AMC12 2013 A

The area of $\triangle ABE$ is $\frac{1}{2} \times AB \times BE = \frac{1}{2} \times 10 \times BE = 40$, so $BE = 8$.

Answer (C): The softball team could only have scored twice as many runs as their opponent when they scored an even number of runs. In those games their opponents scored $\frac{2}{2}+\frac{4}{2}+\frac{6}{2}+\frac{8}{2}+\frac{10}{2}=15$ runs. In the games the softball team lost, their opponents scored $(1+1)+(3+1)+(5+1)+(7+1)+(9+1)=30$ runs. The total number of runs scored by their opponents was $15+30=45$ runs.

Answer (E): Because six tenths of the flowers are pink and two thirds of the pink flowers are carnations, $\frac{6}{10}\cdot\frac{2}{3}=\frac{2}{5}$ of the flowers are pink carnations. Because four tenths of the flowers are red and three fourths of the red flowers are carnations, $\frac{4}{10}\cdot\frac{3}{4}=\frac{3}{10}$ of the flowers are red carnations. Therefore $\frac{2}{5}+\frac{3}{10}=\frac{7}{10}=70\%$ of the flowers are carnations.

Answer (C): Factoring $2^{2012}$ from each of the terms and simplifying gives \[ \frac{2^{2012}(2^2+1)}{2^{2012}(2^2-1)}=\frac{4+1}{4-1}=\frac{5}{3}. \]

Answer (B): The total shared expenses were $105 + 125 + 175 = 405$ dollars, so each traveler's fair share was $\frac{1}{3}\cdot 405 = 135$ dollars. Therefore $t = 135 - 105 = 30$ and $d = 135 - 125 = 10$, so $t - d = 30 - 10 = 20$.

Answer (B): If Shenille attempted $x$ three-point shots and $30-x$ two-point shots, then she scored a total of $\frac{20}{100}\cdot 3 \cdot x+\frac{30}{100}\cdot 2\cdot (30-x)=18$ points. Remark: The given information does not allow the value of $x$ to be determined.

Answer (C): Note that $110=S_9=S_7+S_8=42+S_8$, so $S_8=110-42=68$. Thus $68=S_8=S_6+S_7=S_6+42$, so $S_6=68-42=26$. Similarly, $S_5=42-26=16$, and $S_4=26-16=10$.

Answer (D): Multiplying the given equation by $xy\neq 0$ yields $x^2y+2y=xy^2+2x$. Thus $$ x^2y-2x-xy^2+2y=x(xy-2)-y(xy-2)=(x-y)(xy-2)=0. $$ Because $x-y\neq 0$, it follows that $xy=2$.

Answer (C): Because $\overline{EF}$ is parallel to $\overline{AB}$, it follows that $\triangle FEC$ is similar to $\triangle ABC$ and $\overline{FE}=\overline{FC}$. Thus half of the perimeter of $ADEF$ is $\overline{AF}+\overline{FE}=\overline{AF}+\overline{FC}=\overline{AC}=28$. The entire perimeter is 56.

Answer (D): If $n$ satisfies the equation $\frac{1}{n}=0.\overline{ab}$, then $\frac{100}{n}=ab.\overline{ab}$ and subtracting gives $\frac{99}{n}=ab$. The positive factors of $99$ are $1,3,9,11,33,$ and $99$. Only $n=11,33,$ and $99$ give a number $\frac{99}{n}$ consisting of two different digits, namely $09,03,$ and $01$, respectively. Thus the requested sum is $11+33+99=143$.

Answer (C): Let $x=DE$ and $y=FG$. Then the perimeter of $ADE$ is $x+x+x=3x$, the perimeter of $DFGE$ is $x+(y-x)+y+(y-x)=3y-x$, and the perimeter of $FBCG$ is $y+(1-y)+1+(1-y)=3-y$. Because the perimeters are equal, it follows that $3x=3y-x=3-y$. Solving this system yields $x=\frac{9}{13}$ and $y=\frac{12}{13}$. Thus $DE+FG=x+y=\frac{21}{13}$.

Answer (A): Let the angles of the triangle be $\alpha-\delta$, $\alpha$, and $\alpha+\delta$. Then $3\alpha=\alpha-\delta+\alpha+\alpha+\delta=180^\circ$, so $\alpha=60^\circ$. There are three cases depending on which side is opposite to the $60^\circ$ angle. Suppose that the triangle is $ABC$ with $\angle BAC=60^\circ$. Let $D$ be the foot of the altitude from $C$. The triangle $CAD$ is a $30$-$60$-$90^\circ$ triangle, so $AD=\frac12 AC$ and $CD=\frac{\sqrt3}{2}AC$. There are three cases to consider. In each case the Pythagorean Theorem can be used to solve for the unknown side. If $AB=5$, $AC=4$, and $BC=x$, then $AD=2$, $CD=2\sqrt3$, and $BD=|AB-AD|=3$. It follows that $x^2=BC^2=CD^2+BD^2=21$, so $x=\sqrt{21}$. If $AB=x$, $AC=4$, and $BC=5$, then $AD=2$, $CD=2\sqrt3$, and $BD=|AB-AD|=|x-2|$. It follows that $25=BC^2=CD^2+BD^2=12+(x-2)^2$, and the positive solution is $x=2+\sqrt{13}$. If $AB=x$, $AC=5$, and $BC=4$, then $AD=\frac52$, $CD=\frac{5\sqrt3}{2}$, and $BD=|AB-AD|=|x-\frac52|$. It follows that $16=BC^2=CD^2+BD^2=\frac{75}{4}+(x-\frac52)^2$, which has no solution because $\frac{75}{4}>16$. The sum of all possible side lengths is $2+\sqrt{13}+\sqrt{21}$. The requested sum is $2+13+21=36$.

Answer (B): Let line $AG$ be the required line, with $G$ on $\overline{CD}$. Divide $ABCD$ into triangle $ABF$, trapezoid $BCEF$, and triangle $CDE$, as shown. Their areas are $1$, $5$, and $\frac{3}{2}$, respectively. Hence the area of $ABCD=\frac{15}{2}$, and the area of triangle $ADG=\frac{15}{4}$. Because $AD=4$, it follows that $GH=\frac{15}{8}=\frac{r}{s}$. The equation of $\overline{CD}$ is $y=-3(x-4)$, so when $y=\frac{15}{8}$, $x=\frac{p}{q}=\frac{27}{8}$. Therefore $p+q+r+s=58$.

Answer (B): Because the terms form an arithmetic sequence, $$\log_{12} y=\frac{1}{2}\left(\log_{12}162+\log_{12}1250\right)=\frac{1}{2}\log_{12}(162\cdot1250)$$ $$=\frac{1}{2}\log_{12}\left(2^2 3^4 5^4\right)=\log_{12}\left(2\cdot3^2 5^2\right).$$ Then $$\log_{12} x=\frac{1}{2}\left(\log_{12}162+\log_{12}y\right)=\frac{1}{2}\left(\log_{12}(2\cdot3^4)+\log_{12}(2\cdot3^2 5^2)\right)$$ $$=\frac{1}{2}\log_{12}\left(2^2 3^6 5^2\right)=\log_{12}\left(2\cdot3^3 5\right)=\log_{12}270.$$ Therefore $x=270$.

Answer (D): There are two cases. If Peter and Pauline are given to the same pet store, then there are 4 ways to choose that store. Each of the children must then be assigned to one of the other three stores, and this can be done in $3^3 = 27$ ways. Therefore there are $4 \cdot 27 = 108$ possible assignments in this case. If Peter and Pauline are given to different stores, then there are $4 \cdot 3 = 12$ ways to choose those stores. In this case, each of the children must be assigned to one of the other two stores, and this can be done in $2^3 = 8$ ways. Therefore there are $12 \cdot 8 = 96$ possible assignments in this case. The total number of assignments is $108 + 96 = 204$.

Answer (E): Let $a$, $b$, and $c$ be the number of rocks in piles $A$, $B$, and $C$, respectively. Then $$\frac{40a+50b}{a+b}=43\ \text{and}\ 7b=3a.$$ Because $7$ and $3$ are relatively prime, there is a positive integer $k$ such that $a=7k$ and $b=3k$. Let $\mu_C$ equal the mean weight in pounds of the rocks in $C$ and $\mu_{BC}$ equal the mean weight in pounds of the rocks in $B$ and $C$. Then $$\frac{40\cdot 7k+\mu_C\cdot c}{7k+c}=44,\ \text{so}\ \mu_C=\frac{28k+44c}{c},$$ and $$\mu_{BC}=\frac{50\cdot 3k+(28k+44c)}{3k+c}=\frac{178k+44c}{3k+c}.$$ Clearing the denominator and rearranging yields $(\mu_{BC}-44)c=(178-3\mu_{BC})k$. Because the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds, and the mean weight of the rocks in $B$ is greater than the mean weight of the rocks in $A$, it follows that the mean weight of the rocks in $B$ and $C$ must be greater than $44$ pounds. Thus $(\mu_{BC}-44)c>0$ and therefore $178-3\mu_{BC}$ must be greater than zero. This implies that $\mu_{BC}<\frac{178}{3}=59\frac{1}{3}$. If $k=15c$ and $\mu_C=464$, then $\mu_{BC}=59$. Thus the greatest possible integer value for the weight in pounds of the combined piles $B$ and $C$ is $59$.

Answer (D): For $1\le k\le 11$, the number of coins remaining in the chest before the $k$th pirate takes a share is $\frac{12}{12-k}$ times the number remaining afterward. Thus if there are $n$ coins left for the $12$th pirate to take, the number of coins originally in the chest is $$ \frac{12^{11}\cdot n}{11!} =\frac{2^{22}\cdot 3^{11}\cdot n}{2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7\cdot 11} =\frac{2^{14}\cdot 3^{7}\cdot n}{5^{2}\cdot 7\cdot 11}. $$ The smallest value of $n$ for which this is a positive integer is $5^{2}\cdot 7\cdot 11=1925$. In this case there are $$ 2^{14}\cdot 3^{7}\cdot \frac{11!}{(12-k)!\cdot 12^{k-1}} $$ coins left for the $k$th pirate to take, and note that this amount is an integer for each $k$. Hence the $12$th pirate receives 1925 coins.

Answer (B): Let the vertices of the regular hexagon be labeled in order $A$, $B$, $C$, $D$, $E$, and $F$. Let $O$ be the center of the hexagon, which is also the center of the largest sphere. Let the eighth sphere have center $G$ and radius $r$. Because the centers of the six small spheres are each a distance $2$ from $O$ and the small spheres have radius $1$, the radius of the largest sphere is $3$. Because $G$ is equidistant from $A$ and $D$, the segments $GO$ and $AO$ are perpendicular. Let $x$ be the distance from $G$ to $O$. Then $x+r=3$. The Pythagorean Theorem applied to $\triangle AOG$ gives $(r+1)^2=2^2+x^2=4+(3-r)^2$, which simplifies to $2r+1=13-6r$, so $r=\frac{3}{2}$. Note that this shows that the eighth sphere is tangent to $AD$ at $O$.

Answer (D): By the Power of a Point Theorem, $BC\cdot CX=AC^2-r^2$ where $r=AB$ is the radius of the circle. Thus $BC\cdot CX=97^2-86^2=2013$. Since $BC=BX+CX$ and $CX$ are both integers, they are complementary factors of 2013. Note that $2013=3\cdot 11\cdot 61$, and $CX<BC<AB+AC=183$. Thus the only possibility is $CX=33$ and $BC=61$.

Answer (B): Consider the elements of $S$ as integers modulo 19. Assume $a \succ b$. If $a > b$, then $a - b \le 9$. If $a < b$, then $b - a > 9$; that is $b - a \ge 10$ and so $(a + 19) - b \le 9$. Thus $a \succ b$ if and only if $0 < (a - b)\ (\mathrm{mod}\ 19) \le 9$. Suppose that $(x,y,z)$ is a triple in $S \times S \times S$ such that $x \succ y$, $y \succ z$, and $z \succ x$. There are 19 possibilities for the first entry $x$. Once $x$ is chosen, $y$ can equal $x + i$ for any $i$, $1 \le i \le 9$. Then $z$ is at most $x + 9 + i$ and at least $x + 10$, so once $y$ is chosen, there are $i$ possibilities for the third element $z$. The number of required triples is equal to $19(1 + 2 + \cdots + 9) = 19 \cdot \frac{1}{2} \cdot 9 \cdot 10 = 19 \cdot 45 = 855$.

Answer (A): Let $A_n=\log\!\bigl(n+\log((n-1)+\log(\cdots+\log(3+\log 2)\cdots))\bigr)$. Note that $0<\log 2=A_2<1$. If $0<A_{k-1}<1$, then $k<k+A_{k-1}<k+1$. Hence $0<\log k<\log(k+A_{k-1})=A_k<\log(k+1)\le1$, as long as $\log k>0$ and $\log(k+1)\le1$, which occurs when $2\le k\le9$. Thus $0<A_n<1$ for $2\le n\le9$. Because $0<A_9<1$, it follows that $10<10+A_9<11$, and so $1=\log(10)<\log(10+A_9)=A_{10}<\log(11)<2$. If $1<A_{k-1}<2$, then $k+1<k+A_{k-1}<k+2$. Hence $1<\log(k+1)<\log(k+A_{k-1})=A_k<\log(k+2)\le2$, as long as $\log(k+1)>1$ and $\log(k+2)\le2$, which occurs when $10\le k\le98$. Thus $1<A_n<2$ for $10\le n\le98$. In a similar way, it can be proved that $2<A_n<3$ for $99\le n\le997$, and $3<A_n<4$ for $998\le n\le9996$. For $n=2012$, it follows that $3<A_{2012}<4$, so $2016<2013+A_{2012}<2017$ and $\log 2016<A_{2013}<\log 2017$.

Answer (E): Let $n$ be a 6-digit palindrome, $m=\frac{n}{11}$, and suppose $m$ is a palindrome as well. First, if $m$ is a 4-digit number, then $n=11m<11\cdot 10^5=10^6+10^5$. Thus the first and last digit of $n$ is $1$. Thus the last digit of $m$ is $1$ and then the first digit of $m$ must be $1$ as well. Then $m\le 1991<2000$ and $n=11m<11\cdot 2000=22000$, which is a contradiction. Therefore $m$ is a 5-digit number $abcba$. If $a+b\le 9$ and $b+c\le 9$, then there are no carries in the sum $n=11m=abcba0+abcba$; thus the digits of $n$ in order are $a$, $a+b$, $b+c$, $b+c$, $a+b$, and $a$. Conversely, if $a+b\ge 10$, then the first digit of $n$ is $a+1$ and the last digit $a$; and if $a+b\le 9$ but $b+c\ge 10$, then the second digit of $n$ is $a+b+1$ if $a+b<9$, or $0$ if $a+b=9$, and the previous to last digit is $a+b$. In any case $n$ is not a palindrome. Therefore $n=11m$ is a palindrome if and only if $a+b\le 9$ and $b+c\le 9$. Thus the number of pairs $(m,n)$ is equal to $$ \sum_{b=0}^{9}\sum_{c=0}^{9-b}\sum_{a=1}^{9-b}1=\sum_{b=0}^{9}(10-b)(9-b). $$ Letting $j=10-b$ gives $$ \sum_{j=1}^{10} j(j-1)=\frac{10\cdot 11\cdot 21}{6}-\frac{10\cdot 11}{2}=330. $$ The total number of 6-digit palindromes $abccba$ is determined by 10 choices for each of $b$ and $c$, and 9 choices for $a$, for a total of $9\cdot 10^2=900$. Thus the required probability is $\frac{330}{900}=\frac{11}{30}$.

Answer (C): Assume that the vertices of $ABCD$ are labeled in counterclockwise order. Let $A'$, $B'$, $C'$, and $D'$ be the images of $A$, $B$, $C$, and $D$, respectively, under the rotation. Because $\triangle A'PA$ and $\triangle C'PC$ are isosceles right triangles, points $A'$ and $C'$ are on lines $AB$ and $CD$, respectively. Moreover, because $AP=\sqrt{2}$ and $PC=AC-AP=\sqrt{2}(\sqrt{3}+1)-\sqrt{2}=\sqrt{6}$, it follows that $AA'=\sqrt{2}AP=2$ and $CC'=\sqrt{2}CP=2\sqrt{3}$. By symmetry, points $B'$ and $D'$ are on lines $CD$ and $AB$, respectively. Let $X\ne B$ and $Y\ne D'$ be the intersections of $BC$ and $C'D'$, respectively, with the circle centered at $P$ with radius $PB$. Note that $PD'=PD=PB$, so this circle also contains $D'$. Therefore the required region consists of sectors $APA'$, $BPX$, $CPC'$, and $YPD'$, and triangles $BPA'$, $CPX$, $YPC'$, and $APD'$. Sector $APA'$ has area $\frac14\cdot(\sqrt{2})^2\pi=\frac{\pi}{2}$, and sector $CPC'$ has area $\frac14\cdot(\sqrt{6})^2\pi=\frac{3\pi}{2}$. Let $H$ and $I$ be the midpoints of $AA'$ and $BX$, respectively. Then $PH=AH=\frac{\sqrt{2}}{2}AP=1$, and $PI=HB=AB-AH=\sqrt{3}$. Thus $\triangle BPH$ is a 30-60-90° triangle, implying that $PB=2$ and $\triangle XPB$ is equilateral. Therefore congruent sectors $BPX$ and $YPD'$ each have area $\frac16\cdot2^2\pi=\frac{2\pi}{3}$. Congruent triangles $BPA'$ and $D'PA$ each have altitude $PH=1$ and base $A'B=AB-AH-HA'=\sqrt{3}-1$, so each has area $\frac12(\sqrt{3}-1)$. Congruent triangles $CPX$ and $C'PY$ each have altitude $PI=\sqrt{3}$ and base $XC=BC-BX=\sqrt{3}-1$, so each has area $\frac12(3-\sqrt{3})$. The area of the entire region is $$ \frac{\pi}{2}+\frac{3\pi}{2}+2\cdot\frac{2\pi}{3}+2\left(\frac{\sqrt{3}-1}{2}\right)+2\left(\frac{3-\sqrt{3}}{2}\right)=\frac{10\pi+6}{3}, $$ and $a+b+c=10+6+3=19$.

Answer (E): Assume without loss of generality that the regular 12-gon is inscribed in a circle of radius 1. Every segment with endpoints in the 12-gon subtends an angle of $\frac{360}{12}k = 30k$ degrees for some $1 \le k \le 6$. Let $d_k$ be the length of those segments that subtend an angle of $30k$ degrees. There are 12 such segments of length $d_k$ for every $1 \le k \le 5$ and 6 segments of length $d_6$. Because $d_k = 2\sin(15k^\circ)$, it follows that $d_2 = 2\sin(30^\circ)=1$, $d_3 = 2\sin(45^\circ)=\sqrt2$, $d_4 = 2\sin(60^\circ)=\sqrt3$, $d_6 = 2\sin(90^\circ)=2$, $d_1 = 2\sin(15^\circ)=2\sin(45^\circ-30^\circ)$ $=2\sin(45^\circ)\cos(30^\circ)-2\sin(30^\circ)\cos(45^\circ)=\frac{\sqrt6-\sqrt2}{2},$ and $d_5 = 2\sin(75^\circ)=2\sin(45^\circ+30^\circ)$ $=2\sin(45^\circ)\cos(30^\circ)+2\sin(30^\circ)\cos(45^\circ)=\frac{\sqrt6+\sqrt2}{2}.$ If $a \le b \le c$, then $d_a \le d_b \le d_c$ and the segments with lengths $d_a$, $d_b$, and $d_c$ do not form a triangle with positive area if and only if $d_c \ge d_a + d_b$. Because $d_2=1<\sqrt6-\sqrt2=2d_1<\sqrt2=d_3$, it follows that for $(a,b,c)\in\{(1,1,3),(1,1,4),(1,1,5),(1,1,6)\}$, the segments of lengths $d_a,d_b,d_c$ do not form a triangle with positive area. Similarly, $d_3=\sqrt2<\frac{\sqrt6-\sqrt2}{2}+1=d_1+d_2<\sqrt3=d_4,$ $d_4<d_5=\frac{\sqrt6+\sqrt2}{2}=\frac{\sqrt6-\sqrt2}{2}+\sqrt2=d_1+d_3,$ and $d_5<d_6=2=1+1=2d_2,$ so for $(a,b,c)\in\{(1,2,4),(1,2,5),(1,2,6),(1,3,5),(1,3,6),(2,2,6)\}$, the segments of lengths $d_a,d_b,d_c$ do not form a triangle with positive area. Finally, if $a\ge2$ and $b\ge3$, then $d_a+d_b\ge d_2+d_3=1+\sqrt2>2\ge d_c$, and also if $a\ge3$, then $d_a+d_b\ge 2d_3=2\sqrt2>2=d_c$. Therefore the complete list of forbidden triples $(d_a,d_b,d_c)$ is given by $(a,b,c)\in\{(1,1,3),(1,1,4),(1,1,5),(1,1,6),(1,2,4),(1,2,5),(1,2,6),(1,3,5),(1,3,6),(2,2,6)\}$. For each $(a,b,c)\in\{(1,1,3),(1,1,4),(1,1,5)\}$, there are $\binom{12}{2}$ pairs of segments of length $d_a$ and 12 segments of length $d_c$. For each $(a,b,c)\in\{(1,1,6),(2,2,6)\}$, there are $\binom{12}{2}$ pairs of segments of length $d_a$ and 6 segments of length $d_c$. For each $(a,b,c)\in\{(1,2,4),(1,2,5),(1,3,5)\}$, there are $12^3$ triples of segments with lengths $d_a,d_b,$ and $d_c$. Finally, for each $(a,b,c)\in\{(1,2,6),(1,3,6)\}$, there are $12^2$ pairs of segments with lengths $d_a$ and $d_b$, and 6 segments of length $d_c$. Because the total number of triples of segments equals $\binom{\binom{12}{2}}{3}=\binom{66}{3}$, the required probability equals $1-\frac{3\cdot 12\cdot \binom{12}{2}+2\cdot 6\cdot \binom{12}{2}+3\cdot 12^3+2\cdot 12^2\cdot 6}{\binom{66}{3}}$ $=1-\frac{63}{286}=\frac{223}{286}.$

Answer (A): Let $H=\{z\in\mathbb{C}:\operatorname{Im}(z)>0\}$. If $z_1,z_2\in H$ and $f(z_1)=f(z_2)$, then $$ z_1^2-z_2^2+i(z_1-z_2)=(z_1-z_2)(z_1+z_2+i)=0. $$ Because $\operatorname{Im}(z_1)>0$ and $\operatorname{Im}(z_2)>0$, it follows that $z_1+z_2+i\ne0$. Thus $z_1=z_2$; that is, the function $f$ is one-to-one on $H$. Let $r$ be a positive real number. Note that $f(r)=r^2+1+ir$ describes the top part of the parabola $x=y^2+1$. Similarly, $f(-r)=r^2+1-ir$ describes the bottom part of the parabola $x=y^2+1$. Because $f(i)=-1$, it follows that the image set $f(H)$ equals $\{w\in\mathbb{C}:\operatorname{Re}(w)<(\operatorname{Im}(w))^2+1\}$. Thus the set of complex numbers $w\in f(H)$ with integer real and imaginary parts of absolute value at most $10$ is equal to $$ S=\{w=a+ib\in\mathbb{C}:a,b\in\mathbb{Z},\ |a|\le10,\ |b|\le10,\ \text{and }a<b^2+1\}. $$ Because $f$ is one-to-one, the required answer is $|f^{-1}(S)|=|S|$ and $$ |S|=21^2-\sum_{b=-3}^{3}\sum_{a=b^2+1}^{10}1 =441-\sum_{b=-3}^{3}(10-b^2) =441-(1+6+9+10+9+6+1)=399. $$