Solutions to the Contest Problems of IMO 1965
4.7 Solutions to the Contest Problems of IMO 1965
1. Let us set S =
1 − sin2x
2 =2−2 1 − sin 2 2x =
2. Thus the righthand inequality holds for all x .
2 − 2|cos2x| ≤ 2, implying S ≤
It remains to investigate the left-hand inequality. If π /2 ≤ x ≤ 3 π /2, then cos x ≤ 0 and the inequality trivially holds. Assume now that cosx > 0. Then
≤S 2 = 2 − 2|cos2x|, which is equivalent to cos 2x ≤ 0, i.e., to x ∈ [ π /4, π /2] ∪ [3 π /2, 7 π /4]. Hence the so-
the inequality is equivalent to 2 + 2 cos2x = 4 cos 2 x
lution set is π /4 ≤ x ≤ 7 π /4.
2. Suppose that (x 1 ,x 2 ,x 3 ) is a solution. We may assume w.l.o.g. that |x 1 | ≥ |x 2 |≥ |x 3 |. Suppose that |x 1 | > 0. From the first equation we obtain that
0 = |x 1 |· 11 +a 2 12 3 +a 13 1 | · (a 11 − |a 12 | − |a 13 |) > 0,
which is a contradiction. Hence |x 1 | = 0 and consequently x 1 =x 2 =x 3 = 0.
3. Let d denote the distance between the lines AB and CD. Being parallel to AB and CD, the plane π intersects the faces of the tetrahedron in a parallelogram EFGH . Let X ∈ AB be a point such that HX k DB. Clearly V AEHBFG =V AXEH +V XEHBFG .
D Let MN be the common perpendic- ular to lines AB and CD (M ∈ AB,
N N ∈ CD) and let MN,BN meet the
plane π at Q and R respectively. Then it holds that BR /RN = MQ/QN = k
GQ
C and consequently AX /XB = AE/EC =
A E R AH /HD = BF/FC = BG/GD = k.
Now we have V AXEH /V ABCD =k 3 /(k +
1 ) 3 . Furthermore, if h = 3V ABCD /S ABC B is the height of ABCD from D, then
V XEHBFG = S
2 XBFE k +1 (k + 1) 2 −1−k 2 2k
h and
S XBFE =S ABC −S AXE −S EFC = = . (k + 1) 2 (1 + k) 2
These relations give us V XEHBFG /V ABCD = 3k 2 /(1 + k) 3 . Finally,
V AEHBFG
k 3 + 3k 2
3 V . ABCD (k + 1)
Similarly, V CEFDHG /V ABCD = (3k + 1)/(k + 1) 3 , and hence the required ratio is (k 3 + 3k 2 )/(3k + 1).
358 4 Solutions
4. It is easy to see that all x i are nonzero. Let x 1 x 2 x 3 x 4 = p. The given system of equations can be rewritten as x i + p/x i = 2, i = 1, 2, 3, 4. The equation x+ p/x = 2 has at most two real solutions, say y and z. Then each x i is equal either to y or to z . There are three cases:
(i) x 1 =x 2 =x 3 =x 4 = y. Then y + y 3 = 2 and hence y = 1. (ii) x 1 =x 2 =x 3 = y, x 4 = z. Then z + y 3 =y+y 2 z = 2. It is easy to obtain that the only possibilities for (y, z) are (−1,3) and (1,1). (iii) x 1 =x 2 = y, x 3 =x 4 . In this case the only possibility is y = z = 1. Hence the solutions for (x 1 ,x 2 ,x 3 ,x 4 ) are (1, 1, 1, 1), (−1,−1,−1,3), and the cyclic permutations.
5. (a) Let A ′ and B ′ denote the feet of the perpendiculars from A and B to OB and OA respectively. We claim that H ∈A ′ B ′ . Indeed, since MPHQ is a parallelogram, we have B ′ P /B ′ A = BM/BA = MQ/AA ′ = PH/AA ′ , which implies by Thales’s theorem that H ∈A ′ B ′ . It is easy to see that the locus
of H is the whole segment A ′ B ′ .
(b) In this case the locus of points H is obviously the interior of the triangle OA ′ B ′ .
6. We recall the simple statement that every two diameters of a set must have a common point.
Consider any point B that is an endpoint of k ≥ 2 diameters BC 1 , BC 2 , . . . , BC k . We may assume w.l.o.g. that all the points C 1 , . . . ,C k lie on the arc C 1 C k , whose center is B and measure does not exceed 60 ◦ . We observe that for 1 < i < k any diameter with the endpoint C i has to intersect both the diameters C 1 B and C k B . Hence C i B is the only diameter with an endpoint at C i if i = 2, . . . , k − 1. In other words, with each point that is an endpoint of k ≥ 2 we can associate k − 2 points that are endpoints of exactly one diameter. We now assume that each A i is an endpoint of exactly k i ≥ 0 diameters, and that
k 1 ,...,k s ≥ 2, while k s +1 ,...,k n ≤ 1. The total number D of diameters satisfies the inequality 2D ≤k 1 +k 2 + ··· + k s + (n − s). On the other hand, by the above consideration we have (k 1 − 2) + ··· + (k s − 2) ≤ n − s, i.e., k 1 + ··· + k s ≤ n + s. Hence 2D ≤ (n + s) + (n − s) = 2n, which proves the result.
4.8 Contest Problems 1966 359