Answer: 12.0

Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution:
Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get \[x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0\] Since this is a quartic, there are $4$ total roots (counting multiplicity). We see that $x=0$ always has at least one intersection at $(0,-a)$ (and is in fact a double root).
The other two intersection points have $x$ coordinates $\pm\sqrt{2a-1}$ . We must have $2a-1> 0;$ otherwise we are in the case where the parabola lies entirely above the circle (tangent at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see that $\boxed{12}$

Solution:
Substituting $y = x^2 - a$ gives $x^2 + (x^2 - a)^2 = a^2$ , which simplifies to $x^2 + x^4 - 2x^2a + a^2 = a^2$ . This further simplifies to $x^2(1 + x^2 - 2a) = 0$ . Thus, either $x^2 = 0$ , or $x^2 - 2a + 1 = 0$
Since we care about $a$ , we consider the second case. We solve in terms of $a$ , giving $a = \frac{x^2}{2} + \frac{1}{2}$ . We see that in order to find the range in which $a$ lies, we must find the vertex of this equation, which turns out to be $\left(0, \frac{1}{2}\right)$ . Hence, we know that the minimum is $\frac{1}{2}$ , which further implies that $\boxed{12}$

Solution:
Looking at a graph, it is obvious that the two curves intersect at $(0, -a)$ . We also see that if the parabola goes "in" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.
We have $x^2 - a = -\sqrt{a^2 - x^2}$ . Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$ . Since $x = 0$ is already accounted for, we only need to find one solution for $x^2 = 2a - 1$ , where the right hand side portion is obviously increasing. Since $a = \frac{1}{2}$ begets $x = 0$ (an overcount), we have $\boxed{12}$ as the right answer.

Solution: This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is $2$ , the radius of the circle that matches it has a radius of $\frac{1}{2}$ . This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only $1$ point. When a larger circle is used, it is tangent to $3$ points because the points on either side are now separated from the vertex. Therefore, $\boxed{12}$ is correct.

Solution: Now, let's graph these two equations. We want the blue parabola to be inside this red circle.  Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$ . Expanding, we get $x^4-2ax^2+x^2=0$ . Factoring out the $x^2$ , we get $x^2(x^2-2a+1)=0$ . Then we find that $x=0$ or $x=\pm\sqrt{2a-1}$ . Therefore, $2a-1>0$ , which means $\boxed{12}$

Solution:
In order to solve for the values of $a$ , we need to just count multiplicities of the roots when the equations are set equal to each other: in other words, take the derivative. We know that $\sqrt{a^2 - x^2} = x^2 - a$ . Now, we take square of both sides, and rearrange to obtain $x^4 - (2a - 1)x^2 = 0$ . Now, we may take the second derivative of the equation to obtain $6x^2 - (2a - 1) = 0$ . Now, we must take discriminant. Since we need the roots of that equation to be real and not repetitive (otherwise they would not intersect each other at three points), the discriminant must be greater than zero. Thus,
\[b^2 - 4ac > 0 \rightarrow 0 - 4(6)(-(2a - 1)) > 0 \rightarrow a > \frac{1}{2}\] The answer is $\boxed{12}$ and we are done.

Solution:
Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$ , and they have symmetry across the $y$ -axis, thus, for them to intersect at exactly $3$ points, it suffices to find the $y$ solution.
First, rewrite the second equation to $y=x^2-a\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$ . This value must not be less than $0$ (It is the square root part of the quadratic formula). To find when it is $0$ , we find the roots: \[4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}\] Since $\lim_{a\to \infty}(4a^2-4a+1)=\infty$ , our range is $\boxed{12}$

Solution: We can see that if $a = 1$ , we know that the points where the two curves intersect are $(0, -1), (1, 0)$ and $(-1, 0)$ .Because there are only $3$ intersections and $a > 1/2$ , as well as $a > 1/4$ we know that either $\textbf{(D)}$ or $\textbf{(E)}$ is the correct answer. Then we can test a number from $(1/2, 1/4)$ to eliminate the remaining answer. So $\boxed{12}$ is the correct answer.

Solution: Simply plug in $a = \frac{1}{2}, \frac{1}{4}, 1$ and solve the systems. (This shouldn't take too long.) And then realize that only $a=1$ yields three real solutions for $x$ , so we are done and the answer is $\boxed{12}$

Solution: An ideal solution come to mind is where they intersect at the $x$ -axis at the same time, which is $(a, 0)$ and $(-a, 0).$ Take the root of our $y=x^2-a$ we get $x=\sqrt{a}$ , set them equal we get $a=\sqrt{a}.$ The only answer is $1$ so it only left us with the answer choice $\boxed{12}$


Answer: 12.0

Solution:
Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get \[x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0\] Since this is a quartic, there are $4$ total roots (counting multiplicity). We see that $x=0$ always has at least one intersection at $(0,-a)$ (and is in fact a double root).
The other two intersection points have $x$ coordinates $\pm\sqrt{2a-1}$ . We must have $2a-1> 0;$ otherwise we are in the case where the parabola lies entirely above the circle (tangent at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see that $\boxed{12}$

Solution:
Substituting $y = x^2 - a$ gives $x^2 + (x^2 - a)^2 = a^2$ , which simplifies to $x^2 + x^4 - 2x^2a + a^2 = a^2$ . This further simplifies to $x^2(1 + x^2 - 2a) = 0$ . Thus, either $x^2 = 0$ , or $x^2 - 2a + 1 = 0$
Since we care about $a$ , we consider the second case. We solve in terms of $a$ , giving $a = \frac{x^2}{2} + \frac{1}{2}$ . We see that in order to find the range in which $a$ lies, we must find the vertex of this equation, which turns out to be $\left(0, \frac{1}{2}\right)$ . Hence, we know that the minimum is $\frac{1}{2}$ , which further implies that $\boxed{12}$

Solution:
Looking at a graph, it is obvious that the two curves intersect at $(0, -a)$ . We also see that if the parabola goes "in" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.
We have $x^2 - a = -\sqrt{a^2 - x^2}$ . Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$ . Since $x = 0$ is already accounted for, we only need to find one solution for $x^2 = 2a - 1$ , where the right hand side portion is obviously increasing. Since $a = \frac{1}{2}$ begets $x = 0$ (an overcount), we have $\boxed{12}$ as the right answer.

Solution: This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is $2$ , the radius of the circle that matches it has a radius of $\frac{1}{2}$ . This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only $1$ point. When a larger circle is used, it is tangent to $3$ points because the points on either side are now separated from the vertex. Therefore, $\boxed{12}$ is correct.

Solution: Now, let's graph these two equations. We want the blue parabola to be inside this red circle.  Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$ . Expanding, we get $x^4-2ax^2+x^2=0$ . Factoring out the $x^2$ , we get $x^2(x^2-2a+1)=0$ . Then we find that $x=0$ or $x=\pm\sqrt{2a-1}$ . Therefore, $2a-1>0$ , which means $\boxed{12}$

Solution:
In order to solve for the values of $a$ , we need to just count multiplicities of the roots when the equations are set equal to each other: in other words, take the derivative. We know that $\sqrt{a^2 - x^2} = x^2 - a$ . Now, we take square of both sides, and rearrange to obtain $x^4 - (2a - 1)x^2 = 0$ . Now, we may take the second derivative of the equation to obtain $6x^2 - (2a - 1) = 0$ . Now, we must take discriminant. Since we need the roots of that equation to be real and not repetitive (otherwise they would not intersect each other at three points), the discriminant must be greater than zero. Thus,
\[b^2 - 4ac > 0 \rightarrow 0 - 4(6)(-(2a - 1)) > 0 \rightarrow a > \frac{1}{2}\] The answer is $\boxed{12}$ and we are done.

Solution:
Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$ , and they have symmetry across the $y$ -axis, thus, for them to intersect at exactly $3$ points, it suffices to find the $y$ solution.
First, rewrite the second equation to $y=x^2-a\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$ . This value must not be less than $0$ (It is the square root part of the quadratic formula). To find when it is $0$ , we find the roots: \[4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}\] Since $\lim_{a\to \infty}(4a^2-4a+1)=\infty$ , our range is $\boxed{12}$

Solution: We can see that if $a = 1$ , we know that the points where the two curves intersect are $(0, -1), (1, 0)$ and $(-1, 0)$ .Because there are only $3$ intersections and $a > 1/2$ , as well as $a > 1/4$ we know that either $\textbf{(D)}$ or $\textbf{(E)}$ is the correct answer. Then we can test a number from $(1/2, 1/4)$ to eliminate the remaining answer. So $\boxed{12}$ is the correct answer.

Solution: Simply plug in $a = \frac{1}{2}, \frac{1}{4}, 1$ and solve the systems. (This shouldn't take too long.) And then realize that only $a=1$ yields three real solutions for $x$ , so we are done and the answer is $\boxed{12}$

Solution: An ideal solution come to mind is where they intersect at the $x$ -axis at the same time, which is $(a, 0)$ and $(-a, 0).$ Take the root of our $y=x^2-a$ we get $x=\sqrt{a}$ , set them equal we get $a=\sqrt{a}.$ The only answer is $1$ so it only left us with the answer choice $\boxed{12}$


Answer: 100.0

Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$

Solution: Let our $4$ numbers be $n, n+2, n+4, n+6$ , where $n$ is odd. Then, our sum is $4n+12$ . The only answer choice that cannot be written as $4n+12$ , where $n$ is odd, is $\boxed{100}$

Solution: If the four consecutive odd integers are $2n-3,~ 2n-1, ~2n+1$ and $2n+3$ ; then, the sum is $8n$ . All the integers are divisible by $8$ except $\boxed{100}$

Solution: If the four consecutive odd integers are $a,~ a+2, ~a+4$ and $a+6$ , the sum is $4a+12$ , and $4a+12$ divided by $4$ gives $a+3$ . This means that $a+3$ must be even. The only integer that does not give an even integer when divided by $4$ is $100$ , so the answer is $\boxed{100}$

Solution: From Solution 1, we have the sum of the $4$ numbers to be equal to $4n + 12$ . Taking mod 8 gives us $4n + 4 \equiv b \pmod8$ for some residue $b$ and for some odd integer $n$ . Since $n \equiv 1 \pmod{2}$ , we can express it as the equation $n = 2a + 1$ for some integer $a$ . Multiplying 4 to each side of the equation yields $4n = 8a + 4$ , and taking mod 8 gets us $4n \equiv 4 \pmod{8}$ , so $b = 0$ . All the answer choices except choice D is a multiple of 8, and since 100 satisfies all the conditions the answer is $\boxed{100}$

Solution: Since they want CONSECUTIVE odd numbers, it won't be hard to just list the sums out: \[16=1+3+5+7\] \[40=7+9+11+13\] \[72=15+17+19+21\] \[200=47+49+51+53\] All of the answer choices can be a sum of consecutive odd numbers except $100$ , so the answer is $\boxed{100}$


Answer: 0.13

Which of the following is closest to $\sqrt{65}-\sqrt{63}$
$\textbf{(A)}\ .12 \qquad \textbf{(B)}\ .13 \qquad \textbf{(C)}\ .14 \qquad \textbf{(D)}\ .15 \qquad \textbf{(E)}\ .16$

Solution: We have $\sqrt{65} > 8 > 7.5$ . Also $7.5^2 = (7 + 0.5)^2 = 7^2 + 2 \cdot 7 \cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 < 63$ , so $\sqrt{63} > 7.5$ . Thus $\sqrt{65} + \sqrt{63} > 7.5 + 7.5 = 15$ . Now notice that $\sqrt{65} - \sqrt{63} = \frac{(\sqrt{65} - \sqrt{63})(\sqrt{65} + \sqrt{63})}{\sqrt{65} + \sqrt{63}} = \frac{2}{\sqrt{65} + \sqrt{63}}$ , so $\sqrt{65} - \sqrt{63} < \frac{2}{15} = 0.1333333...$ , so the answer must be $A$ or $B$ . To determine which, we write $\sqrt{65} - \sqrt{63} > 0.125 \iff 65 - 2\sqrt{65 \cdot 63} + 63 > 0.015625 \iff 128 - 0.015625 > 2\sqrt{4095} \iff \sqrt{4095} < 64 - 0.0078125 \iff 4095 < 4096 - 128 \cdot 0.0078125 + 0.0078125^2 = 4096 - 1 + 0.0078125^2$ which is true. Hence as the expression is greater than $0.125$ , and less than or equal to $0.13$ (since we showed it is certainly less than $0.1333333...$ ), it is closest to $0.13$ , which is answer $\boxed{.13}$


Answer: 0.11

Which of the following is closest to the product $(.48017)(.48017)(.48017)$
$\text{(A)}\ 0.011 \qquad \text{(B)}\ 0.110 \qquad \text{(C)}\ 1.10 \qquad \text{(D)}\ 11.0 \qquad \text{(E)}\ 110$

Solution: Clearly, \[.4<.48017<.5\] Since the function $f(x)=x^3$ is strictly increasing, we can say that \[.4^3<.48017^3<.5^3\] from which it follows that $\text{A}$ is much too small and $\text{C}$ is much too large, so $\boxed{0.110}$ is the answer.

Solution: Since $0.48017$ is quite close to $0.5$ , or $\dfrac{1}{2}$ , we can look for the answer choice that is just below $(\dfrac{1}{2})^3=\dfrac{1}{8}=0.125$ , which would be $\boxed{0.110}$


Answer: 12.0

Which of the following is equal to $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}$
$\text{(A) } \frac 14 \qquad \text{(B) } \frac 13 \qquad \text{(C) } \frac 12 \qquad \text{(D) } \frac 23 \qquad \text{(E) } \frac 34$

Solution:
We write both the numerator and denominator with a denominator of $12$ first, since the LCM of $2,3,$ and $4$ is $3\cdot4=12$ .
Next, we multiply both the numerator and denominator by $12$ $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}=\dfrac{\frac{4}{12}-\frac{3}{12}}{\frac{6}{12}-\frac{4}{12}}=\dfrac{\frac{1}{12}}{\frac{2}{12}}=\boxed{12}$ .
-sosiaops


Answer: 12.0

Solution:
Multiplying the numerator and the denominator by the same value does not change the value of the fraction.
We can multiply both by $12$ , getting $\dfrac{4-3}{6-4} = \boxed{12}$


Answer: 0.99

Which of the following numbers is the largest?
$\text{(A)}\ .99 \qquad \text{(B)}\ .9099 \qquad \text{(C)}\ .9 \qquad \text{(D)}\ .909 \qquad \text{(E)}\ .9009$

Solution: We have $.99>.9099>.909>.9009>.9$ , so choice $\boxed{.99}$ is the largest.


Answer: 0.979

Which of the following numbers is the largest?
$\text{(A)}\ 0.97 \qquad \text{(B)}\ 0.979 \qquad \text{(C)}\ 0.9709 \qquad \text{(D)}\ 0.907 \qquad \text{(E)}\ 0.9089$

Solution:
Using the process of elimination:
Tenths digit:
All tenths digits are equal, at $9$
Hundreths digit: $A$ $B$ , and $C$ all have the same hundreths digit of $7$ , and it is greater than the hundredths of $D$ or $E$ (which is $0$ ).  Eliminate both $D$ and $E$
Thousandths digit: $B$ has the largest thousandths digit of the remaining answers, and is the correct answer. $A$ has an "invisible" thousandths digit of $0$ , while $C$ also has a thousdandths digit of $0$
$\boxed{0.979}$


Answer: 2.0

Which of these numbers is less than its reciprocal?
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution: The number $0$ has no reciprocal, and $1$ and $-1$ are their own reciprocals. This leaves only $2$ and $-2$ . The reciprocal of $2$ is $1/2$ , but $2$ is not less than $1/2$ . The reciprocal of $-2$ is $-1/2$ , and $-2$ is less than $-1/2$ , so it is $\boxed{2}$

Solution:
The statement "a number is less than its reciprocal" can be translated as $x < \frac{1}{x}$
Multiplication by $x$ can be done if you do it in three parts: $x>0$ $x=0$ , and $x<0$ .  You have to be careful about the direction of the inequality, as you do not know the sign of $x$
If $x>0$ , the sign of the inequality remains the same.  Thus, we have $x^2 < 1$ when $x > 0$ .  This leads to $0 < x < 1$
If $x=0$ , the inequality $x < \frac{1}{x}$ is undefined.
If $x<0$ , the sign of the inequality must be switched.  Thus, we have $x^2 > 1$ when $x < 0$ .  This leads to $x < -1$
Putting the solutions together, we have $x<-1$ or $0 < x < 1$ , or in interval notation, $(-\infty, -1) \cup(0, 1)$ .  The only answer in that range is $\boxed{2}$

Solution:
Starting again with $x < \frac{1}{x}$ , we avoid multiplication by $x$ .   Instead, move everything to the left, and find a common denominator:
$x < \frac{1}{x}$
$x - \frac{1}{x} < 0$
$\frac{x^2 - 1}{x} < 0$
$\frac{(x+1)(x-1)}{x} < 0$
Divide this expression at $x=-1$ $x=0$ , and $x=1$ , as those are the three points where the expression on the left will "change sign".
If $x<-1$ , all three of those terms will be negative, and the inequality is true.  Therefore, $(-\infty, -1)$ is part of our solution set.
If $-1 < x < 0$ , the $(x+1)$ term will become positive, but the other two terms remain negative.  Thus, there are no solutions in this region.
If $0 < x < 1$ , then both $(x+1)$ and $x$ are positive, while $(x-1)$ remains negative.  Thus, the entire region $(0, -1)$ is part of the solution set.
If $1 < x$ , then all three terms are positive, and there are no solutions.
At all three "boundary points", the function is either $0$ or undefined.  Therefore, the entire solution set is $(-\infty, -1) \cup (0, 1)$ , and the only option in that region is $x=-2$ , leading to $\boxed{2}$

Solution: We can find out all of their reciprocals. Now we compare and see that the answer is $\boxed{2}$


Answer: 1.0

Which positive numbers $x$ satisfy the equation $(\log_3x)(\log_x5)=\log_35$
$\textbf{(A)}\ 3 \text{ and } 5 \text{ only} \qquad  \textbf{(B)}\ 3, 5, \text{ and } 15 \text{ only} \qquad \\ \textbf{(C)}\ \text{only numbers of the form } 5^n \cdot 3^m, \text{ where } n \text{ and } m \text{ are positive integers} \qquad \\ \textbf{(D)}\ \text{all positive } x \neq 1 \qquad  \textbf{(E)}\ \text{none of these}$

Solution:
By the change-of-base formula, we can simplify the left side of the equation: $(\log_3x)(\log_x5) = (\frac{\log_x}{\log_3})(\frac{\log_5}{\log_x}) = \frac{\log_5}{\log_3}$
We see that this in fact simplifies to $\log_35$ , which will always equal the right side of the equation, since they are the same exact expressions.
But we have to be careful because $x \neq 1$ . Plugging in $x=1$ , the left side would equal $(\log_31)(\log_x5) = 0 \cdot \log_x5 = 0$ , and $\log_35$ definitely does not equal $0$
Besides $1$ $x$ can take on any positive value, and the equation would work. Therefore, the answer is $\boxed{1}$ jiang147369


Answer: 10.0

While eating out, Mike and Joe each tipped their server $2$ dollars. Mike tipped $10\%$ of his bill and Joe tipped $20\%$ of his bill. What was the difference, in dollars between their bills?
$\textbf{(A) } 2\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 10\qquad \textbf{(E) } 20$

Solution:
Let $m$ be Mike's bill and $j$ be Joe's bill.
$\frac{10}{100}m=2$
$m=20$
$\frac{20}{100}j=2$
$j=10$
So the desired difference is $m-j=20-10=10 \Rightarrow \boxed{10}$


Answer: 50.0

While exploring a cave, Carl comes across a collection of $5$ -pound rocks worth $$14$ each, $4$ -pound rocks worth $$11$ each, and $1$ -pound rocks worth $$2$ each. There are at least $20$ of each size. He can carry at most $18$ pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?
$\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52$

Solution:
The value of $5$ -pound rocks is $$14\div5=$2.80$ per pound, and the value of $4$ -pound rocks is $$11\div4=$2.75$ per pound. Clearly, Carl should not carry more than three $1$ -pound rocks. Otherwise, he can replace some $1$ -pound rocks with some heavier rocks, preserving the weight but increasing the total value.
We perform casework on the number of $1$ -pound rocks Carl can carry: \[\begin{array}{c|c|c||c} & & & \\ [-2.5ex] \boldsymbol{1}\textbf{-Pound Rocks} & \boldsymbol{4}\textbf{-Pound Rocks} & \boldsymbol{5}\textbf{-Pound Rocks} & \textbf{Total Value} \\ \textbf{(}\boldsymbol{$2}\textbf{ Each)} & \textbf{(}\boldsymbol{$11}\textbf{ Each)} & \textbf{(}\boldsymbol{$14}\textbf{ Each)} & \\ [0.5ex] \hline & & & \\ [-2ex] 0 & 2 & 2 & $50 \\  & & & \\ [-2.25ex] 1 & 3 & 1 & $49 \\ & & & \\ [-2.25ex] 2 & 4 & 0 & $48 \\ & & & \\ [-2.25ex] 3 & 0 & 3 & $48 \end{array}\] Clearly, the maximum value of the rocks Carl can carry is $\boxed{50}$ dollars.

Solution: Since each rock is worth $1$ dollar less than $3$ times its weight (in pounds), the answer is just $3\cdot 18=54$ minus the minimum number of rocks we need to make $18$ pounds. Note that we need at least $4$ rocks (two $5$ -pound rocks and two $4$ -pound rocks) to make $18$ pounds, so the answer is $54-4=\boxed{50}.$


Answer: 900.0

While finding the sine of a certain angle, an absent-minded professor failed to notice that his calculator was not in the correct angular mode. He was lucky to get the right answer. The two least positive real values of $x$ for which the sine of $x$ degrees is the same as the sine of $x$ radians are $\frac{m\pi}{n-\pi}$ and $\frac{p\pi}{q+\pi}$ , where $m$ $n$ $p$ , and $q$ are positive integers. Find $m+n+p+q$

Solution:
Note that $x$ degrees is equal to $\frac{\pi x}{180}$ radians. Also, for $\alpha \in \left[0 , \frac{\pi}{2} \right]$ , the two least positive angles $\theta > \alpha$ such that $\sin{\theta} = \sin{\alpha}$ are $\theta = \pi-\alpha$ , and $\theta = 2\pi + \alpha$
Clearly $x > \frac{\pi x}{180}$ for positive real values of $x$
$\theta = \pi-\alpha$ yields: $x = \pi - \frac{\pi x}{180} \Rightarrow x = \frac{180\pi}{180+\pi} \Rightarrow (p,q) = (180,180)$
$\theta = 2\pi + \alpha$ yields: $x = 2\pi + \frac{\pi x}{180} \Rightarrow x = \frac{360\pi}{180-\pi} \Rightarrow (m,n) = (360,180)$
So, $m+n+p+q = \boxed{900}$


Answer: 90.0

While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next to each other after the break. Find the number of possible seating orders they could have chosen after the break.

Solution:
We proceed with casework based on the person who sits first after the break.
$\textbf{Case 1:}$ A is first. Then the first three people in the row can be ACE, ACF, ADB, ADF, AEB, AEC, AFB, AFC, or AFD, which yield 2, 1, 2, 2, 1, 2, 0, 1, and 1 possible configurations, respectively, implying 2 + 1 + 2 + 2 + 1 + 2 + 0 + 1 + 1 = 12 possible configurations in this case.
$\textbf{Case 2:}$ B is first. Then the first three people in the row can be BDA, BDF, BEA, BEC, BFA, BFC, or BFD, which yield 2, 4, 2, 4, 0, 1, and 2 possible configurations, respectively, implying 2 + 4 + 2 + 4 + 0 + 1 + 2 = 15 possible configurations in this case.
$\textbf{Case 3:}$ C is first. Then the first three people in the row can be CAD, CAE, CAF, CEA, CEB, CFA, CFB, or CFD, which yield 1, 2, 1, 4, 4, 2, 2, and 2 possible configurations, respectively, implying 1 + 2 + 1 + 4 + 4 + 2 + 2 + 2 = 18 possible configurations in this case.
Finally, the number of valid configurations for A and F, B and E, and C and D are equal by symmetry, so our final answer is 2(12 + 15 + 18), which computes to be $\boxed{090}.$ ~peace09

Solution:
We determine the order of A, B, C, relative to each other. Then, we will insert D, E, F into the alignment and calculate the total number of possibilities. This solution can be visualised as standing in lines rather than sitting on chairs.
There are 6 possible alignments for A, B, and C, but we only evaluate 3 because the other 3 cases can be mirrored to overlap these 3 cases.
$\textbf{Case 1: A B C}$ In this case, there must be a person standing between A and B and also between B and C. Also, D cannot be adjacent to C. There are 9 possibilities.
$\textbf{Case 2: A C B}$ In this case, there must be a person standing between B and C. Also, D cannot be adjacent to C. There are 12 possibilities.
$\textbf{Case 3: C A B}$ In this case, there must be a person standing between A and B. Also, D cannot be adjacent to C. There are 24 possibilities.
So the total number of cases is 2(9+12+24)= $\boxed{090}$

Solution:
Consider arrangements of numbers $1, 2, 3, ..., n$
Let $f(n,k)$ be the number of arrangements in which $1$ and $2$ $2$ and $3$ $3$ and $4$ , ..., $k-1$ and $k$ aren't together. We need to find $f(6,6)$
Let $d(n,k)$ be the number of arrangements in which $1$ and $2$ $2$ and $3$ $3$ and $4$ , ..., $k-2$ and $k-1$ aren't together, but $(k-1)$ and $k$ are together.
Then,
\[f(n,k+1) = f(n,k) - d(n,k+1)  ......(1)\]
\[d(n,k+1) = 2f(n-1,k) + d(n-1,k)  ......(2)\]
Hence, \[f(n,k+1)=f(n,k)-f(n-1,k)-f(n-1,k-1)\]
And because $f(n,0) = f(n,1) = n!$ , it's easy to get $f(6,6) = \boxed{090}$


Answer: 23.0

Wind chill is a measure of how cold people feel when exposed to wind outside. A good estimate for wind chill can be found using this calculation \[(\text{wind chill}) = (\text{air temperature}) - 0.7 \times (\text{wind speed}),\] where temperature is measured in degrees Fahrenheit $(^{\circ}\text{F})$ and the wind speed is measured in miles per hour (mph). Suppose the air temperature is $36^{\circ}\text{F}$ and the wind speed is $18$ mph. Which of the following is closest to the approximate wind chill?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 35$

Solution: By substitution, we have \begin{align*} (\text{wind chill}) &= 36 - 0.7 \times 18 \\ &= 36 - 12.6 \\ &= 23.4 \\ &\approx \boxed{23} ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM

Solution: $0.7$ is very close to $\frac{2}{3}$ - therefore, we can substitute $\frac{2}{3}$ into the equation to get $36 - \frac{2}{3} * 18$ , which is $36 - 12 = 24$ . As $\frac{2}{3}$ is slightly less than $0.7$ , the correct answer is slightly less than $24$ . Therefore, the answer is $\boxed{23}$


Answer: 91.0

With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$

Solution: Let $x = \cos 1^\circ + i \sin 1^\circ$ . Then from the identity \[\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},\] we deduce that (taking absolute values and noticing $|x| = 1$ \[|2\sin 1| = |x^2 - 1|.\] But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ - z)$ , if we let our product be $M$ then \[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ\] \[= \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|\] because $\sin$ is positive in the first and second quadrants. Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1$ , and so \[\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.\] It is easy to see that $M = 2^{89}$ and that our answer is $2 + 89 = \boxed{91}$

Solution:
Let $p=\sin1\sin3\sin5...\sin89$
\[p=\sqrt{\sin1\sin3\sin5...\sin177\sin179}\]
\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}\]
\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{(2\sin1\cos1)\cdot(2\sin2\cos2)\cdot(2\sin3\cos3)\cdot....\cdot(2\sin89\cos89)}}\]
\[=\sqrt{\frac{1}{2^{89}}\frac{\sin90\sin91\sin92\sin93...\sin177\sin178\sin179}{\cos1\cos2\cos3\cos4...\cos89}}\]
$=\sqrt{\frac{1}{2^{89}}}$ because of the identity $\sin(90+x)=\cos(x)$
we want $\frac{1}{p^2}=2^{89}$
Thus the answer is $2+89=\boxed{091}$

Solution: Similar to Solution $2$ , so we use $\sin{2\theta}=2\sin\theta\cos\theta$ and we find that: \begin{align*}\sin(4)\sin(8)\sin(12)\sin(16)\cdots\sin(84)\sin(88)&=(2\sin(2)\cos(2))(2\sin(4)\cos(4))(2\sin(6)\cos(6))(2\sin(8)\cos(8))\cdots(2\sin(42)\cos(42))(2\sin(44)\cos(44))\\ &=(2\sin(2)\sin(88))(2\sin(4))\sin(86))(2\sin(6)\sin(84))(2\sin(8)\sin(82))\cdots(2\sin(42)\sin(48))(2\sin(44)\sin(46))\\ &=2^{22}(\sin(2)\sin(88)\sin(4)\sin(86)\sin(6)\sin(84)\sin(8)\sin(82)\cdots\sin(42)\sin(48)\sin(44)\sin(46))\\ &=2^{22}(\sin(2)\sin(4)\sin(6)\sin(8)\cdots\sin(82)\sin(84)\sin(86)\sin(88))\end{align*} Now we can cancel the sines of the multiples of $4$ \[1=2^{22}(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86))\] So $\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)=2^{-22}$ and we can apply the double-angle formula again: \begin{align*}2^{-22}&=(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)\\ &=(2\sin(1)\cos(1))(2\sin(3)\cos(3))(2\sin(5)\cos(5))(2\sin(7)\cos(7))\cdots(2\sin(41)\cos(41))(2\sin(43)\cos(43))\\ &=(2\sin(1)\sin(89))(2\sin(3)\sin(87))(2\sin(5)\sin(85))(2\sin(7)\sin(87))\cdots(2\sin(41)\sin(49))(2\sin(43)\sin(47))\\ &=2^{22}(\sin(1)\sin(89)\sin(3)\sin(87)\sin(5)\sin(85)\sin(7)\sin(83)\cdots\sin(41)\sin(49)\sin(43)\sin(47))\\ &=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\end{align*} Of course, $\sin(45)=2^{-\frac{1}{2}}$ is missing, so we multiply it to both sides: \[2^{-22}\sin(45)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(45))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] \[\left(2^{-22}\right)\left(2^{-\frac{1}{2}}\right)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] \[2^{-\frac{45}{2}}=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] Now isolate the product of the sines: \[\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89)=2^{-\frac{89}{2}}\] And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: \[\csc^2(1)\csc^2(3)\csc^2(5)\csc^2(7)\cdots\csc^2(83)\csc^2(85)\csc^2(87)\csc^2(89)=\left(\frac{1}{2^{-\frac{89}{2}}}\right)^2=\left(2^{\frac{89}{2}}\right)^2=2^{89}\] The answer is therefore $m+n=(2)+(89)=\boxed{091}$

Solution:
Let $p=\prod_{k=1}^{45} \csc^2(2k-1)^\circ$
Then, $\sqrt{\frac{1}{p}}=\prod_{k=1}^{45} \sin(2k-1)^\circ$
Since $\sin\theta=\cos(90^{\circ}-\theta)$ , we can multiply both sides by $\frac{\sqrt{2}}{2}$ to get $\sqrt{\frac{1}{2p}}=\prod_{k=1}^{23} \sin(2k-1)^\circ\cos(2k-1)^\circ$
Using the double-angle identity $\sin2\theta=2\sin\theta\cos\theta$ , we get $\sqrt{\frac{1}{2p}}=\frac{1}{2^{23}}\prod_{k=1}^{23} \sin(4k-2)^\circ$
Note that the right-hand side is equal to $\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \sin(4k)^\circ$ , which is equal to $\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} 2\sin(2k)^\circ\cos(2k)^\circ$ , again, from using our double-angle identity.
Putting this back into our equation and simplifying gives us $\sqrt{\frac{1}{2p}}=\frac{1}{2^{45}}\prod_{k=23}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \cos(2k)^\circ$
Using the fact that $\sin\theta=\cos(90^{\circ}-\theta)$ again, our equation simplifies to $\sqrt{\frac{1}{2p}}=\frac{\sin90^\circ}{2^{45}}$ , and since $\sin90^\circ=1$ , it follows that $2p = 2^{90}$ , which implies $p=2^{89}$ . Thus, $m+n=2+89=\boxed{091}$

Solution:
Recall that $\sin\alpha\cdot \sin(60^{\circ}-\alpha)\cdot \sin(60^{\circ}+\alpha)=\frac{1}{4}\cdot \sin3\alpha$ Since it is in csc, we can write in sin and then take reciprocal.
We can group them by threes, $P=(\sin1^{\circ}\cdot \sin59^{\circ}\cdot \sin61^{\circ})\cdots(\sin29^{\circ}\cdot \sin31^{\circ}\cdot \sin89^{\circ})$ . Thus \begin{align*} P &=\frac{1}{4^{15}}\cdot \sin3^{\circ}\cdot \sin9^{\circ}\cdots\sin87^{\circ}\\ &=\frac{1}{4^{20}}\cdot \sin9^{\circ}\cdot \sin27^{\circ}\cdot \sin45^{\circ}\cdot \sin63^{\circ}\cdot \sin81^{\circ}\\ &=\frac{1}{4^{20}}\cdot \frac{\sqrt{2}}{2}\cdot \sin9^{\circ}\cdot \cos9^{\circ}\cdot \sin27^{\circ}\cdot \cos27^{\circ}\\ &=\frac{1}{4^{21}}\cdot \frac{\sqrt{2}}{2}\cdot \sin18^{\circ}\cdot \cos36^{\circ}=\frac{\sqrt{2}}{2^{45}} \end{align*} So we take reciprocal, $\frac 1P=2^{\frac{89}{2}}$ , the desired answer is $\frac{1}{P^2}=2^{89}$ leads to answer $\boxed{091}$

Solution:
We have
\[\prod_{k=1}^{45} \csc^2(2k-1)^\circ = \left(\frac{1}{\sin1^\circ \cdot \sin3^\circ \cdots \sin89^\circ}\right)^2.\]
Multiplying by $\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}$ gives
\[\left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin1^\circ \sin2^\circ \cdot \sin3^\circ \cdots \sin88^\circ \cdot \sin89^\circ}\right)^2\]
\[= \left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin1^\circ \sin2^\circ \cdot \sin3^\circ \cdots \sin 45^\circ \cdot \cos 44^\circ \cdot \cos 43^\circ \cdots \cos1^\circ}\right)^2.\]
Using $\sin\alpha \cos\alpha = \frac{1}{2}\sin{2\alpha}$ gives
\[\left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\frac{1}{2} \sin2^\circ \cdot \frac{1}{2} \sin4^\circ \cdots \frac{1}{2} \sin88^\circ \cdot \sin45^\circ}\right) ^2\]
\[= \left(\frac{1}{(\frac{1}{2})^{44} \cdot \frac{\sqrt{2}}{2}}\right)^2\]
\[= 2^{89}.\]
Thus, the answer is $2+89 = \boxed{091}.$


Answer: 13.0

With the use of three different weights, namely $1$ lb., $3$ lb., and $9$ lb., how many objects of different weights can be weighed, if the objects is to be weighed and the given weights may be placed in either pan of the scale? $\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 11\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 7$

Solution: The heaviest object that could be weighed with this set weighs $1 + 3 + 9 = 13$ lb., and we can weigh any positive integer weight at most that. This means that $13$ different objects could be weighed, so our answer is $\boxed{13}$ and we are done.


Answer: 34.0

Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?
$\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78$

Solution: Let $x$ represent the distance from home to the stadium, and let $r$ represent the distance from Yan to home. Our goal is to find $\frac{r}{x-r}$ . If Yan walks directly to the stadium, then assuming he walks at a rate of $1$ , it will take him $x-r$ units of time. Similarly, if he walks back home it will take him $r + \frac{x}{7}$ units of time. Because the two times are equal, we can create the following equation: $x-r = r + \frac{x}{7}$ . We get $x-2r=\frac{x}{7}$ , so $\frac{6}{7}x = 2r$ , and $\frac{x}{r} = \frac{7}{3}$ . This minus one is the reciprocal of what we want to find: $\frac{7}{3}-1 = \frac{4}{3}$ , so the answer is $\boxed{34}$

Solution:
 Let $H$ represent Yan's home, $S$ represent the stadium, and $Y$ represent Yan's current position. If Yan walks directly to the stadium, he will reach Point $P$ the same time he will reach $H$ if he is walking home.
Since he bikes $7$ times as fast as he walks and the time is the same, the distance from his home to the stadium must be $7$ times the distance from $P$ to the stadium. If $PS=x$ , then $HS=7x$ and $HP=6x$ . Since Y is the midpoint of $\overline{HP}$ $HY=YP=3x$ . Therefore, the ratio of Yan's distance from his home to his distance from the stadium is $\frac{YH}{YS}=\frac{3x}{4x}=\boxed{34}$


Answer: 34.0

Solution: Let $x$ represent the distance from home to the stadium, and let $r$ represent the distance from Yan to home. Our goal is to find $\frac{r}{x-r}$ . If Yan walks directly to the stadium, then assuming he walks at a rate of $1$ , it will take him $x-r$ units of time. Similarly, if he walks back home it will take him $r + \frac{x}{7}$ units of time. Because the two times are equal, we can create the following equation: $x-r = r + \frac{x}{7}$ . We get $x-2r=\frac{x}{7}$ , so $\frac{6}{7}x = 2r$ , and $\frac{x}{r} = \frac{7}{3}$ . This minus one is the reciprocal of what we want to find: $\frac{7}{3}-1 = \frac{4}{3}$ , so the answer is $\boxed{34}$

Solution:
 Let $H$ represent Yan's home, $S$ represent the stadium, and $Y$ represent Yan's current position. If Yan walks directly to the stadium, he will reach Point $P$ the same time he will reach $H$ if he is walking home.
Since he bikes $7$ times as fast as he walks and the time is the same, the distance from his home to the stadium must be $7$ times the distance from $P$ to the stadium. If $PS=x$ , then $HS=7x$ and $HP=6x$ . Since Y is the midpoint of $\overline{HP}$ $HY=YP=3x$ . Therefore, the ratio of Yan's distance from his home to his distance from the stadium is $\frac{YH}{YS}=\frac{3x}{4x}=\boxed{34}$


Answer: 150.0

Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$

Solution:
Set the time Ian traveled as $I$ , and set Han's speed as $H$ . Therefore, Jan's speed is $H+5.$
We get the following equation for how much Han is ahead of Ian: $H+5I = 70.$
The expression for how much Jan is ahead of Ian is: $2(H+5)+10I.$
This simplifies to: $2H+10+10I.$
However, this is just $2(H+5I)+10.$
Substitute, from the first equation, $H+5I$ as $70.$
Therefore, the answer is $140 + 10$ , which is $150$ , or $\boxed{150}$

Solution:
We let Ian's speed and time equal $I_s$ and $I_t$ , respectively. Similarly, let Han's and Jan's speed and time be $H_s$ $H_t$ $J_s$ $J_t$ . The problem gives us 5 equations
\begin{align} H_s&=I_s+5 \\ H_t&=I_t+1 \\ J_s&=I_s+10 \\ J_t&=I_t+2 \\ H_s \cdot H_t & =I_s \cdot I_t+70 \end{align}
Substituting equations $(1)$ and $(2)$ into $(5)$ gives:
\[(I_s+5)(I_t+1)=I_s I_t+70 \Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \Longrightarrow I_s+5I_t=65 \quad (*)\]
We are asked the difference between Jan's and Ian's distances, or
\[J_s J_t-I_s I_t=x,\]
Where $x$ is the difference between Jan's and Ian's distances and the answer to the problem. Substituting $(3)$ and $(4)$ equations into this equation gives:
\[(I_s+10)(I_t+2)-I_s I_t=x \Longrightarrow I_s I_t+2I_s+10I_t+20-I_s I_t=x \Longrightarrow\]
\[2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x\]
Substituting $(*)$ into this equation gives:
\[2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x\]
Therefore, the answer is $150$ miles or $\boxed{150}$

Solution: Let Ian drive $D$ miles, at a speed of $R$ , for some time $T$ (in hours). Hence, we have $D=RT$ . We can find a similar equation for Han, who drove $D + 70$ miles, at a rate of $R+5$ , for $T+1$ hours, giving us $D + 70 = (R + 5)(T + 1)$ . We can do the same for Jan, giving us $D + x = (R + 10)(T + 2)$ , where $x$ is how much further Jan traveled than Ian. We now have three equations: \[D= RT\] \[D + 70 = (R+5)(T+1) = RT + R + 5T + 5\] \[D + x = (R + 10)(T + 2) = RT + 10 T + 2R + 20.\] Substituting $RT$ for $D$ in the second and third equations and cancelling gives us: \[70 = 5T + R + 5 \Longrightarrow 5T + R = 65\] \[x = 10T + 2R + 20  \Longrightarrow x = 2(5T + R ) + 20  \Longrightarrow x= 2(65) + 20 = 150.\] Since $x = 150$ , our answer is $\boxed{150}$

Solution:
Let Ian drive $d$ miles, $t$ hours, and at speed $s$
Ian's Equation: $d=s \cdot t.$
Han drove 70 more miles, traveled 5 miles per hour faster and traveled 1 more hour than Ian.
Han's Equation: $d+70=(s+5) \cdot (t+1).$
Let Jan have driven $m$ miles. Jan also has driven 10 miles per hour faster and traveled 2 more hours than Ian.
Jan's Equation: $m=(s+10) \cdot (t+2).$
Let's group the equations together:
$(1) \phantom{a} d=s \cdot t$
$(2) \phantom{a} d+70=(s+5) \cdot (t+1)$
$(3) \phantom{a} m=(s+10) \cdot (t+2)$
Let's see what we want to find: We want to find $n$ . The equation is $m=d+n$ where $n$ is the number of more miles traveled by Jan than Ian.
Onto the calculating part.
Expanding the second equation, we get
$d+70=st+5t+s+5.$
Note that $st=d$ by the first equation, so substituting we get
$d+70=d+5t+s+5.$
Simplifying gets us
$(4) \phantom{a} s+5t=65.$
(Note for the above process: You could have substituted $d$ with $st$ but that would lead you to the same result since $d-d=st-st=0.$
Let's look at the third equation. Expanding, we get
$m=st+10t+2s+20.$
Since we want $n$ we want the equation $m=d+n$ . We write the expanded third equation into this form since $st=d.$
$(5) \phantom{a} m=d+(10t+2s+20)$
Let's take a closer look at the $n$ section of the equation:
$(6) \phantom{a} n=10t+2s+20$
This looks very similar to equation 4, if you multiply equation 4 by $2$ you get
$(7) \phantom{a} 10t+2s=130$
Plugging equation 7 into equation 6 we have
$n=(10t+2s)+20 \Rightarrow n=130+20$
Calculating gets us
$(8) \phantom{a} n=150.$
Substituting equation 8 into equation 5 gets us
$m=d+n \Rightarrow m=d+150.$
The $n$ term is $150$ which is what we want to find so the answer is $\boxed{150}.$

Solution: Since Han drove for $1$ hour and drove $70$ miles more than Ian during that hour, we know that Ian's speed is $65$ miles per hour since Han drove $5$ mph faster than him. Now Jan went $10$ mph faster than Ian for $2$ hours, so we can tell that she drove $75 \cdot 2$ miles more than Ian, therefore the answer is $\boxed{150}.$


Answer: 250.0

You and five friends need to raise $1500$ dollars in donations for a charity, dividing the fundraising equally.  How many dollars will each of you need to raise?
$\mathrm{(A) \ } 250\qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 1500 \qquad \mathrm{(D) \ } 7500 \qquad \mathrm{(E) \ } 9000$

Solution: There are $6$ people to split the $1500$ dollars among, so each person must raise $\frac{1500}6=250$ dollars. $\Rightarrow\boxed{250}$


Answer: 0.0

You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2 \times 3 \times 5 \times\ldots \times 61$ of all prime numbers less than or equal to $61$ , and $n$ takes, successively, the values $2, 3, 4,\ldots, 59$ .
Let $N$ be the number of primes appearing in this sequence. Then $N$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 57\qquad \textbf{(E)}\ 58$

Solution:
First, note that $n$ does not have a prime number larger than $61$ as one of its factors.  Also, note that $n$ does not equal $1$
Therefore, since the prime factorization of $n$ only has primes from $2$ to $59$ $n$ and $P$ share at least one common factor other than $1$ .  Therefore $P+n$ is not prime for any $n$ , so the answer is $\Rightarrow{\boxed{0}$


Answer: 4.0

You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$

Solution: First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with $5$ guesses, and one with $4$ . Since the problem is asking for the minimum number, the answer is $\boxed{4}$

Solution: Since the hidden rectangle can only hide two adjacent squares, we may think that we eliminate 8 squares and we're done, but think again. This is the AMC 10, so there must be a better solution (also note that every other solution choice is below 8 so we're probably not done) So, we think again, we notice that we haven't used the adjacent condition, and then it clicks. If we eliminate the four squares with only one edge on the boundary of the 9x9 square. We are left with 5 diagonal squares, since our rectangle cant be diagonal, we can ensure that we find it in 4 moves. So our answer is : $\boxed{4}$

Solution:
We realize that every $2 \times 1$ rectangle must contain an edge and no more than one edge. There are a total of four edges so the answer is $\boxed{4.}$ .
~darrenn.cp


Answer: 4.0

Solution:
The $3 \times 3$ grid can be colored like a checkerboard with alternating black and white squares.
Let the top left square be white, and the rest of the squares alternate colors.
Each $2 \times 1$ rectangle always covers $1$ white square and $1$ black square.
You can ensure that at least one of your guessed squares is covered by the rectangle by choosing either each of the white squares ( $5$ turns) or each of the black squares ( $4$ turns).
Since it is ideal to be the most efficient with our turns, by choosing all the black squares, we guarantee that one of the $\boxed{4}$ squares are of the $2 \times 1$ rectangle.


Answer: 4.0

Solution: First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with $5$ guesses, and one with $4$ . Since the problem is asking for the minimum number, the answer is $\boxed{4}$

Solution: Since the hidden rectangle can only hide two adjacent squares, we may think that we eliminate 8 squares and we're done, but think again. This is the AMC 10, so there must be a better solution (also note that every other solution choice is below 8 so we're probably not done) So, we think again, we notice that we haven't used the adjacent condition, and then it clicks. If we eliminate the four squares with only one edge on the boundary of the 9x9 square. We are left with 5 diagonal squares, since our rectangle cant be diagonal, we can ensure that we find it in 4 moves. So our answer is : $\boxed{4}$

Solution:
We realize that every $2 \times 1$ rectangle must contain an edge and no more than one edge. There are a total of four edges so the answer is $\boxed{4.}$ .
~darrenn.cp


Answer: 4.0

Solution:
The $3 \times 3$ grid can be colored like a checkerboard with alternating black and white squares.
Let the top left square be white, and the rest of the squares alternate colors.
Each $2 \times 1$ rectangle always covers $1$ white square and $1$ black square.
You can ensure that at least one of your guessed squares is covered by the rectangle by choosing either each of the white squares ( $5$ turns) or each of the black squares ( $4$ turns).
Since it is ideal to be the most efficient with our turns, by choosing all the black squares, we guarantee that one of the $\boxed{4}$ squares are of the $2 \times 1$ rectangle.


Answer: 1.0

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

Solution:
Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins.
You must have $1$ more penny.  If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes, and quarters.  But you only have $5$ more coins to assign.
Now you have $61 - 1 = 60$ cents remaining for $4$ coins, which may be nickels, quarters, or dimes.  If you have only one more dime, that leaves $50$ cents in $3$ nickels or quarters, which is impossible.  If you have two dimes, that leaves $40$ cents for $2$ nickels or quarters, which is again impossible.  If you have three dimes, that leaves $30$ cents for $1$ nickel or quarter, which is still impossible.  And all four remaining coins being dimes will not be enough.
Therefore, you must have no more dimes to assign, and the $60$ cents in $4$ coins must be divided between the quarters and nickels.  We quickly see that $2$ nickels and $2$ quarters work.  Thus, the total count is $2$ quarters, $2$ nickels, $1$ penny, plus one more coin of each type that we originally subtracted.  Double-checking, that gives a total $2 + 2 + 1 + 4 = 9$ coins, and a total of $2\cdot 25 + 2\cdot 5 + 1 + (1 + 5 + 10 + 25) = 102$ cents.
There is only $1$ dime in that combo, so the answer is $\boxed{1}$

Solution: We see that there must be 102 cents, so therefore there's at least 2 pennies. That leaves 7 coins. We assume that there are 3 quarters, leaving 25 cents with 4 coins left. If all 4 are nickels, that would only be 20 cents, missing 5. Therefore, one nickel must be changed into 1 dime, so the answer is $\boxed{1}$

Solution: It is clear that there should only be $2$ pennies; any more would take up too many coins, and the limit is $9$ . Now we have $ $1$ left, and $7$ coins to use. Looking at the quarters, we can make methodical guesses. If there is $1$ quarter, then we would have to make $ $0.75$ with $6$ coins. We take a few educated guesses for the nickel and dime combos, and see that $1$ quarter will not work. Trying values for $2$ quarters, we see that this will not work either. When we reach $3$ quarters, the remaining is $ $0.25$ made from $4$ coins. We try with $2$ dimes, which does not work (it only takes $3$ coins) and we try with $1$ dime. After trying $2$ pennies, $3$ quarters, $1$ dime, and $3$ nickels, it is evident that this combo works. Therefore, the answer is $\boxed{1}$


Answer: 12.0

Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution: Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$ , respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each other. If we place $S$ and $T$ next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. $ST\square\square, \square ST\square, \square\square ST$ ). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $TS\square\square, \square TS\square, \square\square TS$ ). Thus there are 6 ways of placing $S$ and $T$ directly next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$ . Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\times 2=12$ ways to place $S$ and $T$ next to each other. Subtracting this from the total number of arrangements gives us $24-12=12$ total arrangements $\implies\boxed{12}$

Solution: Let's try complementary counting. There $4!$ ways to arrange the 4 marbles. However, there are $2\cdot3!$ arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, \[4!-2\cdot3!=\boxed{12}\]

Solution: We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a "super marble." There are $2!$ ways to arrange Steelie and Tiger within this "super marble." Then there are $3!$ ways to arrange the "super marble" and Zara's two other marbles in a row. Since there are $4!$ ways to arrange the marbles without any restrictions, the answer is given by $4!-2!\cdot 3!=\boxed{12}$

Solution:
We will use the following
$\textbf{Georgeooga-Harryooga Theorem:}$ The Georgeooga-Harryooga Theorem states that if you have $a$ distinguishable objects and $b$ of them cannot be together, then there are $\frac{(a-b)!(a-b+1)!}{b!}$ ways to arrange the objects.
$\textit{Proof. (Created by AoPS user RedFireTruck)}$
Let our group of $a$ objects be represented like so $1$ $2$ $3$ , ..., $a-1$ $a$ . Let the last $b$ objects be the ones we can't have together.
Then we can organize our objects like so $\square1\square2\square3\square...\square a-b-1\square a-b\square$
We have $(a-b)!$ ways to arrange the objects in that list.
Now we have $a-b+1$ blanks and $b$ other objects so we have $_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}$ ways to arrange the objects we can't put together.
By fundamental counting principle our answer is $\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$
Proof by RedFireTruck talk ) 12:09, 1 February 2021 (EST)
Back to the problem. By the Georgeooga-Harryooga Theorem , our answer is $\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{12}$


Answer: 97.0

Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution:
For the next five races, Zou wins four and loses one. Let $W$ and $L$ denote a win and a loss, respectively. There are five possible outcome sequences for Zou:
We proceed with casework:
Case (1): Sequences #1-4, in which Zou does not lose the last race.
The probability that Zou loses a race is $\frac13,$ and the probability that Zou wins the next race is $\frac13.$ For each of the three other races, the probability that Zou wins is $\frac23.$
There are four sequences in this case. The probability of one such sequence is $\left(\frac13\right)^2\left(\frac23\right)^3.$
Case (2): Sequence #5, in which Zou loses the last race.
The probability that Zou loses a race is $\frac13.$ For each of the four other races, the probability that Zou wins is $\frac23.$
There is one sequence in this case. The probability is $\left(\frac13\right)^1\left(\frac23\right)^4.$
Answer
The requested probability is \[4\left(\frac13\right)^2\left(\frac23\right)^3+\left(\frac13\right)^1\left(\frac23\right)^4=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},\] from which the answer is $16+81=\boxed{097}.$

Solution: We have $5$ cases, depending on which race Zou lost. Let $\text{W}$ denote a won race, and $\text{L}$ denote a lost race for Zou. The possible cases are $\text{WWWWL, WWWLW, WWLWW, WLWWW, LWWWW}$ . The first case has probability $\left(\frac{2}{3} \right)^4 \cdot \frac{1}{3} = \frac{16}{3^5}$ . The second case has probability $\left( \frac{2}{3} \right)^3 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{8}{3^5}$ . The third has probability $\left( \frac{2}{3} \right)^2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{3^5}$ . The fourth has probability $\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^2 = \frac{8}{3^5}$ . Lastly, the fifth has probability $\frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^3 = \frac{8}{3^5}$ . Adding these up, the total probability is $\frac{16 + 8 \cdot 4}{3^5} = \frac{16 \cdot 3}{3^5} = \frac{16}{81}$ , so $m+n = \boxed{097}$

Solution:
Case 1: Zou loses the first race
In this case, Zou must win the rest of the races. Thus, our probability is $\frac{8}{243}$
Case 2: Zou loses the last race
There is only one possibility for this, so our probability is $\frac{16}{243}$
Case 3: Neither happens
There are three ways that this happens. Each has one loss that is not the last race. Therefore, the probability that one happens is $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{8}{243}$ . Thus, the total probability is $\frac{8}{243} \cdot 3 = \frac{24}{243}$
Adding these up, we get $\frac{48}{243} = \frac{16}{81}$ , so $16+81=\boxed{097}$

Solution: Note that Zou wins one race. The probability that he wins the last race is $\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)=\frac{16}{243}.$ Now, if he doesn't win the last race, then there must be two races where the winner of the previous race loses. We can choose any $4$ of the middle races for Zou to win. So the probability for this case is $4\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2=\frac{32}{243}.$ Thus, the answer is $\frac{16}{243}+\frac{32}{243}=\frac{16}{81}\implies\boxed{097}.$


Answer: 13.0

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,0), B=(16,0), C=(16,16), D=(0,16), E=(32,0), F=(48,0), G=(48,16), H=(32,16), I=(0,8), J=(10,8), K=(10,16), L=(32,6), M=(40,6), N=(40,16); draw(A--B--C--D--A^^E--F--G--H--E^^I--J--K^^L--M--N); label("S", (18,8)); label("S", (50,8)); label("Figure 1", (A+B)/2, S); label("Figure 2", (E+F)/2, S); label("10'", (I+J)/2, S); label("8'", (12,12)); label("8'", (L+M)/2, S); label("10'", (42,11)); label("table", (5,12)); label("table", (36,11)); [/asy]
An $8'\times 10'$ table sits in the corner of a square room, as in Figure $1$ below.
The owners desire to move the table to the position shown in Figure $2$ .
The side of the room is $S$ feet. What is the smallest integer value of $S$ for which the table can be moved as desired without tilting it or taking it apart?
$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15$

Solution:
Small correction: The writer below has maximized the area of the rectangle (with sides parallel to the walls) that fits around the table, but there is a larger single dimension we can find in the table. The height or width is maximized when the diagonal of the table is horizontal or vertical. By the Pythagorean Theorem, this diagonal is $\sqrt{8^2+10^2} = \sqrt{164},$ which is between $\sqrt{144}$ and $\sqrt{169},$ so the answer is still $\textbf{(C)}.$ -hailstone
We begin by thinking about the motion of the table. As it moves, the table will have it's maximum height and width when the rectangle's sides form $45$ degree angles relative to the sides of the square. Therefore, by the Pythagorean Theorem, we have that $S= 4\sqrt{2}+5\sqrt{2}$ , with $4\sqrt{2}$ being the length of the leg formed by the side of the square with length $8$ and $5\sqrt{2}$ being the length of the leg formed by the side of the square with length $10$ . Adding these up yields $9\sqrt{2}$
We have that $\sqrt{2}\approx 1.414\approx 1.4$ . That means that $9\sqrt{2}\approx 12.6$ , which rounds up to $\boxed{13}$


Answer: 18.0

[asy] draw((-2,1)--(2,1)--(2,-1)--(-2,-1)--cycle); draw((0,0)--(0,-1)--(-2,-1)--(-2,0)--cycle); label("$F$",(0,0),E); label("$A$",(-2,1),W); label("$B$",(2,1),E); label("$C$", (2,-1),E); label("$D$",(-2,-1),WSW); label("$E$",(-2,0),W); label("$G$",(0,-1),S); //Credit to TheMaskedMagician for the diagram [/asy]
If rectangle ABCD has area 72 square meters and E and G are the midpoints of sides AD and CD, respectively, then the area of rectangle DEFG in square meters is
$\textbf{(A) }8\qquad \textbf{(B) }9\qquad \textbf{(C) }12\qquad \textbf{(D) }18\qquad \textbf{(E) }24$

Solution:
Solution by e_power_pi_times_i
Since the dimensions of $DEFG$ are half of the dimensions of $ABCD$ , the area of $DEFG$ is $\dfrac{1}{2}\cdot\dfrac{1}{2}$ of $ABCD$ , so the area of $ABCD$ is $\dfrac{1}{4}\cdot72 = \boxed{18}$


Answer: 128.0

[asy] draw((0,0)--(0,3)--(4,0)--cycle,dot); draw((4,0)--(7,0)--(7,10)--cycle,dot); draw((0,3)--(7,10),dot); MP("C",(0,0),SW);MP("A",(0,3),NW);MP("B",(4,0),S);MP("E",(7,0),SE);MP("D",(7,10),NE); [/asy]
Triangle $ABC$ has a right angle at $C, AC=3$ and $BC=4$ . Triangle $ABD$ has a right angle at $A$ and $AD=12$ . Points $C$ and $D$ are on opposite sides of $\overline{AB}$ . The line through $D$ parallel to $\overline{AC}$ meets $\overline{CB}$ extended at $E$ . If \[\frac{DE}{DB}=\frac{m}{n},\] where $m$ and $n$ are relatively prime positive integers, then $m+n=$
$\text{(A) } 25\quad \text{(B) } 128\quad \text{(C) } 153\quad \text{(D) } 243\quad \text{(E) } 256$

Solution:
Solution by e_power_pi_times_i
Let $F$ be the point such that $DF$ and $CF$ are parallel to $CE$ and $DE$ , respectively, and let $DE = x$ and $BE^2 = 169-x^2$ . Then, $[FDEC] = x(4+\sqrt{169-x^2}) = [ABC] + [BED] + [ABD] + [AFD] = 6 + \dfrac{x\sqrt{169-x^2}}{2} + 30 + \dfrac{(x-3)(4+\sqrt{169-x^2})}{2}$ . So, $4x+x\sqrt{169-x^2} = 60 + x\sqrt{169-x^2} - 3\sqrt{169-x^2}$ . Simplifying $3\sqrt{169-x^2} = 60 - 4x$ , and $1521 - 9x^2 = 16x^2 - 480x + 3600$ . Therefore $25x^2 - 480x + 2079 = 0$ , and $x = \dfrac{48\pm15}{5}$ . Checking, $x = \dfrac{63}{5}$ is the answer, so $\dfrac{DE}{DB} = \dfrac{\dfrac{63}{5}}{13} = \dfrac{63}{65}$ . The answer is $\boxed{128}$

Solution:
Solution by Arjun Vikram
Extend lines $AD$ and $CE$ to meet at a new point $F$ . Now, we see that $FAC\sim FDE \sim ACB$ . Using this relationship, we can see that $AF=\frac{15}4$ , (so $FD=\frac{63}4$ ), and the ratio of similarity between $FDE$ and $FAC$ is $\frac{63}{15}$ . This ratio gives us that $\frac{63}5$ . By the Pythagorean Theorem, $DB=13$ . Thus, $\frac{DE}{DB}=\frac{63}{65}$ , and the answer is $63+65=\boxed{128}$


Answer: 36.0

Solution:
We first notice that the paper strips cover up part of the others. Since the width of the overlap is $1$ and the length of the overlap is $1$ , the area of
each of the strips with the overlap is $(10\cdot 1)-1=9$ . Since there are 4 strips, $4\cdot 9=36 \implies \boxed{36}$


Answer: 15.0

[asy] draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); draw((-1.7,0)--(0,0)--(1.7,0),dot); draw((0,0)--(0,.98),dot); MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); [/asy]
A parabolic arch has a height of $16$ inches and a span of $40$ inches. The height, in inches, of the arch at the point $5$ inches from the center $M$ is:
$\text{(A) } 1\quad \text{(B) } 15\quad \text{(C) } 15\tfrac{1}{3}\quad \text{(D) } 15\tfrac{1}{2}\quad \text{(E) } 15\tfrac{3}{4}$

Solution:
Because the arch has a height of $16$ inches, an equation that models the arch is $y = ax^2 + 16$ , where $x$ is the horizontal distance from the center and $y$ is the height.  The arch has a span of $40$ inches, so the arch meets the ground $20$ inches from the center.  That means $0 = 400a + 16$ , so $a = -\frac{1}{25}$
Thus, the equation that models height based on distance from the center $y = -\frac{1}{25}x^2 + 16$ , so the height of the arch $5$ inches from the center is $-\frac{1}{25} \cdot 5^2 + 16 = \boxed{15}$ inches.


Answer: 15.0

[asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$ .
What is the measure of $\measuredangle AED$ in degrees?
$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$

Solution:
Solution by e_power_pi_times_i
Notice that $\measuredangle DAE = 90^\circ+60^\circ = 150^\circ$ and that $AD = AE$ . Then triangle $ADE$ is isosceles, so $\measuredangle AED = \dfrac{180^\circ-150^\circ}{2} = \boxed{15}$

Solution: WLOG, let the side length of the square and the equilateral triangle be $1$ $\angle{DAE}=90^\circ+60^\circ=150^\circ$ . Apply the law of cosines then the law of sines, we find that $\angle{AED}=15^\circ$ . Select $\boxed{15}$


Answer: 4.8

[asy] size(100); real a=4, b=3; // import cse5; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)--(b,a)); D(CP((X+C)/2,C)); D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0)))); //Credit to chezbgone2 for the diagram [/asy]
In $\triangle ABC, AB = 10~ AC = 8$ and $BC = 6$ . Circle $P$ is the circle with smallest radius which passes through $C$ and is tangent to $AB$ . Let $Q$ and $R$ be the points of intersection, distinct from $C$ , of circle $P$ with sides $AC$ and $BC$ , respectively. The length of segment $QR$ is
$\textbf{(A) }4.75\qquad \textbf{(B) }4.8\qquad \textbf{(C) }5\qquad \textbf{(D) }4\sqrt{2}\qquad  \textbf{(E) }3\sqrt{3}$

Solution:
We know that triangle $RCQ$ is similar to triangle $ABC$ . We draw a line to point $D$ on hypotenuse $AB$ such that $\angle QDR$ is $90 ^\circ$ and that $RDQC$ is a rectangle. Since triangle $RCQ$ is similar to triangle $ABC$ , let $RC$ be $4x$ and $RD/CQ$ be $3x$ . Now we have line segment $AQ$ $8-3x$ , and line segment $RB$ $6-4x$ . Since $BD + DA = AB$ , we use simple algebra and Pythagorean Theorem to get $\sqrt {(3x)^2 + (6-4x)^2}$ $\sqrt {(4x)^2 + (8-3x)^2}$ $10$ . Expanding and simplifying gives us $\sqrt {25x^2-48x+36}$ $\sqrt {25x^2-48x+64}$ $10$
Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by $\sqrt {25x^2-48x+36}$ . Now, we can square both sides and simplify to get $0 = 72 - 20 \sqrt{25x^2-48x+36}$ . Dividing both sides by $4$ , we get $18 - 5 \sqrt {25x^2-48x+36}$ $0$ . We then add $5 \sqrt {25x^2-48x+36}$ to both sides to get $18 = 5 \sqrt {25x^2-48x+36}$ . Since this is very messy, let $25x^2 - 48x = y$ . Squaring both sides, we get $324 = 25y + 900, 25y = -576$ . Solving for $y$ , we have $y = -23.04$ . Plugging in $y$ as $25x^2-48x$ , we have $25x^2-48x+23.04 = 0$ . Using the quadratic equation, we get $\frac {48+0}{50}$ . Therefore, $x = \frac {48}{50}$
Remember that our desired answer is the hypotenuse of the triangle $3x - 4x - 5x$ . Since $5x$ is the hypotenuse, our answer is $\boxed{4.8}$


Answer: 15.0

[asy] size(120); path c = Circle((0, 0), 1); pair A = dir(20), B = dir(130), C = dir(240), D = dir(330); draw(c); pair F = 3(A-B) + B; pair G = 3(D-C) + C; pair E = intersectionpoints(B--F, C--G)[0]; draw(B--E--C--A); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, E); //Credit to MSTang for the diagram [/asy]
In the adjoining figure $\measuredangle E=40^\circ$ and arc $AB$ , arc $BC$ , and arc $CD$ all have equal length. Find the measure of $\measuredangle ACD$
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }\left(\frac{45}{2}\right)^\circ\qquad \textbf{(E) }30^\circ$

Solution:
Solution by e_power_pi_times_i
If arcs $AB$ $BC$ , and $CD$ are congruent, then $\measuredangle ACB = \measuredangle BDC = \measuredangle CBD = \theta$ . Because $ABCD$ is cyclic, $\measuredangle CAD = \measuredangle CBD = \theta$ , and $\measuredangle ADB = \measuredangle ACB = \theta$ . Then, $\measuredangle EAD = \measuredangle EDA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ$ $\theta = 55^\circ$ $\measuredangle ACD = 180^\circ - 55^\circ - 110^\circ = \boxed{15}$


Answer: 15.0

[asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\circ$",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$ . Point $A$ lies on the extension of $DC$ past $C$ ;
point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle.
If length $AB$ equals length $OD$ , and the measure of $\measuredangle EOD$ is $45^\circ$ , then the
measure of $\measuredangle BAO$ is
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$

Solution:
Solution by e_power_pi_times_i
Because $AB = OD$ , triangles $ABO$ and $BOE$ are isosceles. Denote $\measuredangle BAO = \measuredangle AOB = \theta$ . Then $\measuredangle ABO = 180^\circ-2\theta$ , and $\measuredangle EBO = \measuredangle OEB = 2\theta$ , so $\measuredangle BOE = 180^\circ-4\theta$ . Notice that $\measuredangle AOB + \measuredangle BOE + 45^\circ = 180^\circ$ . Therefore $\theta+180-4\theta = 135^\circ$ , and $\theta = \boxed{15}$

Solution:
Draw $BO$ . Let $y = \angle BAO$ . Since $AB = OD = BO$ , triangle $ABO$ is isosceles, so $\angle BOA = \angle BAO = y$ . Angle $\angle EBO$ is exterior to triangle $ABO$ , so $\angle EBO = \angle BAO + \angle BOA = y + y = 2y$
Triangle $BEO$ is isosceles, so $\angle BEO = \angle EBO = 2y$ . Then $\angle EOD$ is external to triangle $AEO$ , so $\angle EOD = \angle EAO + \angle AEO = y + 2y = 3y$ . But $\angle EOD = 45^\circ$ , so $\angle BAO = y = 45^\circ/3 = \boxed{15}$


Answer: 1.0

[katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex]

Solution: By the associative property , we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{1}[/katex]

Solution: Notice that the $9 \times 11$ in the denominator of the first fraction cancels with the same term in the second fraction, the $7$ s in the numerator and denominator of the second fraction cancel, and the $3 \times 5$ in the numerator of the first fraction cancels with the same term in the denominator second fraction. Then everything in the expression cancels, leaving us with $\boxed{1}$


Answer: 592.0

chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form $\frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}},$ where $a, b, c, d, e,$ and $f$ are positive integers $a$ and $e$ are relatively prime , and neither $c$ nor $f$ is divisible by the square of any prime . Find the remainder when the product $abcdef$ is divided by 1000.
2004 AIME II Problem 1.png

Solution:
Let $r$ be the length of the radius of the circle.  A right triangle is formed by half of the chord, half of the radius (since the chord bisects it), and the radius. Thus, it is a $30^\circ$ $60^\circ$ $90^\circ$ triangle , and the area of two such triangles is $2 \cdot \frac{1}{2} \cdot \frac{r}{2} \cdot \frac{r\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}$ . The central angle which contains the entire chord is $60 \cdot 2 = 120$ degrees , so the area of the sector is $\frac{1}{3}r^2\pi$ ; the rest of the area of the circle is then equal to $\frac{2}{3}r^2\pi$
The smaller area cut off by the chord is equal to the area of the sector minus the area of the triangle. The larger area is equal to the area of the circle not within the sector and the area of the triangle. Thus, the desired ratio is $\frac{\frac{2}{3}r^2\pi + \frac{r^2\sqrt{3}}{4}}{\frac{1}{3}r^2\pi - \frac{r^2\sqrt{3}}{4}} =  \frac{8\pi + 3\sqrt{3}}{4\pi - 3\sqrt{3}}$
Therefore, $abcdef = 2^53^4 = 2592 \Longrightarrow \boxed{592}$


Answer: 5.0

circle and two distinct lines are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?
$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad {(C)}\ 4 \qquad {(D)}\ 5 \qquad {(E)}\ 6$

Solution: The two lines can both intersect the circle twice, and can intersect each other once, so $2+2+1= \boxed{5}.$


Answer: 647.0

circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle.

Solution:
Call the center of the circle $O$ . By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.
Thus, $\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180$ , or $(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180$
Take the $\tan$ of both sides and use the identity for $\tan(A+B)$ to get \[\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.\]
Use the identity for $\tan(A+B)$ again to get \[\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0.\]
Solving gives $r^2=\boxed{647}$

Solution: Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on ( $a, b, c,$ and $d$ are the tangent lengths, not the side lengths). \[A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}\] $r^2=\frac{A^2}{(a+b+c+d)^2} = \boxed{647}$

Solution: Using the formulas established in solution 2, one notices: \[r^2=\frac{A^2}{(a+b+c+d)^2}\] \[r^2=\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}\] \[r^2=\frac{abc+bcd+acd+abd}{a+b+c+d}\] \[r^2=\boxed{647}\]


Answer: 817.0

circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution:
The location of the center of the circle must be in the $34 \times 13$ rectangle that is one unit away from the sides of rectangle $ABCD$ . We want to find the area of the right triangle with hypotenuse one unit away from $\overline{AC}$ . Let this triangle be $A'B'C'$
Notice that $ABC$ and $A'B'C'$ share the same incenter ; this follows because the corresponding sides are parallel, and so the perpendicular inradii are concurrent, except that the inradii of $\triangle ABC$ extend one unit farther than those of $\triangle A'B'C'$ . From $A = rs$ , we note that $r_{ABC} = \frac{[ABC]}{s_{ABC}} = \frac{15 \cdot 36 /2}{(15+36+39)/2} = 6$ . Thus $r_{A'B'C'} = r_{ABC} - 1 = 5$ , and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, $[A'B'C'] = [ABC] \cdot \left(\frac{r_{A'B'C'}}{r_{ABC}}\right)^2 = \frac{15 \times 36}{2} \cdot \frac{25}{36} = \frac{375}{2}$
The probability is $\frac{2[A'B'C']}{34 \times 13} = \frac{375}{442}$ , and $m + n = \boxed{817}$

Solution:
2004 I AIME-10.png
Let the bisector of $\angle CAD$ be $AE$ , with $E$ on $CD$ . By the angle bisector theorem, $DE = 36/5$ . Since $\triangle AOR \sim \triangle AED$ $O$ is the center of the circle), we find that $AR = 5$ since $OR = 1$ . Also $AT = 35$ so $RT = OQ = 30$
We can apply the same principle again to find that $PT = 27/2$ , and since $QT = 1$ , we find that $PQ = 27/2 - 1 = 25/2$ . The locus of all possible centers of the circle on this "half" of the rectangle is triangle $\triangle OPQ$ . There exists another congruent triangle that is symmetric over $AC$ that has the same area as triangle $\triangle OPQ$ $\triangle APQ$ has area $\frac {1}{2}\cdot OP \cdot PQ = \frac {1}{2}\cdot 30\cdot \frac {25}{2}$ , since $\angle OQP$ is right. Thus the total area that works is $30\cdot \frac {25}{2} = 375$ , and the area of the locus of all centers of any circle with radius 1 is $34\cdot 13 = 442$ . Hence, the desired probability is $\frac {375}{442}$ , and our answer is $\boxed{817}$

Solution:
2004 I AIME-10b.png
Again, the location of the center of the circle must be in the $34 \times 13$ rectangle that is one unit away from the sides of rectangle $ABCD$ . We want to find the area of the right triangle with hypotenuse one unit away from $\overline{AC}$
Let $A$ be at the origin, $B (36,0)$ $C (36,15)$ $D (0,15)$ . The slope of $\overline{AC}$ is $\frac{15}{36} = \frac{5}{12}$ . Let $\triangle A'B'C'$ be the right triangle with sides one unit inside $\triangle ABC$ . Since $\overline{AC} || \overline{A'C'}$ , they have the same slope, and the equation of $A'C'$ is $y = \frac{5}{12}x + c$ . Manipulating, $5x - 12y + 12c = 0$ . We need to find the value of $c$ , which can be determined since $\overline{AC}$ is one unit away from $\overline{A'C'}$ . Since the diagonal contains the origin, we can use the distance from a point to the line formula at the origin:
\[\left|\frac{Ax + By + C}{\sqrt{A^2+B^2}}\right| = 1 \Longrightarrow \left|\frac{(5)(0) + (-12)(0) + 12c}{\sqrt{5^2 + (-12)^2}}\right| = 1\] \[c = \pm \frac{13}{12}\]
The two values of $c$ correspond to the triangle on top and below the diagonal. We are considering $A'B'C'$ which is below, so $c = -\frac{13}{12}$ . Then the equation of $\overline{A'C'}$ is $y = \frac{5}{12}x - \frac{13}{12}$ . Solving for its intersections with the lines $y = 1, x = 35$ (boundaries of the internal rectangle), we find the coordinates of $A'B'C'$ are at $A'\ (5,1)\ B'\ (35,1)\ C'\ (35,\frac{27}{2})$ . The area is $\frac{1}{2}bh = \frac{1}{2}(35-5)\left(\frac{27}{2} - 1\right) = \frac{375}{2}$
Finally, the probability is $\frac{2\cdot \mathrm{area\ of\ triangle}}{34 \times 13} = \frac{375}{442}$ , and $m + n = \boxed{817}$


Answer: 57.0

circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000.

Solution:
We begin as in the first solution.  Once we see that $\triangle EOF$ has side lengths $12$ $20$ , and $24$ , we can compute its area with Heron's formula:
\[K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.\]
Thus, the circumradius of triangle $\triangle EOF$ is $R = \frac{abc}{4K} = \frac{45}{\sqrt{14}}$ .  Looking at $EPFO$ , we see that $\angle OEP = \angle OFP = 90^\circ$ , which makes it a cyclic quadrilateral. This means $\triangle EOF$ 's circumcircle and $EPFO$ 's inscribed circle are the same.
Since $EPFO$ is cyclic with diameter $OP$ , we have $OP = 2R = \frac{90}{\sqrt{14}}$ , so $OP^2 = \frac{4050}{7}$ and the answer is $\boxed{057}$

Solution:
Let $OP=x$
Proceed as the first solution in finding that quadrilateral $EPFO$ has side lengths $OE=20$ $OF=24$ $EP=\sqrt{x^2-20^2}$ , and $PF=\sqrt{x^2-24^2}$ , and diagonals $OP=x$ and $EF=12$
We note that quadrilateral $EPFO$ is cyclic and use Ptolemy's theorem to solve for $x$
\[20\cdot \sqrt{x^2-24^2} + 12\cdot x = 24\cdot \sqrt{x^2-20^2}\]
Solving, we have $x^2=\frac{4050}{7}$ so the answer is $\boxed{057}$

Solution:
Let $M$ be the midpoint of $AB$ and $N$ of $CD$ . As $\angle OMP = \angle ONP$ , quadrilateral $OMPN$ is cyclic with diameter $OP$ . By Cyclic quadrilaterals note that $\angle MPO = \angle MNO$
The area of $\triangle MNP$ can be computed by Herons as \[[MNO] = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.\] The area is also $\frac{1}{2}ON \cdot MN \sin{\angle MNO}$ . Therefore, \begin{align*} \sin{\angle MNO} &= \frac{2[MNO]}{ON \cdot MN}  \\ &= \frac{2}{9}\sqrt{14} \\ \sin{\angle MNO} &= \frac{OM}{OP} \\ &= \frac{2}{9}\sqrt{14} \\ OP &= \frac{90\sqrt{14}}{14} \\ OP^2 &= \frac{4050}{7} \implies \boxed{057}

Solution:
Define $M$ and $N$ as the midpoints of $AB$ and $CD$ , respectively. Because $\angle OMP = \angle ONP = 90^{\circ}$ , we have that $ONPM$ is a cyclic quadrilateral. Hence, $\angle PNM = \angle POM.$ Then, let these two angles be denoted as $\alpha$ .
Now, assume WLOG that $PD = x < 7$ and $PB = y < 15$ (We can do this because one of $PD$ or $PC$ must be less than 7, and similarly for $PB$ and $PA$ ). Then, by Power of a Point on P with respect to the circle with center $O$ , we have that \[(14-x)x = (30-y)y\] \[(7-x)^{2}+176=(15-y)^{2}.\] Then, let $z = (7-x)^{2}$ . From Law of Cosines on $\triangle NMP$ , we have that \[\textrm{cos } \angle MNP = \frac{NP^{2}+MN^{2}-MP^{2}}{2 \cdot NP \cdot MN}\] \[\textrm{cos } \alpha  = \frac{(7-x)^{2} + 12^{2} - (14-x)^{2}}{24 \cdot (7-x)}.\] Plugging in $z$ in gives \[\textrm{cos } \alpha  = \frac{-32}{24 \cdot \sqrt{z}}\] \[\textrm{cos } \alpha  = \frac{-4}{3\sqrt{z}}\] \[\textrm{cos }^{2} \alpha  = \frac{16}{9z}.\] Hence, \[\textrm{tan }^{2} \alpha  = \frac{\frac{9z-16}{9z}}{\frac{16}{9z}} = \frac{9z-16}{16}.\] Then, we also know that \[\textrm{tan } \alpha  = \textrm{tan } \angle MOP  = \frac{MP}{OM} = \frac{14-y}{20}.\] Squaring this, we get \[\textrm{tan }^{2} \alpha  = \frac{z+176}{400}.\] Equating our expressions for $z$ , we get $\frac{z+176}{400} = \frac{9z-16}{16}.$ Solving gives us that $z = \frac{18}{7}$ .
Since $\angle ONP = 90^{\circ}$ , from the Pythagorean Theorem, $OP^{2} = ON^{2}+PN^{2} = 25^{2}-7^{2} + z = 576+z = \frac{4050}{7}$ ,
and thus the answer is $4050+7 = 4057$ , which when divided by a thousand leaves a remainder of $\boxed{57}.$

Solution:
2011 AIMEII Problem 10 CASE 2.png
Let $E$ and $F$ be the midpoints of $\overline{AB}$ and $\overline{CD}$ , respectively, such that $\overline{BE}$ intersects $\overline{CF}$
Since $E$ and $F$ are midpoints, $BE = 15$ and $CF = 7$
$B$ and $C$ are located on the circumference of the circle, so $OB = OC = 25$
Since $\overline{OE}\perp \overline{AB}$ and $\overline{OF}\perp \overline{CD}$ $OE = \sqrt{OB^2-BE^2}=20$ and $OF = \sqrt{OC^2-OF^2}=24$
With law of cosines, $\cos \angle EOF = \frac{OE^2+OF^2-EF^2}{2\cdot OE\cdot OF} = \frac{13}{15}$
Since $EF < OF$ $\angle EOF$ is acute angle. $\sin \angle EOF = \sqrt{1-\cos^2 \angle EOF} = \frac{\sqrt{56}}{15}$ and $\tan \angle EOF = \frac{\sqrt{56}}{13}$
Let $\overline{OF}$ line be $x$ axis.
Line $\overline{DC}$ equation is $x = OF$
Since line $\overline{AB}$ passes point $E$ and perpendicular to $\overline{OD}$ , its equation is $y - E_y = -\frac{1}{\tan \angle EOF} (x - E_x)$
where $E_x = OE\cos{\angle EOF}$ $E_y = OE\sin{\angle EOF}$
Since $P$ is the intersection of $\overline{AB}$ and $\overline{CD}$
$P_x = OF = 24$
$P_y = E_y -\frac{1}{\tan \angle EOF} (OF - E_x) = - \frac{3\sqrt{14}}{7}$ (Negative means point $P$ is between point $F$ and $C$
$OP^2 = P_x^2 + P_y^2 = \frac{4050}{7}$ and the answer is $\boxed{057}$


Answer: 241.0

convex polyhedron $P$ has $26$ vertices, $60$ edges, and $36$ faces, $24$ of which are triangular and $12$ of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $P$ have?

Solution: Every pair of vertices of the polyhedron determines either an edge , a face diagonal or a space diagonal.  We have ${26 \choose 2} = \frac{26\cdot25}2 = 325$ total line segments determined by the vertices.  Of these, $60$ are edges.  Each triangular face has $0$ face diagonals and each quadrilateral face has $2$ , so there are $2 \cdot 12 = 24$ face diagonals.  This leaves $325 - 60 - 24 = \boxed{241}$ segments to be the space diagonals.


Answer: 840.0

convex polyhedron has for its faces 12 squares , 8 regular hexagons , and 6 regular octagons . At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face

Solution:
The polyhedron described looks like this, a truncated cuboctahedron.
The number of segments joining the vertices of the polyhedron is ${48\choose2} = 1128$ . We must now subtract out those segments that lie along an edge or a face.
Since every vertex of the polyhedron lies on exactly one vertex of a square/hexagon/octagon, we have that $V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48$
Each vertex is formed by the intersection of 3 edges. Since every edge is counted twice, once at each of its endpoints, the number of edges $E$ is $\frac{3}{2}V = 72$
Each of the segments lying on a face of the polyhedron must be a diagonal of that face. Each square contributes $\frac{n(n-3)}{2} = 2$ diagonals, each hexagon $9$ , and each octagon $20$ . The number of diagonals is thus $2 \cdot 12 + 9 \cdot 8 + 20 \cdot 6 = 216$
Subtracting, we get that the number of space diagonals is $1128 - 72 - 216 = \boxed{840}$

Solution:
We first find the number of vertices on the polyhedron:
There are 4 corners per square, 6 corners per hexagon, and 8 corners per octagon. Each vertex is where 3 corners coincide, so we count the corners and divide by 3. $\text{vertices} = \frac{12 \cdot 4 + 8 \cdot 6 + 6 \cdot 8}{3}=48$
We know that all vertices look the same (from the problem statement), so we should find the number of line segments originating from a vertex, and multiply that by the number of vertices, and divide by 2 (because each space diagonal is counted twice because it has two endpoints).
Counting the vertices that are on the same face as an arbitrary vertex, we find that there are 13 vertices that aren't possible endpoints of a line originating from the vertex in the middle of the diagram.
You can draw a diagram to count this better: 1988AIME10.png Since 13 aren't possible endpoints, that means that there are 35 possible endpoints per vertex.
The total number of segments joining vertices that aren't on the same face is $48\cdot 35\cdot \frac 12 = 24 \cdot 35 = \boxed{840}$

Solution:
Since at each vertex one square, one hexagon, and one octagon meet, then there are a total of $12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48$ vertices. This means that for each segment we have $48$ choices of vertices for the first endpoint of the segment.
Since each vertex is the meeting point of a square, octagon, and hexagon, then there are $3$ other vertices of the square that are not the first one, and connecting the first point to any of these would result in a segment that lies on a face or edge.
Similarly, there are $5$ points on the adjacent hexagon and $7$ points on adjoining octagon that, when connected to the first point, would result in a diagonal or edge.
However, the square and hexagon share a vertex, as do the square and octagon, and the hexagon and octagon.
Subtracting these from the $47$ vertices we have left to choose from, and adding the $3$ that we counted twice, we get
\[48 \cdot (47 - 3 - 5 - 7 + 3) = 48 \cdot 35 = 1680\]
We over-counted, however, as choosing vertex $A$ then $B$ is the same thing as choosing $B$ then $A$ , so we must divide $1680 / 2 = \boxed{840}$

Solution:
In the same ways as above, we find that there are 48 vertices. Now, notice that there are $\binom{48}{2}$ total possible ways to choose two vertices. However, we must remove the cases where the segments do not lie in the interior of the polyhedron. We get
\[\binom{48}{2}-12\binom{4}{2}-8\binom{6}{2}-6\binom{8}{2}=768\]
We remover all the possible edges of the squares, hexagons, and octagons. However, we have undercounted! We must add back the number of edges because when we subtracted the three binomials from $\binom{48}{2}$ we removed each edge twice (each edge is shared by two polygons). This means that we need to add back the number of edges, 72. Thus, we get $768+72=\boxed{840}$


Answer: 73.0

fair coin is to be tossed $10_{}^{}$ times. Let $\frac{i}{j}^{}_{}$ , in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$

Solution:
Clearly, at least $5$ tails must be flipped; any less, then by the Pigeonhole Principle there will be heads that appear on consecutive tosses.
Consider the case when $5$ tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled $(H)$
There are six slots for the heads to be placed, but only $5$ heads remaining. Thus, using stars-and-bars there are ${6\choose5}$ possible combinations of 6 heads. Continuing this pattern, we find that there are $\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144$ . There are a total of $2^{10}$ possible flips of $10$ coins, making the probability $\frac{144}{1024} = \frac{9}{64}$ . Thus, our solution is $9 + 64 = \boxed{073}$

Solution:
Call the number of ways of flipping $n$ coins and not receiving any consecutive heads $S_n$ . Notice that tails must be received in at least one of the first two flips.
If the first coin flipped is a T, then the remaining $n-1$ flips must fall under one of the configurations of $S_{n-1}$
If the first coin flipped is a H, then the second coin must be a T. There are then $S_{n-2}$ configurations.
Thus, $S_n = S_{n-1} + S_{n-2}$ . By counting, we can establish that $S_1 = 2$ and $S_2 = 3$ . Therefore, $S_3 = 5,\ S_4 = 8$ , forming the Fibonacci sequence . Listing them out, we get $2,3,5,8,13,21,34,55,89,144$ , and the 10th number is $144$ . Putting this over $2^{10}$ to find the probability, we get $\frac{9}{64}$ . Our solution is $9+64=\boxed{073}$

Solution:
We can also split the problem into casework.
Case 1: 0 Heads
There is only one possibility.
Case 2: 1 Head
There are 10 possibilities.
Case 3: 2 Heads
There are 36 possibilities.
Case 4: 3 Heads
There are 56 possibilities.
Case 5: 4 Heads
There are 35 possibilities.
Case 6: 5 Heads
There are 6 possibilities.
We have $1+10+36+56+35+6=144$ , and there are $1024$ possible outcomes, so the probability is $\frac{144}{1024}=\frac{9}{64}$ , and the answer is $\boxed{073}$


Answer: 401.0

function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

Solution:
If $f(2+x)=f(2-x)$ , then substituting $t=2+x$ gives $f(t)=f(4-t)$ . Similarly, $f(t)=f(14-t)$ . In particular, \[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\]
Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\pmod{10}$ are also roots. To see that these may be the only integer roots, observe that the function \[f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}\] satisfies the conditions and has no other roots.
In the interval $-1000\leq x\leq 1000$ , there are $201$ multiples of $10$ and $200$ numbers that are congruent to $4 \pmod{10}$ , therefore the minimum number of roots is $\boxed{401}$

Solution: We notice that the function has reflectional symmetry across both $x=2$ and $x=7$ . We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\pm 10$ as roots. Continuing this shows that the roots are $0 \mod 10$ or $4 \mod 10$ . There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of $\boxed{401}$ $QED \blacksquare$

Solution: Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only counting the total number of x values so ignore the double counted values. This means that $x = 0, \pm 5, \pm 10, \pm 15... \pm 1000$ so the answer is 400 + 1 = $\boxed{401}$

Solution: Let $z$ be an arbitrary zero. If $z=2-x$ , then $x=2-z$ and $2+x=4-z$ . Repeat with other equation to find if $z$ is a zero then so are $4-z$ and $14-z$ . From $0$ , we get $4$ and $14$ . Now note that applying either of these twice will return $z$ , so we must apply them in an alternating fashion for distinct roots. Doing so to $4$ and $14$ returns $10$ and $-10$ , respectively. A pattern will emerge of each path hitting a multiple of $10$ after $2$ moves. Hence, we will reach $\pm 1000$ after $200$ jumps in either direction. Including zero, there are $2\cdot200+1=\boxed{401}$


Answer: 384.0

hexagon is inscribed in a circle . Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$ , has length $31$ . Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$

Solution:
Let $x=AC=BF$ $y=AD=BE$ , and $z=AE=BD$
Ptolemy's Theorem on $ABCD$ gives $81y+31\cdot 81=xz$ , and Ptolemy on $ACDF$ gives $x\cdot z+81^2=y^2$ .
Subtracting these equations give $y^2-81y-112\cdot 81=0$ , and from this $y=144$ . Ptolemy on $ADEF$ gives $81y+81^2=z^2$ , and from this $z=135$ . Finally, plugging back into the first equation gives $x=105$ , so $x+y+z=105+144+135=\boxed{384}$


Answer: 14.0

let $n=x-y^{x-y}$ . Find $n$ when $x=2$ and $y=-2$
$\textbf{(A)}\ -14 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 256$

Solution: Substitute the variables to determine the value of $n$ \[n = 2 - (-2)^{2-(-2)}\] \[n = 2 - (-2)^4\] \[n = 2 - 16\] \[n = -14\] The answer is $\boxed{14}$


Answer: 8.0

line passes through $A\ (1,1)$ and $B\ (100,1000)$ . How many other points with integer coordinates are on the line and strictly between $A$ and $B$
$(\mathrm {A}) \ 0 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 3 \qquad (\mathrm {D}) \ 8 \qquad (\mathrm {E})\ 9$

Solution:
The slope of the line is $\frac{1000-1}{100-1}=\frac{111}{11},$ so all points on the line have the form $(1+11t, 1+111t)$ for some value of $t$ (the rise is 111 and the run is 11). Such a point has integer coordinates if and only if $t$ is an integer, and the point is strictly between $A$ and $B$ if and only if $0<t<9$ . Thus, there are $\boxed{8}$ points with the required property.
-Paixiao


Answer: 1994.0

lucky year is one in which at least one date, when written in the form month/day/year, has the following property: The product of the month times the day equals the last two digits of the year .  For example, 1956 is a lucky year because it has the date 7/8/56 and $7\times 8 = 56$ .  Which of the following is NOT a lucky year?
$\text{(A)}\ 1990 \qquad \text{(B)}\ 1991 \qquad \text{(C)}\ 1992 \qquad \text{(D)}\ 1993 \qquad \text{(E)}\ 1994$

Solution:
We examine only the factors of $90, 91, 92, 93,$ and $94$ that are less than $13$ , because for a year to be lucky, it must have at least one factor between $1$ and $12$ to represent the month.
$90$ has factors of $1, 2, 3, 5, 6, 9,$ and $10$ .  Dividing $90$ by the last number $10$ , gives $9$ .  Thus, $10/9/90$ is a valid date, and $1990$ is a lucky year.
$91$ has factors of $1$ and $7$ .  Dividing $91$ by $7$ gives $13$ , and $7/13/91$ is a valid date.  Thus, $1991$ is a lucky year.
$92$ has factors of $1, 2$ and $4$ .  Dividing $92$ by $4$ gives $23$ , and $4/23/92$ is a valid date.  Thus, $1992$ is a lucky year.
$93$ has factors of $1$ and $3$ .  Dividing $93$ by $3$ gives $31$ , and $3/31/93$ is a valid date.  Thus, $1993$ is a lucky year.
$94$ has factors of $1$ and $2$ .  Dividing $94$ by $2$ gives $47$ , and no month has $47$ days.  Thus, $1994$ is not a lucky year, and the answer is $\boxed{1994}$


Answer: 2.0

palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$
$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

Solution: Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \ldots, 99$ . Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as $110=77+22+11$ . Then, $N = 110$ , and the sum of the digits of $N$ is $1+1+0 = \boxed{2}$

Solution: We already know that two-digit palindromes can only be two-digit multiples of 11; which are: $11, 22, 33, 44, 55, 66, 77, 88,$ and $99$ . Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then, we start counting. $110 \ldots$ Aha! This multiple of 11, 110, not only isn’t a palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of $N$ ’s digits is $1+1+0 = \boxed{2}$

Solution: As stated above, two-digit palindromes can only be two-digit multiples of 11. We can see that if we add anything that are multiples of 11 together, we will again get a multiple of 11. For instance, $11+22=33$ . Since we know this fact and we are finding the smallest value possible, we can start with the first three-digit multiple of 11 which is $110$ . Since this is not a palindrome and can be the sum of 3 two-digit palindromes (see above solutions for more details), $110$ fits the bill. We can see that the sum of $110$ 's digits is $1+1+0 = \boxed{2}$