answer:Let's denote the number of times each team faces another team as "n". Since there are 50 teams, each team will play (50 - 1) = 49 other teams. Now, when team A plays team B, that's one game. But we must remember that this also counts as team B playing team A. So each matchup is counted twice in the total number of games. The total number of games played in the season is given as 4900. So, we can set up the equation: Number of games each team plays * Number of teams / 2 = Total number of games (n * 49) * 50 / 2 = 4900 Now, we solve for n: n * 49 * 50 / 2 = 4900 n * 2450 = 4900 n = 4900 / 2450 n = 2 So, each team faces another team boxed{2} times during the season.

answer:Since 1+x+x^2+x^3=0, then x+x^2+x^3+ldots+x^{2004} = x(1+x+x^2+x^3) + x^5(1+x+x^2+x^3) + x^9(1+x+x^2+x^3) + ldots + x^{1997}(1+x+x^2+x^3) + x^{2001}(1+x+x^2+x^3), = (1+x+x^2+x^3)(x+x^5+x^9+x^{12}+ldots+x^{1997}+x^{2001}), = 0. Therefore, the answer is boxed{0}.

answer:1. Consider the triangle ABC with point D inside the triangle such that the angle angle BDC = angle BAC + 90^circ. 2. We need to show that the circumcircles of triangles ABD and ACD are orthogonal, which means that the tangents to these circles at their common points are perpendicular to each other. 3. Let P and Q be the points where the tangents to these circles at point D intersect the side BC of the triangle. The points P and Q are such that P lies on the tangent from A to the circumcircle of ABD, and Q lies on the tangent from A to the circumcircle of ACD. 4. By the properties of circle tangents and similar triangles, the angles formed by these tangents can be expressed using the angles at point D: [ angle QDP = angle BDC - (angle BDQ + angle CDP) ] 5. Note that because BDQ and CDP are tangents, the following relations hold: [ angle BDQ = angle BAD quad text{and} quad angle CDP = angle CAD ] Therefore, we can rewrite angle QDP as: [ angle QDP = angle BDC - (angle BAD + angle CAD) ] 6. Since angle BDC = angle BAC + 90^circ and angle BAC is the sum of angle BAD and angle CAD, we have: [ angle BAC = angle BAD + angle CAD ] Substituting this in, we get: [ angle QDP = (angle BAC + 90^circ) - (angle BAD + angle CAD) ] Simplifying the right-hand side, we obtain: [ angle QDP = 90^circ ] 7. This shows that the tangents to the circumcircles at D are indeed perpendicular to each other. Thus, the circumcircles of triangles ABD and ACD intersect orthogonally. # Conclusion: boxed{}

answer:1. **Initial Setup**: - Set ( x = 1 ) and ( y = 1 + g(1) ), yielding: [ g(g(1+g(1)) - 1) = 1 + g(1) - g(1) = 1 ]. - Let ( c = g(1+g(1)) - 1 ), then ( g(c) = 1 ). 2. **Further Simplification**: - Set ( y = c ), implying: [ g(xg(c) - x) = cx - g(x) ]. - With ( g(c) = 1 ), it simplifies to: [ g(x - x) = cx - g(x) Rightarrow g(0) = cx - g(x) ]. - Let ( d = g(0) ), then ( g(x) = cx - d ). 3. **Verification and Calculation**: - Substitute ( g(x) = cx - d ) into the original equation: [ g(x(cx - d) - x) = xy - (cx - d) ]. - Simplify and equate coefficients to find ( c ) and ( d ): [ g(x(c^2 - 1) - d) = xy - cx + d ]. - Equating coefficients, ( c^2 - 1 = 0 ) and ( -d = -d ), gives ( c = 1 ) or ( c = -1 ), with ( d = 0 ). - Thus, ( g(x) = x ) or ( g(x) = -x ). - Therefore, ( m = 2 ) and ( t = -2 + 2 = 0 ). Hence, ( m times t = boxed{0} ).