1. Show that among any four points contained in a unit circle, there exist two whose distance is at most√
2.
Solution: If one of the four points lies at the center O of the circle, the statement is trivial. Otherwise, label the points P1, P2, P3, P4
We will use Iverson’s bracket convention: if P is a statement, [P ] is 1 if P is true, 0 if P is false. Note thatb√
kc is equal to the number of positive integers whose squares are at most k, sob√
kc =Pa
Now Pn
k=1[j2 ≤ k] counts the number of k ∈ {1, . . . , n} such that k≥ j2; when j≤ a, j2≤ n, so this number is n + 1 − j2. So
n
X
k=1
b√ kc =
a
X
j=1
n + 1− j2= (n + 1)a− a(a + 1)(2a + 1)/6.
4. [Corrected] Let C be a circle touching the edges of an angle ∠XOY , and let C1 be a circle touching the same edges and passing through the center of C. Let A be the second endpoint of the diameter of C1passing through the center of C, and let B be the intersection of this diameter with C. Prove that the circle centered at A passing through B touches the edges of ∠XOY .
Solution: Let T and T1 be the centers of C and C1, and let r and r1 be their radii. Drop perpendiculars T T0, T1T10, and AA0 to OX; then T T0 = r and T1T10 = r1. But T1 is the midpoint of AT , so T1T10 = (AA0+ T T0)/2; therefore AA0= 2T1T10− T T0 = 2r1− r.
Also AB = AT − BT = 2r1− r, so the circle centered at A with radius AB touches OX at A0. Similarly, this circle touches OY . 5. Find all integers x, y, z satisfying x2+ y2+ z2− 2xyz = 0.
Solution: The only solution of this equation is x = y = z = 0.
First, note that x, y, and z cannot all be odd, as then x2+ y2+ z2− 2xyz would be odd and therefore non-zero. Therefore 2 divides xyz.
But then x2+ y2+ z2= 2xyz is divisible by 4; since all squares are 0 or 1 (mod 4), x, y, and z must all be even. Write x = 2x1, y = 2y1, z = 2z1; then we have 4x21+ 4y21+ 4z21= 16x1y1z1, or x21+ y12+ z21= 4x1y1z1. Since the right-hand side is divisible by 4, x1, y1, and z1 must again be even, so we can write x1= 2x2, y1= 2y2, z1 = 2z2; plugging this in and manipulating we obtain x22+ y22+ z22= 8x2y2z2. In general, if n≥ 1, x2n+ yn2+ zn2= 2n+1xnynznimplies that xn, yn, znare all even, so we can write xn= 2xn+1, yn= 2yn+1, zn= 2zn+1, which satisfy x2n+1+ yn+12 + zn+12 = 2n+2xn+1yn+1zn+1; repeating this argument gives us an infinite sequence of integers (x1, x2, . . .) in which xi = 2xi+1. But then x = 2nxn, so 2n divides x for every n≥ 1; therefore we must have x = 0. Similarly y = z = 0.
Note: The substitution x = yz− w reduces this problem to USAMO 76/3.
6. Find the smallest integer k such that there exist two sequences{ai}, {bi} (i = 1, . . . , k) such that
Solution: The smallest such integer is 1997.
Suppose{ai}, {bi} are two sequences satisfying these four conditions with k ≤ 1996. The second condition tells us that a1 6= b1, so assume without loss of generality that a1< b1. By the first condition, there are 0 ≤ m < n such that a1 = 1996m, b1 = 1996n. Since bi ≥ b1 for all 1 ≤ i ≤ k (by the third condition) and each bi is a power of 1996,Pk
i=1biis divisible by 1996n. Therefore by the fourth conditionPk
i=1ai is divisible by 1996n; letting t denote the number of j’s for which aj = 1996m, we have t· 1996m ≡ Pk of An, and find the sum of all products of two distinct elements of An.
Solution: First we prove a lemma: for any n≥ 1,
β1
The proof is by induction on n; if n = 1, both sides are the set we must have|j| < 2n. Therefore the set on the left is contained in the set on the right. Now if j is odd and|j| < 2n, either (j− 1)/2 or (j + 1)/2 is odd, since these are consecutive integers; let j0 denote the odd one. Then|j0| ≤ (|j| + 1)/2 ≤ 2n−1, as|j| ≤ 2n− 1; since j0 and the proof of the lemma is complete.
We can now conclude from this lemma that
An= all products of two distinct elements of An, we use the formula
X
X
a∈An
a =|An| = 2n.
Now if X, Y are any finite sets of real numbers with P
x∈Xx =
can easily be proven by induction on m. So
X
by considering even n and odd n separately in the last step. So X
8. In an acute triangle ABC with AB6= AC, let V be the intersection of the angle bisector of A with BC, and let D be the foot of the perpendicular from A to BC. If E and F are the intersections of the circumcircle of AV D with CA and AB, respectively, show that the lines AD, BE, CF concur.
Solution: Since ∠ADV = π/2 and A, D, V , E, F are concyclic,
∠BF V = ∠CEV = π/2. Therefore triangles BF V and BDA are similar and triangles CEV and CDA are similar, so BD/BF = AB/V B and CD/CE = AC/V C. But AB/V B = AC/V C by the Angle Bisector theorem, so BD/BF = CD/CE. Also, since
∠F AV = ∠V AE, AE = AF . Therefore BD
DC CE EA
AF F B =BD
BF
CD CE = 1 and AD, BE, CF concur by Ceva’s Theorem.
9. [Corrected] A word is a sequence of 8 digits, each equal to 0 or 1.
Let x and y be two words differing in exactly three places. Show that the number of words differing from each of x and y in at least five places is 38.
Solution: Assume without loss of generality that x = 00000000, y = 00000111. Then a word z differs from each of x and y in at least five places if and only if a + b≥ 5 and a + (3 − b) ≥ 5, where a is the number of 1’s among the first five digits of z and b is the number of 1’s among the last three digits of z. Adding these in-equalities gives 2a ≥ 7, so we must have a ≥ 4; the solutions are (4, 1), (4, 2), and (5, b) for b∈ {0, 1, 2, 3} . The first two solutions give 54 3
1 + 32 = 30 words for z, and the others give 23 = 8, so there are 38 words differing from each of x and y in at least five places.
10. Find all pairs of functions f, g : R→ R such that (a) if x < y, then f (x) < f (y);
(b) for all x, y∈ R, f(xy) = g(y)f(x) + f(y).
Solution: The pairs (f, g) given by
f (x) = t(1− g(x)), g(x) = xm, x≥ 0
−|x|m, x < 0 , where t < 0, m > 0 are the only solutions.
Letting x = 0 in item (b) gives f (0) = f (0)g(y) + f (y), so f (y) = f (0)(1− g(y)). Let t = f(0); then f(y) = t(1 − g(y)). Since f is increasing, we cannot have t = 0. Substituting this formula for f in item (b) gives t(1−g(xy)) = g(y)t(1−g(x))+t(1−g(y)); rearranging we get 1− g(xy) = g(y)(1 − g(x)) + 1 − g(y) = 1 − g(x)g(y), or
g(xy) = g(x)g(y) for all x, y∈ R.
Since g = 1− f/t, g is strictly monotone, so g(1) 6= 0; but g(1) = g(1)2, so we must have g(1) = 1. Therefore g is increasing, so as f is increasing, we must have t < 0. So g(x) > 0 for x >
0; define h : R → R by h = log ◦ g ◦ exp. Then h(x + y) = log g(ex+y) = log(g(ex)g(ey)) = log g(ex) + log g(ey) = h(x) + h(y), h(0) = log g(e0) = 0, and h is strictly increasing. Also h(x + y) = h(x) + h(y) implies h(nx) = nh(x) for n ∈ N and h(−x) = −h(x), so h(αx) = αh(x) for α ∈ Q. Taking sequences of rationals xi ap-proaching x from below and yiapproaching x from above, and using monotonicity, shows that h(x) = xh(1) for all x; let m = h(1). Then we must have m > 0, as h is increasing. So g(x) = xmfor all x > 0.
Now g(−1) < 0, but (g(−1))2= g(1) = 1, so g(−1) = −1; therefore g(−x) = −g(x), so
g(x) = xm, x≥ 0
−|x|m, x < 0 .
Also we have f (x) = t(1− g(x)). It is easy to check that this pair (f, g) is a solution for any m > 0 > t.
11. [Corrected] Let a1, . . . , an be positive numbers, and define A = a1+· · · + an
n G = (a1· · · an)1/n
H = n
a−11 +· · · + a−1n
.
(a) If n is even, show that HA ≤ −1 + 2(AG)n. (b) If n is odd, show that HA ≤ −n−2n +2(nn−1)(AG)n.
Solution: Note that Gn
H = a1· · · an(a−11 +· · · a−1n ) n
= 1
n
n
X
j=1
a1· · · an
aj ≤
1 n
n
X
j=1
aj
n−1
= An−1
by Maclaurin’s inequality, so HA ≤ (AG)n. Since A≥ G, AGn
≥ 1, so HA ≤ (AG)n ≤ −1 + 2(AG)n, proving part (a). For part (b), HA ≤ (GA)n ≤ (AG)n+n−2n ((AG)n− 1) = −n−2n +2(nn−1)(AG)n.