1. For which real values of p does the equation
　　　 ?(x2 - p) + 2 ?(x2 - 1) = x have real roots? What are the roots?
 

 Solution　I must admit to having formed rather a dislike for this type of question which came up in almost every one of the early IMOs. Its sole purpose seems to be to teach you to be careful with one-way implications: the fact that a2 = b2 does not imply a = b.
The lhs is non-negative, so x must be non-negative. Moreover 2?(x2 - 1) <= x, so x <= 2/?3. Also ?(x2 - p) <= x, so p >= 0.
Squaring etc gives that any solution must satisfy x2 = (p - 4)2/(16 - 8p). We require x <= 2/?3 and hence (3p - 4)(p + 4) <= 0, so p <= 4/3.
Substituting back in the original equality we get |3p-4| + 2|p| = |p - 4|, which is indeed true for any p satisfying 0 <= p <= 4/3.
 

 
2. Given a point A and a segment BC, determine the locus of all points P in space for which angle APX = 90 for some X on the segment BC.
 

 Solution　Take the solid sphere on diameter AB, and the solid sphere on diameter AC. Then the locus is the points in one sphere but not the other (or on the surface of either sphere). Given P, consider the plane through P perpendicular to AP and the parallel planes through the other two points of intersection of AP with the two spheres (apart from A) which pass through B and C.
 

 
3. An n-gon has all angles equal and the lengths of consecutive sides satisfy a1 >= a2 >= ... >= an. Prove that all the sides are equal.
 

 Solution　For n odd consider the perpendicular distance of the shortest side from the opposite vertex. This is a sum of terms ai x cosine of some angle. We can go either way round. The angles are the same in both cases, so the inequalities give that a1 = an-1, and hence a1 = ai for all i < n. We get a1 = an by repeating the argument for the next shortest side. The case n even is easier, because we take a line through the vertex with sides a1 and an making equal angles with them and look at the perpendicular distance to the opposite vertex. This gives immediately that a1 = an.
 

 
4. Find all solutions x1, ... , x5 to the five equations xi + xi+2 = y xi+1 for i = 1, ... , 5, where subscripts are reduced by 5 if necessary.
 

 Solution　Successively eliminate variables to get x1(y - 2)(y2 + y - 1)2 = 0. We have the trivial solution xi = 0 for any y. For y = 2, we find xi = s for all i (where s is arbitary). Care is needed for the case y2 + y - 1 = 0, because after eliminating three variables the two remaining equations have a factor y2 + y - 1, and so they are automatically satisfied. In this case, we can take any two xi arbitary and still get a solution. For example, x1 = s, x2 = t, x3 = - s + yt, x4 = - ys - yt, x5 = ys - t.
 

 
5. Prove that cos pi/7 - cos 2pi/7 + cos 3pi/7 = 1/2.
 

 Solution　Consider the roots of x7 + 1 = 0. They are eipi/7, ei3pi/7, ... , ei13pi/7 and must have sum zero since there is no x6 term. Hence, in particular, their real parts sum to zero. But cos7pi/7 = - 1 and the others are equal in pairs, because cos(2pi - x) = cos x. So we get cos pi/7 + cos 3pi/7 + cos 5pi/7 = 1/2. Finally since cos(pi - x) = - cos x, cos 5pi/7 = - cos 2pi/7.
 

 
6. Five students A, B, C, D, E were placed 1 to 5 in a contest with no ties. One prediction was that the result would be the order A, B, C, D, E. But no student finished in the position predicted and to two students predicted to finish consecutively did so. For example, the outcome for C and D was not 1, 2 (respectively), or 2, 3, or 3, 4 or 4, 5. Another prediction was the order D, A, E, C, B. Exactly two students finished in the places predicted and two disjoint pairs predicted to finish consecutively did so. Determine the outcome.
 

 SolutionI cannot see an elegant solution. Start from the second prediction. The disjoint pairs can only be: DA, EC; DC, CB; or AE, CB. The additional requirement of just two correct places means that the only possibilities (in the light of the information about the second predicition) are: DABEC, DACBE, EDACB, AEDCB. The first is ruled out because AB are consecutive. The second is ruled out because C is in the correct place. The fourth is ruled out because A is in the correct place. This leaves EDACB, which is indeed a solution.
 

                        摘自：数学人