answer:Given: - The area of triangle ABC is 15sqrt{3} - angle BAC = 120^circ - Distance from vertex A to the incenter (center of inscribed circle) is 2 We are to find the median of triangle ABC drawn from vertex B. 1. **Define Notations and Apply Trigonometry**: - Let O be the incenter of triangle ABC. - Let r be the radius of the inscribed circle. - Let K, M, N be the points of tangency on sides AC, BC, and AB respectively. 2. **Utilize the Distance Relation**: [ OK = r = AO sin 60^circ = AO cdot frac{sqrt{3}}{2} ] Given AO = 2, we get: [ r = 2 cdot frac{sqrt{3}}{2} = sqrt{3} ] Also, [ AK = AO cos 60^circ = 2 cdot frac{1}{2} = 1 ] 3. **Relate Semi-perimeter p with Area**: Using the formula for the area of a triangle (in terms of its inradius and semi-perimeter): [ S_{Delta ABC} = p cdot r ] Given S_{Delta ABC} = 15sqrt{3} and r = sqrt{3}: [ 15sqrt{3} = p cdot sqrt{3} ] Solving for p: [ p = 15 ] 4. **System of Equations**: Let BM = BN = x and CM = CK = y: [ x + y + 1 = 15 quad text{(sum of tangents and sides equals the semi-perimeter)} ] [ (x + y)^2 = (x + 1)^2 + (y + 1)^2 + (x+1)(y+1) quad text{(Cosine Rule for triangle AOB)} ] Solving for x and y: [ x + y = 14 ] and, [ 14^2 = (x + 1)^2 + (y + 1)^2 + (x+1)(y+1) ] Simplify and solve: [ 196 = x^2 + 2x + 1 + y^2 + 2y + 1 + (x + 1)(y + 1) ] Using x + y = 14, one solution must be tested. Given AC > AB: [ x = 5, y = 9 quad text{(due to condition AC > AB fulfilling x<y)} ] Therefore: [ AB = x + 1 = 6, AC = y + 1 = 10 ] 5. **Median Calculation**: Let P be the midpoint of AC. Using the cosine rule in triangle ABP: [ BP^2 = AB^2 + AP^2 - 2 cdot AB cdot AP cdot cos 120^circ ] Since AP = frac{AC}{2} = 5: [ BP^2 = 6^2 + 5^2 - 2 cdot 6 cdot 5 cdot (-frac{1}{2}) ] Substitution yields: [ BP^2 = 36 + 25 + 2 cdot 6 cdot 5 cdot frac{1}{2} = 36 + 25 + 30 = 91 ] Therefore: [ BP = sqrt{91} ] # Conclusion: [ boxed{sqrt{91}} ]

answer:**Part (a):** 1. Given that a^2 + b^2 = 1 and a + b = 1, we want to show that ab = 0. 2. Start by squaring the second equation: [ (a + b)^2 = 1^2 = 1 ] 3. Expand the left-hand side: [ a^2 + 2ab + b^2 = 1 ] 4. Substitute a^2 + b^2 = 1 into the equation: [ 1 + 2ab = 1 ] 5. Subtract 1 from both sides: [ 2ab = 0 ] 6. Divide both sides by 2: [ ab = 0 ] **Conclusion (a):** Thus, ab = 0. boxed{ab = 0}. **Part (b):** 1. Given that a^3 + b^3 + c^3 = 1, a^2 + b^2 + c^2 = 1, and a + b + c = 1, we want to show that abc = 0. 2. Begin by squaring the sum a + b + c: [ (a + b + c)^2 = 1^2 = 1 ] 3. Expand the left-hand side: [ a^2 + b^2 + c^2 + 2(ab + bc + ca) = 1 ] 4. Substitute a^2 + b^2 + c^2 = 1: [ 1 + 2(ab + bc + ca) = 1 ] 5. Subtract 1 from both sides: [ 2(ab + bc + ca) = 0 ] 6. Divide both sides by 2: [ ab + bc + ca = 0 ] 7. Consider the identity: [ (a^2 + b^2 + c^2)(a + b + c) = a^3 + b^3 + c^3 + (ab + bc + ca)(a + b + c) - 3abc ] 8. Substitute the given values a^2 + b^2 + c^2 = 1, a + b + c = 1, and a^3 + b^3 + c^3 = 1: [ 1 cdot 1 = 1 + (ab + bc + ca)cdot 1 - 3abc ] 9. Simplify using ab + bc + ca = 0: [ 1 = 1 + 0 - 3abc ] [ 1 = 1 - 3abc ] 10. Subtract 1 from both sides: [ 0 = -3abc ] 11. Divide both sides by -3: [ abc = 0 ] **Conclusion (b):** Thus, abc = 0. boxed{abc = 0}.

answer:To solve the problem, we need to determine the maximal number of red points we can obtain on a circle with 800 points, numbered from 1 to 800, using the given dyeing rules. 1. **Understanding the Dyeing Rule:** - If point k is red, then the point located k positions clockwise (i.e., at position (k + k) mod 800 = 2k mod 800) will also turn red. 2. **Forming a Sequence:** - We start with point x as a red point. Following the rule, we generate a sequence of positions: [ x, 2x, 2^2x, 2^3x, ldots ] taken modulo 800. 3. **Periodicity of the Sequence (Modulo 800):** - We're interested in the behavior of the sequence x, 2x, 2^2x, ldots under modulo 800. Note that: [ 800 = 2^5 cdot 5^2 ] 4. **Finding the Length of Distinct Elements:** - The goal is to find the largest number of distinct elements m in the sequence x, 2x, 2^2x, ldots, modulo 800. 5. **Choosing an Appropriate x:** - Select x = 2^5 + 1 = 33. We then consider the sequence: [ 33, 2 cdot 33 = 66, 2^2 cdot 33 = 4 cdot 33 = 132, 2^3 cdot 33 = 8 cdot 33 = 264, ldots ] modulo 800. 6. **Study of Multiplicative Properties:** - We aim to find the distinct powers of 2 modulo 800. Given: [ x, 2x, 2^2x, 2^3x, ldots ] - For n geq 5: [ 2^n cdot 33 equiv 0 pmod{32} quad text{(since 32=2^5)} ] - Additionally, observe the behavior modulo 25: [ 2^{20} equiv 1 pmod{25} ] because the order of 2 modulo 25 is 20. 7. **Combining the Two Moduli Using the Chinese Remainder Theorem:** - We need 2^r equiv 1 pmod{25} and 2^s equiv 0 pmod{32}. - The smallest integer r satisfying 2^r equiv 1 pmod{25} is 20. - For s geq 5, 2^s cdot 33 equiv 0 pmod{32}. - Thus, m geq 25. 8. **Checking for Contradictions:** - If we assume m = 26, then two values would repeat, leading to a contradiction due to periodicity for both moduli (32 and 25). 9. **Conclusion:** - The maximum number of distinct red points under these rules is 25. [ boxed{25} ]

answer:Let's denote the number of bird cages as ( x ). Each bird cage has ( 6 ) parrots and ( 2 ) parakeets, so each cage has a total of ( 6 + 2 = 8 ) birds. The total number of birds in the pet store is ( 48 ), and since each cage has ( 8 ) birds, we can set up the following equation to find the number of cages: ( 8x = 48 ) To solve for ( x ), we divide both sides of the equation by ( 8 ): ( x = frac{48}{8} ) ( x = 6 ) Therefore, the pet store has ( boxed{6} ) bird cages.