SOLUTIONS TO TEAM PROBLEMS 02/95

T1. Let C_{1} be the circle of radius 3 and C_{2} the circle of radius 4. Let A be the center of C_{1}, and let B be the center of C_{2}. Let C be the intersection of ℓ and AB^{⟷}. Set x = AC and note BC = 25 − x. Then ΔAP C is similar to ΔBQC, and we get

\frac{x}{3} = \frac{25 − x}{4}.

Hence, AC = x = (3/7)25 = 75/7 and BC = 25 − x = (4/7)25 = 100/7. Therefore, the distance from P to Q is

\sqrt{(3 × 25/7)^{2} − 9} + \sqrt{(4 × 25/7)^{2} − 16}

= \frac{3}{7}\sqrt{25^{2} − 7^{2}} + \frac{4}{7}\sqrt{25^{2} − 7^{2}} = \sqrt{25^{2} − 7^{2}} = \sqrt{576} = 24.

T2. (1 + 2 + 4 + 8)(1 + 3 + 17 + 19)(1 + 7 + 31 + 61) = 15 × 40 × 100 = 60000

T3. (Use the Euclidean Algorithm.) Let d = gcd(106577, 1071089). Then d divides 1071089 − (10 × 106577) = 5319 and, hence, 106577 − (20 × 5319) = 197. One checks that 197 divides 5319 and therefore both 106577 and 1071089. The answer is 197.

T4. Completing the square twice, f(x, y) = (x − 3y)^{2} + (2y − 1)^{2} − 3. Thus, f(x, y) ≥ −3 for all pairs (x, y) and achieves this value when y = 1/2 and x = 3/2. The answer is (x, y) = (1/2, 3/2).

T5. The sum can be evaluated via

∑_{n=1}^{98} \frac{1}{n(n + 1)(n + 2)(n + 3)} = ∑_{n=1}^{98} \frac{1}{3} (\frac{1}{n(n + 1)(n + 2)} − \frac{1}{(n + 1)(n + 2)(n + 3)})

= \frac{1}{3} (\frac{1}{6} − \frac{1}{99 × 100 × 101}) .

Since 6 divides 99 × 100 × 101, the least common denominator of the difference above is 99×100×101 = 999900. It is easy to see that the resulting numerator will not be divisible by any of the prime divisors of 99×100×101 including 3. The answer is 3×999900 = 2999700.

T6. The answer is 6. After Dave and Michael choose 6, the value of S is 12. No matter what number k Dave chooses after this point, Michael can choose 7 − k. Every turn by both Dave and Michael will increase the sum by 7. After Michael has four more turns, S = 12 + (4 × 7) = 40, so Michael wins with this strategy. No other choice for Michael’s first turn will do (if Michael picks ℓ ∈ {1, 2, 3, 4, 5}, Dave can choose 6 − ℓ for his second move and continue by choosing 7 − k for each integer k Michael chooses; Dave will be able to win).

T7. It suffices to count the number of real roots of

f(x) = (x^{12} + x^{10} + x^{8} + x^{6} + x^{4} + x^{2} + 1) − (x^{5} + x^{3} + x)
= x^{12} + x^{10} + x^{8} + x^{6} − x^{5} + x^{4} − x^{3} + x^{2} − x + 1.

Clearly, f(0) ≠ 0. Since

f(−x) = x^{12} + x^{10} + x^{8} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1,

Descartes Rule of Signs gives that f(x) has no negative roots. Observe that

(x + 1)f(x) = (x + 1) (x^{12} + x^{10} + x^{8} + x^{6} − x^{5} + x^{4} − x^{3} + x^{2} − x + 1)
= x^{13} + x^{12} + x^{11} + x^{10} + x^{9} + x^{8} + x^{7} + 1.

By Descartes Rule of Signs, (x + 1)f(x) and, hence, f(x) has no positive real roots. The answer is 0.

T8. Consider the circle C′ centered at D = (0, −3) and passing through A and B. The measure of angle ∠ADB is π/2, from which it follows that for any point P on the smallest arc (along C′) from A to B, the measure of angle ∠AP B is 3π/4. Similarly, any point P on the smallest arc from A to B along a circle centered at (0, 3) is such that the measure of angle ∠AP B is 3π/4. Let R be the region strictly between these two arcs. Then P is a point in R if and only if the measure of ∠AP B is > 3π/4. The answer is therefore the ratio of the area of C minus the area of R to the area of C. The area of C is 9π. Since DB = 3\sqrt{2} and the measure ∠ADB = π/2, the area of R is

2 × (\frac{1}{2} (\frac{π}{2} ) (3\sqrt{2})^{2} − \frac{1}{2}(3\sqrt{2})^{2}) = 18 (\frac{π}{2} − 1) = 9π − 18.

After a little simplification, we deduce that the probability is 2/π.

T9. Observe that \binom{m}{k} = \binom{m}{m−k}. Thus,

(x+1)^{50}(x+1)^{50} = (∑_{k=0}^{50} \binom{50}{k}x^{k}) (∑_{k=0}^{50} \binom{50}{k}x^{k}) = (∑_{k=0}^{50} \binom{50}{k}x^{k}) (∑_{k=0}^{50} \binom{50}{k}x^{50−k}) .

The coefficient of x^{50} in this last product (when expanded) is

\binom{50}{0}^{2} + \binom{50}{1}^{2} + \binom{50}{2}^{2} + ⋯ + \binom{50}{50}^{2}.

On the other hand, (x + 1)^{50}(x + 1)^{50} = (x + 1)^{100}, so the coefficient of x^{50} must be \binom{100}{50}. An answer is a = 100 and b = 50.

T10. Let t be the positive integer for which 2^{t} divides 1995! and 2^{t+1} does not divide 1995!. So we may write 1995! = 2^{t}m where m is an odd integer. Let

S = ∑_{j=1}^{1995} \frac{1995!}{j}.

Here, S is a sum of integers. Observe that the term 1995!/1024 = 2^{t−10}m is divisible by 2^{t−10} and not by 2^{t−9}. Also, since no other denominator in the sum S above is divisible by 1024, every other term is divisible by 2^{t−9}. Write

S − \frac{1995!}{1024} = 2^{t−9}k,

where k is an integer. Then

\frac{a}{b} = \frac{1}{1995!}S = \frac{1}{2^{t}m} (2^{t−9}k + 2^{t−10}m) = \frac{1}{2^{t}m}2^{t−10} (2k + m) = \frac{2k + m}{2^{10}m}.

Hence,

2^{10}ma = b(2k + m).

Since m is odd, so is 2k + m, and we deduce that b is even. Only one of a or b is even, so a is odd. The above equation implies therefore that r = 10.