IMO 1963
Problem 1 :
For which real values of p does the equation
v(x^2 - p) + 2 v(x^2 - 1) = x have real roots? What are the roots?
Solution 1 :
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 a^2 = b^2 does not imply a = b.
The lhs is non-negative, so x must be non-negative. Moreover 2v(x^2 - 1) <= x, so x <= 2/v3. Also v(x^2
- p) <= x, so p >= 0.
Squaring etc gives that any solution must satisfy x^2 = (p - 4)^2/(16 - 8p). We require x <= 2/v3 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.
Problem 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 2 :
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.
Problem 3 :
An n-gon has all angles equal and the lengths of consecutive sides satisfy . Prove that all the sides are
equal.
Solution 3 :
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 , and hence for all i < n. We get 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 .
Problem 4 :
Find all solutions to the five equations
necessary.
for i = 1, ... , 5, where subscripts are reduced by 5 if
Solution 4 :
Successively eliminate variables to get (y - 2)( + y - 1)^2 = 0. We have the trivial solution = 0 for any y.
For y = 2, we find = s for all i (where s is arbitrary). Care is needed for the case y^2 + y - 1 = 0,
because after eliminating three variables the two remaining equations have a factor y^2 + y - 1, and so
they are automatically satisfied. In this case, we can take any two arbitrary and still get a solution. For
example, x_1 = s, x_2 = t, x_3 = - s + yt, x_4 = - ys - yt, x_5 = ys - t.
Problem 5 :
Prove that cos pi/7 - cos 2pi/7 + cos 3pi/7 = 1/2.
Solution 5 :
Consider the roots of x^7 + 1 = 0. They are e^i/7, e^i3/7, ... , e^i13/7 and must have sum zero since there
is no x^6 term. Hence, in particular, their real parts sum to zero. But cos7/7 = - 1 and the others are equal
in pairs, because cos(2 - x) = cos x. So we get cos/7 + cos3/7 + cos 5/7 = 1/2. Finally since cos( - x) = cos x, cos 5/7 = - cos 2/7.
Problem 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.
Solution 6 :
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.