If the key in mailbox 37 is for mailbox 37, the algorithm is successful immediately. - If not, the scientist uses the key found to open a new mailbox corresponding to the key number found. 3. **Sequence and Convergence:** - The sequence of mailbox numbers opened forms a permutation of the set {1, 2, ..., 80}. - Since there are only 80 mailboxes, this sequence must eventually return to mailbox 37 as it cannot be infinite. - If there was no return to 37 and continued indefinitely, it would imply more than 80 mailboxes exist, which is a contradiction. 4. **Conclusion:** - Hence, the described algorithm will eventually lead the scientist to find key 37 in mailbox 37. [ blacksquare ] Part (b) **Expectation Calculation:** 1. **Graph Representation:** - Construct a directed graph where vertices represent mailboxes. - Draw an edge from vertex (a) to vertex (b) if the key from mailbox (a) opens mailbox (b). 2. **Graph Structure:** - Every vertex (mailbox) has exactly one outgoing and one incoming edge. - The graph is a union of disjoint cycles. 3. **Cycle Calculation:** - Define (I_{1,2,ldots,k}) as the indicator that vertices {1, 2, ..., k} form a cycle. - The expected value of (I_{1,2,ldots,k}) is calculated by the probability of this event: [ mathbb{E}[I_{1,2,ldots,k}] = frac{1}{n} times frac{1}{n-1} times cdots times frac{1}{n-(k-1)} = frac{(n-k)!}{n!} ] 4. **Number of Cycles:** - For any (k), the number of ordered cycles is: [ X_{k} = sum_{text{all ordered cycles of length } k} I_{1,2,ldots,k} ] 5. **Expected Number of Cycles:** - The expected number of cycles of length (k) is: [ mathbb{E}[X_{k}] = frac{n!}{k(n-k)!} times frac{(n-k)!}{n!} = frac{1}{k} ] - The total expected number of cycles in the graph is: [ mathbb{E}[X] = sum_{k=1}^{n} frac{1}{k} = H_{n} ] - Here, (H_{n}) is the (n)-th harmonic number. 6. **Calculation for (n = 80):** - The approximate sum using harmonic series relation: [ H_n approx ln n + gamma + frac{1}{2n} ] - Given (gamma approx 0.577): [ H_{80} approx ln 80 + 0.577 + frac{1}{160} ] [ H_{80} approx 4.382 + 0.577 + 0.00625 approx 4.96525 ] - The expected number of residents who need to use the stick is approximately: [ boxed{4.968} ]

answer:Substituting and expanding, we get [ x^2 + y^2 + z^2 + xyz = left( frac{b}{c} - frac{c}{b} right)^2 + left( frac{a}{c} - frac{c}{a} right)^2 + left( frac{a}{b} - frac{b}{a} right)^2 + left( frac{b}{c} - frac{c}{b} right)left( frac{a}{c} - frac{c}{a} right)left( frac{a}{b} - frac{b}{a} right) ] Expanding each square: [ left( frac{b}{c} - frac{c}{b} right)^2 = frac{b^2}{c^2} - 2 + frac{c^2}{b^2}, quad left( frac{a}{c} - frac{c}{a} right)^2 = frac{a^2}{c^2} - 2 + frac{c^2}{a^2}, quad left( frac{a}{b} - frac{b}{a} right)^2 = frac{a^2}{b^2} - 2 + frac{b^2}{a^2} ] Summing these: [ frac{b^2}{c^2} + frac{c^2}{b^2} + frac{a^2}{c^2} + frac{c^2}{a^2} + frac{a^2}{b^2} + frac{b^2}{a^2} - 6 ] For the product (xyz): [ left( frac{b}{c} - frac{c}{b} right)left( frac{a}{c} - frac{c}{a} right)left( frac{a}{b} - frac{b}{a} right) = left( frac{ab}{c^2} - frac{bc}{a^2} - frac{ca}{b^2} + frac{c^3}{abc} right) ] Expanding this product and simplifying, we find that it is equal to (-4) after considering symmetry and the fact that (frac{a}{b}frac{b}{c}frac{c}{a} = 1). Adding this to the simplified sum: [ frac{b^2}{c^2} + frac{c^2}{b^2} + frac{a^2}{c^2} + frac{c^2}{a^2} + frac{a^2}{b^2} + frac{b^2}{a^2} - 6 - 4 = frac{b^2}{c^2} + frac{c^2}{b^2} + frac{a^2}{c^2} + frac{c^2}{a^2} + frac{a^2}{b^2} + frac{b^2}{a^2} - 10 ] Since each pair of terms like (frac{b^2}{c^2} + frac{c^2}{b^2}) simplifies to at least 2 by AM-GM inequality, we have: [ x^2 + y^2 + z^2 + xyz = 2 + 2 + 2 - 10 = -4 ] Thus, the final answer is (boxed{-4}).

answer:To solve this problem: 1. First, rearrange and complete the square for the x-term and the y-term in the equation: [ 4x^2 - 48x - y^2 + 6y + 50 = 0 ] Completing the square for x, we get: [ 4(x^2 - 12x) = 4[(x-6)^2 - 36] ] [ 4(x-6)^2 - 144 ] Completing the square for y, we get: [ -(y^2 - 6y) = -[(y-3)^2 - 9] ] [ -(y-3)^2 + 9 ] Substituting back, we obtain: [ 4(x-6)^2 - 144 - (y-3)^2 + 9 + 50 = 0 ] [ 4(x-6)^2 - (y-3)^2 - 85 = 0 ] Then, by making this equation resemble the standard hyperbola equation, add 85 to both sides: [ 4(x-6)^2 - (y-3)^2 = 85 ] Divide through by 85: [ frac{(x-6)^2}{21.25} - frac{(y-3)^2}{85} = 1 ] 2. From the above, a^2 = 21.25 so a = frac{1}{3}sqrt{85}. 3. The distance between the vertices is 2a, which is: [ 2 times frac{1}{3}sqrt{85} = frac{2sqrt{85}}{3} ] boxed{frac{2sqrt{85}}{3}}