answer:**Analysis of the problem**: We can analyze each option based on the conditions for determining a circle, the formation of the circumcenter of a triangle, and the properties of the perpendicular diameter theorem. A. A circle can be determined by three non-collinear points. It is essential to emphasize "non-collinear", so this option is incorrect; B. The circumcenter of a triangle is the intersection point of the perpendicular bisectors of any two sides. A point on the perpendicular bisector of a segment is equidistant from the two endpoints of the segment. Therefore, the circumcenter is equidistant from the three vertices of the triangle, making this option incorrect; C. The diameter that bisects a chord is perpendicular to the chord, which is correct; D. In the same or congruent circles, arcs corresponding to equal central angles are equal. It is necessary to emphasize "in the same or congruent circles", so this option is incorrect; Therefore, the correct answer is boxed{C}. **Key Points**: This question mainly examines the conditions for determining a circle, the properties of the circumcenter of a triangle, and the perpendicular diameter theorem.

answer:Given (P, Q) are non-zero polynomials with integer coefficients such that (operatorname{deg}(P) > operatorname{deg}(Q)). We are to show that at least one root of (P) is rational if the polynomial (p cdot P + Q) has a rational root for an infinite number of prime numbers (p). 1. **Identifying Rational Roots**: Let (p) be a prime number, and let (r_p) be a rational root of (p P + Q). Since ( operatorname{deg}(P) > operatorname{deg}(Q) ), the dominant term in (P) will eventually outweigh those in (Q) as (x) increases. 2. **Controlling the magnitude of the roots**: There exists an integer (M) such that for all (x) satisfying (|x| > M), we have [ |P(x)| > |Q(x)| ] Consequently, [ p |P(x)| > |Q(x)| ] Using the triangle inequality, we get [ |p P(x) - Q(x)| > 0 ] This implies that there are no roots for (|x| > M) when (p P(x) - Q(x) = 0). 3. **Bounding the roots**: Hence, the rational root (r_p) must satisfy (|r_p| leq M). 4. **Application of Bolzano-Weierstrass theorem**: By the Bolzano-Weierstrass theorem, there exists a subsequence ( left(r_{phi(p)}right) ) for some strictly increasing function (phi: mathbb{N} to mathbb{P}) such that ( left(r_{phi(p)}right) ) converges to a limit (r). We aim to show that (r) is rational. 5. **Assumptions and goal**: Suppose (r neq 0). We wish to show that the irreducible denominators of ( r_{phi(n)} ) are bounded. If their denominators are bounded by some positive integer (k), then the sequence ( k! r_{phi(n)} ) would be integer-valued and convergent, hence eventually constant. Therefore, ( left(r_{phi(n)} right)_{n geq 0} ) would be stationary and rational. 6. **Denominators not bounded case**: Assume the irreducible denominators of ( r_{phi(n)} ) are unbounded. Without loss of generality, assume the denominator tends to ( +infty ). 7. **Behavior of the denominators**: The denominator of (r_p) must divide (p times a) (where (a neq 0) is the leading coefficient of (P)). Since the denominator tends to (infty), it cannot divide the constant term (b) of (P) for large (n), hence it must be a multiple of (p). 8. **Expressing rational root**: Let (p = phi(n)). Denote ( s_p in mathbb{Z} ) such that [ r_p = frac{s_p}{p f_n} ] Here, ( p f_n ) is the denominator. Since ( r_p ) divides (f_n cdot (b p + c)), there exists ( t_p in mathbb{Z} ) such that [ s_p t_p = (b p + c) f_n ] 9. **Simplifying the expression**: Thus, [ t_{phi(n)} = frac{(b p + c) f_n}{s_p} = frac{(p b + c)}{p r_p} = frac{1}{r_p phi(n)} left( b + frac{c}{phi(n)} right) ] As (n to infty), [ t_{phi(n)} to frac{b}{r} ] Hence, the sequence (left(t_{phi(n)}right)) is an integer-valued convergent sequence, so it is stationary. Therefore, (frac{b}{r}) is an integer, implying (r) is rational. # Conclusion: At least one root of (P) must be rational. blacksquare

answer:1. **Label the Points:** Let the four points be ( A, B, C, ) and ( D ) such that no three of them are collinear. 2. **Consider the Line ( AB ):** Draw the line segment ( AB ). Since no three points are collinear, points ( C ) and ( D ) must lie on the same side of the line ( AB ). 3. **Construct Circle through ( A, B, ) and ( D ):** Construct the circle passing through points ( A, B, ) and ( D ). Denote this circle as ( (ADB) ). 4. **Analyze the Position of Point ( C ):** We need to show that point ( C ) is either on the circumference of the circle ( (ADB) ) or inside it. 5. **Compare Angles ( angle ACB ) and ( angle ADB ):** Suppose without loss of generality (WLOG) that ( angle ACB ge angle ADB ). This assumption is valid because we can always relabel the points if necessary. 6. **Position of Point ( C ):** - If ( angle ACB = angle ADB ), then point ( C ) lies on the circumference of the circle ( (ADB) ). - If ( angle ACB > angle ADB ), then point ( C ) lies inside the circle ( (ADB) ) because the angle subtended by the same chord ( AB ) at the circumference is smaller than the angle subtended at any point inside the circle. 7. **Conclusion:** Therefore, point ( C ) is either on the circumference of the circle ( (ADB) ) or inside it. [ blacksquare ]

answer:Let's assume B's salary is 100 units. If A's salary is 20% less than B's salary, then A's salary is 100 - 20% of 100. 20% of 100 = 0.20 * 100 = 20 units. So, A's salary = 100 - 20 = 80 units. Now, we want to find out by how much percentage B's salary is more than A's. The difference in their salaries is B's salary - A's salary = 100 - 80 = 20 units. To find out by how much percentage B's salary is more than A's, we calculate the percentage increase from A's salary to B's salary: Percentage increase = (Difference in salary / A's salary) * 100 Percentage increase = (20 / 80) * 100 Percentage increase = 0.25 * 100 Percentage increase = 25% So, B's salary is boxed{25%} more than A's salary.