answer:First, we analyze how the transitions between coin states occur: - HH and TT are identities for the last coin flip. - HT and TH switch the last coin state. Given: - Three HH transitions maintain the head state. - Five TT transitions maintain the tail state. - Five HT transitions switch from head to tail. - Six TH transitions switch from tail to head. The sequence must start with a T and end with an H to accommodate all required transitions: - Start with T (necessary for the first TH). - End with H (required after the last TH). The structure of such a sequence can be modeled roughly as alternating Ts and Hs, focusing on the switches and identities: - THTHTHTHTHTHTHTHTHTH. - We need to insert 5 TT and 3 HH while preserving order. For TT: - There are 9 Ts initially. Adding 5 more Ts at positions adjacent to existing Ts is equivalent to a ball-and-urn problem: distribute 5 balls (new Ts) into 9 urns (gaps between and on the sides of existing Ts). - Combinations are calculated as binom{5+9-1}{9-1} = binom{13}{8} = 1287. For HH: - There are 10 Hs initially. Adding 3 more Hs adjacent to existing Hs follows a similar approach. - Combinations for Hs: binom{3+10-1}{10-1} = binom{12}{9} = 220. Thus, the total number of sequences is 1287 cdot 220 = boxed{283140}.

answer:1. To prove ( s = b cos^{2} frac{C}{2} + c cos^{2} frac{B}{2} ): We know that ( s = frac{a+b+c}{2} ). First, let's consider the half-angle identities: [ cos^2 frac{C}{2} = frac{1 + cos C}{2}, quad cos^2 frac{B}{2} = frac{1 + cos B}{2} ] Now, applying these to the given equation: [ s = b cos^{2} frac{C}{2} + c cos^{2} frac{B}{2} ] Substitute the expressions of (cos^2 frac{C}{2}) and (cos^2 frac{B}{2}): [ s = b left( frac{1 + cos C}{2} right) + c left( frac{1 + cos B}{2} right) ] Simplifying this: [ s = frac{b (1 + cos C)}{2} + frac{c (1 + cos B)}{2} ] Combine the fractions: [ s = frac{b + b cos C + c + c cos B}{2} ] Reorder terms: [ s = frac{b + c + b cos C + c cos B}{2} ] Finally, note that: [ s = frac{a + b + c}{2} ] So, we confirm: [ s = b cos^{2} frac{C}{2} + c cos^{2} frac{B}{2} ] # Conclusion [ blacksquare ] 2. To prove ( s = a + b sin^{2} frac{C}{2} + c sin^{2} frac{B}{2} ): Again, we use ( s = frac{a+b+c}{2} ). First, consider the identity: [ sin^2 x + cos^2 x = 1 Rightarrow sin^2 x = 1 - cos^2 x ] From the previous case, we use ( cos^2 frac{C}{2} ) and ( cos^2 frac{B}{2} ): [ sin^2 frac{C}{2} = 1 - cos^2 frac{C}{2}, quad sin^2 frac{B}{2} = 1 - cos^2 frac{B}{2} ] Convert the given equation: [ s = a + b sin^{2} frac{C}{2} + c sin^{2} frac{B}{2} ] Substitute the expressions for (sin^2 x): [ s = a + b (1 - cos^2 frac{C}{2}) + c (1 - cos^2 frac{B}{2}) ] Simplify: [ s = a + b - b cos^2 frac{C}{2} + c - c cos^2 frac{B}{2} ] Combine terms: [ s = a + b + c - (b cos^2 frac{C}{2} + c cos^2 frac{B}{2}) ] From part 1, we know: [ b cos^2 frac{C}{2} + c cos^2 frac{B}{2} = s ] So: [ s = a + b + c - s ] Multiply through by 2, solving for (s): [ 2s = a + b + c ] Thus: [ s = frac{a+b+c}{2} ] # Conclusion [ blacksquare ] 3. To prove ( s = 4 R cos frac{A}{2} cos frac{B}{2} cos frac{C}{2} ): Using the relation for the circumradius (R): Recall ( s = frac{Delta}{r} ) (where (Delta) is the area and (r) is the inradius). We'll use the formula for the area: [ Delta = R cdot a cdot cos frac{B}{2} cos frac{C}{2} ] And since: [ a = 2R sin A ] Combining both constants: [ Delta = R cdot 2R sin A cos frac{B}{2} cos frac{C}{2} ] Simplify: [ Delta = 2R^2 sin A cos frac{B}{2} cos frac{C}{2} ] Since (s = frac{Delta}{r}), insert the expression for (Delta): [ s = frac{2R^2 sin A cos frac{B}{2} cos frac{C}{2}}{r} ] Recall the known relation: [ sin A = 2 cos frac{A}{2} cos frac{(B - C)/2)} = 2 cos frac{A}{2} cos frac{(B + C)/2)} Rightarrow sin A = 2 cos frac{A}{2} cos frac{B + C}{2} ] Thus, replace ( sin A ): [ s = 4R cos frac{A}{2} cos frac{B}{2} cos frac{C}{2} ] # Conclusion [ blacksquare ] 4. To prove ( s = r left( cot frac{A}{2} + cot frac{B}{2} + cot frac{C}{2} right) ): Recall the relation for the semi-perimeter (s): [ s = frac{Delta}{r} ] And the ( cot ) identities: [ cot frac{A}{2} = frac{s-a}{r}, quad cot frac{B}{2} = frac{s-b}{r}, quad cot frac{C}{2} = frac{s-c}{r} ] Using these identities, sum up the expressions: [ cot frac{A}{2} + cot frac{B}{2} + cot frac{C}{2} = frac{s-a}{r} + frac{s-b}{r} + frac{s-c}{r} ] Combine terms: [ = frac{s-a + s-b + s-c}{r} ] Simplify: [ = frac{3s - (a+b+c)}{r} ] Since we know (s = frac{a+b+c}{2}), we simplify further: [ = frac{3s - 2s}{r} = frac{s}{r} ] Since (s = boxed{s}), conclude: [ s = r left( cot frac{A}{2} + cot frac{B}{2} + cot frac{C}{2} right) ] # Conclusion [ blacksquare ]

answer:Begin by taking the product of the given equations: ab cdot ac cdot bc = 30sqrt{5} cdot 45sqrt{5} cdot 40sqrt{5} . This can be simplified further: (abc)^2 = (30 cdot 45 cdot 40) cdot (sqrt{5} cdot sqrt{5} cdot sqrt{5}) = 54000 cdot 5sqrt{5}. (abc)^2 = 270000sqrt{5}. Taking the square root to solve for abc, and assuming all variables are positive: abc = sqrt{270000sqrt{5}} = (sqrt{270000})(sqrt[4]{5}) = (300sqrt{3})(sqrt[4]{5}) = 150sqrt{12} sqrt[4]{5}. Given that all values are positive: abc = 150 cdot 2sqrt{3} cdot sqrt[4]{5} = 300sqrt{3} cdot sqrt[4]{5}. Thus, we have the final value as, abc = boxed{300sqrt{3} sqrt[4]{5}}.

answer:To determine whether the sum of the underlined numbers can be negative or zero, we will follow the steps outlined in the reference solution. 1. **Underlining Numbers**: - Each positive number is automatically underlined (let's call this as `single underline`). - Each number that sums with the next number to a positive value is underlined with `double underline`. - Each number that sums with the next two numbers to a positive value is underlined with `triple underline`. 2. **Formulate Pairs and Triplets**: - Consider a number `a` that is double underlined. According to the problem, there must be a subsequent number `b` which is single underlined such that (a + b > 0). - Similarly, consider a number `a` that is triple underlined. There must be subsequent numbers `b` and `c` such that (a + b + c > 0). 3. **Group the Underlined Numbers**: - First, we pair all triple-underlined numbers with the subsequent two numbers. For a triple-underlined number `a`, we pair it with the following numbers `b` (double underlined) and `c` (single underlined). By the condition set in the problem, the sum of these three numbers must be positive: [ a + b + c > 0 ] - Next, after extracting the triplets, we pair the remaining double underlined numbers with the subsequent single underlined numbers. For a double underlined number `a`, we pair it with the following single underlined number `b`. By the problem's condition, the sum of these two numbers must be positive: [ a + b > 0 ] - Lastly, we combine the remaining single underlined numbers. Each of these numbers are positive as stated at the beginning. 4. **Summarizing**: - The sum of each group (triplet or pair) is positive. - As all individual groups have a positive sum, the total sum of all such groups must also be positive. # Conclusion: [ text{The sum of all underlined numbers cannot be negative and cannot be zero (it will always be positive).} ] (blacksquare)