Solution　Evidently the parallelogram is a red herring, since the circles cover it iff and only if the three circles center A, B, D cover the triangle ABD.
The three circles radius x and centers the three vertices cover an acute-angled triangle ABC iff x is at least R, the circumradius. The circumcenter O is a distance R from each vertex, so the condition is clearly necessary. If the perpendiculars from O to AB, BC, CA are OP, OQ, OR, then the circle center A, radius R covers the quadrilateral APOR, the circle center B, radius R covers the quadrilateral BPOQ, and the circle center C radius R covers the quadrilateral CQOR, so the condition is also sufficient.
We need an expression for R in terms of a and A. We can express BD two ways: 2R sin A, and Ö(a² + 1 - 2a cos A). So a necessary and sufficient condition for the covering is 4 sin²A >= (a² + 1 - 2a cos A), which reduces to a <= cos A + Ö3 sin A, since cos A <= a (the foot of the perpendicular from D onto AB must lie between A and B).

Solution　Let the tetrahedron be ABCD and assume that all edges except AB have length at most 1. The volume is the 1/3 x area BCD x height of A above BCD. The height is at most the height of A above CD, so we maximise the volume by taking the triangles ACD and BCD to be perpendicular. Then the volume is 1/6 x CD x altitude from A to CD x altitude from B to CD. If AC or AD is less than 1, then we can increase the altitude from A to CD whilst keeping CD fixed by taking AC = AD = 1. Similarly for the altitude from B to CD. So if CD = z, the volume is z(1 - z²/4)/6. This is slightly awkward to maximise without using calculus. Factorise as: z(1 - z/2)(1 + z/2)/6 <= z(1 - z/2)(1 + 1/2)/6 = z/4 - z²/8 = 1/8 - (z - 1)²/8 <= 1/8. Checking back, we find that z = 1 gives z(1 - z²/4)/6 = 1/8, so that is indeed the maximum value.

Solution　The key is that c_a - c_b = (a - b)(a + b + 1). Hence the product (c_m+1 - c_k)(c_m+2 - c_k) ... (c_m+n - c_k) is the product of the n consecutive numbers (m - k + 1), ... , (m - k + n), times the product of the n consecutive numbers (m + k + 2), ... , (m + k + n + 1). The first product is just the binomial coefficient (m-k+n)Cn times n!, so it is divisible by n!. The second product is 1/(m + k + 1) x (m + k + 1)(m + k + 2) ... (m + k + n + 1) = 1/(m + k + 1) x (m+k+n+1)C(n+1) x (n+1)!. But m + k + 1 is a prime greater than n + 1, so it has no factors in common with (n+1)!, hence the second product is divisible by (n+1)!. Finally note that c₁c₂ ... c_n= n! (n+1)!.

Solution　Take any triangle similar to A₁B₁C₁ and circumscribing A₀B₀C₀. For example, take an arbitary line through A₀ and then lines through B₀ and C₀ at the appropriate angles to the first line. Label the triangle‘s vertices X, Y, Z so that A₀ lies on YZ, B₀ on ZX, and C₀ on XY. Now any circumscribed ABC (labeled with the same convention) must have C on the circle through A₀, B₀ and Z, because it has angle C = angle Z = angle C₁. Similarly it must have B on the circle through C₀, A₀ and Y, and it must have A on the circle through B₀, C₀ and X.
Consider the side AB. It passes through C₀. Its length is twice the projection of the line joining the centers of the two circles onto AB (because each center projects onto the midpoint of the part of AB that is a chord of its circle). But this projection is maximum when it is parallel to the line joining the two centers. The area is maximised when AB is maximised (because all the triangles are similar), so we take AB parallel to the line joining the centers. [Note, in passing, that this proves that the other sides must also be parallel to the lines joining the respective centers and hence that the three centers form a triangle similar to A₁B₁C₁.]

Solution　Take |a₁| >= |a₂| >= ... >= |a₈|. Suppose that |a₁|, ... , |a_r| are all equal and greater than |a_r+1|. Then for sufficiently large n, we can ensure that |a_s|ⁿ < 1/8 |a₁|ⁿ for s > r, and hence the sum of |a_s|ⁿ for all s > r is less than |a₁|ⁿ. Hence r must be even with half of a₁, ... , a_r positive and half negative.
If that does not exhaust the a_i, then in a similar way there must be an even number of a_i with the next largest value of |a_i|, with half positive and half negative, and so on. Thus we find that c_n = 0 for all odd n.

Solution Let the number of medals remaining at the start of day r be m_r. Then m₁ = m, and 6(m_k - k)/7 = m_k+1 for k < n with m_n = n.
After a little rearrangement, we find that m = 1 + 2(7/6) + 3(7/6)² + ... + n(7/6)^n-1. Summing, we get m = 36(1 - (n + 1)(7/6)ⁿ + n (7/6)ⁿ⁺¹) = 36 + (n - 6)7ⁿ/6^n-1. 6 and 7 are coprime, so 6^n-1 must divide n - 6. But 6^n-1 > n - 6, so n = 6 and m = 36.