Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: $A$ $B$ $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles $B$ and $C$
$\textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59$

Solution: Let the number of rocks in $A$ be $a$ $B$ be $b$ $C$ be $c$ . The total weight of $A$ be $40a$ $B$ be $50b$ $C$ be $kc$
We can write the information given as, $\frac{40a + 50b}{a+b} = 43$ $\frac{40a + kc}{a+c} = 44$ $\frac{50b + kc}{b+c} = ?$
$40a + 50b = 43 a + 43 b$ $3a = 7b$
$40a + kc = 44a + 44 c$ $kc = 4a + 44c = \frac{28}{3}b + 44c$
$\frac{50b + kc}{b+c} = \frac{50 \cdot \frac{3}{7}a + kc}{\frac{3}{7}a+c} = \frac{150a + 7kc}{3a + 7c} = \frac{150a + 28a + 308c}{3a+7c} = \frac{178a + 308c}{3a+7c} = \frac{44(3a+7c)+46a}{3a+7c}$
$= 44 + \frac{46a}{3a+7c} = 44 + \frac{46}{3 + \frac{7}{a}} < 44 + \frac{46}{3} \approx 59.3$
$\frac{50b + kc}{b+c} \le \boxed{59}$

Answer: E 59
Problem: $A$ $B$ $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles $B$ and $C$
$\textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59$

Solution: Let the total number of rocks in pile $A$ be $A_n$ , and the total number of rocks in pile $B$ be $B_n$ . Then, by restriction 3 (the average of $A$ and $B$ ), we can establish the equation: \[\frac{40A_n+50B_n}{A_n+B_n}=43\] .
Cross-multiplying, we get: \[40A_n+50B_n=43A_n+43B_n \implies 3A_n=7B_n\] .
Let's say we have $7k$ rocks in $A$ and $3k$ rocks in $B$ . Hence, we have $280k$ and $150k$ as the total weight of piles $A$ and $B$ , respectively. Let the total weight of $C$ be $m$ , and the total number of rocks in $C$ be $n$ . 
Using the last restriction regarding the average of piles $A$ and $C$ , we have: \[\frac{280k+m}{7k+n}=44 \implies 280k + m=280k + 28k + 44n \implies m=28k+44n\] .
To find the average of piles $B$ and $C$ , we can establish the expression: \[\frac{150k+28k+44n}{3k+n}=\frac{178k+44n}{3k+n}=\frac{132k+44n+46k}{3k+n}=\frac{44(3k+n)+46k}{3k+n}=44+\frac{46k}{3k+n}.\] When we let the final expression equal $59$ , we get: \[44+\frac{46k}{3k+n}=59\implies\frac{46k}{3k+n}=15\] Cross-multiplying, we get: \[46k=45k+14n\implies k=14n\] $k$ is still positive here, so 59 works. As this is the greatest option, we can circle $\textbf{(E)}$ immediately.
To show why $59$ is the greatest, consider the following:
When we let the final expression equal $60$ , we get: \[44+\frac{46k}{3k+n}=60\implies\frac{46k}{3k+n}=16\] Cross-multiplying, we get: \[46k=48k+14n\implies k=-7n\] Since $k$ is positive, the final expression could not equal 60. It further implies that the final expression could not equal any other integer greater than 60. Therefore, we have our final answer $\boxed{59}$

Answer:  59
Problem: A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)} \ 720 \qquad  \textbf{(B)} \ 1296 \qquad  \textbf{(C)} \ 1728 \qquad  \textbf{(D)} \ 1925 \qquad  \textbf{(E)} \ 3850$

Solution: The first pirate takes $\frac{1}{12}$ of the $x$ coins, leaving $\frac{11}{12} x$
The second pirate takes $\frac{2}{12}$ of the remaining coins, leaving $\frac{10}{12}\cdot   \frac{11}{12}*x$
Note that
$12^{11} = (2^2 \cdot 3)^{11} = 2^{22} \cdot 3^{11}$
$11! = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2$
All the $2$ s and $3$ s cancel out of $11!$ , leaving
$11 \cdot 5 \cdot 7 \cdot 5 = 1925$
in the numerator.
We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, $x$ is the denominator, leaving $\boxed{1925}$ coins for the twelfth pirate.

Answer: D 1925
Problem: A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)} \ 720 \qquad  \textbf{(B)} \ 1296 \qquad  \textbf{(C)} \ 1728 \qquad  \textbf{(D)} \ 1925 \qquad  \textbf{(E)} \ 3850$

Solution: The answer cannot be an even number. Here is why:
Consider the highest power of 2 that divides the starting number of coins, and consider how this value changes as each pirate takes their share. At each step, the size of the pile is multiplied by some $\frac{n}{12}$ . This means the highest power of 2 that divides the number of coins is continually decreasing or staying the same (except once, briefly, when we multiply by $\frac{2}{3}$ for pirate 4, but it immediately drops again in the next step).
Therefore, if the 12th pirate's coin total were even, then it can't be the smallest possible value, because we can safely cut the initial pot (and all the intermediate totals) in half. We could continue halving the result until the 12th pirate's total is finally odd.
Only one of the choices given is odd, $\boxed{1925}$

Answer: D 1925
Problem: A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)} \ 720 \qquad  \textbf{(B)} \ 1296 \qquad  \textbf{(C)} \ 1728 \qquad  \textbf{(D)} \ 1925 \qquad  \textbf{(E)} \ 3850$

Solution: Let $x$ be the number of coins the $12$ th pirate takes. Then the number of coins the $k<12$ th pirate takes is $\frac{12}{1} \cdot \frac{12}{2} \cdots \frac{12}{12-k} x$ . For all these to be an integer, we need the denominators to divide into the numerators. Looking at prime factors, obviously there are sufficient $2$ s and $3$ s, so we just need $5^2 \cdot 7 \cdot 11 = \boxed{1925}$ to divide into $x$ . -Frestho

Answer: D 1925
Problem: Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2$ . The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?
$\textbf{(A)} \ \sqrt{2} \qquad \textbf{(B)} \ \frac{3}{2} \qquad \textbf{(C)} \ \frac{5}{3} \qquad \textbf{(D)} \ \sqrt{3} \qquad \textbf{(E)} \ 2$

Solution: It can be seen that the diameter of the eighth sphere is equal to the radius of the seventh sphere by drawing out a diagram of the insides of the seventh sphere. The radius of the seventh sphere is $2+1=3$ , the radius of the eight sphere is $\boxed{32}$

Answer: E 32
Problem: In $\bigtriangleup ABC$ $AB = 86$ , and $AC = 97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)} \ 11 \qquad  \textbf{(B)} \ 28 \qquad  \textbf{(C)} \ 33 \qquad  \textbf{(D)} \ 61 \qquad  \textbf{(E)} \ 72$

Solution: [asy] //Made by samrocksnature size(8cm); pair A,B,C,D,E,X; A=(0,0); B=(-53.4,-67.4); C=(0,-97); D=(0,-86); E=(0,86); X=(-29,-81); draw(circle(A,86)); draw(E--C--B--A--X); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,NE); label("$E$",E,NE); label("$X$",X,dir(250)); dot(A^^B^^C^^D^^E^^X); [/asy]
Let circle $A$ intersect $AC$ at $D$ and $E$ as shown. We apply Power of a Point on point $C$ with respect to circle $A.$ This yields the diophantine equation
\[CX \cdot CB = CD \cdot CE\] \[CX(CX+XB) = (97-86)(97+86)\] \[CX(CX+XB) = 3 \cdot 11 \cdot 61.\]
Since lengths cannot be negative, we must have $CX+XB \ge CX.$ This generates the four solution pairs for $(CX,CX+XB)$ \[(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).\]
However, by the Triangle Inequality on $\triangle ACX,$ we see that $CX>13.$ This implies that we must have $CX+XB= \boxed{61}.$

Answer: D 61
Problem: In $\bigtriangleup ABC$ $AB = 86$ , and $AC = 97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)} \ 11 \qquad  \textbf{(B)} \ 28 \qquad  \textbf{(C)} \ 33 \qquad  \textbf{(D)} \ 61 \qquad  \textbf{(E)} \ 72$

Solution: Let $BX = q$ $CX = p$ , and $AC$ meet the circle at $Y$ and $Z$ , with $Y$ on $AC$ .  Then $AZ = AY = 86$ .  Using the Power of a Point, we get that $p(p+q) = 11(183) = 11 * 3 * 61$ .  We know that $p+q>p$ , and that $p>13$ by the triangle inequality on $\triangle ACX$ .  Thus, we get that $BC = p+q = \boxed{61}$

Answer: D 61
Problem: In $\bigtriangleup ABC$ $AB = 86$ , and $AC = 97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)} \ 11 \qquad  \textbf{(B)} \ 28 \qquad  \textbf{(C)} \ 33 \qquad  \textbf{(D)} \ 61 \qquad  \textbf{(E)} \ 72$

Solution: Let $x$ represent $CX$ , and let $y$ represent $BX$ . Since the circle goes through $B$ and $X$ $AB = AX = 86$ .
Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to $0$ , dividing both sides of the equation by $y$ is allowed.)
$x(x+y) = (97+86)(97-86)$
$x(x+y) = 2013$
The prime factors of $2013$ are $3$ $11$ , and $61$ . Obviously, $x < x+y$ . In addition, by the Triangle Inequality, $BC < AB + AC$ , so $x+y < 183$ . Therefore, $x$ must equal $33$ , and $x+y$ must equal $\boxed{61}$

Answer: D 61
Problem: $ABCD$ is a square of side length $\sqrt{3} + 1$ . Point $P$ is on $\overline{AC}$ such that $AP = \sqrt{2}$ . The square region bounded by $ABCD$ is rotated $90^{\circ}$ counterclockwise with center $P$ , sweeping out a region whose area is $\frac{1}{c} (a \pi + b)$ , where $a$ $b$ , and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a + b + c$
$\textbf{(A)} \ 15 \qquad \textbf{(B)} \ 17 \qquad \textbf{(C)} \ 19 \qquad \textbf{(D)} \ 21 \qquad \textbf{(E)} \ 23$

Solution: We first note that diagonal $\overline{AC}$ is of length $\sqrt{6} + \sqrt{2}$ . It must be that $\overline{AP}$ divides the diagonal into two segments in the ratio $\sqrt{3}$ to $1$ . It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions $2\sqrt{3}$ by $\sqrt{3} + 1$ . The area of the overall region (of the initial and final squares) is therefore twice the area of the original square minus the overlap, or $2 (\sqrt{3} + 1)^2 - 2 (\sqrt{3} + 1) = 2 (4 + 2 \sqrt{3}) - 2 \sqrt{3} - 2 = 6 + 2 \sqrt{3}$
The area also includes $4$ circular segments. Two are quarter-circles centered at $P$ of radii $\sqrt{2}$ (the segment bounded by $\overline{PA}$ and $\overline{PA'}$ ) and $\sqrt{6}$ (that bounded by $\overline{PC}$ and $\overline{PC'}$ ). Assuming $A$ is the bottom-left vertex and $B$ is the bottom-right one, it is clear that the third segment is formed as $B$ swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when $D$ overshoots the final square's left edge. To find these areas, consider the perpendicular from $P$ to $\overline{BC}$ . Call the point of intersection $E$ . From the previous paragraph, it is clear that $PE = \sqrt{3}$ and $BE = 1$ . This means $PB = 2$ , and $B$ swings back inside edge $\overline{BC}$ at a point $1$ unit above $E$ (since it left the edge $1$ unit below). The triangle of the circular sector is therefore an equilateral triangle of side length $2$ , and so the angle of the segment is $60^{\circ}$ . Imagining the process in reverse, it is clear that the situation is the same with point $D$
The area of the segments can be found by subtracting the area of the triangle from that of the sector; it follows that the two quarter-segments have areas $\frac{1}{4} \pi (\sqrt{2})^2 - \frac{1}{2} \sqrt{2} \sqrt{2} = \frac{\pi}{2} - 1$ and $\frac{1}{4} \pi (\sqrt{6})^2 - \frac{1}{2} \sqrt{6} \sqrt{6} = \frac{3 \pi}{2} - 3$ . The other two segments both have area $\frac{1}{6} \pi (2)^2 - \frac{(2)^2 \sqrt{3}}{4} = \frac{2 \pi}{3} - \sqrt{3}$
The total area is therefore \[(6 + 2 \sqrt{3}) + (\frac{\pi}{2} - 1) + (\frac{3 \pi}{2} - 3) + 2 (\frac{2 \pi}{3} - \sqrt{3})\] \[= 2 + 2 \sqrt{3} + 2 \pi + \frac{4 \pi}{3} - 2 \sqrt{3}\] \[= \frac{10 \pi}{3} + 2\] \[= \frac{1}{3} (10 \pi + 6)\]
Since $a = 10$ $b = 6$ , and $c = 3$ , the answer is $a + b + c = 10 + 6 + 3 = \boxed{19}$

Answer: C 19
Problem: Let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = z^2 + iz + 1$ . How many complex numbers $z$ are there such that $\text{Im}(z) > 0$ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $10$
$\textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D)} \ 431 \qquad \textbf{(E)} \ 441$

Solution: Suppose $f(z)=z^2+iz+1=c=a+bi$ . We look for $z$ with $\operatorname{Im}(z)>0$ such that $a,b$ are integers where $|a|, |b|\leq 10$
First, use the quadratic formula:
$z = \frac{1}{2} (-i \pm \sqrt{-1-4(1-c)}) = -\frac{i}{2} \pm \sqrt{ -\frac{5}{4} + c }$
Generally, consider the imaginary part of a radical of a complex number: $\sqrt{u}$ , where $u = v+wi = r e^{i\theta}$
$\operatorname{Im}(\sqrt{u}) = \operatorname{Im}(\pm \sqrt{r} e^{i\theta/2}) = \pm \sqrt{r} \sin(\theta/2) = \pm \sqrt{r}\sqrt{\frac{1-\cos\theta}{2}} = \pm \sqrt{\frac{r-v}{2}}$
Now let $u= -5/4 + c$ , then $v = -5/4 + a$ $w=b$ $r=\sqrt{v^2 + w^2}$
Note that $\operatorname{Im}(z)>0$ if and only if $\pm \sqrt{\frac{r-v}{2}}>\frac{1}{2}$ . The latter is true only when we take the positive sign, and that $r-v > 1/2$
or $v^2 + w^2 > (1/2 + v)^2 = 1/4 + v + v^2$ $w^2 > 1/4 + v$ , or $b^2 > a-1$
In other words, when $b^2 > a-1$ , the equation $f(z)=a+bi$ has unique solution $z$ in the region $\operatorname{Im}(z)>0$ ; and when $b^2 \leq a-1$ there is no solution. Therefore the number of desired solution $z$ is the same as the number of ordered pairs $(a,b)$ such that integers $|a|, |b|\leq 10$ , and that $b^2 \geq a$
When $a\leq 0$ , there is no restriction on $b$ so there are $11\cdot 21 = 231$ pairs;
when $a > 0$ , there are $2(1+4+9+10+10+10+10+10+10+10)=2(84)=168$ pairs.
So there are $231+168=\boxed{399}$ in total.

Answer:  399
Problem: Let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = z^2 + iz + 1$ . How many complex numbers $z$ are there such that $\text{Im}(z) > 0$ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $10$
$\textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D)} \ 431 \qquad \textbf{(E)} \ 441$

Solution: We consider the function $f(z)$ as a mapping from the 2-D complex plane onto itself. We complete the square of $f(z)=z^2+iz+1=(z+\frac{i}{2})^2+\frac{5}{4}$
Now, we must decide the range of $f(z)$ based on the domain of $z$ $\operatorname{Im}(z)>0$ . To do this, we are interested in mapping the boundary line $\operatorname{Im}(z)=0$ . To make the mapping simpler, let $f(z)=g(z)+\frac{5}{4}$ , or $g(z)=(z+\frac{i}{2})^2$
We intend to map of the line $\operatorname{Im}(z)=0$ using the function $g(z)$ . This transformation is equivalent to the polar equation $r=(\frac{1}{2}\csc(\frac{\theta}{2}))^2$ . Using polar and trig identities, we can restate this equation as the rectangular form of a parabola,
$x=y^2-\frac{1}{4}$
where $x=\operatorname{Re}(z)$ and $y=\operatorname{Im}(z)$ . So, we conclude that $f(z)$ maps the line $\operatorname{Im}(z)=0$ to the parabola
$x=y^2-\frac{1}{4}+\frac{5}{4}=y^2+1$
A quick check reveals that the range of $f(z)$ is to the left of the parabola, meaning that any point on or to the right of parabola cannot be reached.
Since the problem requires $|\operatorname{Re}(z)|$ and $|\operatorname{Im}(z)|$ to both be integers and at most 10, all that remains is counting all points with integer coordinates in the range of $f(z), \operatorname{Im}(z)>0$ . To do this, we employ complementary counting.
The points of interest are $|\operatorname{Re}(z)|\leq 10$ and $|\operatorname{Im}(z)|\leq 10$ , resulting in a total of $441$ points. For lattice points on or to the right of the parabola, there are $10$ points for $x=0$ $9$ points for $x=\pm 1$ $6$ points for $x=\pm 2$ , and $1$ point for $x=\pm 3$ . Summing it all together, our answer is $441-(10+2*9+2*6+2*1)=\boxed{399}$

Answer:  399
Problem: What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

Solution: Looking at the prime numbers under $12$ , we see that there are $\left\lfloor\frac{12}{2}\right\rfloor+\left\lfloor\frac{12}{2^2}\right\rfloor+\left\lfloor\frac{12}{2^3}\right\rfloor=6+3+1=10$ factors of $2$ $\left\lfloor\frac{12}{3}\right\rfloor+\left\lfloor\frac{12}{3^2}\right\rfloor=4+1=5$ factors of $3$ , and $\left\lfloor\frac{12}{5}\right\rfloor=2$ factors of $5$ . All greater primes are represented once or none in $12!$ , so they cannot be part of the square. Since we are looking for a perfect square, the exponents on its prime factors must be even, so we can only use $4$ of the $5$ factors of $3$ . Therefore, the prime factorization of the square is $2^{10}\cdot3^4\cdot5^2$ . To find the square root of this, we halve the exponents, leaving $2^5\cdot3^2\cdot5$ . The sum of the exponents is $\boxed{8}$

Answer: C 8
Problem: Cities $A$ $B$ $C$ $D$ , and $E$ are connected by roads $\widetilde{AB}$ $\widetilde{AD}$ $\widetilde{AE}$ $\widetilde{BC}$ $\widetilde{BD}$ $\widetilde{CD}$ , and $\widetilde{DE}$ . How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some cities more than once.)
[asy] unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B); dot (C); dot (D); dot (E); label("$A$",A,S); label("$B$",B,SE); label("$C$",C,E); label("$D$",D,N); label("$E$",E,W); guide squiggly(path g, real stepsize, real slope=45) {  real len = arclength(g);  real step = len / round(len / stepsize);  guide squig;  for (real u = 0; u < len; u += step){  real a = arctime(g, u);  real b = arctime(g, u + step / 2);  pair p = point(g, a);  pair q = point(g, b);  pair np = unit( rotate(slope) * dir(g,a));  pair nq = unit( rotate(0 - slope) * dir(g,b));  squig = squig .. p{np} .. q{nq};  }  squig = squig .. point(g, length(g)){unit(rotate(slope)*dir(g,length(g)))};  return squig; } pen pp = defaultpen + 2.718; draw(squiggly(A--B, 4.04, 30), pp); draw(squiggly(A--D, 7.777, 20), pp); draw(squiggly(A--E, 5.050, 15), pp); draw(squiggly(B--C, 5.050, 15), pp); draw(squiggly(B--D, 4.04, 20), pp); draw(squiggly(C--D, 2.718, 20), pp); draw(squiggly(D--E, 2.718, -60), pp); [/asy]
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18$

Solution: Note that cities $C$ and $E$ can be removed when counting paths because if a path goes in to $C$ or $E$ , there is only one possible path to take out of cities $C$ or $E$ .
So the diagram is as follows:
[asy] unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B);  dot (D);  label("$A$",A,S); label("$B$",B,SE);  label("$D$",D,N);  draw(A--B..D..cycle); draw(A--D); draw(B--D); [/asy]
Now we proceed with casework. Remember that there are two ways to travel from $A$ to $D$ $D$ to $A$ $B$ to $D$ and $D$ to $B$ .:
Case 1 $A \Rightarrow D$ : From $D$ , if the path returns to $A$ , then the next path must go to $B\Rightarrow D \Rightarrow B$ . There are $2 \cdot 1 \cdot 2 = 4$ possibilities of the path $ADABDB$ . If the path goes to $D$ from $B$ , then the path must continue with either $BDAB$ or $BADB$ . There are $2 \cdot 2 \cdot 2 = 8$ possibilities. So, this case gives $4+8=12$ different possibilities.
Case 2 $A \Rightarrow B$ : The path must continue with $BDADB$ . There are $2 \cdot 2 = 4$ possibilities for this case.
Putting the two cases together gives $12+4 = \boxed{16}$

Answer: D 16
Problem: The internal angles of quadrilateral $ABCD$ form an arithmetic progression. Triangles $ABD$ and $DCB$ are similar with $\angle DBA = \angle DCB$ and $\angle ADB = \angle CBD$ . Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of $ABCD$
$\textbf{(A)}\ 210 \qquad \textbf{(B)}\ 220 \qquad \textbf{(C)}\ 230 \qquad \textbf{(D)}\ 240 \qquad \textbf{(E)}\ 250$

Solution: Since the angles of Quadrilateral $ABCD$ form an arithmetic sequence, we can assign each angle with the value $a$ $a+d$ $a+2d$ , and $a+3d$ . Also, since these angles form an arithmetic progression, we can reason out that $(a)+(a+3d)=(a+d)+(a+2d)=180$
For the sake of simplicity, lets rename the angles of each similar triangle. Let $\angle ADB = \angle CBD = \alpha$ $\angle DBA = \angle DCB = \beta$ $\angle CDB = \angle BAD = \gamma$
Now the four angles of $ABCD$ are $\beta$ $\alpha + \beta$ $\gamma$ , and $\alpha + \gamma$
As for the similar triangles, we can name their angles $y$ $y+b$ , and $y+2b$ . Therefore $y+y+b+y+2b=180$ and $y+b=60$ . Because these 3 angles are each equal to one of $\alpha, \beta, \gamma$ , we know that one of these three angles is equal to 60 degrees.
Now we we use trial and error. Let $\alpha = 60^{\circ}$ . Then the angles of ABCD are $\beta$ $60^{\circ} + \beta$ $\gamma$ , and $60^{\circ} + \gamma$ . Since these four angles add up to 360, then $\beta + \gamma= 120$ . If we list them in increasing value, we get $\beta$ $\gamma$ $60^{\circ} + \beta$ $60^{\circ} + \gamma$ . Note that this is the only sequence that works because the common difference is less than 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, $\alpha, \beta, \gamma$ also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice.
If we apply the same reasoning to $\beta$ and $\gamma$ , we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, $\boxed{240}$ is the correct answer.

Answer: D 240