AMC12 2014 B

Leah has 7 pennies and 6 nickels, which are worth $7 \cdot 1 + 6 \cdot 5 = 37$ cents.

The special allows Orvin to purchase balloons at $\frac{1 + \frac{2}{3}}{2} = \frac{5}{6}$ times the regular price. Because Orvin had just enough money to purchase 30 balloons at the regular price, he may now purchase $30 \cdot \frac{6}{5} = 36$ balloons.

The fraction of Randy’s trip driven on pavement was $1 - \frac{1}{3} - \frac{1}{5} = \frac{7}{15}$. Therefore the entire trip was $20 \div \frac{7}{15} = 20 \cdot \frac{15}{7} = \frac{300}{7}$ miles.

Let $m$ be the cost of a muffin and $b$ the cost of a banana in dollars. Then $2m + 16b = 2(4m + 3b)$, which simplifies to $m = \frac{5}{3}b$.

Answer (A): Denote the height of a pane by $5x$ and the width by $2x$. Then the square window has height $2\cdot 5x+6$ inches and width $4\cdot 2x+10$ inches. Solving $2\cdot 5x+6=4\cdot 2x+10$ gives $x=2$. The side length of the square window is $26$ inches.

Answer (D): Let $a$ be the amount in a regular lemonade. Then a large lemonade holds $\frac{3}{2}a$, and Ann had $\frac{1}{4}\cdot\frac{3}{2}a-\frac{3}{8}a$ lemonade left right before she gave Ed part of her drink. She gave him $\frac{1}{3}\cdot\frac{3}{8}a+2=\frac{1}{8}a+2$ ounces. Because Ann and Ed drank the same amount of lemonade, it follows that $a+\left(\frac{1}{8}a+2\right)=\frac{3}{2}a-\left(\frac{1}{8}a+2\right)$, and $4=\frac{1}{4}a$. Thus $a=16$ ounces, $\frac{3}{2}a=24$ ounces, and together they drank $16+24=40$ ounces.

Answer (D): Let $x=\frac{n}{30-n}$ so that $n=\frac{30x}{x+1}$. Because $x$ and $x+1$ are relatively prime, it follows that $x+1$ must be a factor of 30. Because $n$ is positive and less than 30 it follows that $x+1\ge 2$. Thus $x+1$ equals 2, 3, 5, 6, 10, 15, or 30. Hence there are 7 possible values for $n$, namely 15, 20, 24, 25, 27, 28, and 29.

English (OCR): Answer (C): As indicated by the leftmost column $A+B\le 9$. Then both the second and fourth columns show that $C=0$. Because $A$, $B$, and $C$ are distinct digits, $D$ must be at least $3$. The following values for $(A,B,C,D)$ show that $D$ may be any of the $7$ digits that are at least $3$: $(1,2,0,3)$, $(1,3,0,4)$, $(2,3,0,5)$, $(2,4,0,6)$, $(2,5,0,7)$, $(2,6,0,8)$, $(2,7,0,9)$.

Answer (B): By the Pythagorean Theorem, $AC = 5$. Because $5^2 + 12^2 = 13^2$, the converse of the Pythagorean Theorem applied to $\triangle DAC$ implies that $\angle DAC = 90^\circ$. The area of $\triangle ABC$ is $6$ and the area of $\triangle DAC$ is $30$. Thus the area of the quadrilateral is $6 + 30 = 36$.

Answer (D): Let $m$ be the total mileage of the trip. Then $m$ must be a multiple of $55$. Also, because $m=cba-abc=99(c-a)$, it is a multiple of $9$. Therefore $m$ is a multiple of $495$. Because $m$ is at most a 3-digit number and $a$ is not equal to $0$, $m=495$. Therefore $c-a=5$. Because $a+b+c\le 7$, the only possible $abc$ is $106$, so $a^2+b^2+c^2=1+0+36=37$.

Answer (E): The numbers in the list have a sum of $11\cdot 10=110$. The value of the 11th number is maximized when the sum of the first ten numbers is minimized subject to the following conditions. - If the numbers are arranged in nondecreasing order, the sixth number is 9. - The number 8 occurs either 2, 3, 4, or 5 times, and all other numbers occur fewer times. If 8 occurs 5 times, the smallest possible sum of the first 10 numbers is $8+8+8+8+8+9+9+9+9+10=86$. If 8 occurs 4 times, the smallest possible sum of the first 10 numbers is $1+8+8+8+8+9+9+9+10+10=80$. If 8 occurs 3 times, the smallest possible sum of the first 10 numbers is $1+1+8+8+8+9+9+10+10+11=75$. If 8 occurs 2 times, the smallest possible sum of the first 10 numbers is $1+2+3+8+8+9+10+11+12+13=77$. Thus the largest possible value of the 11th number is $110-75=35$.

Answer (B): Denote a triangle by the string of its side lengths written in nonincreasing order. Then $S$ has at most one equilateral triangle and at most one of the two triangles 442 and 221. The other possible elements of $S$ are 443, 441, 433, 432, 332, 331, and 322. All other strings are excluded by the triangle inequality. Therefore $S$ has at most 9 elements.

Answer (C): There is a triangle with side lengths $1$, $a$, and $b$ if and only if $a>b-1$. There is a triangle with side lengths $\frac1b$, $\frac1a$, and $1$ if and only if $\frac1a>1-\frac1b$, that is, $a<\frac{b}{b-1}$. Therefore there are no such triangles if and only if $b-1\ge a\ge \frac{b}{b-1}$. The smallest possible value of $b$ satisfies $b-1=\frac{b}{b-1}$, or $b^2-3b+1=0$. The solution with $b>1$ is $\frac12(3+\sqrt5)$. The corresponding value of $a$ is $\frac12(1+\sqrt5)$.

Answer (D): Denote the edge lengths by $x$, $y$, and $z$. The surface area is $2(xy+yz+zx)=94$ and the sum of the lengths of the edges is $4(x+y+z)=48$. Therefore $144=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=x^2+y^2+z^2+94$, so $x^2+y^2+z^2=50$. By the Pythagorean Theorem applied twice, each of the 4 internal diagonals has length $\sqrt{50}$, and their total length is $4\sqrt{50}=20\sqrt{2}$. A right rectangular prism with edge lengths 3, 4, and 5 satisfies the conditions of the problem.

Answer (C): Because $k\ln k=\ln(k^k)$ and the log of a product is the sum of the logs, $p=\ln\prod_{k=1}^6 k^k$. Therefore $e^p$ is the integer $1^1\cdot2^2\cdot3^3\cdot4^4\cdot5^5\cdot6^6=2^{16}\cdot3^9\cdot5^5$, and the largest power of $2$ dividing $e^p$ is $2^{16}$.

Answer (E): Because $P(0)=k$, it follows that $P(x)=ax^3+bx^2+cx+k$. Thus $P(1)=a+b+c+k=2k$ and $P(-1)=-a+b-c+k=3k$. Adding these equations gives $2b=3k$. Hence $$ P(2)+P(-2)=(8a+4b+2c+k)+(-8a+4b-2c+k) =8b+2k=12k+2k=14k. $$

Answer (E): The line passing through point $Q=(20,14)$ with slope $m$ has equation $y-14=m(x-20)$. The requested values for $m$ are those for which the system \[ \left\{ \begin{array}{l} y-14=m(x-20)\\ y=x^2 \end{array} \right. \] has no solutions. Solving for $y$ in the first equation and substituting into the second yields $m(x-20)+14=x^2$, which reduces to $x^2-mx+(20m-14)=0$. This equation has no solution for $x$ when the discriminant is negative, that is, when $m^2-4\cdot(20m-14)=m^2-80m+56<0$. This quadratic in $m$ is negative between its two roots $40\pm\sqrt{40^2-56}$, which are the required values of $r$ and $s$. The requested sum is $r+s=2\cdot40=80$.

Answer (B): The circular arrangement 14352 is bad because the sum 6 cannot be achieved with consecutive numbers, and the circular arrangement 23154 is bad because the sum 7 cannot be so achieved. It remains to show that these are the only bad arrangements. Given a circular arrangement, sums 1 through 5 can be achieved with a single number, and if the sum $n$ can be achieved, then the sum $15-n$ can be achieved using the complementary subset. Therefore an arrangement is not bad as long as sums 6 and 7 can be achieved. Suppose 6 cannot be achieved. Then 1 and 5 cannot be adjacent, so by a suitable rotation and/or reflection, the arrangement is $1bc5e$. Furthermore, $\{b,c\}$ cannot equal $\{2,3\}$ because $1+2+3=6$; similarly $\{b,c\}$ cannot equal $\{2,4\}$. It follows that $e=2$, which then forces the arrangement to be 14352 in order to avoid consecutive 213. This arrangement is bad. Next suppose that 7 cannot be achieved. Then 2 and 5 cannot be adjacent, so again without loss of generality the arrangement is $2bc5e$. Reasoning as before, $\{b,c\}$ cannot equal $\{3,4\}$ or $\{1,4\}$, so $e=4$, and then $b=3$ and $c=1$, to avoid consecutive 421; therefore the arrangement is 23154, which is also bad. Thus there are only two bad arrangements up to rotation and reflection.

Answer (E): Assume without loss of generality that the radius of the top base of the truncated cone (frustum) is $1$. Denote the radius of the bottom base by $r$ and the radius of the sphere by $a$. The figure on the left is a side view of the frustum. Applying the Pythagorean Theorem to the triangle on the right yields $r=a^2$. The volume of the frustum is $$\frac{1}{3}\pi(r^2+r\cdot1+1^2)\cdot2a=\frac{1}{3}\pi(a^4+a^2+1)\cdot2a.$$ Setting this equal to twice the volume of the sphere, $\frac{4}{3}\pi a^3$, and simplifying gives $a^4-3a^2+1=0$, or $r^2-3r+1=0$. Therefore $r=\frac{3+\sqrt{5}}{2}$.

Answer (B): The domain of $\log_{10}(x-40)+\log_{10}(60-x)$ is $40<x<60$. Within this domain, the inequality $\log_{10}(x-40)+\log_{10}(60-x)<2$ is equivalent to each of the following: $\log_{10}((x-40)(60-x))<2$, $(x-40)(60-x)<10^2=100$, $x^2-100x+2500>0$, and $(x-50)^2>0$. The last inequality is true for all $x\ne50$. Thus the integer solutions to the original inequality are $41,42,\ldots,49,51,52,\ldots,59$, and their number is 18.

Answer (C): Let $x = BE = GH = CF$, and let $\theta = \angle DHG = \angle AGJ = \angle FKH$. Note that $AD = GJ = HK = 1$. In right triangle $GDH$, $x\sin\theta = DG = 1 - AG = 1 - \cos\theta$, so $x = \dfrac{1-\cos\theta}{\sin\theta}$. Then $1 = CD = CF + FH + HD = x + \sin\theta + x\cos\theta$. Substituting for $x$ gives $$ 1=\frac{1-\cos\theta}{\sin\theta}+\sin\theta+\frac{1-\cos\theta}{\sin\theta}\cdot\cos\theta $$ $$ =\frac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta}+\sin\theta $$ $$ =\frac{\sin^2\theta}{\sin\theta}+\sin\theta=2\sin\theta. $$ It follows that $\sin\theta=\dfrac12$, so $\theta=30^\circ$, and $$ x=\frac{1-\frac{\sqrt3}{2}}{\frac12}=2-\sqrt3. $$

Answer (C): First note that once the frog is on pad 5, it has probability $\frac{1}{2}$ of eventually being eaten by the snake, and a probability $\frac{1}{2}$ of eventually exiting the pond without being eaten. It is therefore necessary only to determine the probability that the frog on pad 1 will reach pad 5 before being eaten. Consider the frog’s jumps in pairs. The frog on pad 1 will advance to pad 3 with probability $\frac{9}{10}\cdot\frac{8}{10}=\frac{72}{100}$, will be back at pad 1 with probability $\frac{9}{10}\cdot\frac{2}{10}=\frac{18}{100}$, and will retreat to pad 0 and be eaten with probability $\frac{1}{10}$. Because the frog will eventually make it to pad 3 or make it to pad 0, the probability that it ultimately makes it to pad 3 is $\frac{72}{100}\div\left(\frac{72}{100}+\frac{10}{100}\right)=\frac{36}{41}$, and the probability that it ultimately makes it to pad 0 is $\frac{10}{100}\div\left(\frac{72}{100}+\frac{10}{100}\right)=\frac{5}{41}$. Similarly, in a pair of jumps the frog will advance from pad 3 to pad 5 with probability $\frac{7}{10}\cdot\frac{6}{10}=\frac{42}{100}$, will be back at pad 3 with probability $\frac{7}{10}\cdot\frac{4}{10}+\frac{3}{10}\cdot\frac{8}{10}=\frac{52}{100}$, and will retreat to pad 1 with probability $\frac{3}{10}\cdot\frac{2}{10}=\frac{6}{100}$. Because the frog will ultimately make it to pad 5 or pad 1 from pad 3, the probability that it ultimately makes it to pad 5 is $\frac{42}{100}\div\left(\frac{42}{100}+\frac{6}{100}\right)=\frac{7}{8}$, and the probability that it ultimately makes it to pad 1 is $\frac{6}{100}\div\left(\frac{42}{100}+\frac{6}{100}\right)=\frac{1}{8}$. The sequences of pairs of moves by which the frog will advance to pad 5 without being eaten are $1\to3\to5,\;1\to3\to1\to3\to5,\;1\to3\to1\to3\to1\to3\to5,$ and so on. The sum of the respective probabilities of reaching pad 5 is then $\frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{1}{8}\cdot\frac{36}{41}\cdot\frac{7}{8}+\cdots$ $=\frac{63}{82}\left(1+\frac{9}{82}+\left(\frac{9}{82}\right)^2+\cdots\right)$ $=\frac{63}{82}\div\left(1-\frac{9}{82}\right)$ $=\frac{63}{73}.$ Therefore the requested probability is $\frac{1}{2}\cdot\frac{63}{73}=\frac{63}{146}.$

Answer (C): Let $n=\binom{2014}{k}$. Note that $2016\cdot2015\equiv(-1)(-2)=2\pmod{2017}$ and $2016\cdot2015\cdots(2015-k)\equiv(-1)(-2)\cdots(-(k+2))=(-1)^k (k+2)!\pmod{2017}$. Because $n\cdot k!\cdot(2014-k)!=2014!$, it follows that $\,n\cdot k!\cdot(2014-k)!\cdot\big((2015-k)\cdots2015\cdot2016\big)\cdot2 \equiv 2014!\cdot2015\cdot2016\cdot(-1)^k (k+2)!\pmod{2017}.$ Thus $2n\cdot k!\cdot2016!\equiv(-1)^k (k+2)!\cdot2016!\pmod{2017}.$ Dividing by $2016!\cdot k!$, which is relatively prime to $2017$, gives $2n\equiv(-1)^k (k+2)(k+1)\pmod{2017}.$ Thus $n\equiv(-1)^k\binom{k+2}{2}\pmod{2017}$. It follows that $S=\sum_{k=0}^{62}(-1)^k\binom{k+2}{2}=1+\sum_{k=1}^{31}\left(\binom{2k+2}{2}-\binom{2k+1}{2}\right)$ $=1+\sum_{k=1}^{31}(2k+1)=32^2=1024\pmod{2017}.$

Answer (D): Let $x=AC$, $y=AD$, and $z=BE$. Because the arcs $ABC$, $BCD$, and $CDE$ are congruent, it follows that $AC=BD=CE=x$. By Ptolemy’s Theorem applied to the quadrilaterals $ABCD$, $ABDE$, and $BCDE$, it follows that $$10y+9=x^2,\quad 30+14x=yz,\quad \text{and}\quad 100+3z=x^2.$$ Solving for $y$ and $z$ in the first and third equations and substituting in the second equation gives $$30+14x=\left(\frac{x^2-9}{10}\right)\left(\frac{x^2-100}{3}\right)=\frac{x^4-109x^2+900}{30},$$ which implies that $$900+420x=x^4-109x^2+900.$$ Thus $x^3-109x-420=0$. This equation factors as $(x-12)(x+5)(x+7)=0$. Because $x>0$ it follows that $x=12$, $y=\frac{1}{10}(x^2-9)=\frac{135}{10}=\frac{27}{2}$, and $z=\frac{1}{3}(x^2-100)=\frac{44}{3}$. The required sum of diagonals equals $3x+y+z=\frac{385}{6}$, so $m+n=385+6=391$.

Answer (D): If $x=\frac{1}{2}\pi y$, then the given equation is equivalent to $$2\cos(\pi y)\left(\cos(\pi y)-\cos\left(\frac{4028\pi}{y}\right)\right)=\cos(2\pi y)-1.$$ Dividing both sides by $2$ and using the identity $\frac{1}{2}(1-\cos(2\pi y))=\sin^2(\pi y)$ yields $$\cos^2(\pi y)-\cos(\pi y)\cos\left(\frac{4028\pi}{y}\right)=\frac{1}{2}(\cos(2\pi y)-1)=-\sin^2(\pi y).$$ This is equivalent to $$1=\cos(\pi y)\cos\left(\frac{4028\pi}{y}\right).$$ Thus either $\cos(\pi y)=\cos\left(\frac{4028\pi}{y}\right)=1$ or $\cos(\pi y)=\cos\left(\frac{4028\pi}{y}\right)=-1$. It follows that $y$ and $\frac{4028}{y}$ are both integers having the same parity. Therefore $y$ cannot be odd or a multiple of $4$. Finally, let $y=2a$ with $a$ a positive odd divisor of $4028=2^2\cdot 19\cdot 53$, that is $a\in\{1,19,53,19\cdot 53\}$. Then $\cos(\pi y)=\cos(2a\pi)=1$ and $\cos\left(\frac{4028\pi}{y}\right)=\cos\left(\frac{2014\pi}{a}\right)=1$. Therefore the sum of all solutions $x$ is $$\pi(1+19+53+19\cdot 53)=\pi(19+1)(53+1)=1080\pi.$$