AMC10 2008 B

The number of points could be any integer between $5 \cdot 2 = 10$ and $5 \cdot 3 = 15$, inclusive. The number of possibilities is $15 - 10 + 1 = 6$.

The two sums are $1 + 10 + 17 + 22 = 50$ and $4 + 9 + 16 + 25 = 54$, so the positive difference between the sums is $54 - 50 = 4$.

The properties of exponents imply that $$ \sqrt[3]{x\sqrt{x}} = (x \cdot x^{1/2})^{1/3} = (x^{3/2})^{1/3} = x^{1/2}. $$

A single player can receive the largest possible salary only when the other 20 players on the team are each receiving the minimum salary of $15,000$. Thus the maximum salary for any player is $700,000 - 20 \cdot 15,000 = 400,000$.

Answer (A): Note that $(y-x)^2=(x-y)^2$, so $$(x-y)^2\$(y-x)^2=(x-y)^2\$(x-y)^2=\left((x-y)^2-(x-y)^2\right)^2=0^2=0.$$

Answer (C): Because $AB + BD = AD$ and $AB = 4BD$, it follows that $BD = \frac{1}{5}\cdot AD$. By similar reasoning, $CD = \frac{1}{10}\cdot AD$. Thus \[ BC = BD - CD = \frac{1}{5}\cdot AD - \frac{1}{10}\cdot AD = \frac{1}{10}\cdot AD. \]

The side length of the large triangle is 10 times the side length of each small triangle, so the area of the large triangle is $10^2 = 100$ times the area of each small triangle.

The total number of dollars spent is the even number 50, so the number of roses purchased must also be even. In addition, the number of roses purchased cannot exceed $\frac{50}{3}$. Therefore the number of roses purchased must be one of the even integers between 0 and 16, inclusive. This gives 9 possibilities for the number of bouquets.

Answer (A): The quadratic formula implies that the two solutions are $x_1=\dfrac{2a+\sqrt{4a^2-4ab}}{2a}$ and $x_2=\dfrac{2a-\sqrt{4a^2-4ab}}{2a}$, so the average is $\dfrac{1}{2}(x_1+x_2)=\dfrac{1}{2}\left(\dfrac{2a}{2a}+\dfrac{2a}{2a}\right)=1.$

Answer (A): Let $O$ be the center of the circle, and let $D$ be the intersection of $OC$ and $AB$. Because $OC$ bisects minor arc $AB$, $OD$ is a perpendicular bisector of chord $AB$. Hence $AD=3$, and applying the Pythagorean Theorem to $\triangle ADO$ yields $OD=\sqrt{5^2-3^2}=4$. Therefore $DC=1$, and applying the Pythagorean Theorem to $\triangle ADC$ yields $AC=\sqrt{3^2+1^2}=\sqrt{10}$.

Answer (B): Note that $u_5 = 2u_4 + 9$ and $128 = u_6 = 2u_5 + u_4 = 5u_4 + 18$. Thus $u_4 = 22$, and it follows that $u_5 = 2 \cdot 22 + 9 = 53$.

Answer (A): During the year Pete takes $44\times 10^{5}+5\times 10^{4}=44.5\times 10^{5}$ steps. At 1800 steps per mile, the number of miles Pete walks is $\dfrac{44.5\times 10^{5}}{18\times 10^{2}}=\dfrac{44.5}{18}\times 10^{3}\approx 2.5\times 10^{3}=2500.$

Because the mean of the first $n$ terms is $n$, their sum is $n^2$. Therefore the $n$th term is $n^2 - (n-1)^2 = 2n - 1$, and the 2008th term is $2 \cdot 2008 - 1 = 4015$.

Answer (B): Because $\triangle OAB$ is a $30-60-90^\circ$ triangle, we have $BA=\frac{5\sqrt{3}}{3}$. Let $A'$ and $B'$ be the images of $A$ and $B$, respectively, under the rotation. Then $B'=(0,5)$, $B'A'$ is horizontal, and $B'A'=BA=\frac{5\sqrt{3}}{3}$. Hence $A'$ is in the second quadrant and $$ A'=\left(-\frac{5}{3}\sqrt{3},5\right). $$

Answer (A): By the Pythagorean Theorem we have $a^2+b^2=(b+1)^2$, so $a^2=(b+1)^2-b^2=2b+1$. Because $b$ is an integer with $b<100$, $a^2$ is an odd perfect square between 1 and 201, and there are six of these, namely, 9, 25, 49, 81, 121, and 169. Hence $a$ must be 3, 5, 7, 9, 11, or 13, and there are 6 triangles that satisfy the given conditions.

Answer (A): If one die is rolled, 3 of the 6 possible numbers are odd. If two dice are rolled, 18 of the 36 possible outcomes have odd sums. In each of these cases, the probability of an odd sum is $\frac{1}{2}$. If no die is rolled, the sum is 0, which is not odd. The probability that no die is rolled is equal to the probability that both coin tosses are tails, which is $\left(\frac{1}{2}\right)^2=\frac{1}{4}$. Thus the requested probability is \[ \left(1-\frac{1}{4}\right)\cdot\frac{1}{2}=\frac{3}{8}. \]

Answer (B): The responses on these three occasions, in order, must be YNN, NYN, or NNY, where Y indicates approval and N indicates disapproval. The probability of each of these is $(0.7)(0.3)(0.3)=0.063$, so the requested probability is $3(0.063)=0.189$.

Answer (B): Let $n$ be the number of bricks in the chimney. Then the number of bricks per hour Brenda and Brandon can lay working alone is $\frac{n}{9}$ and $\frac{n}{10}$, respectively. Working together they can lay $\left(\frac{n}{9}+\frac{n}{10}-10\right)$ bricks in an hour, or $5\left(\frac{n}{9}+\frac{n}{10}-10\right)$ bricks in 5 hours to complete the chimney. Thus $5\left(\frac{n}{9}+\frac{n}{10}-10\right)=n,$ and the number of bricks in the chimney is $n=900$.

Answer (E): The portion of each end of the tank that is under water is a circular sector with two right triangles removed as shown. The hypotenuse of each triangle is 4, and the vertical leg is 2, so each is a $30-60-90^\circ$ triangle. Therefore the sector has a central angle of $120^\circ$, and the area of the sector is $$\frac{120}{360}\cdot \pi(4)^2=\frac{16}{3}\pi.$$ The area of each triangle is $\frac12(2)(2\sqrt3)$, so the portion of each end that is underwater has area $\frac{16}{3}\pi-4\sqrt3$. The length of the cylinder is 9, so the volume of the water is $$9\left(\frac{16}{3}\pi-4\sqrt3\right)=48\pi-36\sqrt3.$$

Answer (B): Of the 36 possible outcomes, the four pairs (1, 4), (2, 3), (2, 3), and (4, 1) yield a sum of 5. The six pairs (1, 6), (2, 5), (2, 5), (3, 4), (3, 4), and (4, 3) yield a sum of 7. The four pairs (1, 8), (3, 6), (3, 6), and (4, 5) yield a sum of 9. Thus the probability of getting a sum of 5, 7, or 9 is $(4+6+4)/36 = 7/18$. Note: The dice described here are known as Sicherman dice. The probability of obtaining each sum between 2 and 12 is the same as that on a pair of standard dice.

Answer (C): Let the women be seated first. The first woman may sit in any of the 10 chairs. Because men and women must alternate, the number of choices for the remaining women is 4, 3, 2, and 1. Thus the number of possible seating arrangements for the women is $10\cdot 4! = 240$. Without loss of generality, suppose that a woman sits in chair 1. Then this woman's spouse must sit in chair 4 or chair 8. If he sits in chair 4 then the women sitting in chairs 7, 3, 9, and 5 must have their spouses sitting in chairs 10, 6, 2, and 8, respectively. If he sits in chair 8 then the women sitting in chairs 5, 9, 3, and 7 must have their spouses sitting in chairs 2, 6, 10, and 4, respectively. So for each possible seating arrangement for the women there are two arrangements for the men. Hence, there are $2\cdot 240 = 480$ possible seating arrangements.

Answer (C): There are $6!/(3!2!1!)=60$ distinguishable orders of the beads on the line. To meet the required condition, the red beads must be placed in one of four configurations: positions 1, 3, and 5, positions 2, 4, and 6, positions 1, 3, and 6, or positions 1, 4, and 6. In the first two cases, the blue bead can be placed in any of the three remaining positions. In the last two cases, the blue bead can be placed in either of the two adjacent remaining positions. In each case, the placement of the white beads is then determined. Hence there are $2\cdot 3+2\cdot 2=10$ orders that meet the required condition, and the requested probability is $\frac{10}{60}=\frac{1}{6}$.

Answer (B): Because the area of the border is half the area of the floor, the same is true of the painted rectangle. The painted rectangle measures $a-2$ by $b-2$ feet. Hence $ab=2(a-2)(b-2)$, from which $0=ab-4a-4b+8$. Add 8 to each side of the equation to produce $$8=ab-4a-4b+16=(a-4)(b-4).$$ Because the only integer factorizations of 8 are $$8=1\cdot 8=2\cdot 4=(-4)\cdot(-2)=(-8)\cdot(-1),$$ and because $b>a>0$, the only possible ordered pairs satisfying this equation for $(a-4,b-4)$ are $(1,8)$ and $(2,4)$. Hence $(a,b)$ must be one of the two ordered pairs $(5,12)$, or $(6,8)$.

Answer (C): Let $M$ be on the same side of line $BC$ as $A$, such that $\triangle BMC$ is equilateral. Then $\triangle ABM$ and $\triangle MCD$ are isosceles with $\angle ABM=10^\circ$ and $\angle MCD=110^\circ$. Hence $\angle AMB=85^\circ$ and $\angle CMD=35^\circ$. Therefore $\angle AMD=360^\circ-\angle AMB-\angle BMC-\angle CMD$ $=360^\circ-85^\circ-60^\circ-35^\circ=180^\circ.$ It follows that $M$ lies on $AD$ and $\angle BAD=\angle BAM=85^\circ$.

Answer (B): Number the pails consecutively so that Michael is presently at pail 0 and the garbage truck is at pail 1. Michael takes $200/5 = 40$ seconds to walk between pails, so for $n \ge 0$ he passes pail $n$ after $40n$ seconds. The truck takes 20 seconds to travel between pails and stops for 30 seconds at each pail. Thus for $n \ge 1$ it leaves pail $n$ after $50(n-1)$ seconds, and for $n \ge 2$ it arrives at pail $n$ after $50(n-1) - 30$ seconds. Michael will meet the truck at pail $n$ if and only if $50(n-1) - 30 \le 40n \le 50(n-1)$ or, equivalently, $5 \le n \le 8$. Hence Michael first meets the truck at pail 5 after 200 seconds, just as the truck leaves the pail. He passes the truck at pail 6 after 240 seconds and at pail 7 after 280 seconds. Finally, Michael meets the truck just as it arrives at pail 8 after 320 seconds. These conditions imply that the truck is ahead of Michael between pails 5 and 6 and that Michael is ahead of the truck between pails 7 and 8. However, the truck must pass Michael at some point between pails 6 and 7, so they meet a total of five times.