Answer: 568.0

Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255$

Solution:
It should first be noted that given any quadrilateral of fixed side lengths, there is exactly one way to manipulate the angles so that the quadrilateral becomes cyclic.
Proof. Given a quadrilateral $ABCD$ where all sides are fixed (in a certain order), we can construct the diagonal $\overline{BD}$ . When $BD$ is the minimum allowed by the triangle inequality, one of the angles $\angle DAB$ or $\angle BCD$ will be degenerate and measure $0^\circ$ , so opposite angles will sum to less than $180^\circ$ . When $BD$ is the maximum allowed, one of the angles will be degenerate and measure $180^\circ$ , so opposite angles will sum to more than $180^\circ$ . Thus, since the sum of opposite angles increases continuously as $BD$ is lengthened from the minimum to the maximum values, there is a unique value of $BD$ somewhere in the middle such that the sum of opposite angles is exactly $180^\circ$
Denote $a$ $b$ $c$ , and $d$ as the integer side lengths of the quadrilateral. Without loss of generality, let $a\ge b \ge c \ge d$
Since $a+b+c+d = 32$ , the Triangle Inequality implies that $a \le 15$
We will now split into $5$ cases.
Case $1$ $a = b = c = d$ $4$ side lengths are equal)
Clearly there is only $1$ way to select the side lengths $(8,8,8,8)$ , and no matter how the sides are rearranged only $1$ unique quadrilateral can be formed.
Case $2$ $a = b = c > d$ or $a > b = c = d$ $3$ side lengths are equal)
If $3$ side lengths are equal, then each of those side lengths can only be integers from $6$ to $10$ except for $8$ (because that is counted in the first case). Obviously there is still only $1$ unique quadrilateral that can be formed from one set of side lengths, resulting in a total of $4$ quadrilaterals.
Case $3$ $a = b > c = d$ $2$ pairs of side lengths are equal)
$a$ and $b$ can be any integer from $9$ to $15$ , and likewise $c$ and $d$ can be any integer from $1$ to $7$ . However, a single set of side lengths can form $2$ different cyclic quadrilaterals (a rectangle and a kite), so the total number of quadrilaterals for this case is $7\cdot{2} = 14$
Case $4$ $a = b > c > d$ or $a > b = c > d$ or $a > b > c = d$ $2$ side lengths are equal)
If the $2$ equal side lengths are each $1$ , then the other $2$ sides must each be $15$ , which we have already counted in an earlier case. If the equal side lengths are each $2$ , there is $1$ possible set of side lengths. Likewise, for side lengths of $3$ there are $2$ sets. Continuing this pattern, we find a total of $1+2+3+4+4+5+7+5+4+4+3+2+1 = 45$ sets of side lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are $3$ possible quadrilaterals that can be formed, so the total number of quadrilaterals for this case is $3\cdot{45} = 135$
Case $5$ $a > b > c > d$ (no side lengths are equal)
Using the same counting principles starting from $a = 15$ and eventually reaching $a = 9$ , we find that the total number of possible side lengths is $69$ . There are $4!$ ways to arrange the $4$ side lengths, but there is only $1$ unique quadrilateral for $4$ rotations, so the number of quadrilaterals for each set of side lengths is $\frac{4!}{4} = 6$ . The total number of quadrilaterals is $6\cdot{69} = 414$
And so, the total number of quadrilaterals that can be made is $414 + 135 + 14 + 4 + 1 = \boxed{568}$

Solution:
As with solution $1$ we would  like to note that given any quadrilateral we can change its angles to make a cyclic one.
Let $a \ge b \ge c\ge d$ be the sides of the quadrilateral.
There are $\binom{31}{3}$ ways to partition $32$ . However, some of these will not be quadrilaterals since they would have one side bigger than the sum of the other three.  This occurs when $a \ge 16$ .  For $a=16$ $b+c+d=16$ . There are $\binom{15}{2}$ ways to partition $16$ . Since $a$ could be any of the four sides, we have counted $4\binom{15}{2}$ degenerate quadrilaterals. Similarly, there are $4\binom{14}{2}$ $4\binom{13}{2} \cdots 4\binom{2}{2}$ for other values of $a$ .  Thus, there are $\binom{31}{3} - 4\left(\binom{15}{2}+\binom{14}{2}+\cdots+\binom{2}{2}\right) = \binom{31}{3} - 4\binom{16}{3} = 2255$ non-degenerate partitions of $32$ by the hockey stick theorem.  We then account for symmetry. If all sides are congruent (meaning the quadrilateral is a square), the quadrilateral will be counted once. If the quadrilateral is a rectangle (and not a square), it will be counted twice. In all other cases, it will be counted 4 times. Since there is $1$ square case, and $7$ rectangle cases, there are $2255-1-2\cdot7=2240$ quadrilaterals counted 4 times. Thus there are $1+7+\frac{2240}{4} = \boxed{568}$ total quadrilaterals.

Solution:
As with solution $1$ we find that there are $2255$ ways to form a quadrilateral if we don't account for rotations. We now apply Burnside's_Lemma . There are four types of actions in the group acting on the set of quadrilaterals. We will consider each individually:
Identity: maps a quadrilateral with sides $a,b,c,d$ in that order to $a,b,c,d$ . Obviously all members of the set of quadrilaterals are fixed points, for a total of $2255$ .
Rotation by one: maps a quadrilateral from $a,b,c,d$ to $b,c,d,a$ . For this to have a fixed point we need $a=b,b=c,c=d,d=a$ , so the only quadrilateral that is a fixed point is the square with side length $8$ , for a total of $1$ .
Rotation by two: maps a quadrilateral from $a,b,c,d$ to $c,d,a,b$ . For this to be a fixed point we need $a=c$ and $b=d$ . Thus the quadrilateral is of the form $x,y,x,y$ —a rectangle. We can count that there are $15$ rectangle cases, namely $(a, b, c, d) = \{ (1, 15, 1, 15), (2, 14, 2, 14), \cdots, (15, 1, 15, 1)  \}$ .
Roration by three: maps a quadrilateral from $a,b,c,d$ to $d,a,b,c$ . Similarly to the rotation by one case, there is one fixed point here.
Summing up, we get that the total number of groups is $2255+1+15+1=2272$ . Since there are $4$ members of the group our final answer is $\frac{2272}{4}=\boxed{568}$


Answer: 41.0

Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
Noting that $\angle OQP$ and $\angle ORP$ are right angles, we realize that we can draw a semicircle with diameter $\overline{OP}$ and points $Q$ and $R$ on the semicircle. Since the radius of the semicircle is $100$ , if $\overline{QR} \leq 100$ , then $\overarc{QR}$ must be less than or equal to $60^{\circ}$
This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:
Given $a, b$ such that $0<a, b<75$ , what is the probability that $|a-b| \leq 30$ ?
Through simple geometric probability, we get that $P = \frac{16}{25}$
The answer is $16+25=\boxed{041}$

Solution:
Put $\triangle POQ$ and $\triangle POR$ with $O$ on the origin and the triangles on the $1^{st}$ quadrant.
The coordinates of $Q$ and $P$ is $(200 \cos^{2}a,200 \cos a\sin a )$ $(200\cos^{2}b,200\cos(b)\sin b)$ . So $PQ^{2}$ $(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}$ , which we want to be less then $100^{2}$ .
So $(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} <= 100^{2}$ \[(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4}\] \[\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(\cos^{2} b+\sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-\cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[(\cos a\sin b)^{2} +(\cos b\sin a)^{2} - 2 (\cos a \sin b)(\cos b \sin a)\le \frac{1}{4}\] \[(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4}\] \[\sin^{2} (b-a) \le \frac{1}{4}\] So we want $-\frac{1}{2} \le \sin (b-a) \le \frac{1}{2}$ , which is equivalent to $-30 \le b-a \le 30$ or $150 \le b-a \le 210$ . The second inequality is impossible so we only consider what the first inequality does to our $75$ by $75$ box in the $ab$ plane. This cuts off two isosceles right triangles from opposite corners with side lengths $45$ from the $75$ by $75$ box. Hence the probability is $1-\frac{45^2}{75^2} = 1- \frac{9}{25}=\frac{16}{25}$ and the answer is $16+25 = \boxed{41}$

Solution: Let $QR=x.$ Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: $OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.$ Now observe that quadrilateral $OQRP$ is a cyclic quadrilateral . Thus, we are able to apply Ptolemy's Theorem to it: \[200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),\] \[x + 200 (\cos a \sin b) = 200 (\sin a \cos b),\] \[x = 200(\sin a \cos b - \sin b \cos a),\] \[x = 200 \sin(a-b).\] We want $|x| \le 100$ (the absolute value comes from the fact that $a$ is not necessarily greater than $b,$ so we cannot assume that $Q$ is to the right of $R$ as in the diagram), so we substitute: \[|200 \sin(a-b)| \le 100,\] \[|\sin(a-b)| \le \frac{1}{2},\] \[|a-b| \le 30 ^\circ,\] \[-30 \le a-b \le 30.\] By simple geometric probability (see Solution 2 for complete explanation), $\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},$ so $m+n = \boxed{041}.$

Solution:
Impose a coordinate system as follows:
Let the midpoint of $\overline{OP}$ be the origin, and let $\overline{OP}$ be the x-axis. We construct a circle with center at the origin with radius 100. Since $\angle OQP$ and $\angle ORP$ are both right angles, points $Q$ and $R$ are on our circle. Place $Q$ and $R$ in the first quadrant of the Cartesian Plane. Suppose we construct $Q'$ and $R'$ such that they are clockwise rotations of $Q$ and $R$ , respectively by an angle of $2b$ degrees. Thus, we see that $\overline{QR}=100\sqrt{2}\sqrt{\cos(2|a-b|)}$ . We want this quantity to be less than $100$ . This happens when $\cos(2|a-b|) \ge 1/2,$ or when $|a-b|\le 30^{\circ}$ . The probability that the last inequality is satisfied is $16/25$ . Therefore, the probability that $QR$ is less than $100$ is $16/25$ . Hence, $m+n=\boxed{41}$

Solution:
WLOG, let $b\ge a$ . It does not actually matter, but it is necessary for this particular setup. It should be apparent that $\Delta RAQ\sim\Delta OAP$ . We write the equation
\[\dfrac{RA}{AO}=\dfrac{RQ}{OP}.\]
If we examine right triangle $\Delta ROA$ , we can see that $\sin(b-a)=\dfrac{RA}{AO}$ . Also, we are given $OP=200$ , so now we have
\[\sin(b-a)=\dfrac{QR}{200}.\]
We want $QR$ to be less than or equal to $100$ ; this is equivalent to $\dfrac{QR}{200}\le\dfrac12.$ We solve from there:
\begin{align*}
\dfrac{QR}{200}&\le\dfrac12 \\
\sin(b-a)&\le\dfrac12 \\
\arcsin(\sin(b-a))&\le\arcsin\left(\dfrac12\right) \\
b-a&\le30^\circ. \\
\end{align*}
(Notice that if $a>b$ , then this would become $a-b\le30^circ.$ As in Solution 1, we can write $|a-b|\le30$ .) One can now proceed as in Solution 1, but let us tackle the geometric probability for completeness.
We now have transformed this problem into another problem asking for the probability of two uniformly, randomly, and independently chosen real numbers between $0$ and $75$ being no more than $30$ from each other.
If the first number (let this be $x$ ) is between $30$ and $45$ , then the other number can be from $x-30$ to $x+30$ - a range of $60$ . Thus, the probability that this contributes is $\dfrac{45-30}{75}\cdot\dfrac{60}{75}=\dfrac4{25}$
If $x$ is between $0$ and $30$ or $45$ to $75$ (these two cases are equivalent), the chance is the same as that of the average value since the ranges are uniform. For $x=15$ (the average), the second number can be from $0$ to $15+30=45$ - a range of $45$ . The total range is $(30-0)+(75-45)=30+30=60$ . Thus, this case contributes $\dfrac{45}{75}\cdot\dfrac{60}{75}=\dfrac{12}{25}$
Adding the two, we get $\dfrac{16}{25}$ for an answer of $16+25=\boxed{041}$


Answer: 59.0

Two real numbers are selected independently at random from the interval $[-20, 10]$ . What is the probability that the product of those numbers is greater than zero?
$\textbf{(A)}\ \frac{1}{9} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{4}{9} \qquad\textbf{(D)}\ \frac{5}{9} \qquad\textbf{(E)}\ \frac{2}{3}$

Solution:
We will use complementary counting. The probability that the product is negative can be found by finding the probability that one number is positive and the other number is negative. The probability of a positive number being selected is $\frac13$ , and the probability of a negative number being selected is $\frac23$ . So, we see that the probability of the product being negative is \[2 \cdot \frac23 \cdot \frac13 = \frac49.\]
We multiply by $2$ because you can either pick the negative or positive number first.
Thus, the probability of the product being positive is \[1-\frac49 = \boxed{59}\]


Answer: 4.0

Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("$3$",(-0.9,h1*s),N,fontsize(10));  real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1); draw(ellipse((d,0),r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]
$\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1$

Solution: The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\boxed{4}$ times as much.


Answer: 4.0

Solution: The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\boxed{4}$ times as much.


Answer: 21.0

Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?
$\textbf{(A)}\  \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\  \text{The first height is } 10\% \text{ more than the second.}\\ \textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}\\ \textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}$

Solution: Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \[r_2=\frac{11r_1}{10}\] \[\pi r_1^2h_1=\pi r_2^2h_2\] Substituting the first equation into the second and dividing both sides by $\pi$ , we get \[r_1^2h_1=\frac{121r_1^2}{100}h_2\implies h_1=\frac{121h_2}{100}.\] Therefore, $\boxed{21}$


Answer: 21.0

Solution: Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \[r_2=\frac{11r_1}{10}\] \[\pi r_1^2h_1=\pi r_2^2h_2\] Substituting the first equation into the second and dividing both sides by $\pi$ , we get \[r_1^2h_1=\frac{121r_1^2}{100}h_2\implies h_1=\frac{121h_2}{100}.\] Therefore, $\boxed{21}$


Answer: 12.0

Two sides of a triangle have lengths $10$ and $15$ .  The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides.  How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$

Solution:
The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,
therefore the length of the side perpendicular to that altitude will be between $10$ and $15$ . The only answer choice that meets this requirement is $\boxed{12}$

Solution:
Let the height to the side of length $15$ be $h_{1}$ , the height to the side of length 10 be $h_{2}$ , the area be $A$ , and the height to the unknown side be $h_{3}$
Because the area of a triangle is $\frac{bh}{2}$ , we get that $15(h_{1}) = 2A$ and $10(h_{2}) = 2A$ , so, setting them equal, $h_{2} = \frac{3h_{1}}{2}$ .  From the problem, we know that $2h_{3} = h_{1} + h_{2}$ .  Substituting, we get that \[h_{3} = 1.25h_{1}.\] Thus, the side length is going to be $\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{12}$


Answer: 12.0

Solution:
Let $x$ be the height of triangle when the base is $10$ and $y$ is the height of the triangles when the base is $15$ . This means the height for when the triangles has the $3$ rd side length, the height would be $\dfrac{(x+y)}{2}$ giving us the following equation:
$\dfrac{10x}{2}=\dfrac{15y}{2}=\dfrac{n(x+y)}{2}$ For $n$ , we can plug in the answer choices and check when the ratio of $\dfrac{x}{y}$ for $\dfrac{10x}{2}=\dfrac{n(x+y)}{2}$ and $\dfrac{15y}{2}=\dfrac{n(x+y)}{2}$ is the same because we can observe that only for one value, the ratio remains constant. From brute force we get that $\boxed{12}$ meaning that $12$ is the $3$ rd base.
~ Batmanstark


Answer: 160.0

Two skaters, Allie and Billie, are at points $A$ and $B$ , respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$ . At the same time Allie leaves $A$ , Billie leaves $B$ at a speed of $7$ meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?

Solution:
Label the point of intersection as $C$ . Since $d = rt$ $AC = 8t$ and $BC = 7t$ . According to the law of cosines
\begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60^\circ\\ 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\ t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100}{3}.\end{align*}
Since we are looking for the earliest possible intersection, $20$ seconds are needed. Thus, $8 \cdot 20 = \boxed{160}$ meters is the solution.

Solution:
Let $P$ be the point of intersection between the skaters, Allie and Billie. We can draw a line that goes through $P$ and is parallel to $\overline{AB}$ . Letting this line be the $x$ -axis, we can reflect $B$ over the $x$ -axis to get $B'$ . As reflections preserve length, $B'X = XB$
We then draw lines $BB'$ and $PB'$ . We can let the foot of the perpendicular from $P$ to $BB'$ be $X$ , and we can let the foot of the perpendicular from $P$ to $AB$ be $Y$ . In doing so, we have constructed rectangle $PXBY$
By $d=rt$ , we have $AP = 8t$ and $PB = PB' = 7t$ , where $t$ is the number of seconds it takes the skaters to meet. Furthermore, we have $30-60-90$ triangle $PAY$ , so $AY = 4t$ , and $PY = 4t\sqrt{3}$ . Since we have $PY = XB = B'X$ $B'X = 4t\sqrt{3}$ . By Pythagoras, $PX = t$
As $PXBY$ is a rectangle, $PX = YB$ . Thus $AY + YB = AB \Rightarrow AY + PX = AB$ , so we get $4t + t = 100$ . Solving for $t$ , we find $t = 20$
Our answer, $AP$ , is equivalent to $8t$ . Thus, $AP = 8 \cdot 20 = \boxed{160}$

Solution:
We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$ . Then we can produce the following diagram:

Now, if we drop an altitude from point $M$ , we get :

We know this from the $30-60-90$ triangle that is formed. From this we get that: \[(7x)^2 = (4 \sqrt{3} x)^2 + (100-4x)^2\]
\[\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2\]
\[\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)\]
Therefore, we get that $x = \frac{100}{3}$ or $x = 20$ . Since $20< \frac{100}{3}$ , we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \cdot x = 8 \cdot 20 = \boxed{160}$ meters.


Answer: 335.0

Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
This solution refers to the Diagram section.
As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$ ) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\mathcal{P}.$ Let $\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is the foot of the perpendicular from $O_3$ to line $\ell,$ so $\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\overline{O_1O_2}.$ We wish to find $T_3A.$  Note that:
Now, we focus on cross-sections $O_1O_3T_3T_1$ and $\mathcal{R}:$
We have the following diagram:  In cross-section $O_1O_3T_3T_1,$ since $\overline{O_1T_1}\parallel\overline{O_3T_3}$ as discussed, we obtain $\triangle O_1T_1B\sim\triangle O_3T_3B$ by AA, with the ratio of similitude $\frac{O_1T_1}{O_3T_3}=\frac{36}{13}.$ Therefore, we get $\frac{O_1B}{O_3B}=\frac{49+O_3B}{O_3B}=\frac{36}{13},$ or $O_3B=\frac{637}{23}.$
In cross-section $\mathcal{R},$ note that $O_1O_3=49$ and $DO_3=\frac{O_1O_2}{2}=36.$ Applying the Pythagorean Theorem to right $\triangle O_1DO_3,$ we have $O_1D=\sqrt{1105}.$ Moreover, since $\ell\perp\overline{O_1C}$ and $\overline{DO_3}\perp\overline{O_1C},$ we obtain $\ell\parallel\overline{DO_3}$ so that $\triangle O_1CB\sim\triangle O_1DO_3$ by AA, with the ratio of similitude $\frac{O_1B}{O_1O_3}=\frac{49+\frac{637}{23}}{49}.$ Therefore, we get $\frac{O_1C}{O_1D}=\frac{\sqrt{1105}+DC}{\sqrt{1105}}=\frac{49+\frac{637}{23}}{49},$ or $DC=\frac{13\sqrt{1105}}{23}.$
Finally, note that $\overline{O_3T_3}\perp\overline{T_3A}$ and $O_3T_3=13.$ Since quadrilateral $DCAO_3$ is a rectangle, we have $O_3A=DC=\frac{13\sqrt{1105}}{23}.$ Applying the Pythagorean Theorem to right $\triangle O_3T_3A$ gives $T_3A=\frac{312}{23},$ from which the answer is $312+23=\boxed{335}.$

Solution:
The centers of the three spheres form a $49$ $49$ $72$ triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the $72$ side of this triangle. Take its midpoint $M$ , which is $36$ away from the midpoint $A$ of the $72$ side, and connect these two midpoints.
Now consider the point at which the plane is tangent to the small sphere, and connect $M$ with the small sphere's tangent point $B$ . Extend $\overline{MB}$ through $B$ until it hits the ray from $A$ through the center of the small sphere (convince yourself that these two intersect). Call this intersection $D$ , the center of the small sphere $C$ , we want to find $BD$
By Pythagoras, $AC=\sqrt{49^2-36^2}=\sqrt{1105}$ , and we know that $MA=36$ and $BC=13$ . We know that $\overline{MA}$ and $\overline{BC}$ must be parallel, using ratios we realize that $CD=\frac{13}{23}\sqrt{1105}$ . Apply the Pythagorean theorem to $\triangle BCD$ $BD=\frac{312}{23}$ , so $312 + 23 = \boxed{335}$

Solution:
This solution refers to the Diagram section.  The isosceles triangle of centers $O_1 O_2 O$ $O$ is the center of sphere of radii $13$ ) has sides $O_1 O = O_2 O = 36 + 13 = 49,$ and $O_1 O_2 = 36 + 36 = 72.$
Let $N$ be the midpoint $O_1 O_2$
The isosceles triangle of points of tangency $T_1 T_2 T$ has sides $T_1 T = T_2 T = 2 \sqrt{13 \cdot 36} = 12 \sqrt{13}$ and $T_1 T_2 = 72.$
Let $M$ be the midpoint $T_1 T_2.$
The height $TM$ is $\sqrt {12^2 \cdot 13 - 36^2} = 12 \sqrt {13-9} = 24.$
The tangents of the half-angle between the planes is $\frac {TO}{AT} = \frac {MN - TO}{TM},$ so $\frac {13}{AT} = \frac {36 - 13}{24},$ \[AT = \frac{24\cdot 13}{23} = \frac {312}{23} \implies  312 + 23 = \boxed{335}.\] vladimir.shelomovskii@gmail.com, vvsss


Answer: 300.0

Two squares of a $7\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible?

Solution:
There are ${49 \choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards.
Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\frac{49-1}{2}=24$ such pairs. There are then ${49 \choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each of the ${49 \choose 2}-24$ pairs, there are three other pairs that yield an equivalent board.
Thus, the number of inequivalent boards is $\frac{{49 \choose 2} - 24}{4} + \frac{24}{2} = \boxed{300}$ . For a $(2n+1) \times (2n+1)$ board, this argument generalizes to $n(n+1)(2n^2+2n+1)$ inequivalent configurations.

Solution:
There are 4 cases:
1. The center square is occupied, in which there are $12$ cases.
2. The center square isn't occupied and the two squares that are opposite to each other with respect to the center square, in which there are $12$ cases.
3. The center square isn't occupied and the two squares can rotate to each other with a $90^{\circ}$ rotation with each other and with respect to the center square, in which case there are $12$ cases.
4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are $\dbinom{12}{2} \cdot \frac{16}{4} = 264$ cases.
Add up all the values for each case to get $\boxed{300}$ as your answer.


Answer: 40.0

Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$

Solution:
First, choose the two letters to be repeated in each set. $\dbinom{5}{2}=10$ .  Now we have three remaining elements that we wish to place into two separate subsets. There are $2^3 = 8$ ways to do so because each of the three remaining letters can be placed either into the first or second subset. Both of those subsets contain the two chosen elements, so their intersection is the two chosen elements). Unfortunately, we have over-counted  (Take for example $S_{1} = \{a,b,c,d \}$ and $S_{2} = \{a,b,e \}$ ). Notice how $S_{1}$ and $S_{2}$ are interchangeable. Division by two will fix this problem. Thus we have:
$\dfrac{10 \times 8}{2} = 40 \implies \boxed{40}$

Solution:
Another way of looking at this problem is to break it down into cases.
First, our two subsets can have 2 and 5 elements. The 5-element subset (aka the set) will contain the 2-element subset. There are $\dbinom{5}{2}=10$ ways to choose the 2-element subset. Thus, there are $10$ ways to create these sets.
Second, the subsets can have 3 and 4 elements. $3+4=7$ non-distinct elements. $7-5=2$ elements in the intersection. There are $\dbinom{5}{3}=10$ ways to choose the 3-element subset. For the 4-element subset, two of the elements must be the remaining elements (not in the 3-element subset). The other two have to be a subset of the 3-element subset. There are $\dbinom{3}{2}=3$ ways to choose these two elements, which means there are 3 ways to choose the 4-element subset. Therefore, there are $10\cdot3=30$ ways to choose these sets.
This leads us to the answer:
$10+30=40 \implies \boxed{40}$

Solution: There are $\dbinom{5}{2}=10$ ways to choose the 2 shared elements. We now must place the 3 remaining elements into the subsets. Using stars and bars, we can notate this as: $I I I X \rightarrow \dbinom{4}{3}=4$ . Thus, $4*10=\boxed{40}$


Answer: 20.0

Two swimmers, at opposite ends of a $90$ -foot pool, start to swim the length of the pool,
one at the rate of $3$ feet per second, the other at $2$ feet per second.
They swim back and forth for $12$ minutes. Allowing no loss of times at the turns, find the number of times they pass each other.
$\textbf{(A)}\ 24\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 18$

Solution:
First, note that it will take $30$ seconds for the first swimmer to reach the other side and $45$ seconds for the second swimmer to reach the other side.  Also, note that after $180$ seconds (or $3$ minutes), both swimmers will complete an even number of laps, essentially returning to their starting point.

At this point, find the number of meeting points in the first $3$ minutes, then multiply by four to get the answer.  From the graph (where the x-axis is the time in seconds and the y-axis is distance from one side of the pool), there are five meeting points, so the two swimmers will pass each other $20$ times, which is answer choice $\boxed{20}$


Answer: 40.0

Two tangents are drawn to a circle from an exterior point $A$ ; they touch the circle at points $B$ and $C$ respectively.
A third tangent intersects segment $AB$ in $P$ and $AC$ in $R$ , and touches the circle at $Q$ . If $AB=20$ , then the perimeter of $\triangle APR$ is
$\textbf{(A)}\ 42\qquad \textbf{(B)}\ 40.5 \qquad \textbf{(C)}\ 40\qquad \textbf{(D)}\ 39\frac{7}{8} \qquad \textbf{(E)}\ \text{not determined by the given information}$

Solution: Since $Q$ can be anywhere on the circle between $A$ and $B$ , it can basically be "on top" of $A$ . Then $R$ will be at the same point as $A$ , so $APR$ form a degenerate triable with side length $AB=20$ . So its perimeter will be $40$ . Since $BP=PQ$ and $QR=CR$ by power of a point, as $AP$ and $AR$ decrease in length, $PR=PQ+QR$ will "grow" to compensate, so the perimeter will stay constant with a value of $\boxed{40}$


Answer: 36.0

Two tangents to a circle are drawn from a point $A$ .  The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$ .  What is the degree measure of $\angle{BAC}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60$

Solution:
In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).
In order to utilize this theorem, the degree measures of the arcs must be found.  First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d.
Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.
Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:
1/2 (216°-144°) = 1/2 (72°) $=\boxed{36}.$

Solution: Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is $\frac{3x-2x}2 = \frac {x} 2$ by the arc length formula. Also note that $3x + 2x = 360$ , which simplifies to $x= 72.$ Hence the angle formed by the tangents is equal to $\boxed{36}$

Solution:
Let the center of the circle be $O$ . The radii extending from $O$ to the points of tangency $B$ and $C$ are both perpendicular to lines $AB$ and $AC$ . Thus $\angle ABO = \angle ACO = 90^{\circ}$ , and quadrilateral $ABOC$ is cyclic, since the opposite angles add to $180^{\circ}$
Since the ratio of the arc lengths is $2 : 3$ , the shorter arc (the arc cut off by $\angle BOC$ ) is $144^{\circ}$
$\angle BOC$ and $\angle BAC$ must add up to $180^{\circ}$ since quadrilateral $ABOC$ is cyclic, so $\angle BAC = 180 - 144 = \boxed{36}$


Answer: 118.0

Two thousand points are given on a circle . Label one of the points $1$ . From this point, count $2$ points in the clockwise direction and label this point $2$ . From the point labeled $2$ , count $3$ points in the clockwise direction and label this point $3$ . (See figure.) Continue this process until the labels $1,2,3\dots,1993\,$ are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as $1993$
AIME 1993 Problem 9.png

Solution:
The label $1993$ will occur on the $\frac12(1993)(1994) \pmod{2000}$ th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$
Simplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{2000}$ . Therefore, one of $1993 - n$ or $1994 + n$ is odd, and each of them must be a multiple of $125$ or $16$
For $1993 - n$ to be a multiple of $125$ and $1994 + n$ to be a multiple of $16$ $n \equiv 118 \pmod {125}$ and $n\equiv 6 \pmod {16}$ . The smallest $n$ for this case is $118$
In order for $1993 - n$ to be a multiple of $16$ and $1994 + n$ to be a multiple of $125$ $n\equiv 9\pmod{16}$ and $n\equiv 6\pmod{125}$ . The smallest $n$ for this case is larger than $118$ , so $\boxed{118}$ is our answer.

Solution:
Two labels $a$ and $b$ occur on the same point if $\ a(a+1)/2\equiv \ b(b+1)/2\pmod{2000}$ . If we assume the final answer be $n$ , then we have $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$
Multiply $2$ on both side we have $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{4000}$ . As they have different parities, the even one must be divisible by $32$ . As $(1993 - n)+(1994 + n)\equiv 2\pmod{5}$ , one of them is divisible by $5$ , which indicates it's divisible by $125$
Which leads to four different cases: $1993-n\equiv 0\pmod{4000}$ $1994+n\equiv 0\pmod{4000}$ $1993-n\equiv 0\pmod{32}$ and $1994+n\equiv 0\pmod{125}$ $1993-n\equiv 0\pmod{125}$ and $1994+n\equiv 0\pmod{32}$ . Which leads to $n\equiv 1993,2006,3881$ and $118\pmod{4000}$ respectively, and only $n=118$ satisfied.Therefore answer is $\boxed{118}$ .(by ZJY)


Answer: 532.0

Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$ , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a^{}_{}$ when it should have been a $b^{}_{}$ , or as a $b^{}_{}$ when it should be an $a^{}_{}$ . However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let $S_a^{}$ be the three-letter string received when $aaa^{}_{}$ is transmitted and let $S_b^{}$ be the three-letter string received when $bbb^{}_{}$ is transmitted. Let $p$ be the probability that $S_a^{}$ comes before $S_b^{}$ in alphabetical order. When $p$ is written as a fraction in lowest terms , what is its numerator

Solution:
Let us make a chart of values in alphabetical order, where $P_a,\ P_b$ are the probabilities that each string comes from $aaa$ and $bbb$ multiplied by $27$ , and $S_b$ denotes the partial sums of $P_b$ (in other words, $S_b = \sum_{n=1}^{b} P_b$ ): \[\begin{array}{|r||r|r|r|} \hline \text{String}&P_a&P_b&S_b\\ \hline aaa & 8 & 1 & 1 \\ aab & 4 & 2 & 3 \\ aba & 4 & 2 & 5 \\ abb & 2 & 4 & 9 \\ baa & 4 & 2 & 11 \\ bab & 2 & 4 & 15 \\ bba & 2 & 4 & 19 \\ bbb & 1 & 8 & 27 \\ \hline \end{array}\]
The probability is $p=\sum P_a \cdot (27 - S_b)$ , so the answer turns out to be $\frac{8\cdot 26 + 4 \cdot 24 \ldots 2 \cdot 8 + 1 \cdot 0}{27^2} = \frac{532}{729}$ , and the solution is $\boxed{532}$

Solution:
Let $S(a,n)$ be the $n$ th letter of string $S(a)$ .
Compare the first letter of the string $S(a)$ to the first letter of the string $S(b)$ .
There is a $(2/3)^2=4/9$ chance that $S(a,1)$ comes before $S(b,1)$ .
There is a $2(1/3)(2/3)=4/9$ that $S(a,1)$ is the same as $S(b,1)$
If $S(a,1)=S(b,1)$ , then you do the same for the second letters of the strings.  But you have to multiply the $4/9$ chance that $S(a,2)$ comes before $S(b,2)$ as there is a $4/9$ chance we will get to this step.
Similarly, if $S(a,2)=S(b,2)$ , then there is a $(4/9)^3$ chance that we will get to comparing the third letters and that $S(a)$ comes before $S(b)$
So we have $p=(4/9)+(4/9)^2+(4/9)^3=4/9+16/81+64/729=532/729$ . Therefore, the answer is $\boxed{532}$

Solution: Consider $n$ letter strings instead. If the first letters all get transmitted correctly, then the $a$ string will be first. Otherwise, the only way is for both of the first letters to be the same, and then we consider the next $n-1$ letter string following the first letter. This easily leads to a recursion: $p_n=\frac23\cdot\frac23+2\cdot\frac23\cdot\frac13p_{n-1}=\frac49+\frac49p_{n-1}$ . Clearly, $p_0=0\implies p_1=\frac49\implies p_2=\frac{52}{81}\implies p_3=\frac{532}{729}$ . Therefore, the answer is $\boxed{532}$

Solution: The probability that $S_a$ will take the form $a$ _ _ and that $S_b$ will take the form $b$ _ _ is $\frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}$ . Then, the probability that both $S_a$ and $S_b$ will share the same first digit is $2\cdot\frac{2}{3}\cdot\frac{1}{3} = \frac{4}{9}$ . Now if the first digits of either sequence are the same, then we must now consider these same probabilities for the second letter of each sequence. The probability that when the first two letters of both sequences are the same, that the second letter of $S_a$ is $a$ and that the second letter of $S_b$ is $b$ is $\frac{2}{3}\cdot\frac{2}{3}=\frac{4}{9}$ . Similarly, the probability that when the first two letters of both sequences are the same, that the second set of letters in both sets of sequences are the same is $2\cdot\frac{2}{3}\cdot\frac{1}{3} = \frac{4}{9}$ . Now, if the last case is true then the probability that $S_a$ precedes $S_b$ is $\frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}$ . Therefore the total probability would be: $\frac{4}{9} + \frac{4}{9}\left(\frac{4}{9} + \frac{4}{9}\left(\frac{4}{9}\right)\right) = \frac{4}{9}+\frac{4}{9}\left(\frac{52}{81}\right) = \frac{4}{9} + \frac{208}{729} = \frac{532}{729}$ . Therefore the answer is $\boxed{532}$


Answer: 90.0

Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$

Solution: Call the given grid "Grid A". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions $(n-1) \times (n-1)$ . There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of edges in Grid B is $2n(n-1)$ , and the number of ways to pick two squares out of Grid A is $\dbinom{n^2}{2}$ . So, the probability that the two chosen squares are adjacent is $\frac{2n(n-1)}{\binom{n^2}{2}} = \frac{2n(n-1)}{\frac{n^2(n^2-1)}{2}} = \frac{4}{n(n+1)}$ . We wish to find the smallest positive integer $n$ such that $\frac{4}{n(n+1)} < \frac{1}{2015}$ , and by inspection the first such $n$ is $\boxed{090}$

Solution:
Consider a $3 \times 3$ grid, where there are $4$ corner squares, $4$ edge squares, and $1$ center square. A $4 \times 4$ grid has $4$ corner squares, $8$ edge squares, and $4$ center squares. By examining simple cases, we can conclude that for a grid that is $n \times n$ , there are always $4$ corners squares, $4(n-2)$ edge squares, and $n^2-4n+4$ center squares.
Each corner square is adjacent to $2$ other squares, edge squares to $3$ other squares, and center squares to $4$ other squares. In the problem, the second square can be any square that is not the first, which means there are $n^2-1$ to choose from. With this information, we can conclude that the probability that second unit square is adjacent to the first is $\frac{2}{n^2-1}(\frac{4}{n^2}) +\frac{3}{n^2-1}(\frac{4(n-2)}{n^2}) +\frac{4}{n^2-1}(\frac{n^2-4n+4}{n^2})$
Simplifying, we get $\frac{4}{n(n+1)}$ which we can set to be less than $\frac{1}{2015}$ . By inspection, we find that the first such $n$ is $\boxed{090}$

Solution: There are 3 cases in this problem. Number one, the center squares. Two, the edges, and three, the corners. The probability of getting one center square and an adjacent square is $\frac{(n-2)(n-2)}{n^2}$ multiplied by $\frac{4}{n^2 -1}$ . Add that to the probability of an edge and an adjacent square( $\frac{4n-8}{n^2}$ multiplied by $\frac{3}{n^2-1}$ ) and the probability of a corner and an adjacent square( $\frac{4}{n^2}$ multiplied by $\frac{2}{n^2-1}$ ) to get $\frac{4n^2-4n}{n^4-n^2}$ . Simplify to get $\frac{4}{n^2+n}$ . With some experimentation, we realize that the smallest value of n is $\boxed{090}$


Answer: 4.0

Two walls and the ceiling of a room meet at right angles at point $P.$ A fly is in the air one meter from one wall, eight meters from the other wall, and nine meters from point $P$ . How many meters is the fly from the ceiling?
$\text{(A) }\sqrt{13} \qquad \text{(B) }\sqrt{14} \qquad \text{(C) }\sqrt{15} \qquad \text{(D) }4 \qquad \text{(E) }\sqrt{17}$

Solution: We can use the formula for the diagonal of the rectangle, or $d=\sqrt{a^2+b^2+c^2}$ The problem gives us $a=1, b=8,$ and $c=9.$ Solving gives us $9=\sqrt{1^2 + 8^2 + c^2} \implies c^2=9^2-8^2-1^2 \implies c^2=16  \implies c=\boxed{4}.$


Answer: 4.0

Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ $1$
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$

Solution:
This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.
The first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$
To solve the system of equations:
$p=3c-4$
$p=4c-12$
$3c-4=4c-12$
$c=8$
$p=20.$
Let $x$ be the number of years until Pete is twice as old as his cousin.
$20+x=2(8+x)$
$20+x=16+2x$
$x=4$
The answer is $\boxed{4}$


Answer: 4.0

Solution:
This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.
The first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$
To solve the system of equations:
$p=3c-4$
$p=4c-12$
$3c-4=4c-12$
$c=8$
$p=20.$
Let $x$ be the number of years until Pete is twice as old as his cousin.
$20+x=2(8+x)$
$20+x=16+2x$
$x=4$
The answer is $\boxed{4}$


Answer: 27.0

Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are $6$ empty chairs, how many people are in the room?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

Solution: Working backwards, if $3/4$ of the chairs are taken and $6$ are empty, then there are three times as many taken chairs as empty chairs, or $3 \cdot 6 = 18$ . If $x$ is the number of people in the room and $2/3$ are seated, then $\frac23 x = 18$ and $x = \boxed{27}$


Answer: 72.0

Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose?
$\text{(A)}\ 4 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 72 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 144$

Solution: There are $3$ possibilities for the meat and $4$ possibilites for the dessert, for a total of $4\times3=12$ possibilities for the meat and the dessert. There are $4$ possibilities for the first vegetable and $3$ possibilities for the second, but order doesn't matter, so we overcounted by a factor of $2$ . For example, we counted 'baked beans and corn' and 'corn and baked beans' as $2$ different possibilities, so the total possibilites for the two vegetables is $\frac{4\times3}{2}=6$ , and the total number of possibilites is $12\times6=\boxed{72}.$


Answer: 25.0

Tyrone had $97$ marbles and Eric had $11$ marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 18 \qquad \mathrm{(D)}\ 25 \qquad \mathrm{(E)}\ 29$

Solution: Let $x$ be the number of marbles Tyrone gave to Eric. Then, $97-x = 2\cdot(11+x)$ . Solving for $x$ yields $75=3x$ and $x = 25$ . The answer is $\boxed{25}$

Solution: Since the number of balls Tyrone and Eric have a specific ratio when Tyrone gives some of his balls to Eric, we can divide the total amount of balls by their respective ratios. Altogether, they have $97+11$ balls, or $108$ balls, and it does not change when Tyrone gives some of his to Eric, since none are lost in the process. After Tyrone gives some of his balls away, the ratio between the balls is $2:1$ , and added together is $108$ . The two numbers add up to $3$ and we divide $108$ by $3$ , and the outcome is $36$ . This is the number of balls Eric has, and so doubling that results in the number of balls Tyrone has, which is $72$ $97-72$ is $25$ , and $36-11$ is $25$ , thus proving our statement true, and the answer is $\boxed{25}$


Answer: 56.0

Usain is walking for exercise by zigzagging across a $100$ -meter by $30$ -meter rectangular field, beginning at point $A$ and ending on the segment $\overline{BC}$ . He wants to increase the distance walked by zigzagging as shown in the figure below $(APQRS)$ . What angle $\theta = \angle PAB=\angle QPC=\angle RQB=\cdots$ will produce in a length that is $120$ meters? (This figure is not drawn to scale. Do not assume that he zigzag path has exactly four segments as shown; there could be more or fewer.)
[asy] import olympiad; draw((-50,15)--(50,15)); draw((50,15)--(50,-15)); draw((50,-15)--(-50,-15)); draw((-50,-15)--(-50,15)); draw((-50,-15)--(-22.5,15)); draw((-22.5,15)--(5,-15)); draw((5,-15)--(32.5,15)); draw((32.5,15)--(50,-4.090909090909)); label("$\theta$", (-41.5,-10.5)); label("$\theta$", (-13,10.5)); label("$\theta$", (15.5,-10.5)); label("$\theta$", (43,10.5)); dot((-50,15)); dot((-50,-15)); dot((50,15)); dot((50,-15)); dot((50,-4.09090909090909)); label("$D$",(-58,15)); label("$A$",(-58,-15)); label("$C$",(58,15)); label("$B$",(58,-15)); label("$S$",(58,-4.0909090909)); dot((-22.5,15)); dot((5,-15)); dot((32.5,15)); label("$P$",(-22.5,23)); label("$Q$",(5,-23)); label("$R$",(32.5,23)); [/asy]
$\textbf{(A)}~\arccos\frac{5}{6}\qquad\textbf{(B)}~\arccos\frac{4}{5}\qquad\textbf{(C)}~\arccos\frac{3}{10}\qquad\textbf{(D)}~\arcsin\frac{4}{5}\qquad\textbf{(E)}~\arcsin\frac{5}{6}$

Solution: Drop an altitude from $P$ to $AB$ and let its base be $x$ . Note that if we repeat this for $Q$ and $R$ , all four right triangles (including $\triangle{RSC}$ ) will have the same trig ratios. By proportion, the hypotenuse $AP$ is $\frac{x}{100}(120) = \frac65 x$ , so $\cos\theta = \frac{x}{(\frac65x)} = \frac56 \Rightarrow \theta = \boxed{56}$


Answer: 30.0

Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1.  Length $r$ is found to be $32$ inches.  After rearranging the blocks as in Figure 2, length $s$ is found to be $28$ inches.  How high is the table?
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13pt)); path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle; path block = origin--(3,0)--(3,1.5)--(0,1.5)--cycle; path rotblock = origin--(1.5,0)--(1.5,3)--(0,3)--cycle; draw(table^^shift((14,0))*table); filldraw(shift((7,0))*block^^shift((5.5,7))*rotblock^^shift((21,0))*rotblock^^shift((18,7))*block,gray); draw((7.25,1.75)--(8.5,3.5)--(8.5,8)--(7.25,9.75),Arrows(size=5)); draw((21.25,3.25)--(22,3.5)--(22,8)--(21.25,8.25),Arrows(size=5)); unfill((8,5)--(8,6.5)--(9,6.5)--(9,5)--cycle); unfill((21.5,5)--(21.5,6.5)--(23,6.5)--(23,5)--cycle); label("$r$",(8.5,5.75)); label("$s$",(22,5.75)); [/asy]
$\textbf{(A) }28\text{ inches}\qquad\textbf{(B) }29\text{ inches}\qquad\textbf{(C) }30\text{ inches}\qquad\textbf{(D) }31\text{ inches}\qquad\textbf{(E) }32\text{ inches}$

Solution: Let $h$ $l$ , and $w$ represent the height of the table and the length and width of the wood blocks, respectively, in inches. From Figure 1, we have $l+h-w=32$ , and from Figure 2, $w+h-l=28$ . Adding the equations gives $2h=60 \implies h=30$ , which is $\boxed{30}$


Answer: 10.0

Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99$

Solution: You need $2$ dimes, $1$ nickel, and $4$ pennies for the first $25$ cents. From $26$ cents to $50$ cents, you only need to add $1$ quarter. From $51$ cents to $75$ cents, you also only need to add $1$ quarter. The same for $76$ cents to $99$ cents. Notice that instead of $100$ , it is $99$ . We are left with $3$ quarters, $1$ nickel, $2$ dimes, and $4$ pennies. Thus, the correct answer is $3+2+1+4=\boxed{10}$


Answer: 6.0

Solution:
There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN.
Since A can only go to C or D, which each only have 1 way to get to N each, there are $1+1=2$ ways to get from A to N.
Since B can only go to A, C or N, and A only has 2 ways to get to N, C only has 1 way and to get from B to N is only 1 way, there are $2+1+1=4$ ways to get from B to N.
M can only go to either B or A, A has 2 ways and B has 4 ways, so M has $4+2=6$ ways to get to N.
6 is $\boxed{6}$


Answer: 190.0

Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?
$\text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190$

Solution: Since a multiple-digit prime number is not divisible by either 2 or 5, it must end with 1, 3, 7, or 9 in the units place. The remaining digits given must therefore appear in the tens place. Hence our answer is $20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\Rightarrow\boxed{190}$


Answer: 115.0

Using the letters $A$ $M$ $O$ $S$ , and $U$ , we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" $USAMO$ occupies position
$\mathrm{(A) \ } 112\qquad \mathrm{(B) \ } 113\qquad \mathrm{(C) \ } 114\qquad \mathrm{(D) \ } 115\qquad \mathrm{(E) \ } 116$

Solution: There are $4!\cdot 4$ "words" beginning with each of the first four letters alphabetically. From there, there are $3!\cdot 3$ with $U$ as the first letter and each of the first three letters alphabetically. After that, the next "word" is $USAMO$ , hence our answer is $4\cdot 4!+3\cdot 3!+1=\boxed{115}$

Solution: Let $A = 1$ $M = 2$ $O = 3$ $S = 4$ , and $U = 5$ . Then counting backwards, $54321, 54312, 54231, 54213, 54132, 54123$ , so the answer is $\boxed{115}$


Answer: 90.0

Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$
$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$

Solution: Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn't matter. Then $C>A$ and $B>D$ . Therefore the number of ways to choose the four integers is $\tbinom{6}{2}\tbinom{4}{2}=90$ , and the answer is $\boxed{90}$

Solution: Setting $y = Ax^2+B = Cx^2+D$ , we find that $Ax^2-Cx^2 = x^2(A-C) = D-B$ , so $x^2 = \frac {D-B}{A-C} \ge 0$ by the trivial inequality. This implies that $D-B$ and $A-C$ must both be positive or negative. If two distinct values are chosen for $(A, C)$ and $(B, D)$ respectively, there are $2$ ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by $2$ at the end, however, since the $2$ curves aren't considered distinct. Calculating, we get \[\frac {1}{2} \cdot \binom {6}{2} \binom {4}{2} \cdot 2 = \boxed{90}.\] ~ike.chen

Solution:
Like in Solution 2 , we find $\frac {D-B}{A-C} \ge 0$ . Notice that, since $D \ne B$ , this expression can never equal $0$ , and since $A \ne C$ , there won't be a divide-by- $0$ . This means that every choice results in either a positive or a negative value.
For every choice of $(A,B,C,D)$ that results in a positive value, we can flip $B$ and $D$ to obtain a corresponding negative value. This is a bijection (we could flip $B$ and $D$ again to obtain the original choice (injectivity) and we could flip $B$ and $D$ from any negative choice to obtain the corresponding positive choice (surjectivity)), so half of the choices are positive (where the curves intersect) and half are negative (where they don't).
This means that of the $\frac{6\cdot5\cdot4\cdot3}{2} = 180$ total choices (dividing by $2$ because the order of the curves does not matter), half of them, or $\frac{180}{2} = \boxed{90}$ , lead to intersecting curves.


Answer: 4.2

Viswam walks half a mile to get to school each day. His route consists of $10$ city blocks of equal length and he takes $1$ minute to walk each block. Today, after walking $5$ blocks, Viswam discovers he has to make a detour, walking $3$ blocks of equal length instead of $1$ block to reach the next corner. From the time he starts his detour, at what speed, in mph, must he walk, in order to get to school at his usual time? [asy] // Diagram by TheMathGuyd size(13cm); // this is an important stickman to the left of the origin pair C=midpoint((-0.5,0.5)--(-0.6,0.05)); draw((-0.5,0.5)--(-0.6,0.05)); // Head to butt draw((-0.64,0.16)--(-0.7,0.2)--C--(-0.47,0.2)--(-0.4,0.22)); // LH-C-RH draw((-0.6,0.05)--(-0.55,-0.1)--(-0.57,-0.25)); draw((-0.6,0.05)--(-0.68,-0.12)--(-0.8,-0.20));  filldraw(circle((-0.5,0.5),0.1),white,black);  int i; real d,s; // gap and side d=0.2; s=1-2*d; for(i=0; i<10; i=i+1) {   //dot((i,0), red); //marks to start   filldraw((i+d,d)--(i+1-d,d)--(i+1-d,1-d)--(i+d,1-d)--cycle, lightgrey, black);   filldraw(conj((i+d,d))--conj((i+1-d,d))--conj((i+1-d,1-d))--conj((i+d,1-d))--cycle,lightgrey,black); }  fill((5+d,-d/2)--(6-d,-d/2)--(6-d,d/2)--(5+d,d/2)--cycle,lightred);  draw((0,0)--(5,0)--(5,1)--(6,1)--(6,0)--(10.1,0),deepblue+linewidth(1.25)); //Who even noticed label("School", (10,0),E, Draw()); [/asy] $\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.2 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 4.8 \qquad \textbf{(E)}\ 5$
(NOTE: THE FOLLOWING DIAGRAM WAS NOT SHOWN DURING THE ACTUAL EXAM, BUT IS NOW HERE TO GUIDE STUDENTS IN PICTURING THE PROBLEM)

Solution: Note that Viswam walks at a constant speed of $60$ blocks per hour as he takes $1$ minute to walk each block. After walking $5$ blocks, he has taken $5$ minutes, and he has $5$ minutes remaining, to walk $7$ blocks. Therefore, he must walk at a speed of $7 \cdot 60 \div 5 = 84$ blocks per hour to get to school on time, from the time he starts his detour. Since he normally walks $\frac{1}{2}$ mile, which is equal to $10$ blocks, $1$ mile is equal to $20$ blocks. Therefore, he must walk at $84 \div 20 = 4.2$ mph from the time he starts his detour to get to school on time, so the answer is $\boxed{4.2}$

Solution: Viswam walks $10$ blocks, or half a mile, in $10$ minutes. Therefore, he walks at a rate of $3$ mph. From the time he takes his detour, he must travel $7$ blocks instead of $5$ . Our final equation is $\frac{7}{5} \times 3 = \frac{21}{5} = \boxed{4.2}$

Solution:
We can cheese this problem.
Notice that Viswam will need to walk $7$ blocks during the second half as opposed to his normal $5$ blocks. Since rate is equal to distance over time, this implies that the final answer will likely be a multiple of $7$ , since you will need to convert $7$ blocks to miles. The only answer choice that is a multiple of $7$ is $\boxed{4.2}$

Solution:
To travel $\frac{1}{2}$ of a mile in total, each block must be $\frac{1}{20}$ of a mile long. Since Viswam takes $1$ minute to walk along each block, it would take him $10$ minutes normally.
Viswam has already travelled for $5$ mins by the time he encounters the detour, so he must travel $7$ block lengths in the remaining $5$ minutes. The distance he has to travel is $7\cdot \frac{1}{20} = \frac{7}{20}$ of a mile. Therefore,
\[\frac{7}{20} \mathrm{\ miles} = r\cdot 5\mathrm{\ mins}.\] \[\frac{7\mathrm{\ miles}}{100\mathrm{\ mins}} = r.\]
As $60$ mins equals $1$ hour, we set up the following proportion:
\[r = \frac{7\mathrm{\ miles}}{100\mathrm{\ mins}} = \frac{m\mathrm{\ miles}}{60\mathrm{\ mins}}.\]
Cross multiplying and cancelling units yields
\[m = \frac{7\cdot 60}{100},\]
or $\boxed{4.2}$

Solution:
When calculating the speed of his normal route, we get:
$\frac{0.5 \mathrm{miles}}{10 \mathrm{minutes}}= \frac{3 \mathrm{miles}}{60 \mathrm{minutes}}$
Keep in mind that his route is $10$ blocks long. Since we count $7$ blocks when the detour starts, than this would mean that he has $\frac{7}{10}$ miles left to walk, considering the normal amount of blocks that he usually walks. Multiplying by $6$ to get the rate, we get, $\frac{42}{10}=4.2$
Therefore, the answer is $\boxed{4.2}$

Solution:
We can start by splitting the walk into two sections: The 5-block section before the detour and the 7-block section during and after the detour.
Since Viswam is walking at his normal rate for the first 5, it took him 5 minutes to walk that much. Since it must take him a total of 10 min in order for him to reach on time, he has 5 min for the second section of his walk.
We can figure out that 1 block equals 0.05 mi by dividing 10 by 0.5. Now, we can convert 7 blocks per 5 min into 4.2 miles per 1 hour. So, we arrive at $\boxed{4.2}$


Answer: 60.0

Walter gets up at 6:30 a.m., catches the school bus at 7:30 a.m., has 6 classes that last 50 minutes each, has 30 minutes for lunch, and has 2 hours additional time at school.  He takes the bus home and arrives at 4:00 p.m.  How many minutes has he spent on the bus?
$\text{(A)}\ 30 \qquad \text{(B)}\ 60 \qquad \text{(C)}\ 75 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 120$

Solution:
There are $4\frac{1}{2}$ hours from 7:30 a.m. to noon, and $4$ hours from noon to 4 p.m., meaning Walter was away from home for a total of $8\frac{1}{2}$ hours, or $8.5\times 60 = 510$ minutes.
Tallying up the times he was not on the bus, he has $6\times 50 = 300$ minutes in classes, $30$ minutes at lunch, and $2\times 60 = 120$ minutes of "additional time".  That is a total of $300 + 30 + 120 = 450$ minutes that Walter was away from home, but not on the bus.
Therefore, Walter spent $510 - 450 = 60$ minutes on the bus, and the answer is $\boxed{60}$


Answer: 41.0

Walter has exactly one penny, one nickel, one dime and one quarter in his pocket.  What percent of one dollar is in his pocket?
$\text{(A)}\ 4\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 41\% \qquad \text{(E)}\ 59\%$

Solution: Walter has $1 + 5 + 10 + 25 = 41$ cents in his pocket.  There are $100$ cents in a dollar.  Therefore, Walter has $\frac{41}{100}\cdot 100\% = 41\%$ of a dollar in his pocket, and the right answer is $\boxed{41}$


Answer: 84.0

Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?
$\textbf{(A) }42\qquad\textbf{(B) }84\qquad\textbf{(C) }126\qquad\textbf{(D) }178\qquad\textbf{(E) }252$

Solution:
We need to find the least common multiple of the four numbers given.
$\textrm{LCM}(3, 4, 6, 7) = \textrm{LCM}(3, 2^2, 2 \cdot 3, 7) = 2^2 \cdot 3 \cdot 7 = 84$
So the answer is $\boxed{84}$


Answer: 5.0

What are the sign and units digit of the product of all the odd negative integers strictly greater than $-2015$ $\textbf{(A) }$ It is a negative number ending with a 1. $\textbf{(B) }$ It is a positive number ending with a 1. $\textbf{(C) }$ It is a negative number ending with a 5. $\textbf{(D) }$ It is a positive number ending with a 5. $\textbf{(E) }$ It is a negative number ending with a 0.

Solution:
Since $-5>-2015$ , the product must end with a $5$
The multiplicands are the odd negative integers from $-1$ to $-2013$ . There are $\frac{|-2013+1|}2+1=1006+1$ of these numbers. Since $(-1)^{1007}=-1$ , the product is negative.
Therefore, the answer must be $\boxed{5.}$


Answer: 0.0

Solution:
Since $-1$ raised to an odd integer is $-1$ and $-1$ raised to an even integer exponent is $1$
$(-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = \boxed{0}.$


Answer: 64.0

Solution: We can synchronously multiply ${2^3}$ to the expresions both above and below the fraction bar. Thus, \[\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.\] Hence, the fraction equals to $\boxed{64}$

Solution: We have \[\frac{2^3+2^3}{2^{-3}+2^{-3}} = \frac{8 + 8}{\frac{1}{8} + \frac{1}{8}} = \frac{16}{\frac{1}{4}} = 16 \cdot 4 = 64,\] so our answer is $\boxed{64}$


Answer: 1010000.0

Solution: Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{j=1}i + \sum^{100}_{i=1} \sum^{100}_{j=1}j \\ &= \sum^{100}_{i=1} (100i) + 100 \sum^{100}_{j=1}j \\ &= 100 \sum^{100}_{i=1}i + 100 \sum^{100}_{j=1}j \\ &= 100\cdot5050 + 100\cdot5050 \\ &= \boxed{1010000} ~MRENTHUSIASM

Solution: Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \left(\sum^{100}_{j=1} i + \sum^{100}_{j=1} j\right) \\ &= \sum^{100}_{i=1} (100i+5050) \\ &= 100\sum^{100}_{i=1} i + \sum^{100}_{i=1} 5050 \\ &= 100\cdot5050+5050\cdot100 \\ &= \boxed{1010000} ~Vfire ~MRENTHUSIASM

Solution: Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ Since the nested summation is symmetric with respect to $i$ and $j,$ it follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{i=1} (2i) \\ &= 2\sum^{100}_{i=1} \sum^{100}_{i=1} i \\ &= 2\sum^{100}_{i=1} 5050 \\ &= 2\cdot(5050\cdot100) \\ &= \boxed{1010000} ~Vfire ~MRENTHUSIASM

Solution: The sum contains $100\cdot100=10000$ terms, and the average value of both $i$ and $j$ is $\frac{101}{2}.$ Therefore, the sum becomes \[10000\cdot\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{1010000}.\] ~Rejas ~MRENTHUSIASM

Solution:
We start by writing out the first few terms: \[\begin{array}{ccccccccc} (1+1) &+ &(1+2) &+ &(1+3) &+ &\cdots &+ &(1+100) \\ (2+1) &+ &(2+2) &+ &(2+3) &+ &\cdots &+ &(2+100) \\ (3+1) &+ &(3+2) &+ &(3+3) &+ &\cdots &+ &(3+100) \\ [-1ex] &&&&\vdots&&&& \\ (100+1) &+ &(100+2) &+ &(100+3) &+ &\cdots &+ &(100+100). \end{array}\] From the first terms in the parentheses, the sum $1+2+3+\cdots+100$ occurs $100$ times vertically.
From the second terms in the parentheses, the sum $1+2+3+\cdots+100$ occurs $100$ times horizontally.
Recall that the sum of the first $100$ positive integers is $1+2+3+\cdots+100=\frac{101\cdot100}{2}=5050.$ Therefore, the answer is \[2\cdot\left(5050\cdot100\right)=\boxed{1010000}.\] ~RandomPieKevin ~MRENTHUSIASM

Solution:
When we expand the nested summation as shown in Solution 5, note that:
Together, the nested summation becomes \begin{align*} \sum^{100}_{k=1}\left[(k+1)k\right] + \sum^{99}_{k=1}\left[(k+101)(100-k)\right] &= \sum^{100}_{k=1}\left[k^2+k\right] + \sum^{99}_{k=1}\left[-k^2-k+10100\right] \\ &= \sum^{100}_{k=1}k^2 + \sum^{100}_{k=1}k - \sum^{99}_{k=1}k^2 - \sum^{99}_{k=1}k + \sum^{99}_{k=1}10100 \\ &= \left(\sum^{100}_{k=1}k^2 - \sum^{99}_{k=1}k^2\right) + \left(\sum^{100}_{k=1}k - \sum^{99}_{k=1}k\right) + \sum^{99}_{k=1}10100 \\ &= 100^2+100+10100\cdot99 \\ &= \boxed{1010000} ~MRENTHUSIASM


Answer: 397.0

What is the $100\text{th}$ number in the arithmetic sequence $1,5,9,13,17,21,25,...$
$\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405$

Solution: To get from the $1^\text{st}$ term of an arithmetic sequence to the $100^\text{th}$ term, we must add the common difference $99$ times.  The first term is $1$ and the common difference is $5-1=9-5=13-9=\cdots = 4$ , so the $100^\text{th}$ term is \[1+4(99)=397 \rightarrow \boxed{397}\]

Solution:
Alternatively you could create an equation for the arithmetic sequence: $a_{n}=a_{1}+4(n-1)=1+4n-4=4n-3$
$a_{100}=4(100)-3=397$ , or $\boxed{397}$


Answer: 1.0

What is the $100^\text{th}$ digit to the right of the decimal point in the decimal form of $4/37$
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution:
$\frac{4}{37}=\frac{12}{111}=\frac{108}{999}=0.108108108...$
Since this repeats every three digits, digit number x = digit number (x mod 3), and the 100th digit = (100 mod 3)th digit = 1st digit = $\boxed{1}$


Answer: 24.0

What is the area enclosed by the graph of $|3x|+|4y|=12$
$\mathrm{(A)}\ 6      \qquad \mathrm{(B)}\ 12      \qquad \mathrm{(C)}\ 16      \qquad \mathrm{(D)}\ 24      \qquad \mathrm{(E)}\ 25$

Solution:
If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if $|a|=b$ , then $a$ is either $b$ or $-b$ ):
\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}
We can then put these equations in slope-intercept form in order to graph them.
\begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*}
Now you can graph the lines to find the shape of the graph:

We can easily see that it is a rhombus with diagonals of $6$ and $8$ . The area is $\dfrac{1}{2}\times 6\times8$ , or $\boxed{24}$

Solution:
The graph is symmetric with respect to both coordinate axes, and in the first quadrant it coincides with the graph of the line $3x + 4y = 12.$ Therefore the region is a rhombus, and the area is
\[\text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{24}\]


Answer: 2.0

What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2} \qquad\textbf{(B)}\ \pi+2 \qquad\textbf{(C)}\ \pi+2\sqrt{2} \qquad\textbf{(D)}\ 2\pi+\sqrt{2} \qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$

Solution:
Consider the case when $x \geq 0$ $y \geq 0$ \[x^2+y^2=x+y\] \[(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}\] Notice the circle intersect the axes at points $(0, 1)$ and $(1, 0)$ . Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of $\frac{\sqrt{2}}{2}$ and a triangle: \[A = \frac{\pi}{4} +\frac{1}{2}\]  Because of symmetry, the area is the same in all four quadrants.
The answer is $\boxed{2}$


Answer: 2.0

What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$

Solution:
Another way to solve this problem is using cases.
Though this may seem tedious, we only have to do one case since the area enclosed is symmetrical.
The equation for this figure is $x^2+y^2=|x|+|y|$ To make this as easy as possible,
we can make both $x$ and $y$ positive. Simplifying the equation for $x$ and $y$ being positive,
we get the equation $x^{2} +y^{2} -x-y = 0.$
Using the complete the square method, we get $\left(x-\frac{1}{2}\right)^{2} + \left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$
Therefore, the origin of this section of the shape is at $\left(\frac{1}{2}, \frac{1}{2}\right).$
Using the equation we can also see that the radius has a length of $\frac{\sqrt{2}}{2}$
With this shape we see that this shape can be cut into a right triangle and a semicircle.
The length of the hypotenuse of the triangle is $\sqrt{2}$ so using special right triangles, we see that
the area of the triangle is $\frac{1}{2}$ . The semicircle has the area of $\frac{1}{4}\pi$
But this is only $1$ case. There are $4$ cases in total so we have to multiply $\frac{1}{2}+\frac{1}{4}\pi$ by $4$
After multiplying, our answer is: \[\boxed{2}.\]

Solution:
We solve with cases. The cases are:
Case 1: $x\geq0, y\geq0.$ Case 2: $x\geq0, y<0.$ Case 3: $x<0, y\geq0.$ Case 4: $x<0, y<0.$
We can quickly realize that the whole figure is symmetrical; so when you figure out the first case, you get the first part is $\left(x-\dfrac12\right)^2+\left(x-\dfrac12\right)^2$ you can figure out the whole figure: (scaled 8x).  The way we figure out the area is by splitting it the following way:
We know each of the red lines is a diameter of the circle which is $\sqrt2$ . So the area of the red is 2. We know that the area of each semicircle is $\dfrac14 \pi$ so the area of the semicircles combines is $\pi$ . Thus we get $\boxed{2}.$


Answer: 8.0

What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$
$\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12$

Solution:
First consider, $|x-1|+|y-1| \le 1.$ We can see that it is a square with a radius of $1$ (diagonal $\sqrt{2}$ ). The area of the square is $\sqrt{2}^2 = 2.$
Next, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \le 1.$ This will double the square reflecting over x-axis.
So now we have $2$ squares.
Finally, we add one more absolute value and obtain $||x|-1|+||y|-1| \le 1.$ This will double the squares as we reflect the $2$ squares we already have over the y-axis.
Concluding, we have $4$ congruent squares.  The total area is $4\cdot2 =$ $\boxed{8}$

Solution:
We first consider the lattice points that satisfy $||x|-1| = 0$ and $||y|-1| = 1$ . The lattice points satisfying these equations
are $(1,0), (1,2), (1,-2), (-1,0), (-1,2),$ and $(-1,-2).$ By symmetry, we also have points $(0,1), (2,1), (-2,1), (0,-1), (2,-1),$ and $(-2,-1)$ when $||x|-1| = 1$ and $||y|-1| = 0$ .  Graphing and connecting these points, we form 5 squares. However,
we can see that any point within the square in the middle does not satisfy the given inequality (take $(0,0)$ , for instance). As
noted in the above solution, each square has a diagonal $2$ for an area of $\frac{2^2}{2} = 2$ , so the total area is $4\cdot2 =$ $\boxed{8}.$

Solution: The value of $|x|$ and $|y|$ can be a maximum of 1 when the other is 0. Therefore the value of $x$ and $y$ range from -2 to 2. This forms a diamond shape which has area $4 \times \frac{2^2}{2}$ which is $\boxed{8}.$

Solution:
We start by considering the graph of $|x|+|y|\leq 1$ . To get from this graph to $||x|-1|+||y|-1| \leq 1$ we have to translate it by $\pm 1$ on the $x$ axis and $\pm 1$ on the $y$ axis.
Graphing $|x|+|y|\leq 1$ we get a square with side length of $\sqrt{2}$ , so the area of one of these squares is just $2$
We have to multiply by $4$ since there are $4$ combinations of shifting the $x$ and $y$ axis.
So we have $2\times 4$ which is $\boxed{8}$


Answer: 8.0

Solution:
First consider, $|x-1|+|y-1| \le 1.$ We can see that it is a square with a radius of $1$ (diagonal $\sqrt{2}$ ). The area of the square is $\sqrt{2}^2 = 2.$
Next, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \le 1.$ This will double the square reflecting over x-axis.
So now we have $2$ squares.
Finally, we add one more absolute value and obtain $||x|-1|+||y|-1| \le 1.$ This will double the squares as we reflect the $2$ squares we already have over the y-axis.
Concluding, we have $4$ congruent squares.  The total area is $4\cdot2 =$ $\boxed{8}$

Solution:
We first consider the lattice points that satisfy $||x|-1| = 0$ and $||y|-1| = 1$ . The lattice points satisfying these equations
are $(1,0), (1,2), (1,-2), (-1,0), (-1,2),$ and $(-1,-2).$ By symmetry, we also have points $(0,1), (2,1), (-2,1), (0,-1), (2,-1),$ and $(-2,-1)$ when $||x|-1| = 1$ and $||y|-1| = 0$ .  Graphing and connecting these points, we form 5 squares. However,
we can see that any point within the square in the middle does not satisfy the given inequality (take $(0,0)$ , for instance). As
noted in the above solution, each square has a diagonal $2$ for an area of $\frac{2^2}{2} = 2$ , so the total area is $4\cdot2 =$ $\boxed{8}.$

Solution: The value of $|x|$ and $|y|$ can be a maximum of 1 when the other is 0. Therefore the value of $x$ and $y$ range from -2 to 2. This forms a diamond shape which has area $4 \times \frac{2^2}{2}$ which is $\boxed{8}.$

Solution:
We start by considering the graph of $|x|+|y|\leq 1$ . To get from this graph to $||x|-1|+||y|-1| \leq 1$ we have to translate it by $\pm 1$ on the $x$ axis and $\pm 1$ on the $y$ axis.
Graphing $|x|+|y|\leq 1$ we get a square with side length of $\sqrt{2}$ , so the area of one of these squares is just $2$
We have to multiply by $4$ since there are $4$ combinations of shifting the $x$ and $y$ axis.
So we have $2\times 4$ which is $\boxed{8}$


Answer: 6.0

Solution:
The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$
Therefore, the area of the shaded figure is \[2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{6}.\] ~MRENTHUSIASM ~Wilhelm Z

Solution: To find the area of the shaded figure, we subtract the area of the smaller triangle (base $4$ and height $2$ ) from the area of the larger triangle (base $4$ and height $5$ ): \[\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=\boxed{6}.\] ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)

Solution: The consecutive vertices of the shaded figure are $(1,0),(3,2),(5,0),$ and $(3,5).$ By the Shoelace Theorem , the area is \[\frac12\cdot|(1\cdot2+3\cdot0+5\cdot5+3\cdot0)-(0\cdot3+2\cdot5+0\cdot3+5\cdot1)|=\frac12\cdot12=\boxed{6}.\] ~Taco12 ~I-AM-DA-KING

Solution: We have $4$ lattice points in the interior and $6$ lattice points on the boundary. By Pick's Theorem , the area of the shaded figure is \[4+\frac{6}{2}-1 = 4+3-1 = \boxed{6}.\] ~danprathab


Answer: 6.0

Solution:
The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$
Therefore, the area of the shaded figure is \[2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{6}.\] ~MRENTHUSIASM ~Wilhelm Z

Solution: To find the area of the shaded figure, we subtract the area of the smaller triangle (base $4$ and height $2$ ) from the area of the larger triangle (base $4$ and height $5$ ): \[\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=\boxed{6}.\] ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)

Solution: The consecutive vertices of the shaded figure are $(1,0),(3,2),(5,0),$ and $(3,5).$ By the Shoelace Theorem , the area is \[\frac12\cdot|(1\cdot2+3\cdot0+5\cdot5+3\cdot0)-(0\cdot3+2\cdot5+0\cdot3+5\cdot1)|=\frac12\cdot12=\boxed{6}.\] ~Taco12 ~I-AM-DA-KING

Solution: We have $4$ lattice points in the interior and $6$ lattice points on the boundary. By Pick's Theorem , the area of the shaded figure is \[4+\frac{6}{2}-1 = 4+3-1 = \boxed{6}.\] ~danprathab


Answer: 6.0

Solution: The area of the square around the pinwheel is 25. The area of the pinwheel is equal to $\text{the square } - \text{ the white space.}$ Each of the four triangles have a base of 3 units and a height of 2.5 units, and so their combined area is 15 units squared. Then the unshaded space consists of the four triangles with total area of 15, and there are four white corner squares. Therefore the area of the pinwheel is $25-(15+4)$ which is $\boxed{6}$

Solution:
We'd like to use Pick's Theorem on one of the kites, except it doesn't immediately apply since there is a single vertex (in the middle of the diagram) of each kite that does not lie on a lattice point.
We can remedy this be pretending the figure is twice as big:

Now we can safely use Pick's Theorem:
\[A=\frac{b}{2}+i-1=\frac{6}{2}+4-1=6\]
However since we scaled the figure's dimensions by $2$ , we scaled its area by $4$ (since the area of similar shapes scales quadratically with the scaling factor).  Therefore the area of each kite is $\frac{6}{4}$ and the area of all four kites combined is $\boxed{6}$

Solution:
The area of any kite (concave OR convex) with diagonals $p$ $q$ is $\frac{1}{2}pq$ .  Let $p$ be the smaller diagonal and $q$ be the longer diagonal.  Then by Pythagorean Theorem $p=\sqrt{2}$ . Similarly, $q$ is $\sqrt{2}$ less than half of the diagonal of the $5 \times 5$ grid, or $q=\frac{5\sqrt{2}}{2}-\sqrt{2}=\frac{3\sqrt{2}}{2}$ .  Therefore the area of the four kites is just:
\[A=4\cdot\frac{1}{2}pq=4\cdot\frac{1}{2}\cdot\sqrt{2}\cdot\frac{3\sqrt{2}}{2}=\boxed{6}\]


Answer: 16.0

What is the area of the triangle formed by the lines $y=5$ $y=1+x$ , and $y=1-x$
$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$

Solution:
First, we need to find the coordinates where the graphs intersect.
We want the points x and y to be the same. Thus, we set $5=x+1,$ and get $x=4.$ Plugging this into the equation, $y=1-x,$ $y=5$ , and $y=1+x$ intersect at $(4,5)$ , we call this line x.
Doing the same thing, we get $x=-4.$ Thus, $y=5$ . Also, $y=5$ and $y=1-x$ intersect at $(-4,5)$ , and we call this line y.
It's apparent the only solution to $1-x=1+x$ is $0.$ Thus, $y=1.$ $y=1-x$ and $y=1+x$ intersect at $(0,1)$ , we call this line z.
Using the Shoelace Theorem we get: \[\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}\] $=$ So, our answer is $\boxed{16.}$

Solution: Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get $\frac{4\cdot8}{2}$ which is equal to $\boxed{16}$

Solution: $y = x + 1$ and $y = -x + 1$ have $y$ -intercepts at $(0, 1)$ and slopes of $1$ and $-1$ , respectively. Since the product of these slopes is $-1$ , the two lines are perpendicular. From $y = 5$ , we see that $(-4, 5)$ and $(4, 5)$ are the other two intersection points, and they are $8$ units apart. By symmetry, this triangle is a $45-45-90$ triangle, so the legs are $4\sqrt{2}$ each and the area is $\frac{(4\sqrt{2})^2}{2} = \boxed{16}$


Answer: 53332.8

What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?
$\textbf{(A) }48000\qquad\textbf{(B) }49999.5\qquad\textbf{(C) }53332.8\qquad\textbf{(D) }55555\qquad\textbf{(E) }56432.8$

Solution: We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are $4! = 24$ ways to arrange the other numbers, so each number appears in each spot $24$ times. Therefore, the sum of all such numbers is $24 \times (1+3+5+7+8) \times (11111) = 24 \times 24 \times 11111 = 6399936.$ Since there are $5! = 120$ such numbers, we divide $6399936 \div 120$ to get $\boxed{53332.8}$

Solution: We can first solve for the mean for the digits $1, 3, 5, 7,$ and $9$ since each is $2$ away from each other. The mean of the numbers than can be solved using these digits is $55555$ . The total amount of numbers that can be formed using these digits is $5! =120$ . The sum of these numbers is $55555(120) = 6666600$ . Now we can find out the total value that was gained by replacing the $8$ with a $9$ . We can start how be calculating the gain when the $8$ was in the ones digit. Since there are $4! = 24$ numbers with the $8$ in the ones digit and $1$ was gain from each of them, $24$ is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of $24+240+2400+24000+240000=266664$ as the total amount that was gained. Subtract this amount from the sum of the digits using the $9$ instead of the $8$ to get $6666600-266664=6399936$ . Finally, we divide this by $120$ to get the average. $\frac{6399936}{120}= \boxed{53332.8}$

Solution:
The average value of the digits is $(1 + 3 + 5 + 7 + 8)/5 = 4.8$ .
Values occur in every place so $4.8 * 11111 = \boxed{53332.8}$


Answer: 45.0

What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$

Solution: Remind that $\text{slope}=\dfrac{\Delta y}{\Delta x}=\tan \theta$ where $\theta$ is the angle between the slope and $x$ -axis. $k_1=2=\tan \alpha$ $k_2=\dfrac{1}{3}=\tan \beta$ . The angle formed by the two lines is $\alpha-\beta$ $\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1$ . Therefore, $\alpha-\beta=\boxed{45}$

Solution:
Follow Solution 2 up until the lattice points section. Let's use $(0,0)$ $(2,4)$ , and $(9,3)$ . The distance between the origin and $(2,4)$ is $\sqrt{20}$ . The distance between the origin and $(9,3)$ is $\sqrt{90}$ . The distance between $(2,4)$ and $(9,3)$ is $\sqrt{50}$ . Using the Law of Cosines, we see the $50 = 90 + 20 - 2\times\sqrt{20}$ $\times\sqrt{90}$ $\times\cos(\theta)$ , where $\theta$ is the angle we are looking for.
Simplifying, we get $-60 = -2\times(\sqrt{20}) \times(\sqrt{90}) \times\cos(\theta)$
$30 =  \sqrt{1800} \times\cos(\theta)$
$30 =  30\sqrt{2} \times\cos(\theta)$
$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}= \cos(\theta)$
Thus, $\theta = \boxed{45}$

Solution: We can set up vectors $\vec{a} = <1,2>$ and $\vec{b} = <3,1>$ to represent the two lines. We know that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \cos \theta$ . Plugging the vectors in gives us $\cos \theta = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$ .  From this we get that $\theta = \boxed{45}$

Solution:
Let $Z_1 = 3 + i$ and $Z_2 = 1 + 2i$ \begin{align*} Z_2 &= Z_1 \cdot re^{i\theta} \\ 1+2i&=(3+i) \cdot re^{i\theta} \\ 1+2i&=(3 + i) \cdot r(\cos\theta + i\sin\theta) \\ 1+2i&=3r\cos\theta - r\sin\theta  + 3ri\sin\theta + ri\cos\theta \\ \end{align*}
From this we have: \begin{align} 1 &= 3r\cos\theta - r\sin\theta \\ 2 &= r\cos\theta + 3r\sin\theta  \end{align}
To solve this we must compute $r$ \begin{align*} r &= \frac{|Z_2|}{|Z_1|} \\ &= \frac{\sqrt{5}}{\sqrt{10}} \\ &= \frac{\sqrt{2}}{2} \end{align*}
Using elimination we have: $3\cdot(2) - (1)$ \begin{align*} 5 &= 10r\sin\theta \\ \frac{1}{2r} &= \sin\theta \\ \frac{1}{2\frac{\sqrt{2}}{2}} &= \sin\theta \\ \frac{\sqrt{2}}{2} &= \sin\theta \\ \theta &= \boxed{45}

Solution:
The lines $y = 2x, y = \frac {1}{3}x$ , and $x = 3$ form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line $y = 2x$ $\alpha$ , and call the angle that is formed by the x-axis and the line $y = \frac {1}{3}x$ $\beta$ . We try to find $\sin (\alpha - \beta)$ first, and then try to see if any of the answer choices match up.
$\sin (\alpha - \beta)$ $\sin \alpha$ $\cos \beta$ $\sin \beta$ $\cos \alpha$
Using soh-cah-toa, we find that $\sin \alpha = \frac {2}{\sqrt 5}, \sin \beta = \frac {1}{\sqrt 10}, \cos \alpha = \frac {1}{\sqrt 5},$ and $\cos \beta = \frac {3}{\sqrt 10}$
Plugging it all in, we find that $\sin (\alpha - \beta) = \frac {5}{\sqrt {50}}$ , which is equivalent to $\frac {\sqrt 2}{2}$ . Since $\sin (\alpha - \beta) = \frac {\sqrt 2}{2}$ , we get that $\alpha - \beta = 45^{\circ}$ . Therefore, the answer is $\boxed{45}$


Answer: 45.0

Solution: We can take any two lines of this form, since the angle between them will always be the same. Let's take $y=2x$ for the line with slope of 2 and $y=\frac{1}{3}x$ for the line with slope of 1/3. Let's take 3 lattice points and create a triangle. Let's use $(0,0)$ $(1,2)$ , and $(3,1)$ . The distance between the origin and $(1,2)$ is $\sqrt{5}$ . The distance between the origin and $(3,1)$ is $\sqrt{10}$ . The distance between $(1,2)$ and $(3,1)$ is $\sqrt{5}$ . We notice that we have a triangle with 3 side lengths: $\sqrt{5}$ $\sqrt{5}$ , and $\sqrt{10}$ . This forms a 45-45-90 triangle, meaning that the angle is $\boxed{45}$


Answer: 60.0

Solution:
At $10:00$ , the hour hand will be on the $10$ while the minute hand on the $12$
This makes them $\frac{1}{6}$ th of a circle apart, and $\frac{1}{6}\cdot360^{\circ}=\boxed{60}$

Solution:
We know that the full clock is a circle, and therefore has 360 degrees. Considering that there are $12$ numbers, the distance between one number will be $360\div 12=30$ .
If the time is $10:00$ , then the hour hand will be on $10$ , and the minute hand will be on, $12$ , making them $2$ numbers apart, so they will be $60$ degrees apart, or $\boxed{60}$


Answer: 2003.0

What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$

Solution:
The first $2003$ even counting numbers are $2,4,6,...,4006$
The first $2003$ odd counting numbers are $1,3,5,...,4005$
Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$
$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$
$= 1+1+1+...+1 = \boxed{2003}$

Solution: Using the sum of an arithmetic progression formula, we can write this as $\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{2003}$

Solution:
The formula for the sum of the first $n$ even numbers, is $S_E=n^{2}+n$ , (E standing for even).
Sum of first $n$ odd numbers, is $S_O=n^{2}$ , (O standing for odd).
Knowing this, plug $2003$ for $n$
$S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow$ $\boxed{2003}$

Solution: In the case that we don't know if $0$ is considered an even number, we note that it doesn't matter! The sum of odd numbers is $O=1+3+5+...+4005$ . And the sum of even numbers is either $E_1=0+2+4...+4004$ or $E_2=2+4+6+...+4006$ . When compared to the sum of odd numbers, we see that each of the $n$ th term in the series of even numbers differ by $1$ . For example, take series $O$ and $E_1$ . The first terms are $1$ and $0$ . Their difference is $|1-0|=1$ . Similarly, take take series $O$ and $E_2$ . The first terms are $1$ and $2$ . Their difference is $|1-2|=1$ . Since there are $2003$ terms in each set, the answer $\boxed{2003}$

Solution: We can pair each term of the sums - the first even number with the first odd number, the second with the second, and so forth. Then, there are 2003 pairs with a difference of 1 in each pair - 2-1 is 1, 4-3 is 1, 6-5 is 1, and so on. Then, the solution is $1 \cdot 2003$ , and the answer is $\boxed{2003}$


Answer: 2003.0

Solution:
The first $2003$ even counting numbers are $2,4,6,...,4006$
The first $2003$ odd counting numbers are $1,3,5,...,4005$
Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$
$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$
$= 1+1+1+...+1 = \boxed{2003}$

Solution: Using the sum of an arithmetic progression formula, we can write this as $\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{2003}$

Solution:
The formula for the sum of the first $n$ even numbers, is $S_E=n^{2}+n$ , (E standing for even).
Sum of first $n$ odd numbers, is $S_O=n^{2}$ , (O standing for odd).
Knowing this, plug $2003$ for $n$
$S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow$ $\boxed{2003}$

Solution: In the case that we don't know if $0$ is considered an even number, we note that it doesn't matter! The sum of odd numbers is $O=1+3+5+...+4005$ . And the sum of even numbers is either $E_1=0+2+4...+4004$ or $E_2=2+4+6+...+4006$ . When compared to the sum of odd numbers, we see that each of the $n$ th term in the series of even numbers differ by $1$ . For example, take series $O$ and $E_1$ . The first terms are $1$ and $0$ . Their difference is $|1-0|=1$ . Similarly, take take series $O$ and $E_2$ . The first terms are $1$ and $2$ . Their difference is $|1-2|=1$ . Since there are $2003$ terms in each set, the answer $\boxed{2003}$

Solution: We can pair each term of the sums - the first even number with the first odd number, the second with the second, and so forth. Then, there are 2003 pairs with a difference of 1 in each pair - 2-1 is 1, 4-3 is 1, 6-5 is 1, and so on. Then, the solution is $1 \cdot 2003$ , and the answer is $\boxed{2003}$


Answer: 80.0

What is the greatest integer less than or equal to \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?\]
$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$

Solution: We write \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot\frac{3^{100}}{3^{96}}+\frac{2^{96}}{3^{96}+2^{96}}\cdot\frac{2^{100}}{2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot 81+\frac{2^{96}}{3^{96}+2^{96}}\cdot 16.\] Hence we see that our number is a weighted average of 81 and 16, extremely heavily weighted toward 81. Hence the number is ever so slightly less than 81, so the answer is $\boxed{80}$

Solution:
\[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{2^{96}\left(\frac{3^{100}}{2^{96}}\right)+2^{96}\left(2^{4}\right)}{2^{96}\left(\frac{3}{2}\right)^{96}+2^{96}(1)}=\frac{\frac{3^{100}}{2^{96}}+2^{4}}{\left(\frac{3}{2}\right)^{96}+1}=\frac{\frac{3^{100}}{2^{100}}\cdot2^{4}+2^{4}}{\left(\frac{3}{2}\right)^{96}+1}=\frac{2^{4}\left(\frac{3^{100}}{2^{100}}+1\right)}{\left(\frac{3}{2}\right)^{96}+1}.\]
We can ignore the 1's on the end because they won't really affect the fraction. So, the answer is very very very close but less than the new fraction.
\[\frac{2^{4}\left(\frac{3^{100}}{2^{100}}+1\right)}{\left(\frac{3}{2}\right)^{96}+1}<\frac{2^{4}\left(\frac{3^{100}}{2^{100}}\right)}{\left(\frac{3}{2}\right)^{96}},\]
\[\frac{2^{4}\left(\frac{3^{100}}{2^{100}}\right)}{\left(\frac{3}{2}\right)^{96}}=\frac{3^{4}}{2^{4}}*2^{4}=3^{4}=81.\]
So, our final answer is very close but not quite 81, and therefore the greatest integer less than the number is $\boxed{80}$

Solution:
Let $x=3^{96}$ and $y=2^{96}$ . Then our fraction can be written as $\frac{81x+16y}{x+y}=\frac{16x+16y}{x+y}+\frac{65x}{x+y}=16+\frac{65x}{x+y}$ .
Notice that $\frac{65x}{x+y}<\frac{65x}{x}=65$ .
So , $16+\frac{65x}{x+y}<16+65=81$ .
And our only answer choice less than 81 is $\boxed{80}$ (RegularHexagon)

Solution: Dividing by $2^{96}$ in both numerator and denominator, this fraction can be rewritten as \[\frac{81 \times (1.5)^{96} + 16}{(1.5)^{96} + 1}.\] Notice that the $+1$ and the $+16$ will be so insignificant compared to a number such as $(1.5)^{96},$ and that thereby the fraction will be ever so slightly less than $81$ . Thereby, we see that the answer is $\boxed{80}.$

Solution: If you multiply $(3^{96} + 2^{96})$ by $(3^{4} + 2^{4})$ (to get the exponent up to 100), you'll get $(3^{100} + 2^{100}) + 3^{96} \cdot 2^{4} + 2^{96} \cdot 3^{4}$ . Thus, in the numerator, if you add and subtract by $3^{96} \cdot 2^{4}$ and $2^{96} \cdot 3^{4}$ , you'll get $\frac{(3^{4} + 2^{4})(3^{100} + 2^{100}) - 3^{96} \cdot 2^{4} - 2^{96} \cdot 3^{4}}{3^{96}+2^{96}}$ . You can then take out out the first number to get $3^{4} + 2^{4} - \frac{3^{96} \cdot 2^{4} + 2^{96} \cdot 3^{4}}{3^{96}+2^{96}}$ . This can then be written as $87 - \frac{16 \cdot 3^{96} + 16 \cdot 2^{96} + 75 \cdot 2^{96}}{3^{96}+2^{96}}$ , factoring out the 16 and splitting the fraction will give you $87 - 16 - \frac{65 \cdot 2^{96}}{3^{96}+2^{96}}$ , giving you $81 - \frac{65 \cdot 2^{96}}{3^{96}+2^{96}}$ . While you can roughly say that $\frac{65 \cdot 2^{96}}{3^{96}+2^{96}} < 1$ you can also notice that the only answer choice less than 81 is 80, thus the answer is $\boxed{80}.$

Solution: If you factor out $3^{100}$ from the numerator and $3^{96}$ from the denominator, you will get $\frac{3^{100}\left(1+(\frac{2}{3}\right)^{100})}{3^{96}\left(1+(\frac{2}{3}\right)^{96})}$ . Divide the numerator and denominator by $3^{96}$ to get $\frac{81\left(1+(\frac{2}{3}\right)^{100})}{\left(1+(\frac{2}{3}\right)^{96})}$ . We see that every time we multiply $\frac{2}{3}$ by itself, it slightly decreases, so $1+(\frac{2}{3})^{100}$ will be ever so slightly smaller than $1+(\frac{2}{3})^{96}$ . Thus, the decimal representation of $\frac{\left(1+(\frac{2}{3}\right)^{100})}{\left(1+(\frac{2}{3}\right)^{96})}$ will be extremely close to $1$ , so our solution will be the largest integer that is less than $81$ . Thus, the answer is $\boxed{80}.$


Answer: 90.0

What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$

Solution: We might at first think that the answer would be $9$ , because $1+2+3 \dots +n = 45$ when $n = 9$ . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence $-44, -43, \cdots, 44, 45$ cancels out except $45$ .  Thus, the answer is, intuitively, $\boxed{90}$ integers.

Solution: To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be $\frac12$ if the middle two numbers are $0$ and $1$ , so the answer is $\frac{45}{\frac12}=\boxed{90}$


Answer: 90.0

Solution: We might at first think that the answer would be $9$ , because $1+2+3 \dots +n = 45$ when $n = 9$ . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence $-44, -43, \cdots, 44, 45$ cancels out except $45$ .  Thus, the answer is, intuitively, $\boxed{90}$ integers.

Solution: To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be $\frac12$ if the middle two numbers are $0$ and $1$ , so the answer is $\frac{45}{\frac12}=\boxed{90}$


Answer: 22.0

What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than $2019$
$\textbf{(A) } 11 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 23 \qquad\textbf{(E) } 27$

Solution: Observe that $2019_{10} = 5613_7$ . To maximize the sum of the digits, we want as many $6$ s as possible (since $6$ is the highest value in base $7$ ), and this will occur with either of the numbers $4666_7$ or $5566_7$ . Thus, the answer is $4+6+6+6 = 5+5+6+6 = \boxed{22}$

Solution: Note that all base $7$ numbers with $5$ or more digits are in fact greater than $2019$ . Since the first answer that is possible using a $4$ digit number is $23$ , we start with the smallest base $7$ number that whose digits sum to $23$ , namely $5666_7$ . But this is greater than $2019_{10}$ , so we continue by trying $4666_7$ , which is less than 2019. So the answer is $\boxed{22}$

Solution: Again note that you want to maximize the number of $6$ s to get the maximum sum.  Note that $666_7=342_{10}$ , so you have room to add a thousands digit base $7$ .  Fix the $666$ in place and try different thousands digits, to get $4666_7$ as the number with the maximum sum of digits.  The answer is $\boxed{22}$