answer:In this problem, we are given a triangle (ABC) where (angle B = 2angle C) and a point (D) on the ray (BA) such that (AC = BD). We need to prove that (AB + BC > CD). 1. Let's denote (angle ACB = alpha). Then, (angle ABC = 2alpha). 2. The sum of the angles in triangle (ABC) is: [ angle BAC + angle ABC + angle ACB = 180^circ ] Substituting the known angles, we get: [ angle BAC + 2alpha + alpha = 180^circ ] This simplifies to: [ angle BAC = 180^circ - 3alpha ] 3. Let (K) be a point on the bisector of (angle B), such that (BK = BC). Therefore, we place segment (BK) equal in length to segment (BC). 4. Triangles (ACB) and (DBK) are congruent by the Side-Angle-Side (SAS) criterion: - (AC = BD) by construction, - (BC = BK) by construction, - (angle ACB = angle DBK = alpha), since (BK) is on the bisector of (angle ABC). Consequently, we have (AB = DK). 5. Now, consider the isosceles triangle (BKC). Here, (angle BKC = alpha) (isosceles triangles have equal base angles). 6. We observe that in the isosceles triangle (BKC), (angle BKC = alpha) is the smallest angle. Hence, it follows that the side opposite to (angle BKC) is the smallest. This implies: [ BC > CK ] 7. From the congruence and inequality above, we add (BC) to both sides of the equation (AB = DK): [ AB + BC = DK + BC ] Since (BC > CK), we have: [ DK + BC > DK + CK ] Therefore: [ DK + BC > DC quad (text{because } DK + CK = DC) ] 8. Substituting (AB = DK) back, it results: [ AB + BC > DC ] Thus, we have demonstrated that: [ AB + BC > CD ] [boxed{}]

answer:First, we compute the square of the given complex number: z=(2-i)^{2}=(2^2-2times2times i+i^2)=4-4i-1=3-4i Next, we observe the signs of the real and imaginary parts of z. The real part is positive (+3), and the imaginary part is negative (-4i). Therefore, the corresponding point (3,-4) lies in the fourth quadrant of the complex plane. Hence, the answer is: boxed{D. text{Fourth quadrant}}

answer:To find the total number of bales of hay in the barn at the end of the week, we need to add the initial number of bales to the total number of bales stacked each day. Initial bales: 28 Day 1: 10 bales Day 2: 15 bales Day 3: 8 bales Day 4: 20 bales Day 5: 12 bales Day 6: 4 bales Day 7: 18 bales Total bales stacked during the week = 10 + 15 + 8 + 20 + 12 + 4 + 18 Total bales stacked during the week = 87 Now, add the initial bales to the total bales stacked: Total bales in the barn = Initial bales + Total bales stacked Total bales in the barn = 28 + 87 Total bales in the barn = 115 At the end of the week, there are boxed{115} bales of hay in the barn.

answer:1. **Double Counting the Triplets:** We want to count the number of ordered triplets of points ((X, Y, Z)) from (S) such that (XY = YZ). This condition implies that (Y) is equidistant from (X) and (Z). 2. **Fixing Two Random Points and Finding Possible Third Points:** Fix (X neq Z) (which can be done in (n(n-1)) ways as there are (n) points and choosing 2 different points from (n) points). The point (Y) must lie on the perpendicular bisector of the segment ([XZ]). Since no three points in (S) are collinear, any line in the plane intersects the set (S) in at most two points. Thus, there are at most 2 choices for (Y). 3. **Bound on the Total Triplets:** Hence, the total number of such triplets ((X, Y, Z)) where (Y) is the midpoint is at most (2n(n-1)). 4. **Using the Given Hypothesis:** For each point (P) in (S), there is a radius (r) such that at least (k) points in (S) are at distance (r) from (P). For a fixed point (Y), there are at least (k) points (X) and at least (k) points (Z) (at distance (r) from (Y), including (P)). 5. **Counting the Required Triplets:** Given (Y), there are at least (k(k-1)) pairs ((X, Z)) such that both (X) and (Z) are exactly distance (r) from (Y). So, we have at least (n cdot k(k-1)) such triplets. 6. **Inequality Relating the Two Counts:** Since we counted triplets both ways (the loose upper bound of triplets and the guaranteed quantity from the hypothesis): [ n cdot k(k-1) leq 2n(n-1) ] 7. **Simplifying the Inequality:** Divide both sides by (n) (assuming (n neq 0)): [ k(k-1) leq 2(n-1) ] 8. **Solving the Quadratic Inequality:** To find (k), rewrite the inequality: [ k^2 - k leq 2(n-1) ] Completing the square on the left side: [ k^2 - k + frac{1}{4} leq 2(n-1) + frac{1}{4} ] Simplifying: [ left(k - frac{1}{2}right)^2 leq 2n - frac{3}{2} ] Thus, [ left(k - frac{1}{2}right)^2 < 2n ] 9. **Taking the Square Root and Final Bound:** Taking the square root on both sides gives: [ k - frac{1}{2} < sqrt{2n} ] Therefore, [ k < frac{1}{2} + sqrt{2n} ] # Conclusion: Finally, we have shown: [ boxed{k < frac{1}{2} + sqrt{2n}} ]