Solutions to 1994 TEAM PROBLEMS

T1. The sum of the grades in the class is 20 × 72.45 = 1449. This is also the sum of the 18 remaining students’ grades. Thus, their average is 1449/18 = 80.5.

T2. The Rational Root Test implies that the only “possible” rational roots are of the form a/b where a divides 1155 and b divides 1. Since 1155 = 3×5×7×11, this means that there are 32 possible rational roots (±1, ±3, ±5, ±7, ±11, ±15, ±21, …). Observe that all 32 of these numbers are odd. But if x is an odd number, then x^{8} − 24x^{7} − 18x^{5} + 39x^{2} + 1155 is odd and, therefore, not 0. It follows that there are no rational roots to the given equation. The answer is 0.

T3. Use the equation 2^{log_{2} x} = x to obtain

f(x) = x^{g(h(x))} = x^{1/ log_{2} x} = (2^{log_{2} x})^{1/ log_{2} x} = 2.

Thus, f(x) is 2 for all values of x, and the answer is 2.

T4. The answer is A = 999. The set S = 0, 1, 2, …, 997 satisfies (i) since it has 998 = A − 1 elements and since 1 ≤ a + b ≤ 996 + 997 = 1993 for any two elements a and b from this set S. To see that (ii) holds, suppose we have a set T consisting of A = 999 consecutive integers. We can write T in the form T = {u, u + 1, u + 2, …, u + 998} for some integer u. Observe that u + (u + j) = 2u + j for 1 ≤ j ≤ 998 so each of the numbers 2u + 1, 2u + 2, 2u + 3, …, 2u + 998 can be written in the form a + b for some distinct a and b from T. Also, (u + j) + (u + 998) = 2u + 998 + j for 1 ≤ j ≤ 997 so 2u + 999, 2u + 1000, 2u + 1001, …, 2u + 1995 can also be written in the form a + b for some distinct a and b from T. In other words, there are 1995 consecutive integers from 2u + 1 to 2u + 1995 each of which can be written in the form a + b for some distinct a and b from T. At least one of these must be divisible by 1994, so (ii) holds.

T5. An important observation here is that (a, b) is visible from the origin if and only if the greatest common divisor of a and b is 1. Take b = 2 × 3 × 5 × 7 = 210. Then each of 2, 3, 4, …, 8 has a common factor > 1 with b, so b satisfies the condition in the problem. (Note: 210 is the only answer to this problem.)

T6. Descartes’ Rule of Signs gives that there is exactly one positive real root and exactly one negative real root of the given polynomial. Observe that ±\sqrt{2} are roots of the given polynomial. Hence, the only real roots to the given polynomial are ±\sqrt{2}, and the answer is (\sqrt{2})^{2} + (−\sqrt{2})^{2} = 2 + 2 = 4.

T7. Observe that

−\sqrt{2501} + \sqrt{2502} = \frac{1}{\sqrt{2501} + \sqrt{2502}} < \frac{1}{50 + 50} = \frac{1}{100}.

Similarly, each of −\sqrt{2503} + \sqrt{2504}, −\sqrt{2505} + \sqrt{2506}, …, −\sqrt{2999} + \sqrt{3000} is < 1/100. We get that

\sqrt{2500} − \sqrt{2501}+\sqrt{2502} − \sqrt{2503} + ⋯ − \sqrt{2999} + \sqrt{3000}

< \sqrt{2500} + \frac{1}{100} + ⋯ + \frac{1}{100} = 50 + \frac{250}{100} = 52.5.

We claim the answer is 52. We need only show now that

\sqrt{2500} − \sqrt{2501} + \sqrt{2502} − \sqrt{2503} + ⋯ − \sqrt{2999} + \sqrt{3000} > 52.

In other words, we want

\sqrt{3000} > 52 − \sqrt{2500} + \sqrt{2501} − \sqrt{2502} + \sqrt{2503} − ⋯ + \sqrt{2999}.

An argument as above gives

52 − \sqrt{2500} + \sqrt{2501} − ⋯ + \sqrt{2999} < 52 + 250 (\frac{1}{100}) = 54.5,

so we need only show that \sqrt{3000} > 54.5 (this is correct – do you understand why?) which follows since 3000 > 2970.25 = 54.5^{2}.

T8. The number a_{2} is a power of 2. Observe that the last digits of 2, 2^{2}, 2^{3}, 2^{4}, … are 2, 4, 8, 6, 2, 4, 8, 6, . . .. In other words, 2^{4k+1}, 2^{4k+2}, 2^{4k+3}, 2^{4k+4} end with 2, 4, 8, 6, respectively, for each non-negative integer k. Since a_{2} = 2a^{3}, we want to know in which of the forms 4k+1, 4k+2, 4k+3, 4k+4 we can write a_{3}. Since a_{3} is a power of 3, we consider successive powers of 3. Note that 3, 3^{2}, 3^{3}, 3^{4}, … are in the form 4k+3, 4k+1, 4k+3, 4k+1, … (for different k’s). In other words, if 3 is raised to an odd power, then the result is of the form 4k + 3; and if 3 is raised to an even power, then the result is of the form 4k + 1. Since a_{3} = 3a^{4} and a_{4} = 4a^{5}, a_{3} is 3 raised to an even power and, hence, of the form 4k + 1 for some positive integer k. We get that the final digit of a_{2} = 2^{a_{3}} is 2.

T9. If two of the points are (x_{1}, y_{1}) and (x_{2}, y_{2}), then the slope of the line through the two points is (y_{2} − y_{1})/(x_{2} − x_{1}). Also,

y_{1} = 2x_{1}^{3} − 4x_{1} + 2, y_{2} = 2x_{2}^{3} − 4x_{2} + 2,
y_{1} = x_{1}^{3} + 2x_{1} − 1, and y_{2} = x_{2}^{3} + 2x_{2} − 1.

Taking the difference of the first two of these, we get that

y_{2} − y_{1} = 2(x_{2}^{3} − x_{1}^{3}) − 4(x_{2} − x_{1}) = (x_{2} − x_{1}) (2(x_{2}^{2} + x_{2}x_{1} + x_{1}^{2}) − 4) .

Hence, the slope of the line through the two points is

\frac{y_{2} − y_{1}}{x_{2} − x_{1}} = 2(x_{2}^{2} + x_{2}x_{1} + x_{1}^{2}) − 4.

Taking another difference of equations above gives

y_{2} − y_{1} = x_{2}^{3} − x_{1}^{3} + 2(x_{2} − x_{1}) = (x_{2} − x_{1}) (x_{2}^{2} + x_{2}x_{1} + x_{1}^{2} + 2) .

Hence, the slope of the line through the two points is also

\frac{y_{2} − y_{1}}{x_{2} − x_{1}} = x_{2}^{2} + x_{2}x_{1} + x_{1}^{2} + 2.

This means that

2(x_{2}^{2} + x_{2}x_{1} + x_{1}^{2}) − 4 = x_{2}^{2} + x_{2}x_{1} + x_{1}^{2} + 2

so that x_{2}^{2} + x_{2}x_{1} + x_{1}^{2} = 6. Substituting this into either of the two formulas for the slope above, we get that the slope is 8. (Comment: Since the answer does not depend on which two points we consider, we can conclude that the three points of intersection all lie on a straight line.)

T10. We use the notation P Q to represent the distance between two points P and Q. Let O represent the origin. Let E be a point on \overleftrightarrow{OA} such that \overleftrightarrow{CE} is perpendicular to \overleftrightarrow{OA}. Observe that ∠CEA = ∠DBA = 90^{∘}. One easily gets that ΔCEA is similar to ΔDBA. Hence, AB/AD = AE/AC so that AB × AC = AE × AD. The given information implies that AB × AC = 1 and AD = 2. Therefore, AE = 1/2 which implies OE = 1/2. It follows that the x−coordinate of C is 1/2. (Note that the measurements in the drawing are not correct. Also, 17 has nothing to do with this answer.)