AMC10 2006 B

Because $$(-1)^k = \begin{cases} 1, & \text{if } k \text{ is even,} \\ -1, & \text{if } k \text{ is odd,} \end{cases}$$ the sum can be written as $$(-1+1) + (-1+1) + \cdots + (-1+1) = 0 + 0 + \cdots + 0 = 0.$$

(A) Because $4\spadesuit 5=(4+5)(4-5)=-9$, it follows that $3\spadesuit(4\spadesuit 5)=3\spadesuit(-9)=(3+(-9))(3-(-9))=(-6)(12)=-72.$

(A) Let $c$ and $p$ represent the number of points scored by the Cougars and the Panthers, respectively. The two teams scored a total of 34 points, so $c+p=34$. The Cougars won by 14 points, so $c-p=14$. The solution is $c=24$ and $p=10$, so the Panthers scored 10 points.

The circle with diameter 3 has area $\pi (\frac{3}{2})^2$. The circle with diameter 1 has area $\pi (\frac{1}{2})^2$. Therefore the ratio of the blue-painted area to the red-painted area is $$\frac{\pi (\frac{3}{2})^2 - \pi (\frac{1}{2})^2}{\pi (\frac{1}{2})^2} = 8.$$

The side length of the square is at least equal to the sum of the smaller dimensions of the rectangles, which is $2 + 3 = 5$. If the rectangles are placed as shown, it is in fact possible to contain them within a square of side length 5. Thus the smallest possible area is $5^2 = 25$.

Since the square has side length $2/\pi$, the diameter of each circular section is $2/\pi$. The boundary of the region consists of 4 semicircles, whose total perimeter is twice the circumference of a circle having diameter $2/\pi$. Hence the perimeter of the region is $$2 \cdot \left( \pi \cdot \frac{2}{\pi} \right) = 4.$$

(A) We have $\sqrt{\dfrac{x}{1-\dfrac{x-1}{x}}}=\sqrt{\dfrac{x}{\dfrac{x-x+1}{x}}}=\sqrt{\dfrac{x}{\dfrac{1}{x}}}=\sqrt{x^2}=|x|.$ When $x<0$, the given expression is equivalent to $-x$.

The square has side length $\sqrt{40}$. Let $r$ be the radius of the semicircle. Then $$r^2 = (\sqrt{40})^2 + \left(\frac{\sqrt{40}}{2}\right)^2 = 40 + 10 = 50,$$ so the area of the semicircle is $\frac{1}{2}\pi r^2 = 25\pi$.

Francesca's 600 grams of lemonade contains $25 + 386 = 411$ calories, so 200 grams of her lemonade contains $411/3 = 137$ calories.

Let the sides of the triangle have lengths $x$, $3x$, and $15$. The Triangle Inequality implies that $3x < x + 15$, so $x < 7.5$. Because $x$ is an integer, the greatest possible perimeter is $7 + 21 + 15 = 43$.

Since $n!$ contains the product $2 \cdot 5 \cdot 10 = 100$ whenever $n \ge 10$, it suffices to determine the tens digit of $$7! + 8! + 9! = 7!(1 + 8 + 8 \cdot 9) = 5040(1 + 8 + 72) = 5040 \cdot 81.$$ This is the same as the units digit of $4 \cdot 1$, which is $4$.

Substituting $x = 1$ and $y = 2$ into the equations gives $$1 = \frac{2}{4} + a \quad \text{and} \quad 2 = \frac{1}{4} + b.$$ It follows that $$a + b = \left(1 - \frac{2}{4}\right) + \left(2 - \frac{1}{4}\right) = 3 - \frac{3}{4} = \frac{9}{4}.$$

Joe has 2 ounces of cream in his cup. JoAnn has drunk 2 ounces of the 14 ounces of coffee-cream mixture in her cup, so she has only $12/14 = 6/7$ of her 2 ounces of cream in her cup. Therefore the ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee is $$\frac{2}{\frac{6}{7} \cdot 2} = \frac{7}{6}.$$

Since $a$ and $b$ are roots of $x^2 - mx + 2 = 0$, we have $ab = 2$. In a similar manner, the constant term of $x^2 - px + q$ is the product of $a + (1/b)$ and $b + (1/a)$, so $$q = \left(a + \frac{1}{b}\right) \left(b + \frac{1}{a}\right) = ab + 1 + 1 + \frac{1}{ab} = 2 + 2 + \frac{1}{2} = \frac{9}{2}.$$

Since ∠BAD = 60°, isosceles △BAD is also equilateral. As a consequence, △AEB, △AED, △BED, △BFD, △BFC, and △CFD are congruent. These six triangles have equal areas and their union forms rhombus ABCD, so each has area 24/6 = 4. Rhombus BFDE is the union of △BED and △BFD, so its area is 8.

In the years from 2004 through 2020, Each Leap Day occurs $3 \cdot 365 + 366 = 1461$ days after the preceding Leap Day. When 1461 is divided by 7 the remainder is 5. So the day of the week advances 5 days for each 4-year cycle. In the four cycles from 2004 to 2020, the Leap Day will advance 20 days. So Leap Day in 2020 will occur one day of the week earlier than in 2004, that is, on a Saturday.

After Alice puts the ball into Bob's bag, his bag will contain six balls: two of one color and one of each of the other colors. After Bob selects a ball and places it into Alice's bag, the two bags will have the same contents if and only if Bob picked one of the two balls in his bag that are the same color. Because there are six balls in the bag when Bob makes his selection, the probability of selecting one of the same colored pair is $\frac{2}{6} = \frac{1}{3}$.

18. (E) Note that the first several terms of the sequence are: $2,\ 3,\ \frac{3}{2},\ \frac{1}{2},\ \frac{1}{3},\ \frac{2}{3},\ 2,\ 3,\ \ldots,$ so the sequence consists of a repeating cycle of 6 terms. Since $2006=334\cdot 6+2$, we have $a_{2006}=a_2=3$.

(A) Since $OC=1$ and $OE=2$, it follows that $\angle EOC=60^\circ$ and $\angle EOA=30^\circ$. The area of the shaded region is the area of the $30^\circ$ sector $DOE$ minus the area of congruent triangles $OBD$ and $OBE$. First note that $Area(\text{Sector }DOE)=\frac{1}{12}(4\pi)=\frac{\pi}{3}.$ In right triangle $OCE$, we have $CE=\sqrt{3}$, so $BE=\sqrt{3}-1$. Therefore $Area(\triangle OBE)=\frac{1}{2}(\sqrt{3}-1)(1).$ The required area is consequently $\frac{\pi}{3}-2\left(\frac{\sqrt{3}-1}{2}\right)=\frac{\pi}{3}+1-\sqrt{3}.$

(E) The slope of line $AB$ is $(178-(-22))/(2006-6)=1/10$. Since the line $AD$ is perpendicular to the line $AB$, its slope is $-10$. This implies that \[ -10=\frac{y-(-22)}{8-6}, \] so $y=-10(2)-22=-42$, and $D=(8,-42)$. As a consequence, \[ AB=\sqrt{2000^2+200^2}=200\sqrt{101}\ \text{ and }\ AD=\sqrt{2^2+20^2}=2\sqrt{101}. \] Thus \[ \text{Area}(ABCD)=AB\cdot AD=400\cdot 101=40,400. \]

(C) On each die the probability of rolling $k$, for $1 \le k \le 6$, is \[ \frac{k}{1+2+3+4+5+6}=\frac{k}{21}. \] There are six ways of rolling a total of 7 on the two dice, represented by the ordered pairs $(1,6)$, $(2,5)$, $(3,4)$, $(4,3)$, $(5,2)$, and $(6,1)$. Thus the probability of rolling a total of 7 is \[ \frac{1\cdot 6+2\cdot 5+3\cdot 4+4\cdot 3+5\cdot 2+6\cdot 1}{21^2} =\frac{56}{21^2} =\frac{8}{63}. \]

(D) The total cost of the peanut butter and jam is $N(4B+5J)=253$ cents, so $N$ and $4B+5J$ are factors of $253=11\cdot23$. Because $N>1$, the possible values of $N$ are $11$, $23$, and $253$. If $N=253$, then $4B+5J=1$, which is impossible since $B$ and $J$ are positive integers. If $N=23$, then $4B+5J=11$, which also has no solutions in positive integers. Hence $N=11$ and $4B+5J=23$, which has the unique positive integer solution $B=2$ and $J=3$. So the cost of the jam is $11(3)(5\text{¢})=\$1.65$.

gles be $R$ and $S$ as shown. Then the required area is $T = R + S$. Let $a$ and $b$, respectively, be the bases of the triangles with areas $R$ and $3$, as indicated. If two triangles have the same altitude, then the ratio of their areas is the same as the ratio of their bases. Thus $\dfrac{a}{b}=\dfrac{R}{3}=\dfrac{R+S+7}{3+7},\ \text{so}\ \dfrac{R}{3}=\dfrac{T+7}{10}.$ Similarly, $\dfrac{S}{7}=\dfrac{S+R+3}{7+7},\ \text{so}\ \dfrac{S}{7}=\dfrac{T+3}{14}.$ Thus $T=R+S=3\left(\dfrac{T+7}{10}\right)+7\left(\dfrac{T+3}{14}\right).$ From this we obtain $10T=3(T+7)+5(T+3)=8T+36,$ and it follows that $T=18$.

(B) Through $O$ draw a line parallel to $\overline{AD}$ intersecting $\overline{PD}$ at $F$. Then $AOFD$ is a rectangle and $OPF$ is a right triangle. Thus $DF=2$, $FP=2$, and $OF=4\sqrt{2}$. The area of trapezoid $AOPD$ is $12\sqrt{2}$, and the area of hexagon $AOBCPD$ is $2\cdot 12\sqrt{2}=24\sqrt{2}$.

(B) The 4-digit number on the license plate has the form $aabb$ or $abab$ or $baab$, where $a$ and $b$ are distinct integers from 0 to 9. Because Mr. Jones has a child of age 9, the number on the license plate is divisible by 9. Hence the sum of the digits, $2(a+b)$, is also divisible by 9. Because of the restriction on the digits $a$ and $b$, this implies that $a+b=9$. Moreover, since Mr. Jones must have either a 4-year-old or an 8-year-old, the license plate number is divisible by 4. These conditions narrow the possibilities for the number to 1188, 2772, 3636, 5544, 6336, 7272, and 9900. The last two digits of 9900 could not yield Mr. Jones’s age, and none of the others is divisible by 5, so he does not have a 5-year-old. Note that 5544 is divisible by each of the other eight non-zero digits.