Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In rectangle $ABCD$ , angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$ , where $E$ is on $\overline{AB}$ $F$ is on $\overline{AD}$ $BE=6$ and $AF=2$ . Which of the following is closest to the area of the rectangle $ABCD$ [asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("$A$", A, dir((5, 3.5)--A)); label("$B$", B, dir((5, 3.5)--B)); label("$C$", C, dir((5, 3.5)--C)); label("$D$", D, dir((5, 3.5)--D)); label("$E$", E, dir((5, 3.5)--E)); label("$F$", F, dir((5, 3.5)--F)); label("$2$", (0,1), dir(0)); label("$6$", (7.5,0), N);[/asy] $\text{(A)}\ 110\qquad\text{(B)}\ 120\qquad\text{(C)}\ 130\qquad\text{(D)}\ 140\qquad\text{(E)}\ 150$

Solution: Since $\angle C = 90^\circ$ , each of the three smaller angles is $30^\circ$ , and $\triangle BEC$ and $\triangle CDF$ are both $30-60-90$ triangles.
[asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("$A$", A, dir((5, 3.5)--A)); label("$B$", B, dir((5, 3.5)--B)); label("$C$", C, dir((5, 3.5)--C)); label("$D$", D, dir((5, 3.5)--D)); label("$E$", E, dir((5, 3.5)--E)); label("$F$", F, dir((5, 3.5)--F)); label("$2$", (0,1), plain.E, fontsize(10)); label("$x$", (9,3.5), E, fontsize(10)); label("$x-2$", (0,5), plain.E, fontsize(10)); label("$y$", (5,7), N, fontsize(10)); label("$6$", (7.5,0), S, fontsize(10));[/asy]
Defining the variables as illustrated above, we have $x = 6\sqrt{3}$ from $\triangle BEC$
Then $x-2 = 6\sqrt{3} - 2$ , and $y = \sqrt{3} (6 \sqrt{3} - 2) = 18 - 2\sqrt{3}$
The area of the rectangle is thus $xy = 6\sqrt{3}(18 - 2\sqrt{3}) = 108\sqrt{3} - 36$
Using the approximation $\sqrt{3} \approx 1.7$ , we get an area of just under $147.6$ , which is closest to answer $\boxed{3} > 1.7$ ).

Answer: E 3
Problem: In rectangle $ABCD$ , angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$ , where $E$ is on $\overline{AB}$ $F$ is on $\overline{AD}$ $BE=6$ and $AF=2$ . Which of the following is closest to the area of the rectangle $ABCD$ [asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("$A$", A, dir((5, 3.5)--A)); label("$B$", B, dir((5, 3.5)--B)); label("$C$", C, dir((5, 3.5)--C)); label("$D$", D, dir((5, 3.5)--D)); label("$E$", E, dir((5, 3.5)--E)); label("$F$", F, dir((5, 3.5)--F)); label("$2$", (0,1), dir(0)); label("$6$", (7.5,0), N);[/asy] $\text{(A)}\ 110\qquad\text{(B)}\ 120\qquad\text{(C)}\ 130\qquad\text{(D)}\ 140\qquad\text{(E)}\ 150$

Solution: Use the process above, but use $\sqrt{3} \approx 1.73$ . You should get $[ABCD]=150.84$ , which then you select $\boxed{150}$ . Notice that the actual area, when plugged into a calculator, yields about $151.0614872$

Answer: E 150
Problem: The sum of the lengths of the twelve edges of a rectangular box is $140$ , and 
the distance from one corner of the box to the farthest corner is $21$ . The total surface area of the box is
$\text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812$

Solution: Let $x, y$ , and $z$ be the unique lengths of the edges of the box.  Each box has $4$ edges of each length, so: \[4x + 4y + 4z = 140 \ \Longrightarrow \ x + y + z = 35.\] The spacial diagonal (longest distance) is given by $\sqrt{x^2 + y^2 + z^2}$ .  Thus, we have $\sqrt{x^2 + y^2 + z^2} = 21$ , so $x^2 + y^2 + z^2 = 21^2$
Our target expression is the surface area of the box:
\[S = 2xy + 2xz + 2yz.\]
Since $S$ is a symmetric polynomial of degree $2$ , we try squaring the first equation to get:
\[35^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy +2yz + 2xz = 35^2.\]
Substituting in our long diagonal and surface area expressions, we get: $21^2 + S = 35^2$ , so $S = (35 + 21)(35 - 21) = 56\cdot 14 = 784$ , which is option $\boxed{784}$

Answer: B 784
Problem: The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$ ’s separated by blocks of $2$ ’s with $n$ $2$ ’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$

Solution: The sum of the first $1$ numbers is $1$
The sum of the next $2$ numbers is $2 + 1$
The sum of the next $3$ numbers is $2 + 2 + 1$
In general, we can write "the sum of the next $n$ numbers is $1 + 2(n-1)$ ", where the word "next" follows the pattern established above.
Thus, we first want to find what triangular numbers $1234$ is between.  By plugging in various values of $n$ into $f(n) = \frac{n(n+1)}{2}$ , we find:
$f(50) = 1275$
$f(49) = 1225$
Thus, we want to add up all those sums from "next $1$ number" to the "next $49$ numbers", which will give us all the numbers up to and including the $1225^{th}$ number.  Then, we can manually tack on the remaining $2$ s to hit $1234$
We want to find:
$\sum_{n=1}^{49} 1 + 2(n-1)$
$\sum_{n=1}^{49} 2n - 1$
$\sum_{n=1}^{49} 2n - \sum_{n=1}^{49} 1$
$2 \sum_{n=1}^{49} n - 49$
$2\cdot \frac{49\cdot 50}{2} - 49$
$49^2$
$2401$
Thus, the sum of the first $1225$ terms is $2401$ .  We have to add $9$ more $2$ s to get to the $1234^{th}$ term, which gives us $2419$ , or option $\boxed{2419}$

Answer: B 2419
Problem: The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$ ’s separated by blocks of $2$ ’s with $n$ $2$ ’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$

Solution: The $k$ th appearance of 1 is at position $1 + 2 + \dots + k = \frac{k(k + 1)}{2}$ . Then there are $k$ 1's and $\frac{k(k + 1)}{2} - k = \frac{k(k - 1)}{2}$ 2's among the first $\frac{k(k + 1)}{2}$ numbers, so the sum of these $\frac{k(k + 1)}{2}$ terms is $k + k(k - 1) = k^2$
When $k = 49$ $\frac{k(k + 1)}{2} = 1225$ , and when $k = 50$ $\frac{k(k + 1)}{2} = 1275$
The sum of the first 1225 terms is $49^2 = 2401$ . The numbers in positions 1226 through 1234 are all 2's, so their sum is $(1234 - 1226 + 1) \cdot 2 = 18$ . Therefore, the sum of the first 1234 terms is $2401 + 18 = \boxed{2419}$

Answer:  2419
Problem: Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$

Solution: Complete the square to get \[(x-7)^2 + (y-3)^2 = 64.\] Applying Cauchy-Schwarz directly, \[64\cdot25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2.\] \[40 \ge 3x+4y-33\] \[3x+4y \le 73.\] Thus our answer is $\boxed{73}$

Answer: B 73
Problem: Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$

Solution: The first equation is a circle , so we find its center and radius by completing the square $x^2 - 14x + y^2 - 6y = 6$ , so \[(x-7)^2 + (y-3)^2 = (x^2- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9 = 64.\]
So we have a circle centered at $(7,3)$ with radius $8$ , and we want to find the max of $3x + 4y$
The set of lines $3x + 4y = A$ are all parallel , with slope $-\frac{3}{4}$ .  Increasing $A$ shifts the lines up and/or to the right.
We want to shift this line up high enough that it's tangent to the circle, but not so high that it misses the circle altogether.  This means $3x + 4y = A$ will be tangent to the circle.
Imagine that this line hits the circle at point $(a,b)$ .  The slope of the radius connecting the center of the circle, $(7,3)$ , to tangent point $(a,b)$ will be $\frac{4}{3}$ , since the radius is perpendicular to the tangent line.
So we have a point, $(7,3)$ , and a slope of $\frac{4}{3}$ that represents the slope of the radius to the tangent point.  Let's start at the point $(7,3)$ .  If we go $4k$ units up and $3k$ units right from $(7,3)$ , we would arrive at a point that's $5k$ units away.  But in reality we want $5k = 8$ to reach the tangent point, since the radius of the circle is $8$
Thus, $k = \frac{8}{5}$ , and we want to travel $4\cdot \frac{8}{5}$ up and $3\cdot \frac{8}{5}$ over from the point $(7,3)$ to reach our maximum.  This means the maximum value of $3x + 4y$ occurs at $\left(7 +3\cdot \frac{8}{5}, 3 + 4\cdot \frac{8}{5}\right)$ , which is $\left(\frac{59}{5}, \frac{47}{5}\right).$
Plug in those values for $x$ and $y$ , and you get the maximum value of $3x + 4y = 3\cdot\frac{59}{5} + 4\cdot\frac{47}{5} = \boxed{73}$

Answer: B 73
Problem: Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$

Solution: Let the tangent point be $P$ , and the tangent line's x-intercept be $Q$ . Consider the horizontal line starting from center of circle (O) meeting the tangent line at K. Now triangle $OPK$ is 3-4-5, $OP=8$ , so $OK = \frac{5}{3}*8 = \frac{40}{3}$ . Note that the horizontal distance from $O$ to the origin is $7$ , and the horizontal distance from K to Q is 4, ( $\frac{4}{3}$ of its y coordinate), so the x-intercept is $7+4+OK = 73/3$ . The value of $3x+4y$ is 73 at point $Q$ . Note that this value is constant on the tangent line, so there is no need to calculate the coordinate of $P$ $\boxed{73}$

Answer: B 73
Problem: Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$

Solution: Let $z = 3x + 4y$ . Solving for $y$ , we get $y = (z - 3x)/4$ . Substituting into the given equation, we get \[x^2 + \left( \frac{z - 3x}{4} \right)^2 = 14x + 6 \cdot \frac{z - 3x}{4} + 6,\] which simplifies to \[25x^2 - (6z + 152)x + (z^2 - 24z - 96) = 0.\]
This quadratic equation has real roots in $x$ if and only if its discriminant is nonnegative, so \[(6z + 152)^2 - 4 \cdot 25 \cdot (z^2 - 24z - 96) \ge 0,\] which simplifies to \[-64z^2 + 4224z + 32704 \ge 0,\] which can be factored as \[-64(z + 7)(z - 73) \ge 0.\] The largest value of $z$ that satisfies this inequality is $\boxed{73}$

Answer: B 73
Problem: Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$

Solution: First, we move all the non-constant terms of the constraint to one side and assign it to the function $g(x,y)$ \[g(x,y)=x^2+y^2-14x-6y.\] Since we are trying to maximize $f(x,y)=3x+4y$ , we need to solve for $x$ and $y$ in the system \[\begin{cases}x^2+y^2-14x-6y=6,\\\nabla g(x,y)=\lambda\nabla f(x,y).\end{cases}\] We have that \begin{align*}\nabla f(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}3x+4y\\\dfrac{\partial}{\partial y}3x+4y\end{pmatrix}\\&=\begin{pmatrix}3\\4\end{pmatrix},\end{align*} and \begin{align*}\nabla g(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}x^2+y^2-14x-6y\\\dfrac{\partial}{\partial y}x^2+y^2-14x-6y\end{pmatrix}\\&=\begin{pmatrix}2x-14\\2y-6\end{pmatrix}\end{align*}. To solve the original system, we can solve for $x$ and $y$ in terms of $\lambda$ using our equations from the gradients, then substitute them into the first equation. We have that $x=\frac{3}{2}\lambda+7$ and $y=2\lambda+3$ . Substituting into the first equation, we have that \begin{align*}\left(\frac{3}{2}\lambda+7\right)^2+(2\lambda+3)^2-14\left(\frac{3}{2}\lambda+7\right)-6(2\lambda+3)&=6\\\frac{25}{4}\lambda^2&=64\\\lambda&=\pm\frac{16}{5}\end{align*} Using the solutions of $x$ and $y$ in terms of $\lambda$ that we found earlier, we have that \[3x+4y=\frac{25}{2}\lambda+33.\] Because we are trying to maximize this function, we will use the positive solution for $\lambda$ . Therefore, after substituting, we have that the largest value of $g(x,y)$ that satisfies $f(x,y)=6$ is $\boxed{73}$

Answer: B 73
Problem: Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$

Solution: Completing the square gives the equation of a circle, $(x-7)^2+(y-3)^2=64.$ Seeing that we would like to maximize $3x+4y,$ we parameterize the circle using polar coordinates: \begin{align*} x&=7+8\cos\theta&y&=3+8\sin\theta. \end{align*} Then, we have $3x+4y=3(7+8\cos\theta)+4(3+8\sin\theta)=33+24\cos\theta+32\sin\theta.$ Since $a\cos\theta+b\sin\theta\le\sqrt{a^2+b^2},$ the desired answer is $33+\sqrt{24^2+32^2}=33+40=\boxed{73}.$ - Ultroid999OCPN

Answer:  73
Problem: Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$

Solution: The equation $x^2 + y^2 = 14x + 6y + 6$ can be written\[
(x-7)^2 + (y-3)^2 = 8^2,
\]which defines a circle of radius 8 centered at $(7,3)$ . If $k$ is a possible value of $3x + 4y$ for $(x,y)$ on the circle, then the line $3x + 4y = k$ must intersect the circle in at least one point. The largest value of $k$ occurs when the line is tangent to the circle, and is therefore perpendicular to the radius at the point of tangency. Because the slope of the tangent line is $-3/4$ the slope of the radius is $\ 4/3$ . It follows that the point on the circle that yields the maximum value of $3x + 4y$ is one of the two points of tangency,\[
x = 7 + \frac{3 \cdot 8}{5} = \frac{59}{5}, \hspace{.3in} y = 3 +
\frac{4 \cdot 8}{5} = \frac{47}{5},
\]or\[
x = 7 - \frac{3 \cdot 8}{5} = \frac{11}{5}, \hspace{.3in} y = 3 -
\frac{4 \cdot 8}{5} = - \frac{17}{5}.
\][asy]
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8));
draw((0,15)--origin--(15,0));
dot(" $(7,3)$ ",(7,3),S);
draw(Circle((7,3),8));
line a = line((0,9),(12,0));
line b = line((0,12),(16,0));
line c = line((0,18.3),(18.3*4/3,0));
draw(a^^b^^c);
[/asy]
The first point of tangency gives\[
3x + 4y = 3 \cdot \frac{59}{5} + 4 \cdot \frac{47}{5} =
\frac{177}{5}+\frac{188}{5}= 73,
\]and the second one gives \[\, 3x + 4y = \frac{33}{5}-\frac{68}{5} = -7.\] Thus, $\boxed{73}$ is the desired maximum, while $-7$ is the minimum.

Answer:  73
Problem: An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:
(a) the selection of four red marbles;
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbles; and
(d) the selection of one marble of each color.
What is the smallest number of marbles satisfying the given condition?
$\text{(A)}\ 19\qquad\text{(B)}\ 21\qquad\text{(C)}\ 46\qquad\text{(D)}\ 69\qquad\text{(E)}\ \text{more than 69}$

Solution: Let the bag contain $n$ marbles total, with $r, w, b, g$ representing the number of red, white, blue, and green marbles, respectively.  Note that $r + w + b + g = n$
The number of ways to select four red marbles out of the set of marbles without replacement is:
\[\binom{r}{4} = \frac{r!}{24\cdot (r -4)!}\]
The number of ways to select one white and three red marbles is:
\[\binom{w}{1}\binom{r}{3} = \frac{w\cdot r!}{6\cdot (r - 3)!}\]
The number of ways to select one white, one blue, and two red marbles is:
\[\binom{w}{1}\binom{b}{1} \binom{r}{2} = \frac{wb\cdot r!}{2(r-2)!}\]
The number of ways to select one marble of each colors is:
\[\binom{w}{1}\binom{b}{1} \binom{g}{1}\binom{r}{1} = wbgr\] Setting the first and second statements equal, we find:
\[\frac{r!}{24\cdot (r -4)!}  = \frac{w\cdot r!}{6\cdot (r - 3)!}\]
\[r - 3  = 4w\]
Setting the first and third statements equal, we find:
\[\frac{r!}{24\cdot (r -4)!} = \frac{wb\cdot r!}{2(r-2)!}\]
\[(r-3)(r-2) = 12wb\]
Setting the last two statements equal, we find:
\[\frac{wb\cdot r!}{2(r-2)!} = wbgr\]
\[r - 1 = 2g\]
These are all the "linking equations" that are needed; the transitive property of equality makes the other three equalities unnecessary.
From the first equation, we know that $r$ must be $3$ more than a multiple of $4$ , or that $r \equiv 3 \mod 4$
Putting the first equation into the second equation, we find $r-2 = 3b$ .  Therefore, $r \equiv 2 \mod 3$ .  Using the Chinese Remainder Theorem , we find that $r \equiv 11 \mod 12$
The third equation gives no new restrictions on $r$ ; it is already odd by the first equation.
Thus, the minimal positive value of $r$ is $11$ .  This requires $g=\frac{r - 1}{2} = 5$ by the third equation, and $w = \frac{r-3}{4} = 2$ by the first equation.  Finally, the second equation gives $b = \frac{(r-3)(r-2)}{12w} = 3$
The minimal total number of marbles is $11 + 5 + 2 + 3 = \boxed{21}$

Answer: B 21
Problem: Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$ , and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$ . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$

Solution: The two equations of the balls are
\[x^2 + y^2 + \left(z - \frac{21}{2}\right)^2 \le 36\]
\[x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}\]
Note that along the $z$ axis, the first ball goes from $10.5 \pm 6$ , and the second ball goes from $1 \pm 4.5$ .  The only integer value that $z$ can be is $z=5$
Plugging that in to both equations, we get:
\[x^2 + y^2 \le \frac{23}{4}\]
\[x^2 + y^2 \le \frac{17}{4}\]
The second inequality implies the first inequality, so the only condition that matters is the second inequality.
From here, we do casework, noting that $|x|, |y| \le 3$
For $x=\pm 2$ , we must have $y=0$ .  This gives $2$ points.
For $x = \pm 1$ , we can have $y\in \{-1, 0, 1\}$ .  This gives $2\cdot 3 = 6$ points.
For $x = 0$ , we can have $y \in \{-2, -1, 0, 1, 2\}$ .  This gives $5$ points.
Thus, there are $\boxed{13}$

Answer: D 13
Problem: Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$ , and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$ . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$

Solution: Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a 2-dimensional plane (the previous z-axis is the new x-axis while the y-axis remains the same).
The spheres now become circles with centers at $(1,0)$ and $(\frac{21}{2},0)$ .  They have radii $\frac{9}{2}$ and $6$ , respectively.
Let circle $A$ be the circle centered on $(1,0)$ and circle $B$ be the one centered on $(\frac{21}{2},0)$
The point on circle $A$ closest to the center of circle $B$ is $(\frac{11}{2},0)$ .  The point on circle B closest to the center of circle $A$ is $(\frac{11}{2},0)$
Taking a look back at the 3-dimensional coordinate grid with the spheres, we can see that their intersection appears to be a circle with congruent "dome" shapes on either end.  Because the tops of the "domes" are at $(0,0,\frac{9}{2})$ and $(0,0,\frac{11}{2})$ , respectively, the lattice points inside the area of intersection must have z-value $5$ (because $5$ is the only integer between $\frac{9}{2}$ and $\frac{11}{2}$ ).  Thus, the lattice points in the area of intersection must all be on the 2-dimensional circle.  The radius of the circle will be the distance from the z-axis.
Now, looking at the 2-dimensional coordinate plane, we see that the radius of the circle (now the distance from the x-axis, because there is no more z-axis) is the altitude of a triangle with two points on centers of circles $A$ and $B$ and third point at the first quadrant intersection of the circles.  Let's call that altitude $h$
We know all three side lengths of this triangle: $\frac{9}{2}$ (the radius of circle $A$ ), $6$ (the radius of circle $B$ ), and $\frac{19}{2}$ (the distance between the centers of circles $A$ and $B$ ).  We can now find the area of the triangle using Heron's formula:
\[s=\frac{\frac{9}{2}+6+\frac{19}{2}}{2}=20\]
\[A=\sqrt{(20)(20-\frac{9}{2})(20-6)(20-\frac{19}{2})}=\sqrt{110}\]
Using the area of the triangle, we can find that altitude $h$ from the x-axis:
\[\sqrt{110}=\frac{h*\frac{19}{2}}{2}\]
\[h=\frac{4*\sqrt{110}}{19}\]
Remember, the altitude $h$ is also the radius of the circle containing all the solutions to the problem.
Going back to the 3-dimensional grid and looking at the circle, we can again make the figure 2-dimensional.
The radius $h$ of the circle in a 2-dimensional plane is $\frac{4\sqrt{110}}{19}$ , a little greater than $2$ .  We know that $h>2$ because $4\sqrt{110}>4*10$ , and $\frac{40}{19}>2$
Finally, looking at a circle with radius slightly larger than $2$ , we see that there are $\boxed{13}$

Answer: D 13
Problem: Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$ , and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$ . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$

Solution: Note that the spheres are on the same $x$ and $y$ axis. Therefore, we can draw the spheres so that only the $x$ and $z$ axis are featured.
$A = (1, 0, 5)$ $\sqrt{1^2+4^2} = \sqrt{17}<\frac92$ $A$ is inside the smaller sphere. $\sqrt{1^2+(\frac{11}{2})^2} = \frac{\sqrt{126}}{2}<6$ $A$ is inside the larger sphere. Point $A$ is inside the intersection.
$B = (2, 0, 5)$ $\sqrt{2^2+4^2} = \sqrt{20}<\frac92$ $B$ is inside the smaller sphere. $\sqrt{2^2+(\frac{11}{2})^2} = \frac{\sqrt{137}}{2}<6$ $B$ is inside the larger sphere. Point $B$ is inside the intersection.
$C = (3, 0, 5)$ $\sqrt{3^2+4^2} = \sqrt{25}>\frac92$ $C$ is outside the smaller sphere. $\sqrt{3^2+(\frac{11}{2})^2} = \frac{\sqrt{157}}{2}>6$ $C$ is outside the larger sphere. Point $C$ is outside the intersection.
The intersection of $2$ spheres is a circle, by drawing the circle flat we can see that there are $4$ more points within the intersection.
We can see that points with the same $x$ -coordinates with $A$ are still inside the circle. $B$ is very close to the circle, the points with the same $x$ -coordinates with $B$ are not necessarily inside the circle.
For example, $D = (2,1,5)$ $\sqrt{2^2+1^2+4^2}=\sqrt{21}>\frac92$ $D$ is outside the smaller sphere. $\sqrt{2^2 + 1^2 + (\frac{11}{2})^2} = \frac{\sqrt{141}}{2}<6$ $D$ is inside the larger sphere. Point $D$ is outside the intersection.
Therefore, the answer is $9+4 = \boxed{13}$

Answer: D 13
Problem: On a $4\times 4\times 3$ rectangular parallelepiped, vertices $A$ $B$ , and $C$ are adjacent to vertex $D$ . The perpendicular distance from $D$ to the plane containing $A$ $B$ , and $C$ is closest to
[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, S); label("$D$", D, E); label("$4$", (4,2,0), SW); label("$4$", (2,4,0), SE); label("$3$", (0, 4, 1.5), E); [/asy]
$\text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9$

Solution: By placing the cube in a coordinate system such that $D$ is at the origin, $A(0,0,3)$ $B(4,0,0)$ , and $C(0,4,0)$ , we find that the equation of plane $ABC$ is:
\[\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1,\] so $3x + 3y + 4z - 12 = 0.$ The equation for the distance of a point $(a,b,c)$ to a plane $Ax + By + Cz + D = 0$ is given by:
\[\frac{|Aa + Bb + Cc + D|}{\sqrt{A^2 + B^2 + C^2}}.\]
Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where $a=b=c=0$ ) to the plane is given by:
\[\frac{D}{\sqrt{A^2 + B^2 + C^2}} = \frac{12}{\sqrt{9 + 9 + 16}} = \frac{12}{\sqrt{34}}.\]
Since $\sqrt{34} < 6$ , this number should be just a little over $2$ , and the correct answer is $\boxed{2.1}$

Answer: C 2.1