AMC10 2012 B

There are $18-2 = 16$ more students than rabbits per classroom.\nAltogether there are $4 \cdot 16 = 64$ more students than rabbits.

The width of the rectangle is the diameter of the circle, so the\nwidth is $2 \cdot 5 = 10$. The length of the rectangle is $2 \cdot 10 = 20$. Therefore the area\nof the rectangle is $10 \cdot 20 = 200$.

The given point is 12 units above the horizontal line $y = 2000$. The reflected point will be 12 units below the line, and 24 units below the given point. The coordinates of the reflected point are (1000, 1988).

Answer (A): The 7 marbles left over will fill one more bag of 6 marbles leaving 1 marble left over.

Tax is 10% and tip is 15%, so her total cost is 100% + 10% +\n15% = 125% of her meal. Thus her meal costs $\frac{27.50}{1.25} = \$22$.

Consider $x$ and $y$ as points on the real number line, with $x$\nnecessarily to the right of $y$.\nThen $x - y$ is the distance between $x$ and $y$.\nXiaoli’s rounding moved $x$ to the right and moved $y$ to the left. Therefore the\ndistance between them increased, and her estimate is larger than $x - y$.\n\nTo see that the other answer choices are not correct, let $x = 2.9$ and $y = 2.1$,\nand round each by 0.1. Then $x - y = 0.8$ and Xiaoli’s estimated difference is\n$(2.9 + 0.1) - (2.1 - 0.1) = 1.0$.

Let $h$ be the number of holes dug by the chipmunk. Then the\nchipmunk hid $3h$ acorns, while the squirrel hid $4(h - 4)$ acorns. Since they hid\nthe same number of acorns, $3h = 4(h - 4)$. Solving gives $h = 16$. Thus the\nchipmunk hid $3 \cdot 16 = 48$ acorns.

Answer (B): If $x-2>0$, then the given inequality is equivalent to $1<x-2<5$, or $3<x<7$. The integer solutions in this case are $4,5,$ and $6$. If $x-2<0$, then the given inequality is equivalent to $-5<x-2<-1$, or $-3<x<1$. The integer solutions in this case are $-2,-1,$ and $0$. The sum of all integer solutions is $12$.

The sum of two integers is even if they are both even or both\nodd. The sum of two integers is odd if one is even and one is odd. Only the\nmiddle two integers have an odd sum, namely $41 - 26 = 15$. Hence at least one\ninteger must be even. A list satisfying the given conditions in which there is\nonly one even integer is 1, 25, 1, 14, 1, 15.

Multiplying the given equation by $6N$ gives $MN = 36$. The\ndivisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Each of these divisors can be\npaired with a divisor to make a product of 36. Hence there are 9 ordered pairs\n$(M, N)$.

Answer (A): There are 3 choices for Saturday (anything except cake) and for the same reason 3 choices for Thursday. Similarly there are 3 choices for Wednesday, Tuesday, Monday, and Sunday (anything except what was to be served the following day). Therefore there are $3^6=729$ possible dessert menus.

. Answer (B): Note that $\angle ABC = 90^\circ$, so $\triangle ABC$ is a $45$-$45$-$90^\circ$ triangle. Because hypotenuse $AC = 10\sqrt{2}$, the legs of $\triangle ABC$ have length $10$. Therefore $AB = 10$ and $BD = BC + CD = 10 + 20 = 30$. By the Pythagorean Theorem, $AD = \sqrt{10^2 + 30^2} = \sqrt{1000}.$ Because $31^2 = 961$ and $32^2 = 1024$, it follows that $31 < AD < 32$.

Answer (B): Let $x$ be Clea's rate of walking and $r$ be the rate of the moving escalator. Because the distance is constant, $24(x+r)=60x$. Solving for $r$ yields $r=\frac{3}{2}x$. Let $t$ be the time required for Clea to make the escalator trip while just standing on it. Then $rt=60x$, so $\frac{3}{2}xt=60x$. Therefore $t=40$ seconds.

Answer (D): Construct the altitude for one of the equilateral triangles to its base on the square. Label the vertices of one of the resulting $30-60-90^\circ$ triangles $A$, $B$, and $C$, as shown. Then $AB=\sqrt{3}$ and $BC=3$. Label one of the intersection points of the two equilateral triangles $D$ and the center of the square $E$. Then $\triangle CDE$ is a $30-60-90^\circ$ triangle, $CE=3-\sqrt{3}$, and $DE=\frac{3-\sqrt{3}}{\sqrt{3}}=\sqrt{3}-1$. The area of $\triangle CDE$ is $\frac{1}{2}\cdot(3-\sqrt{3})\cdot(\sqrt{3}-1)=2\sqrt{3}-3$. Hence the area of the rhombus is $4\cdot(2\sqrt{3}-3)=8\sqrt{3}-12$.

A total of 15 games are played, so all 6 teams could not be\ntied for the most wins as this would require $15/6 = 2.5$ wins per team. However,\nit is possible for 5 teams to be tied, each with 3 wins and 2 losses. One such\noutcome can be constructed by labeling 5 of the teams A, B, C, D, and E, and\nplacing these labels at distinct points on a circle. If each of these teams beat\nthe 2 labeled teams clockwise from its respective labeled point, as well as the\nremaining unlabeled team, all 5 would tie with 3 wins and 2 losses.

Connect the centers of the three circles to form an equilateral\ntriangle with side length 4. Then the requested area is equal to the area of this\ntriangle and a $300^\circ$ sector of each circle. The equilateral triangle has base 4 and\naltitude $2\sqrt{3}$, so its area is\n\n$\frac{1}{2} \cdot 4 \cdot 2\sqrt{3} = 4\sqrt{3}$.\n\nThe area of each sector is $\frac{300}{360} \cdot \pi \cdot 2^2 = \frac{10\pi}{3}$. Hence the total area is $3 \cdot \frac{10\pi}{3} + 4\sqrt{3} = 10\pi + 4\sqrt{3}$.

Answer (C): Each sector forms a cone with slant height 12. The circumference of the base of the smaller cone is \$\frac{120}{360}\cdot 2\cdot 12\cdot \pi=8\pi\$. Hence the radius of the base of the smaller cone is 4 and its height is \$\sqrt{12^2-4^2}=8\sqrt{2}\$. Similarly, the circumference of the base of the larger cone is \$16\pi\$. Hence the radius of the base of the larger cone is 8 and its height is \$4\sqrt{5}\$. The ratio of the volume of the smaller cone to the volume of larger cone is \[ \frac{\frac{1}{3}\pi\cdot 4^2\cdot 8\sqrt{2}}{\frac{1}{3}\pi\cdot 8^2\cdot 4\sqrt{5}}=\frac{\sqrt{10}}{10}. \]

On average for every 500 people tested, 1 will test positive because he or she has the disease, while 2% · 499 ≈ 10 will test positive even though they do not have the disease. In other words, of approximately 11 people who test positive, only 1 has the disease, so the probability is approximately$\frac{1}{11}$.

Answer (C): Because $\triangle EBF$ is similar to $\triangle EAD$, it follows that $\dfrac{BF}{AD}=\dfrac{BE}{AE}$, or $\dfrac{BF}{30}=\dfrac{2}{8}$, giving $BF=\dfrac{15}{2}$. The area of trapezoid $BFDG$ is $\dfrac{1}{2}h(b_1+b_2)=\dfrac{1}{2}\cdot AB\cdot(BF+GD)=\dfrac{1}{2}\cdot 6\cdot\left(\dfrac{15}{2}+15\right)=\dfrac{135}{2}$.

Answer (A): The smallest initial number for which Bernardo wins after one round is the smallest integer solution of $2n + 50 \ge 1000$, which is 475. The smallest initial number for which he wins after two rounds is the smallest integer solution of $2n + 50 \ge 475$, which is 213. Similarly, the smallest initial numbers for which he wins after three and four rounds are 82 and 16, respectively. There is no initial number for which Bernardo wins after more than four rounds. Thus $N = 16$, and the sum of the digits of $N$ is 7.

Answer (A): Since 4 of the 6 segments have length $a$, some 3 of the points (call them $A$, $B$, and $C$) must form an equilateral triangle of side length $a$. The fourth point $D$ must be a distance $a$ from one of $A$, $B$, or $C$, and without loss of generality it can be assumed to be $A$. Thus $D$ lies on a circle of radius $a$ centered at $A$. The distance from $D$ to one of the other 2 points (which can be assumed to be $B$) is $2a$, so $BD$ is a diameter of this circle and therefore is the hypotenuse of right triangle $DCB$ with legs of lengths $a$ and $b$. Thus $b^2=(2a)^2-a^2=3a^2$, and the ratio of $b$ to $a$ is $\sqrt{3}$.

Answer (B): If $a_1=1$, then the list must be an increasing sequence. Otherwise let $k=a_1$. Then the numbers 1 through $k-1$ must appear in increasing order from right to left, and the numbers from $k$ through 10 must appear in increasing order from left to right. For $2\le k\le 10$ there are $\binom{9}{k-1}$ ways to choose positions in the list for the numbers from 1 through $k-1$, and the positions of the remaining numbers are then determined. The number of lists is therefore $$ 1+\sum_{k=2}^{10}\binom{9}{k-1}=\sum_{k=0}^{9}\binom{9}{k}=2^9=512. $$ OR If $a_{10}$ is not 1 or 10, then numbers larger than $a_{10}$ must appear in reverse order in the list, and numbers smaller than $a_{10}$ must appear in order. However, 1 and 10 cannot both appear first in the list, so the placement of either 1 or 10 would violate the given conditions. Hence $a_{10}=1$ or 10. By similar reasoning, when reading the list from right to left the number that appears next must be the smallest or largest unused integer from 1 to 10. This gives 2 choices for each term until there is one number left. Hence there are $2^9=512$ choices.

Answer (D): The discarded tetrahedron can be viewed as having an isosceles right triangle of side 1 as its base, with an altitude of 1. Therefore its volume is $\frac{1}{6}$. It can also be viewed as having an equilateral triangle of side length $\sqrt{2}$ as its base, in which case its altitude $h$ must satisfy $\frac{1}{3}\cdot\frac{\sqrt{3}}{4}\cdot(\sqrt{2})^2\cdot h=\frac{1}{6},$ which implies that $h=\frac{\sqrt{3}}{3}$. The height of the remaining solid is the long diagonal of the cube minus $h$, which is $\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$.

Answer (B): There are two cases to consider. Case 1 Each song is liked by two of the girls. Then one of the three pairs of girls likes one of the six possible pairs of songs, one of the remaining pairs of girls likes one of the remaining two songs, and the last pair of girls likes the last song. This case can occur in $3\cdot 6\cdot 2=36$ ways. Case 2 Three songs are each liked by a different pair of girls, and the fourth song is liked by at most one girl. There are $4!=24$ ways to assign the songs to these four categories, and the last song can be liked by Amy, Beth, Jo, or no one. This case can occur in $24\cdot 4=96$ ways. The total number of possibilities is $96+36=132$.

Answer (E): Label the columns having arrows as $c_1, c_2, c_3, \ldots, c_7$ according to the figure. Call those segments that can be traveled only from left to right forward segments. Call the segments $s_1$, $s_2$, and $s_3$, in columns $c_2$, $c_4$, and $c_6$, respectively, which can be traveled only from right to left, back segments. Denote $S$ as the set of back segments traveled for a path. First suppose that $S=\emptyset$. Because it is not possible to travel a segment more than once, it follows that the path is uniquely determined by choosing one forward segment in each of the columns $c_j$. There are $2, 2, 4, 4, 4, 2,$ and $2$ choices for the forward segment in columns $c_1, c_2, c_3, c_4, c_5, c_6,$ and $c_7$, respectively. This gives a total of $2^{10}$ total paths in this case. Next suppose that $S=\{s_1\}$. The two forward segments in $c_2$, together with $s_1$, need to be part of the path, and once the forward segment from $c_1$ is chosen, the order in which the segments of $c_2$ are traveled is determined. Moreover, there are only $2$ choices for possible segments in $c_3$ depending on the last segment traveled in $c_2$, either the bottom $2$ or the top $2$. For the rest of the columns, the path is determined by choosing any forward segment. Thus the total number of paths in this case is $2\cdot 1\cdot 2\cdot 4\cdot 4\cdot 2\cdot 2=2^8$, and by symmetry this is also the total for the number of paths when $S=\{s_3\}$. A similar argument gives $2\cdot 1\cdot 2\cdot 4\cdot 2\cdot 1\cdot 2=2^6$ trips for the case when $S=\{s_1,s_3\}$. Suppose $S=\{s_2\}$. Because $s_2$ is traveled, it follows that $2$ forward segments in $c_4$ need to belong to the path, one of them above $s_2$ ($2$ choices) and the other below it ($2$ choices). Once these are determined, there are $2$ possible choices for the order in which these segments are traveled: the bottom forward segment first, then $s_2$, then the top forward segment, or vice versa. Next, there are only $2$ possible forward segments that can be selected in $c_3$ and also only $2$ possible forward segments that can be selected in $c_5$. The forward segments in $c_1, c_2, c_6,$ and $c_7$ can be freely selected ($2$ choices each). This gives a total of $(2^3\cdot 2\cdot 2)\cdot 2^4=2^9$ paths. If $S=\{s_1,s_2\}$, then the analysis is similar, except for the last step, where the forward segments of $c_1$ and $c_2$ are determined by the previous choices. Thus there are $(2^3\cdot 2\cdot 2)\cdot 2^2=2^7$ possibilities, and by symmetry the same number when $S=\{s_2,s_3\}$. Finally, if $S=\{s_1,s_2,s_3\}$, then in the last step, all forward segments of $c_1, c_2, c_6,$ and $c_7$ are determined by the previous choices and hence there are $2^3\cdot 2\cdot 2=2^5$ possible paths. Altogether the total number of paths is $2^{10}+2\cdot 2^8+2^6+2^9+2\cdot 2^7+2^5=2400$.