answer:1. **Define Variables:** Let (x) be the number of points scored by the Eagles after the deduction, and (y) be the points scored by the Hawks. 2. **Set Up Equations:** [ x + y = 50 quad text{(total points)} ] [ (x + 4) - y = 12 quad text{(margin of victory before deduction)} ] 3. **Modify the second equation for the deduction:** [ x + 4 - y = 12 ] Simplifying, we have: [ x - y = 8 ] 4. **Add the equations to solve for (x):** [ (x + y) + (x - y) = 50 + 8 ] [ 2x = 58 quad Rightarrow quad x = 29 ] 5. **Substitute (x) back into the total points equation:** [ 29 + y = 50 quad Rightarrow quad y = 21 ] 6. **Conclusion:** The number of points scored by the Hawks is (21). boxed{The final answer is (boxed{C})}

answer:1. **Assign Variables**: Let the side length of the second square be ( s_2 ). The diagonal of this square, denoted as ( d_2 ), and the side length of the first square, ( s_1 ), are related by: [ s_1 = 2d_2 ] 2. **Calculate the Diagonal of the Second Square**: Using the Pythagorean theorem, the diagonal of the second square ( d_2 ) is: [ d_2 = s_2 sqrt{2} ] 3. **Express ( s_1 ) in terms of ( s_2 )**: Substitute ( d_2 ) in the expression for ( s_1 ): [ s_1 = 2(s_2 sqrt{2}) = 2s_2 sqrt{2} ] 4. **Calculate Areas**: - The area of the first square ( A_1 ) with side ( s_1 ) is: [ A_1 = s_1^2 = (2s_2 sqrt{2})^2 = 8s_2^2 ] - The area of the second square ( A_2 ) with side ( s_2 ) is: [ A_2 = s_2^2 ] 5. **Calculate the Ratio of the Areas**: [ text{Ratio} = frac{A_1}{A_2} = frac{8s_2^2}{s_2^2} = 8 ] 6. **Conclusion**: The ratio of the area of the first square to that of the second square is ( 8 ). The final answer is boxed{textbf{(B)} 8}

answer:**Step 1:** The problem is asking us to find the value of the real number m that makes the sets A and B equal. We are given that A=B. **Step 2:** Since A=B, we can write the elements of both sets as equal: 3=3m. **Step 3:** Solving for m, we find that m=frac{3}{3}=1. However, this does not satisfy the given condition A=B, since 3 neq 3cdot 1. **Step 4:** Re-examining the problem, we notice that there is another possible solution for m: m=0. This would make 3m=0, which is an element of both sets A and B. **Step 5:** With m=0, both sets become A=B={3,0,}, ensuring that A=B is true. Thus, the value of the real number m that satisfies A=B is boxed{m=0}.

answer:1. **Introduction of Conic (mathcal{P})**: We consider the conic (mathcal{P}) which is tangent to the lines ((AB)), ((BD)), ((DC)), ((CA)), and ((XY)). Our goal is to show that this conic is a parabola. 2. **Projective Transformation (mathcal{T})**: This conic (mathcal{P}) induces a projective transformation (mathcal{T}) from the line ((AB)) to the line ((DC)). Specifically, (mathcal{T}) sends: - The point (B) to (D), - The point (A) to (C), - The midpoint (M) of ([AB]) to the midpoint (N) of ([CD]). 3. **Tangency at Infinity**: Since the ratios (r_{ABM}) and (r_{DCN}) are preserved, the projective transformation (mathcal{T}) must send the point at infinity on line ((AB)) to the point at infinity on line ((CD)). This implies that the line at infinity is tangent to the conic (mathcal{P}). Hence, (mathcal{P}) is a parabola. 4. **Conclusion That (mathcal{P}) is a Parabola**: Therefore, we conclude: mathcal{P} text{ is a parabola.} 5. **Focal Properties of the Parabola**: Let (T) denote the focus of the parabola (mathcal{P}). It is known that the circumcircles of triangles whose sides are tangent to the parabola pass through its focus. 6. **Circumcircles of Specific Triangles**: Specifically, the circumcircles of the following triangles pass through the focus (T): - (triangle EXY): - (triangle FBD): - (triangle BMX): - (triangle YCN). 7. **Required Angle Equality**: To show tangency of the circumcircles of (triangle EXY) and (triangle FMN), it suffices to prove the angle equality: widehat{YTN} = widehat{MTX}. 8. **Angle Argument**: We analyze the following chain of equalities: [ begin{aligned} widehat{NTY} &= widehat{NCY} && text{(since (YNCT) is cyclic)} &= widehat{MXB} && text{(since (ABCD) is cyclic)} &= widehat{MTX} && text{(since (BMXT) is cyclic)}. end{aligned} ] 9. **Final Conclusion**: The above angle equality confirms the tangency of the circumcircles of (triangle EXY) and (triangle FMN). Therefore, we conclude that the circumcircles of triangles (EXY) and (FMN) are indeed tangent. boxed{}