answer:# Step-by-Step Solution Part 1: Finding the values of a, b, and c Given the equation a^{2}+b^{2}+c^{2}-6a-8b-10c+50=0, we can rewrite it by completing the squares for a, b, and c respectively. This process involves rearranging and grouping terms: [ a^{2} - 6a + b^{2} - 8b + c^{2} - 10c + 50 = 0 ] Adding and subtracting the necessary squares (i.e., (3)^2, (4)^2, and (5)^2) inside the equation gives us: [ (a^{2} - 6a + 9) + (b^{2} - 8b + 16) + (c^{2} - 10c + 25) = 0 ] This simplifies to: [ left(a-3right)^{2}+left(b-4right)^{2}+left(c-5right)^{2}=0 ] Given that a square of a real number is always non-negative, and the sum of these squares equals zero, each term must be zero. Therefore: [ begin{align*} a-3 &= 0 b-4 &= 0 c-5 &= 0 end{align*} ] Solving these equations gives us: [ begin{align*} a &= 3 b &= 4 c &= 5 end{align*} ] Thus, the values of a, b, and c are 3, 4, and 5 respectively. We encapsulate this part of the answer as: [ boxed{a = 3, b = 4, c = 5} ] Part 2: Finding the radius of the circumcircle of triangle ABC Given that a=3, b=4, and c=5, and recognizing that 3^{2}+4^{2}=5^{2}, we identify that triangle ABC is a right triangle with the right angle opposite to side c. In a right triangle, the hypotenuse is the diameter of the circumcircle. Therefore, the radius R of the circumcircle is half the length of the hypotenuse: [ R = frac{1}{2}c = frac{1}{2} times 5 = 2.5 ] Thus, the radius of the circumcircle of triangle ABC is 2.5. We encapsulate this part of the answer as: [ boxed{R = 2.5} ]

answer:Solution: ① By the definition of the new operation "ast" and setting c=0 in property (3), we have (a ast b) ast 0 = 0 ast (ab) + (a ast 0) + (0 ast b) = ab + a + b, which means a ast b = ab + a + b. Therefore, f(x) = x ast frac{1}{x} = 1 + x + frac{1}{x}. When x > 0, f(x) = 1 + x + frac{1}{x} geq 1 + 2sqrt{x cdot frac{1}{x}} = 1 + 2 = 3, equality holds if and only if x = frac{1}{x}, i.e., x = 1. Therefore, the minimum value of the function f(x) on (0, +infty) is 3; hence statement ① is correct. ② The domain of the function is (-infty, 0) cup (0, +infty), since f(1) = 1 + 1 + 1 = 3 and f(-1) = 1 - 1 - 1 = -1, thus f(-1) neq -f(1) and f(-1) neq f(1), the function f(x) is neither odd nor even, so statement ② is incorrect. ③ The derivative of the function f'(x) = 1 - frac{1}{x^2}. Setting f'(x) = 0 gives x = pm 1, since f'(x) > 0 for x in (-infty, -1) or (1, +infty), the intervals of monotonic increase for the function f(x) are (-infty, -1) and (1, +infty). Hence statement ③ is correct; Therefore, the correct statements are ① and ③, thus the answer is: boxed{C} By setting c=0 in condition (3), we obtain the expression for a ast b = ab + a + b, which leads to the expression for f(x). The problem can be solved by analyzing the function's odd-even nature, monotonicity, and properties of extremum. This problem introduces a new operation and examines properties such as the extremum, odd-even nature, and monotonicity of functions. Finding the expression for the function by setting c=0 is key to solving this problem. It is comprehensive and somewhat challenging.

answer:1. **Introduction to Key Concepts:** - Consider a triangle ( ABC ) where ( C ) and ( B ) have their respective foot heights intersecting circles. - We need to demonstrate that points ( M, N, P, Q ) are concyclic. 2. **Setup Orthocenter ( H ):** - Let ( H ) be the orthocenter of ( triangle ABC ). - Denote ( H_{A} ) as the foot of the altitude from ( A ) to ( BC ). 3. **Identification of Circles:** - We are dealing with two circles: - Circle with diameter ([AB]), - Circle with diameter ([AC]). 4. **Orthogonality and Intersections:** - From orthogonality, ( H_A ) lies on both circles due to the properties of perpendiculars drawn from corresponding vertices. - Specifically, ( H_A ) lies on: - The circle with diameter ([AB]), - The circle with diameter ([AC]). 5. **Power of Point Reasoning for Circles:** - By considering the power of point ( H_A ) with respect to the two circles: - The power of ( H_A ) relative to the circle with diameter ([AB]) is given by: [ H_M cdot H_N = HA cdot HH_A ] - Similarly, for the circle with diameter ([AC]): [ H_P cdot H_Q = HA cdot HH_A ] 6. **Equating the Products:** - By equating the respective power of points derived from the above conditions: [ H_M cdot H_N = H_A cdot HH_A ] and [ H_P cdot H_Q = H_A cdot HH_A ] - We find that: [ H_M cdot H_N = H_P cdot H_Q ] 7. **Conclusion Using Cyclic Points Theorem:** - According to the converse of the power of a point theorem, if four points ( M, N, P, Q ) satisfy this condition, then they must be concyclic. - Hence, ( M, N, P, Q ) are cocyclic. [ boxed{} ]

answer:To solve this, we shift the graph of y=cos2x to the right by frac{pi}{8} units. This gives us y=cos2(x- frac{pi}{8})=cos(2x- frac{pi}{4}). Therefore, the answer is: boxed{frac{pi}{8}}. This conclusion can be drawn by using the transformation rules of the graph of the function y=Acos(omega x+varphi). This question mainly examines the transformation rules of the graph of the function y=Acos(omega x+varphi), and is considered a basic question.