1. An equilateral triangle ABC is given. Points K and L are chosen on its sides AB and AC, respectively, such that|BK| = |AL|. Let P be the intersection of the segments BL and CK. Determine the ratio|AK| : |KB| if it is known that the segments AP and CK are perpendicular.
Solution: Let M be the point on BC with BK = AL = CM , and put P = BL∩ CK, Q = CK ∩ AM, R = AM ∩ BL. Then the rotation about the center of the triangle taking A to B to C to A takes P to Q to R to P , so P QR is equilateral. By assumption,
∠AP Q = 90◦, so ∠P AR = 30◦ = ∠RP A. Hence QR = RA, and likewise BP = P R. By two applications of Menelaos’ theorem,
QA
2. In a community of more than six people, each member exchanges letters with precisely three other members of the community. Prove that the community can be divided into two nonempty groups so that each member exchanges letters with at least two members of the group he belongs to.
Solution: Let n be the number of people. Consider a graph whose vertices correspond to the people, with edges between people who exchange letters. Since each vertex has degree greater than 1, a cycle must exist. Find a cycle of minimal length and let it have x people. Put the people in the cycle into group A and the others into group B. Note that each person in A corresponds with at least 2 other people in A.
If x ≥ 5, then A and B satisfy the condition of the problem. If a member of B exchanged letters with two people in A, he and at most x/2 + 1 people in A would form a cycle, contradicting our choice of the shortest cycle.
If x≤ (n + 1)/2, which in particular holds for x ≤ 4, the following algorithm produces satisfactory groups: as long as there exists a
person in B corresponding with at least two people in A, transfer him into A. It is clear that this gives what we need provided that B does not end up empty. However, in the original groups, there were at most x edges between A and B, and each transfer reduces this number by at least 1. So B ends up with at least n− 2x > 0 members, and the proof is complete.
3. Find all functions f : R→ R such that the equality f (f (x) + y) = f (x2− y) + 4f(x)y holds for all pairs of real numbers x, y.
Solution: Clearly, f (x) = x2 satisfies the functional equation.
Now assume that there is a nonzero value a such that f (a) 6= a2. Let y = (x2− f(x))/2 in the functional equation to find that
f (f (x)/2 + x2/2) = f (f (x)/2 + x2/2) + 2f (x)(x2− f(x)) or 0 = 2f (x)(x2 − f(x)). Thus, for each x either f(x) = 0 or f (x) = x2. In both cases, f (0) = 0.
Setting x = a from above, either f (a) = 0 or f (a) = 0 or f (a) = a2. The latter is false, so f (a) = 0. Now, let x = 0 and then x = a in the functional equation to find that
f (y) = f (−y), f (y) = f (a2− y)
and so f (y) = f (−y) = f(a2+ y); that is, the function is peri-odic with nonzero period a2. Let y = a2 in the original functional equation to obtain
f (f (x)) = f (f (x) + a2) = f (x2− a2) + 4a2f (x) = f (x2) + 4a2f (x).
However, putting y = 0 in the functional equation gives f (f (x)) = f (x2) for all x. Thus, 4a2f (x) = 0 for all x. Since a is nonzero, f (x) = 0 for all x. Therefore, either f (x) = x2 or f (x) = 0.
4. Is it possible to place 100 solid balls in space so that no two of them have a common interior point, and each of them touches at least one-third of the others?
Solution: Find a sphere S with minimum radius r, and form the sphere X with the same center but with three times the radius.
Any sphere with radius r and touching S is enclosed within X, so the intersection of X and any such sphere is (4/3)πr3. Any sphere with a radius larger than r and touching S completely encloses a sphere with radius r touching S at the same point; thus, the intersection of X and any sphere touching S is at least (4/3)πr3. Since the volume of X is 27 times this volume, at most 26 spheres can touch S, which is less than 33, one-third of the other spheres. Therefore, it is not possible to have such an arrangement.
5. Several integers are given (some of them may be equal) whose sum is equal to 1492. Decide whether the sum of their seventh powers can equal
(a) 1996;
(b) 1998.
Solution:
(a) Consider a set of 1492 1’s, 4 2’s, and 8 -1’s. Their sum is 1492, and the sum of their seventh powers is 1492(1) + 4(128) + 8(−1) = 1996.
(b) By Fermat’s Little Theorem, x7≡ x (mod 7). Thus, the sum of the numbers’ seventh powers must be congruent to the sum of the numbers, modulo 7. But 19986≡ 1492 (mod 7), so the numbers’ seventh powers cannot add up to 1998.
6. In a certain language there are only two letters, A and B. The words of this language obey the following rules:
(a) The only word of length 1 is A.
(b) A sequence of letters X1X2· · · XnXn+1, where Xi ∈ {A, B} for each i, is a word if and only if it contains at least one A but is not of the form X1X2· · · XnA where X1X2· · · Xn is a word.
Show that there are precisely 39951997 − 1 words which do not begin with AA and which are composed of 1998 A’s and 1998 B’s.
Solution: Any word with at least two A’s is either a sequence ending with a B or a word followed by two A’s (and any such is a word). Call these type-one words and type-two words, respectively.
Let f (x) be the number of x-letter words containing 1998 B’s and not starting with two A’s. Then f (1998) = 0 because any such words would have 0 A’s, violating condition (b). We find a formula for f (x + 2) for x≥ 1998. Note that there are at least two A’s in the word.
TYPE-ONE WORDS: Of the first x + 1 letters, 1997 are B’s. So, there are 1997x+1 type-one words with 1998 B’s. If the first two letters are A, then of the next x− 1 letters, 1997 are B’s. So, f(x + 2) only counts 1997x+1 − 1997x−1 type-one words.
TYPE-TWO WORDS: A type-two word is counted by f (x + 2) if and only if the first x letters form a word counted by f (x). So, there are f (x) type-two words counted by f (x + 2).
Therefore, for x≥ 1998,
f (x + 2) =x + 1 1997
−x − 1 1997
+ f (x).
Then
f (3996) =3995 1997
−3993 1997
+· · · +1999 1997
−1997 1997
+ f (1998).
Since 19971997
= 1 and f (1998) = 0, the sum telescopes to give f (3996) = 39951997 − 1.