Answer: 75.0

Solution:
Using the properties of logarithms, $\log_{10}xyz = 81$ by taking the log base 10 of both sides, and $(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}z)= 468$ by using the fact that $\log_{10}ab = \log_{10}a + \log_{10}b$
Through further simplification, we find that $\log_{10}x+\log_{10}y+\log_{10}z = 81$ . It can be seen that there is enough information to use the formula $\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$ , as we have both $\ a+b+c$ and $\ 2ab+2ac+2bc$ , and we want to find $\sqrt {a^2 + b^2 + c^2}$
After plugging in the values into the equation, we find that $\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2$ is equal to $\ 6561 - 936 = 5625$
However, we want to find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$ , so we take the square root of $\ 5625$ , or $\boxed{075}$

Solution:
Let $a=\log_{10}x$ $b=\log_{10}y,$ and $c=\log_{10}z$
We have $a+b+c=81$ and $a(b+c)+bc=ab+ac+bc=468$ . Since these two equations look a lot like Vieta's for a cubic, create the polynomial $x^3-81x^2+468x=0$ (leave the constant term as $0$ to make things easy). Dividing by $x$ yields $x^2-81x+468=0$
Now we use the quadratic formula:
$x=\frac{81\pm\sqrt{81^2-4\cdot468}}{2}$
$x=\frac{81\pm\sqrt{4689}}{2}$
$x=\frac{81+3\sqrt{521}}{2}$ $x=\frac{81-3\sqrt{521}}{2}$
Since the question asks for $\sqrt{a^2+b^2+c^2}$ (remember one of the values was the solution $x=0$ that we divided out in the beginning), we find:
$\sqrt{\left(\frac{81+3\sqrt{521}}{2}\right)^2+\left(\frac{81-3\sqrt{521}}{2}\right)^2}$
$=\sqrt{2\cdot\frac{81^2+(3\sqrt{521})^2}{4}}$
$=\sqrt{\frac{11250}{2}}$
$=\boxed{075}$


Answer: 20.0

Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$

Solution:
Let $\log_2{x} = \log_y{16}=k$ , so that $2^k=x$ and $y^k=16 \implies y=2^{\frac{4}{k}}$ . Then we have $(2^k)(2^{\frac{4}{k}})=2^{k+\frac{4}{k}}=2^6$
We therefore have $k+\frac{4}{k}=6$ , and deduce $k^2-6k+4=0$ . The solutions to this are $k = 3 \pm \sqrt{5}$
To solve the problem, we now find \begin{align*} (\log_2\tfrac{x}{y})^2&=(\log_2 x - \log_2 y)^2\\ &=(k-\tfrac{4}{k})^2=(3 \pm \sqrt{5} - \tfrac{4}{3 \pm \sqrt{5}})^2 \\ &= (3 \pm \sqrt{5} - [3 \mp \sqrt{5}])^2\\ &= (3 \pm \sqrt{5} - 3 \pm \sqrt{5})^2\\ &=(\pm 2\sqrt{5})^2 \\ &= \boxed{20} ~Edits by BakedPotato66

Solution: After obtaining $k + \frac{4}{k} = 6$ , notice that the required answer is $\left(k - \frac{4}{k}\right)^{2} = k^2 - 8 + \frac{16}{k^2} = \left(k^2 + 8 + \frac{16}{k^2}\right) - 16 = \left(k+\frac{4}{k}\right)^2 - 16 = 6^2 - 16 = \boxed{20}$ , as before.

Solution:
From the given data, $\log_2(x) = \frac{1}{\log_{16}(y)}$ , or $\log_2(x) = \frac{4}{{\log_{2}(y)}}$
We know that $xy=64$ , so $x= \frac{64}{y}$
Thus $\log_2\left(\frac{64}{y}\right) = \frac{4}{{\log_{2}(y)}}$ , so $6-\log_2(y) = \frac{4}{{\log_{2}(y)}}$ , so $6(\log_2(y))-(\log_2(y))^2=4$
Solving for $\log_2(y)$ , we obtain $\log_2(y)=3+\sqrt{5}$
Easy resubstitution further gives $\log_2(x)=\frac{4}{3+\sqrt{5}}$ . Simplifying, we obtain $\log_2(x)= 3-\sqrt{5}$
Looking back at the original problem, we have What is $(\log_2{\tfrac{x}{y}})^2$
Deconstructing this expression using log rules, we get $(\log_2{x}-\log_2{y})^2$
Plugging in our known values, we get $((3-\sqrt{5})-(3+\sqrt{5}))^2$ or $(-2\sqrt{5})^2$
Our answer is $\boxed{20}$


Answer: 36.0

Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution: Because $y^3=x^2$ , set $x=a^3$ $y=a^2$ $a\neq 0$ ). Put them in $(y-x)^2=4y^2$ we get $(a^2(a-1))^2=4a^4$ which implies $a^2-2a+1=4$ . Solve the equation to get $a=3$ or $-1$ . Since $x$ and $y$ are positive, $a=3$ and $x+y=3^3+3^2=\boxed{36}$

Solution:
Since $a^2 = |a|^2$ , we can rewrite the second equation as $(x-y)^2=4y^2$
Let $u=x+y$ . The second equation becomes
\[(u-2y)^2 = 4y^2\] \[u^2 - 4uy = 0\] \[u = 4y\] \[x+y = 4y\] \[x = 3y.\]
Substituting this into the first equation, we have
\[y^3 = (3y)^2,\] so $x = 9$
Hence $x = 27$ and $x + y = \boxed{36}.$

Solution:
We will use the difference of squares in the second equation.
\[(y-x)^2=4y^2\] \[(y-x)^2-(2y)^2=0\] \[(y-x-2y)(y-x+2y)=0\] \[-(x+y)(3y-x)=0\]
Since x and y are positive, x+y is non-zero. Thus, \[3y=x\]
Substituting into the first equation:
\[y^3=x^2\] \[y^3=9y^2\] \[y=9, x=27 \rightarrow x+y=\boxed{36}\]


Answer: 36.0

Solution: Let's take the second equation and square root both sides. This will obtain $y-x = \pm2y$ . Solving the case where $y-x=+2y$ , we'd find that $x=-y$ . This is known to be false because both $x$ and $y$ have to be positive, and $x=-y$ implies that at least one of the variables is not positive. So we instead solve the case where $y-x=-2y$ . This means that $x=3y$ . Inputting this value into the first equation, we find: \[y^3 = (3y)^2\] \[y^3 = 9y^2\] \[y=9\] This means that $x=3y=3(9)=27$ . Therefore, $x+y=9+27=\boxed{36}$

Solution:
first expand
$(y-x)^2 = 4y^2$
$y^2-2xy+x^2 = 4y^2$
$y^2-2yx+x^2 = 4y^2$
$x^2-2xy-3y^2 = 0$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
consider
a=1
b=-2y
c=-3y^2
$x=\frac{2y\pm\sqrt{(-2y)^2-4(1)(-3y^2)}}{2a}$
$x=\frac{2y\pm\sqrt{16y^2}}{2a}$
$\frac{2y+4y^{2}}{2}$ or $\frac{2y-4y^{2}}{2}$
$x=y+2y$ and $x=y-2y$
$x=3y$ and $x=-y$
we can see both x and y will be postive in $x=3y$
now do same as solution 2 \[y^3 = (3y)^2\] \[y^3 = 9y^2\] \[y=9\] This means that $x=3y=3(9)=27$ . Therefore, $x+y=9+27=\boxed{36}$


Answer: 2500.0

Postman Pete has a pedometer to count his steps. The pedometer records up to $99999$ steps, then flips over to $00000$ on the next step. Pete plans to determine his mileage for a year. On January $1$ Pete sets the pedometer to $00000$ . During the year, the pedometer flips from $99999$ to $00000$ forty-four times. On December $31$ the pedometer reads $50000$ . Pete takes $1800$ steps per mile. Which of the following is closest to the number of miles Pete walked during the year?
$\textbf{(A)}\ 2500 \qquad \textbf{(B)}\ 3000 \qquad \textbf{(C)}\ 3500 \qquad \textbf{(D)}\ 4000 \qquad \textbf{(E)}\ 4500$

Solution: Every time the pedometer flips, Pete has walked $100,000$ steps. Therefore, Pete has walked a total of $100,000 \cdot 44 + 50,000 = 4,450,000$ steps, which is $4,450,000/1,800 = 2472.2$ miles, which is the closest to the answer choice $\boxed{2500}$


Answer: 2500.0

Postman Pete has a pedometer to count his steps. The pedometer records up to 99999 steps, then flips over to 00000 on the next step. Pete plans to determine his mileage for a year. On January 1 Pete sets the pedometer to 00000. During the year, the pedometer flips from 99999 to 00000 forty-four
times. On December 31 the pedometer reads 50000. Pete takes 1800 steps per mile. Which of the following is closest to the number of miles Pete walked during the year?
$\mathrm{(A)}\ 2500\qquad\mathrm{(B)}\ 3000\qquad\mathrm{(C)}\ 3500\qquad\mathrm{(D)}\ 4000\qquad\mathrm{(E)}\ 4500$

Solution:
Every time the pedometer flips from $99999$ to
$00000$ Pete has walked $100000$ steps.
So, if the pedometer flipped $44$ times
Pete walked $44*100000+50000=4450000$ steps.
Dividing by $1800$ steps per mile gives $2472.\overline{2}$
This is closest to answer $\boxed{2500}$


Answer: 11400.0

Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same $760$ -page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in $45$ seconds and Chandra reads a page in $30$ seconds.
If Bob and Chandra both read the whole book, Bob will spend how many more seconds reading than Chandra?
$\textbf{(A)}\ 7,600\qquad\textbf{(B)}\ 11,400\qquad\textbf{(C)}\ 12,500\qquad\textbf{(D)}\ 15,200\qquad\textbf{(E)}\ 22,800$

Solution:
The information is the same for Problems 14,15, and 16. Therefore, we shall only use the information we need. All we need for this problem is that there's 760 pages, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds. A lot of people will find how long it takes Bob to read the book, how long it takes Chandra to read the book, and then find the seconds. However, if we just set up the expression, we can find an easier way.
$760\cdot45 - 760\cdot30 = 760(45-30) = 760(15) = \boxed{11,400}$


Answer: 7200.0

Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?
$\textbf{(A)}\ 6400\qquad\textbf{(B)}\ 6600\qquad\textbf{(C)}\ 6800\qquad\textbf{(D)}\ 7000\qquad\textbf{(E)}\ 7200$

Solution: The amount of pages Bob, Chandra, and Alice will read is in the ratio 4:6:9. Therefore, Bob, Chandra, and Alice read 160, 240, and 360 pages respectively. They would also be reading for the same amount of time because the ratio of the pages read was based on the time it takes each of them to read a page. Therefore, the amount of seconds each person reads is simply $160 \cdot 45 = \boxed{7200}$

Solution:
We can solve this using equations. Let Bob read $x$ pages and Chandra read $y$ pages. Then Alice would read $760-x-y$ pages. Then we can set up the equations using the time it takes them to read a page: $45x=30y=20(760-x-y)$ . Since we only have two variables, we only need two equations. Let's use $45x=30y$ and $45x=20(760-x-y)$ . We can simplify $45x=30y$ down to $3x=2y$ and $45x=20(760-x-y)$ down to $9x=4(760-x-y)$ . We can then solve the equations accordingly:
\[9x=3040-4x-4y\] \[13x=3040-4y\]
From the equation $3x=2y$ , we know that $4y=6x$
\[13x=3040-6x\] \[19x=3040\] \[x=160\]
Now that we know $x$ , we can multiply it by $45$ (Bob's speed) to get the answer: $160\times45$ $\boxed{7200}$


Answer: 456.0

Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?
$\textbf{(A)}\ 425\qquad\textbf{(B)}\ 444\qquad\textbf{(C)}\ 456\qquad\textbf{(D)}\ 484\qquad\textbf{(E)}\ 506$

Solution:
Same as the previous problem, we only use the information we need. Note that it's not just Chandra reads half of it and Bob reads the rest since they have different reading rates. In this case, we set up an equation and solve.
Let $x$ be the number of pages that Chandra reads.
$30x = 45(760-x)$ Distribute the $45$
$30x = 45(760) - 45x$ Add $45x$ to both sides
$75x = 45(760)$ Divide both sides by $15$ to make it easier to solve
$5x = 3(760)$ Divide both sides by $5$
$x = 3(152) = \boxed{456}$

Solution: Chandra and Bob read at a rate of $30:45$ seconds per page, respectively. Simplifying that gets us Bob reads $2$ pages for every $3$ pages that Chandra reads. Therefore Chandra should read $\frac{3}{2+3}=\frac{3}{5}$ of the book. $\frac{3}{5}\cdot760$ $\boxed{456}$


Answer: 40.0

Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.
$\circ$ Art's cookies are trapezoids: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label("5 in", (2.5,0), S); label("3 in", (5,1.5), E); label("3 in", (3.5,3), N);[/asy]
$\circ$ Roger's cookies are rectangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label("4 in", (2,0), S); label("2 in", (4,1), E);[/asy]
$\circ$ Paul's cookies are parallelograms: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label("3 in", (1.5,0), S); label("2 in", (2.5,1), W);[/asy]
$\circ$ Trisha's cookies are triangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label("3 in", (1.5,0), S); label("4 in", (3,2), E);[/asy]
Each friend uses the same amount of dough, and Art makes exactly $12$ cookies. Art's cookies sell for $60$ cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90$

Solution: The area of one of Art's cookies is $3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12$ . As he has $12$ cookies in a batch, the amount of dough each person used is $12 \cdot 12=144$ . Roger's cookies have an area of $\frac{144}{2 \cdot 4}=\frac{144}{8}= 18$ cookies in  a batch. In total, the amount of money Art will earn is $12 \cdot 60=720$ . Thus, the amount Roger would need to charge per cookie is $\frac{720}{18}=\boxed{40}$


Answer: 24.0

Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.
$\circ$ Art's cookies are trapezoids: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label("5 in", (2.5,0), S); label("3 in", (5,1.5), E); label("3 in", (3.5,3), N);[/asy]
$\circ$ Roger's cookies are rectangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label("4 in", (2,0), S); label("2 in", (4,1), E);[/asy]
$\circ$ Paul's cookies are parallelograms: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label("3 in", (1.5,0), S); label("2 in", (2.5,1), W);[/asy]
$\circ$ Trisha's cookies are triangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label("3 in", (1.5,0), S); label("4 in", (3,2), E);[/asy]
How many cookies will be in one batch of Trisha's cookies?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24$

Solution: Art's cookies have areas of $3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12$ . There are 12 cookies in one of Art's batches so everyone used $12 \cdot 12=144 \text{ in}^2$ of dough. Trisha's cookies have an area of $\frac{3 \cdot 4}{2}=6$ so she has $\frac{144}{6}=\boxed{24}$ cookies per batch.


Answer: 5.5

Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
The average price of his '70s stamps is closest to
$\text{(A)}\ 3.5 \text{ cents} \qquad \text{(B)}\ 4 \text{ cents} \qquad \text{(C)}\ 4.5 \text{ cents} \qquad \text{(D)}\ 5 \text{ cents} \qquad \text{(E)}\ 5.5 \text{ cents}$

Solution:
The price of all the stamps in the '70s together over the total number of stamps is equal to the average price.
\[\frac{(12)(0.06)+(12)(0.06)+(6)(0.04)+(13)(0.05)}{12+12+6+13}\\ = \frac{0.72+0.72+0.24+0.65}{43}\\ = \frac{2.33}{43} \approx \boxed{5.5}\]


Answer: 24.0

Solution: France and Spain are European countries. The number of '80s stamps from France is $15$ and the number of '80s stamps from Spain is $9$ . The total number of stamps is $15+9=\boxed{24}$


Answer: 1.06

Solution: Brazil 50s and 60s total 11 stamps with each 6 cents, Peru 50s and 60s total 10 stamps with each 4 cents. So total $11*0.06+10*0.04 = \boxed{1.06}$


Answer: 5760.0

Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?
$\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880$

Solution: Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $5$ factorial. Now, we multiply this product by $2!$ because there are $2!$ ways to arrange the Arabic books within themselves, and $4!$ ways to arrange the Spanish books within themselves. Multiplying all these together, we have $2! \cdot 4! \cdot 5!=\boxed{5760}$


Answer: 2.935363332e+18

Prove that there exists a positive integer $k$ such that $k\cdot2^n+1$ is composite for every integer $n$

Solution: Indeed, $\boxed{2935363331541925531}$ has the requisite property.


Answer: 5.0

Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m_{}$ and $n_{}$ are integers, find $m+n.$

Solution:
The angle $\theta$ is the angle formed by two perpendiculars drawn to $BO$ , one on the plane determined by $OAB$ and the other by $OBC$ . Let the perpendiculars from $A$ and $C$ to $\overline{OB}$ meet $\overline{OB}$ at $P.$ Without loss of generality , let $AP = 1.$ It follows that $\triangle OPA$ is a $45-45-90$ right triangle , so $OP = AP = 1,$ $OB = OA = \sqrt {2},$ and $AB = \sqrt {4 - 2\sqrt {2}}.$ Therefore, $AC = \sqrt {8 - 4\sqrt {2}}.$
From the Law of Cosines $AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\cos \theta,$ so
\[8 - 4\sqrt {2} = 1 + 1 - 2\cos \theta \Longrightarrow \cos \theta = - 3 + 2\sqrt {2} = - 3 + \sqrt{8}.\]
Thus $m + n = \boxed{005}$

Solution:
Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that $A = (1,0,0),$ $B = (0,1,0),$ $C = ( - 1,0,0),$ $D = (0, - 1,0),$ and $O = (0,0,z),$ where $z$ is unknown.
We first find $z.$ Note that
\[\overrightarrow{OA}\cdot \overrightarrow{OB} = \parallel \overrightarrow{OA}\parallel \parallel \overrightarrow{OB}\parallel \cos 45^\circ.\]
Since $\overrightarrow{OA} =\, <1,0, - z>$ and $\overrightarrow{OB} =\, <0,1, - z> ,$ this simplifies to
\[z^{2}\sqrt {2} = 1 + z^{2}\implies z^{2} = 1 + \sqrt {2}.\]
Now let's find $\cos \theta.$ Let $\vec{u}$ and $\vec{v}$ be normal vectors to the planes containing faces $OAB$ and $OBC,$ respectively. From the definition of the dot product as $\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta$ , we will be able to solve for $\cos \theta.$ A cross product yields (alternatively, it is simple to find the equation of the planes $OAB$ and $OAC$ , and then to find their normal vectors)
\[\vec{u} = \overrightarrow{OA}\times \overrightarrow{OB} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & - z \\ 0 & 1 & - z \end{array}\right| =\, < z,z,1 > .\]
Similarly,
\[\vec{v} = \overrightarrow{OB}\times \overrightarrow{OC} - \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & - z \\ - 1 & 0 & - z \end{array}\right| =\, < - z,z,1 > .\]
Hence, taking the dot product of $\vec{u}$ and $\vec{v}$ yields
\[\cos \theta = \frac{ \vec{u} \cdot \vec{v} }{ \parallel \vec{u} \parallel \parallel \vec{v} \parallel } = \frac{- z^{2} + z^{2} + 1}{(\sqrt {1 + 2z^{2}})^{2}} =  \frac {1}{3 + 2\sqrt {2}} = 3 - 2\sqrt {2} = 3 - \sqrt {8}.\]
Flipping the signs (we found the cosine of the supplement angle) yields $\cos \theta = - 3 + \sqrt {8},$ so the answer is $\boxed{005}$

Solution:
Similar to Solution 1, $\angle APC$ is the dihedral angle we want. WLOG, we will let $AB=1,$ meaning $AC=\sqrt{2}$
Because $\triangle OAB,\triangle OBC$ are isosceles, $\angle ABP = 67.5^{\circ}$ $PC=PA=\cos(\angle PAB)=\cos(22.5^{\circ})$
Thus by the half-angle identity,
\[PA=\cos\left(\frac{45}{2}\right) = \sqrt{\frac{1+\cos(45^{\circ})}{2}}\] \[= \sqrt{\frac{2+\sqrt{2}}{4}}.\]
Now looking at triangle $\triangle PAC,$ we drop the perpendicular from $P$ to $AC$ , and call the foot $H$ . Then $\angle CPH = \theta / 2.$ By Pythagoreas, \[PH=\sqrt{\frac{2+\sqrt{2}}{4}-\frac{1}{2}}=\frac{\sqrt[4]{2}}{2}.\]
We have that \[\cos\left(\frac{\theta}{2}\right)=\frac{\sqrt[4]{2}}{\sqrt{2+\sqrt{2}}},\text{ so}\] \[\cos(\theta)=2\cos^{2}\left(\frac{\theta}{2}\right)-1\] \[=2\left(\frac{\sqrt{2}}{2+\sqrt{2}}\right)-1\] \[=2(\frac{2\sqrt{2}-2}{2})-1\] \[=-3+\sqrt{8}.\]
Because $m$ and $n$ can be negative integers, our answer is $(-3)+8=\boxed{005.}$


Answer: 110.0

Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?
$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$

Solution:
The only option that is easily divisible by $55$ is $110$ , which gives 2 hours of travel. And, the formula is $\frac{15}{30} + \frac{110}{55} = \frac{5}{2}$
And, $\text{Average Speed}$ $\frac{\text{Total Distance}}{\text{Total Time}}$
Thus, $\frac{125}{50} = \frac{5}{2}$
Both are equal and thus our answer is $\boxed{110}.$

Solution: To calculate the average speed, simply evaluate the total distance over the total time. Let the number of additional miles he has to drive be $x.$ Therefore, the total distance is $15+x$ and the total time (in hours) is \[\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.\] We can set up the following equation: \[\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.\] Simplifying the equation, we get \[15+x=25+\frac{10x}{11}.\] Solving the equation yields $x=110,$ so our answer is $\boxed{110}$

Solution: If he travels $15$ miles at a speed of $30$ miles per hour, he travels for 30 min. Average rate is total distance over total time so $(15+d)/(0.5 + t) = 50$ , where d is the distance left to travel and t is the time to travel that distance. Solve for $d$ to get $d = 10+50t$ . You also know that he has to travel $55$ miles per hour for some time, so $d=55t$ . Plug that in for d to get $55t = 10+50t$ and $t=2$ and since $d=55t$ $d = 2\cdot55 =110$ , the answer is $\boxed{110}$

Solution:
Let $h$ be the amount of hours Qiang drives after his first 15 miles. Average speed, which we know is $50$ mph, means total distance over total time. For 15 miles at 30 mph, the time taken is $\frac{1}{2}$ hour, so the total time for this trip would be $\frac{1}{2} + h$ hours. For the total distance, 15 miles are traveled in the first part and $55h$ miles in the second. This gives the following equation:
\[\dfrac{15+55h}{\frac{1}{2}+h} = 50.\]
Cross multiplying, we get that $15 + 55h = 50h + 25$ , and simple algebra gives $h=2$ . In 2 hours traveling at 55 mph, the distance traveled is $\frac{2 \hspace{0.05 in} \text{hours}}{1} \cdot \frac{55 \hspace{0.05 in} \text{miles}}{1 \hspace{0.05 in} \text{hour}} = 2 \cdot 55 \hspace{0.05 in} \text{miles} = 110 \hspace{0.05 in} \text{miles}$ , which is choice $\boxed{110}$


Answer: 116.0

Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$

Solution:
Let $R(x)=P(x)+Q(x).$ Since the $x^2$ -terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.
Note that \begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*} so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$
It follows that the equation of $R(x)$ is \[R(x)=-\frac12x+c\] for some constant $c,$ and we wish to find $R(0)=c.$
We substitute $x=20$ into this equation to get $106=-\frac12\cdot20+c,$ from which $c=\boxed{116}.$

Solution:
Let \begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\ Q(x) &= &\hspace{1mm}-2x^2 + cx + d, \end{alignat*} for some constants $a,b,c$ and $d.$
We are given that \begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\ Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\ P(20) &= &800 + 20a + b &= 53,  &&(3) \\ Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) \end{alignat*} and we wish to find \[P(0)+Q(0)=b+d.\] We need to cancel $a$ and $c.$ Since $\operatorname{lcm}(16,20)=80,$ we subtract $4\cdot[(3)+(4)]$ from $5\cdot[(1)+(2)]$ to get \[b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.\] ~MRENTHUSIASM

Solution:
Like Solution 2, we can begin by setting $P$ and $Q$ to the quadratic above, giving us \begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\ Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\ P(20) &= &800 + 20a + b &= 53,  &&(3) \\ Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) \end{alignat*} We can first add $(1)$ and $(2)$ to obtain $16(a-c) + (b+d) = 108.$
Next, we can add $(3)$ and $(4)$ to obtain $20(a-c) + (b+d) = 106.$ By subtracting these two equations, we find that $4(a-c) = -2,$ so substituting this into equation $[(1) + (2)],$ we know that $4 \cdot (-2) + (b+d) = 108,$ so therefore $b+d = \boxed{116}.$

Solution:
Let \begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\ Q(x) &= &\hspace{1mm}-2x^2 + cx + d, \end{alignat*} By substituting $(16, 54)$ and $(20, 53)$ into these equations, we can get: \begin{align*} 2(16)^2 + 16a + b &= 54, \\ 2(20)^2 + 20a + b &= 53. \end{align*} Hence, $a = -72.25$ and $b = 698.$
Similarly, \begin{align*} -2(16)^2 + 16c + d &= 54, \\ -2(20)^2 + 20c + d &= 53. \end{align*} Hence, $c = 71.75$ and $d = -582.$
Notice that $b = P(0)$ and $d = Q(0).$ Therefore \[P(0) + Q(0) = 698 + (-582) = \boxed{116}.\] ~Littlemouse

Solution:
Add the equations of the polynomials $y=2x^2+ax+b$ and $y=-2x^2+cx+d$ to get $2y=(a+c)x+(b+d)$ . This equation must also pass through the two points $(16,54)$ and $(20,53)$
Let $m=a+c$ and $n=b+d$ . We then have two equations: \begin{align*} 108&=16m+n, \\ 106&=20m+n. \end{align*} We are trying to solve for $n=P(0)$ . Using elimination: \begin{align*} 540&=80m+5n, \\ 424&=80m+4n. \end{align*} Subtracting both equations, we find that $n=\boxed{116}$


Answer: 85.0

Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$

Solution:
Let the unknown $\angle BAD$ be $x$
First, we draw diagonal $BD$ and $AC$ $I$ is the intersection of the two diagonals. The diagonals each form two isosceles triangles, $\triangle BCD$ and $\triangle ABC$
Using this, we find: $\angle DBC = \angle CDB = 5^\circ$ and $\angle BAC = \angle BCA = 55^\circ$ . Expanding on this, we can fill in a couple more angles. $\angle ABD = 70^\circ - 5^\circ = 65^\circ$ $\angle ACD = 170^\circ - 55^\circ = 115^\circ$ $\angle BIA = \angle CID = 180^\circ - (65^\circ + 55^\circ) = 60^\circ$ $\angle BIC =   \angle AID = 180^\circ - 60^\circ = 120^\circ$
We can rewrite $\angle CAD$ and $\angle BDA$ in terms of $x$ $\angle CAD = x - 55^\circ$ and $\angle BDA = 180^\circ - (120^\circ + x - 55^\circ) = 115^\circ - x$
Let us relabel $AB = BC = CD = a$ and $AD = b$
By Rule of Sines on $\triangle ACD$ and $\triangle ABD$ respectively, $\frac{\sin(\angle CAD)}{a} = \frac{\sin(\angle ACD)}{b}$ , and $\frac{\sin(\angle ABD)}{b} = \frac{\sin(\angle BDA)}{a}$
In a more convenient form, $\frac{\sin(x-55^\circ)}{a} = \frac{\sin(115^\circ)}{b} \implies \frac{a}{b} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$
and $\frac{\sin(65^\circ)}{b} = \frac{\sin(115^\circ-x)}{a} \implies \frac{a}{b} = \frac{\sin(115^\circ-x)}{\sin(65^\circ)}$
$\implies \frac{\sin(115^\circ-x)}{\sin(65^\circ)} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$
Now, by identity $\sin(\theta) = \sin(180^\circ-\theta)$ $\sin(65^\circ) = \sin(115^\circ)$
Therefore, $\sin(115^\circ-x) = \sin(x-55^\circ).$ This equation is only satisfied by option $\boxed{85}$

Solution: Using a protractor and rule, draw an accurate diagram ( Example Diagram ). $\angle BAD$ looks slightly less than $90$ degrees. Therefore the answer is $\boxed{85}$ as $85$ is slightly less than $90$


Answer: 85.0

Solution:
To start off, draw a diagram like in solution two and label the points. Create lines $\overline{AC}$ and $\overline{BD}$ . We can call their intersection point $Y$ . Note that triangle $BCD$ is an isosceles triangle so angles $CDB$ and $CBD$ are each $5$ degrees. Since $AB$ equals $BC$ , angle $BAC$ equals $55$ degrees, thus making angle $AYB$ equal to $60$ degrees. We can also find out that angle $CYB$ equals $120$ degrees.
Extend $\overline{CD}$ and $\overline{AB}$ and let their intersection be $E$ . Since angle $BEC$ plus angle $CYB$ equals $180$ degrees, quadrilateral $YCEB$ is a cyclic quadrilateral.
Next, draw a line from point $Y$ to point $E$ . Since angle $YBC$ and angle $YEC$ point to the same arc, angle $YEC$ is equal to $5$ degrees. Since $EYD$ is an isosceles triangle (based on angle properties) and $YAE$ is also an isosceles triangle, we can find that $YAD$ is also an isosceles triangle. Thus, each of the other angles is $\frac{180-120}{2}=30$ degrees. Finally, we have angle $BAD$ equals $30+55=\boxed{85}$ degrees.

Solution:
First, connect the diagonal $DB$ , then, draw line $DE$ such that it is congruent to $DC$ and is parallel to $AB$ . Because triangle $DCB$ is isosceles and angle $DCB$ is $170^\circ$ , the angles $CDB$ and $CBD$ are both $\frac{180-170}{2} = 5^\circ$ . Because angle $ABC$ is $70^\circ$ , we get angle $ABD$ is $65^\circ$ . Next, noticing parallel lines $AB$ and $DE$ and transversal $DB$ , we see that angle $BDE$ is also $65^\circ$ , and subtracting off angle $CDB$ gives that angle $EDC$ is $60^\circ$
Now, because we drew $ED = DC$ , triangle $DEC$ is equilateral. We can also conclude that $EC=DC=CB$ meaning that triangle $ECB$ is isosceles, and angles $CBE$ and $CEB$ are equal.
Finally, we can set up our equation. Denote angle $BAD$ as $x^\circ$ . Then, because $ABED$ is a parallelogram, the angle $DEB$ is also $x^\circ$ . Then, $CEB$ is $(x-60)^\circ$ . Again because $ABED$ is a parallelogram, angle $ABE$ is $(180-x)^\circ$ . Subtracting angle $ABC$ gives that angle $CBE$ equals $(110-x)^\circ$ . Because angle $CBE$ equals angle $CEB$ , we get \[x-60=110-x\] , solving into $x=\boxed{85}$


Answer: 120.0

Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$
[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); [/asy]
$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$

Solution:
A rhombus has sides of equal length. Because the perimeter of the rhombus is $52$ , each side is $\frac{52}{4}=13$ . In a rhombus, diagonals are perpendicular and bisect each other, which means $\overline{AE}$ $12$ $\overline{EC}$
Consider one of the right triangles:

$\overline{AB}$ $13$ , and $\overline{AE}$ $12$ . Using the Pythagorean theorem, we find that $\overline{BE}$ $5$ .
You know the Pythagorean triple, (5, 12, 13).
Thus the values of the two diagonals are $\overline{AC}$ $24$ and $\overline{BD}$ $10$ .
The area of a rhombus is = $\frac{d_1\cdot{d_2}}{2}$ $\frac{24\cdot{10}}{2}$ $120$
$\boxed{120}$

Solution: Right off the bat, we can see that the perimeter of the figure is 52. Dividing this by four, we can get that each side is equal to 13. By drawing a line perpendicular to the one given, we can split the figure into four right triangles. 12 (24/2) is equal to the height of one small right triangle, and 13 is the slanted side. Using the Pythagorean theorem we can find that 169 (13 squared) - 144 (12 squared) = 25 (five squared). With this, we can determine that each small right triangle equals 30. Multiplying that by four we can get $\boxed{120}$


Answer: 750.0

Solution:
If you draw altitudes from $A$ and $B$ to $CD,$ the trapezoid will be divided into two right triangles and a rectangle. You can find the values of $a$ and $b$ with the Pythagorean theorem
\[a=\sqrt{15^2-12^2}=\sqrt{81}=9\]
\[b=\sqrt{20^2-12^2}=\sqrt{256}=16\]
$ABYX$ is a rectangle so $XY=AB=50.$
\[CD=a+XY+b=9+50+16=75\]
The area of the trapezoid is
\[12\cdot \frac{(50+75)}{2} = 6(125) = \boxed{750}\]


Answer: 6.0

Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,$ and $BC=6$ . What is $AC$
$\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7$

Solution: $\angle ADB$ and $\angle ACB$ are both subtended by segment $AB$ , hence $\angle ACB = \angle ADB = 40^\circ$ . By considering $\triangle ABC$ , it follows that $\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ$ . Hence $\triangle ABC$ is isosceles, and $AC = BC = \boxed{6}.$


Answer: 17.0

Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XF\cdot XG$
$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$

Solution:
Using the given ratios, note that $\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.$
By AA Similarity, $\triangle AXD \sim \triangle EXY$ with a ratio of $\frac{DX}{XY} = \frac{9}{16}$ and $\triangle ACX \sim \triangle EFX$ with a ratio of $\frac{AX}{XE} = \frac{DX}{XY} = \frac{9}{16}$ , so $\frac{XF}{CX} = \frac{16}{9}$
Now we find the length of $BD$ . Because the quadrilateral is cyclic, we can simply use the Law of Cosines. \[BD^2=3^2+8^2-48\cos\angle BAD=2^2+6^2-24\cos (180-\angle BAD)=2^2+6^2+24\cos\angle BAD\] \[\rightarrow \cos\angle BAD = \frac{11}{24}\] \[\rightarrow BD=\sqrt{51}\] By Power of a Point, $CX\cdot XG = DX\cdot XB = \frac{\sqrt{51}}{4} \frac{3\sqrt{51}}{4}$ . Thus $XF\cdot XG = \frac{XF}{CX} CX\cdot XG = \frac{51}{3} = \boxed{17}.$

Solution:
We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point.
Let $Z$ be the intersection of $AC$ and $BD$ . First, from $ABCD$ being a cyclic quadrilateral, we have that $\triangle BCZ \sim \triangle AZD$ $\triangle BZA \sim \triangle CDZ$ . Therefore, $\frac{2}{BZ} = \frac{8}{AZ}$ $\frac{6}{CZ} = \frac{3}{BZ}$ , and $\frac{2}{CZ} = \frac{8}{DZ}$ , so we have $BZ = \frac{1}{2}CZ$ $AZ = 2CZ$ , and $DZ = 4CZ$ . By Ptolemy's Theorem, \[(AB)(CD) + (BC)(DA) = (AC)(BD) = (AZ + ZC)(BZ + ZD)\] \[\rightarrow 3 \cdot 6 + 2 \cdot 8 = 34 = \left(2CZ + ZC\right)\left(\frac{1}{2}CZ + 4CZ\right) = \frac{27}{2}CZ^2.\] Thus, $CZ^2 = \frac{68}{27}$ . Then, by Power of a Point, $GX \cdot XC = BX \cdot XD = \frac{3}{4} \cdot \frac{1}{4}BD^2 = \frac{3}{16} \cdot \left(\frac{9}{2}CZ\right)^2 = \frac{9 \cdot 17}{16}$ . So, $XG = \frac{9 \cdot 17}{16XC}$ .
Next, observe that $\triangle ACX \sim \triangle EFX$ , so $\frac{XE}{XF} = \frac{AX}{XC}$ . Also, $\triangle{AXD} \sim \triangle{EXY}$ , so $\frac{8}{AX} = \frac{EY}{XE}$ . We can compute $EY = \frac{128}{9}$ after noticing that $XY = BD - BY - DX = BD - \frac{11}{36}BD - \frac{1}{4}BD = \frac{4}{9}BD$ and that $\frac{8}{DX} = \frac{32}{BD} = \frac{EY}{XY} = \frac{EY}{\frac{4}{9}BD}$ . So, $\frac{8}{AX} = \frac{128}{9XE}$ . Then, $\frac{XE}{AX} = \frac{XF}{XC} = \frac{16}{9} \rightarrow XF = \frac{16}{9}XC$
Multiplying our equations for $XF$ and $XG$ yields that $XF \cdot XG = \frac{9 \cdot 17}{16XC} \cdot \frac{16}{9}XC = \boxed{17}.$

Solution:
Denote $P$ to be the intersection between line $AE$ and circle $O$ . Note that $\angle GFE = \angle ACG = \angle APG = 180 - \angle GPE$ , making $\angle GFE + \angle GDE = 180$ . Thus, $PEFG$ is a cyclic quadrilateral. Using Power of a Point on $X$ gives $XP \cdot XE = XG \cdot XF$
Since $\triangle ADX \sim \triangle EYX$ and $\triangle ACX \sim \triangle EFX$ $AX/XE = XD/YX = 9/16$ . Using Power of a Point on $X$ again, $(AX)(PX) = (BX)(DX)$ . Plugging in $AX=9/16 XE$ gives: \[\dfrac{9}{16}(XE)(PX) = (BX)(DX) = \dfrac{9}{16}(FX)(GX)\] By Law of Cosines , we can find $BD = \sqrt{51}$ , as in Solution 1. Now, $BX = 3/4 (\sqrt{51})$ and $DX = 1/4 (\sqrt{51})$ , making $\dfrac{9}{16}(FX)(GX) = \left( \dfrac{\sqrt{51}}{4}\right)\left( \dfrac{3\sqrt{51}}{4}\right) = \dfrac{3(51)}{16}$ . This gives us $FX \cdot GX = \boxed{17}$ as a result.

Solution:
$\because$ $AC 	\parallel EF$ $\quad \therefore$ $\triangle ACX \sim \triangle EFX$ $\quad \frac{XF}{XC} = \frac{XE}{XA}$
By Power of a Point, $XG \cdot XC = XD \cdot XB$
By multiplying the $2$ equations we get $XF \cdot XG = \frac{XE}{XA} \cdot XD \cdot XB$
$\because$ $YE 	\parallel AD$ $\quad \therefore$ $\triangle EYX \sim \triangle ADX$ $\quad \frac{XD}{XY} = \frac{XA}{XE}, \quad XD \cdot XE = XA \cdot XY, \quad XD = \frac{XA \cdot XY}{XE}$
By substitution, $XF \cdot XG = \frac{XE}{XA} \cdot \frac{XA \cdot XY}{XE} \cdot XB = XY \cdot XB = \frac{4}{9} BD \cdot \frac{3}{4} BD = \frac{BD^2}{3}$
Let $a = AB$ $b = BC$ $c = CD$ $d = AD$ $p = AC$ , and $q = BD$
By Ptolemy's theorem, $p \cdot q = a \cdot c + b \cdot d$
\[[ABD] = \frac12 \cdot ad \cdot \sin A, \quad [BCD] = \frac12 \cdot bc \cdot \sin C = \frac12 \cdot bc \cdot \sin A\]
\[[ABC] = \frac12 \cdot ab \cdot \sin B, \quad [ACD] = \frac12 \cdot cd \cdot \sin D = \frac12 \cdot cd \cdot \sin B\]
\[[ABCD] = [ABD] + [BCD] = \frac12 \cdot ad \cdot \sin A + \frac12 \cdot bc \cdot \sin A = \frac12 (ad + bc) \sin A\]
\[[ABCD] = [ABC] + [ACD] = \frac12 \cdot ab \cdot \sin B + \frac12 \cdot cd \cdot \sin B = \frac12 (ab + cd) \sin B\]
\[\frac{ab + cd}{ad + bc} = \frac{ \sin A }{ \sin B} = \frac{ \frac{q}{2R} }{ \frac{p}{2R} } = \frac{q}{p}, \quad p = \frac{q(ad + bc)}{ab + cd}\]
\[\frac{q(ad + bc)}{ab + cd} \cdot q = ac + bd\]
\[BD^2 = q^2 = \frac{ (ac + bd)(ab + cd) }{ad + bc} = \frac{(3 \cdot 6 + 2 \cdot 8)(3 \cdot 2 + 6 \cdot 8)}{3 \cdot 8 + 2 \cdot 6} = 51\]
\[XF \cdot XG = \frac{51}{3} = \boxed{17}\]


Answer: 360.0

Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$

Solution:
It's crucial to draw a good diagram for this one. Since $AC=20$ and $CD=30$ , we get $[ACD]=300$ . Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$ . Let $FE=x$ . Since $AE=5$ , then $AF=5-x$
By dropping this altitude, we can also see two similar triangles, $\triangle BFE \sim \triangle DCE$ . Since $EC$ is $20-5=15$ , and $DC=30$ , we get that $BF=2x$
Now, if we redraw another diagram just of $ABC$ , we get that $(2x)^2=(5-x)(15+x)$ because of the altitude geometric mean theorem which states that in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into.
Expanding, simplifying, and dividing by the GCF, we get $x^2+2x-15=0$ . This factors to $(x+5)(x-3)$ , which has roots of $x=-5, 3$ . Since lengths cannot be negative, $x=3$ . Since $x=3$ , that means the altitude $BF=2\cdot3=6$ , or $[ABC]=60$ . Thus $[ABCD]=[ACD]+[ABC]=300+60=\boxed{360}$

Solution: Let the points be $A(-10,0)$ $\:B(x,y)$ $\:C(10,0)$ $\:D(10,30)$ ,and $\:E(-5,0)$ , respectively. Since $B$ lies on line $DE$ , we know that $y=2x+10$ . Furthermore, since $\angle{ABC}=90^\circ$ $B$ lies on the circle with diameter $AC$ , so $x^2+y^2=100$ . Solving for $x$ and $y$ with these equations, we get the solutions $(0,10)$ and $(-8,-6)$ . We immediately discard the $(0,10)$ solution as $y$ should be negative. Thus, we conclude that $[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{360}$

Solution: Let $\angle C = \angle{ACB}$ and $\angle{B} = \angle{CBE}.$ Using Law of Sines on $\triangle{BCE}$ we get \[\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}\] and LoS on $\triangle{ABE}$ yields \[\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.\] Divide the two to get $\tan{B} = 3 \tan{C}.$ Now, \[\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}\] and solve the quadratic, taking the positive solution (C is acute) to get $\tan{C} = \frac{1}{3}.$ So if $AB = a,$ then $BC = 3a$ and $[ABC] = \frac{3a^2}{2}.$ By Pythagorean Theorem, $10a^2 = 400 \iff \frac{3a^2}{2} = 60$ and the answer is $300 + 60 \iff \boxed{360}.$

Solution:
Denote $EB$ as $x$ . By the Law of Cosines: \[AB^2 = 25 + x^2 - 10x\cos(\angle DEC)\] \[BC^2 = 225 + x^2 + 30x\cos(\angle DEC)\]
Adding these up yields: \[400 = 250 + 2x^2 + 20x\cos(\angle DEC) \Longrightarrow x^2 + \frac{10x}{\sqrt{5}} - 75 = 0\] By the quadratic formula, $x = 3\sqrt5$
Observe: \[[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DEC) = (3)(20) = 60\]
Thus the desired area is $\frac{1}{2}(30)(20) + 60 = \boxed{360}$

Solution:
Let $C = (0, 0)$ and $D = (0, 30)$ . Then $E = (-15, 0), A = (-20, 0),$ and $B$ lies on the line $y=2x+30.$ So the coordinates of $B$ are \[(x, 2x+30).\]
We can make this a vector problem. $\overrightarrow{\mathbf{B}} = \begin{pmatrix} x \\ 2x+30 \end{pmatrix}.$ We notice that point $B$ forms a right angle, meaning vectors $\overrightarrow{\mathbf{BC}}$ and $\overrightarrow{\mathbf{BA}}$ are orthogonal, and their dot-product is $0$
We determine $\overrightarrow{\mathbf{BC}}$ and $\overrightarrow{\mathbf{BA}}$ to be $\begin{pmatrix} -x \\ -2x-30 \end{pmatrix}$ and $\begin{pmatrix} -20-x \\ -2x-30 \end{pmatrix}$ , respectively. (To get this, we use the fact that $\overrightarrow{\mathbf{BC}} = \overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}}$ and similarly, $\overrightarrow{\mathbf{BA}} = \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}}.$
Equating the cross-product to $0$ gets us the quadratic $-x(-20-x)+(-2x-30)(-2x-30)=0.$ The solutions are $x=-18, -10.$ Since $B$ clearly has a more negative x-coordinate than $E$ , we take $x=-18$ . So $B = (-18, -6).$
From here, there are multiple ways to get the area of $\Delta{ABC}$ to be $60$ , and since the area of $\Delta{ACD}$ is $300$ , we get our final answer to be \[60 + 300 = \boxed{360}.\]

Solution:
Let $F$ be the midpoint of $AC$ , and draw $FG // CD$ where $G$ is on $BD$ . We have $EF=5,FC=10$
$\Delta EFG \sim \Delta ECD \implies FG=10=FA=FC$ . Therefore $ABCG$ is a cyclic quadrilateral.
Notice that $\angle EFG=90^\circ, EG=\sqrt{5^2+10^2}=5\sqrt{5} \implies BE=\frac{AE\cdot EC}{EG}=\frac{5\cdot 15}{5\sqrt{5}}=3\sqrt{5}$ via Power of a Point.
The altitude from $B$ to $AC$ is then equal to $GF\cdot \frac{BE}{GE}=10\cdot \frac{3\sqrt 5}{5 \sqrt 5}=6$
Finally, the total area of $ABCD$ is equal to $\frac 12 \cdot 20 \left(30+6 \right) =\boxed{360}.$

Solution:
Let $AB = x$ $BC = y$
Looking at the diagram we have $x^2 + y^2 = 20^2$ $DE = \sqrt{30^2+15^2} = 15\sqrt{5}$ $[ACD] = \frac{1}{2} \cdot 20 \cdot 30 = 300$
Because $\triangle CEF \sim \triangle CAB$ $EF = AB \cdot \frac{CE}{CA} = x \cdot \frac{15}{20} = \frac{3x}{4}$
$BF = BC - CF = BC - BC \cdot \frac{CE}{CA} = \frac{1}{4} \cdot BC = \frac{y}{4}$
$BE = \sqrt{ \left( \frac{3x}{4} \right) ^2 + \left( \frac{y}{4} \right) ^2 } = \frac{ \sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$ ,  we get $BE = \frac{ \sqrt{8x^2 + 400} }{4} = \frac{ \sqrt{2x^2 + 100} }{2}$
$[ABC] = \frac{1}{2} \cdot x \cdot y$
Because $\triangle ABC$ and $\triangle ACD$ share the same base, $\frac{[ABC]}{[ACD]} = \frac{BE}{DE}$
$[ABC] = [ACD] \cdot \frac{BE}{DE} = 300 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ 15 \sqrt{5} }$
$\frac{1}{2} \cdot x \cdot y = 20 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ \sqrt{5} }$
$xy = 4 \sqrt{10x^2 + 500}$
By $x^2 + y^2 = 400$ $y = \sqrt{400 - x^2}$ . So, $x \cdot  \sqrt{400 - x^2} = 4 \sqrt{10x^2 + 500}$
$x^2 (400 - x^2) = 16 (10x^2 + 500)$
Let $x^2 = a$ $a (400 - a) = 16 (10a + 500)$ $400a - a^2 = 160a + 8000$ $a^2 - 240a + 8000 = 0$ $(a-200)(a-40) = 0$
Because $x < 20$ $a$ can only equal 40. $a = 40$ $x = 2 \sqrt{10}$ $y = 6 \sqrt{10}$
$[ABC] = \frac{1}{2} \cdot 2 \sqrt{10} \cdot 6 \sqrt{10} = 60$
$[ABCD] = [ABC] + [ACD] = 60 + 300 = \boxed{360}$

Solution:
Drop perpendiculars $\overline{AF}$ and $\overline{CG}$ to $\overline{BD}.$ Notice that since $\angle AEF=\angle CEG$ (since they are vertical angles) and $\angle AFE=\angle CGE=90^\circ,$ triangles $AEF$ and $CEG$ are similar. Therefore, we have
\[x/EF=CE/AE=15/5=3,\]
where $EG=x.$ Therefore, $EF=x/3.$
Additionally, angle chasing shows that triangles $CEG$ and $DCG$ are also similar. This gives $CG/x=DC/CE=30/15=2,$ so $CG=2x.$ Thus, applying the Pythagorean Theorem to triangle $CEG$ gives
\[x^2+(2x)^2=15^2,\]
so $EG=x=3\sqrt 5.$ Our pairs of similar triangles then allow us to fill in the following lengths (in this order):
\[EF=x/3=\sqrt 5, CG=2x=6\sqrt 5, AF=CG/3=2\sqrt 5, DG=2\cdot CG=12\sqrt 5.\]
Now, let $BF=y.$ Angle chasing shows that triangle $ABF$ and $BCG$ are similar, so $BG/AF=CG/BF.$ Plugging in known lengths gives
\[\dfrac{y+4\sqrt 5}{2\sqrt 5}=\dfrac{6\sqrt 5}{y}.\]
This gives $y=2\sqrt 5.$ Now we know all the lengths that make up $BD,$ which allows us to find
\[BD=2\sqrt 5+\sqrt 5+3\sqrt 5+12\sqrt 5=18\sqrt 5.\]
Therefore,
\begin{align*} [ABCD] &= [ABD]+[CBD] \\ &= (BD)(AF)/2+(BD)(CG)/2 \\ &= (18\sqrt 5)(2\sqrt 5)/2+(18\sqrt 5)(6\sqrt 5)/2 \\ &= \boxed{360}


Answer: 360.0

Solution:
It's crucial to draw a good diagram for this one. Since $AC=20$ and $CD=30$ , we get $[ACD]=300$ . Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$ . Let $FE=x$ . Since $AE=5$ , then $AF=5-x$
By dropping this altitude, we can also see two similar triangles, $\triangle BFE \sim \triangle DCE$ . Since $EC$ is $20-5=15$ , and $DC=30$ , we get that $BF=2x$
Now, if we redraw another diagram just of $ABC$ , we get that $(2x)^2=(5-x)(15+x)$ because of the altitude geometric mean theorem which states that in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into.
Expanding, simplifying, and dividing by the GCF, we get $x^2+2x-15=0$ . This factors to $(x+5)(x-3)$ , which has roots of $x=-5, 3$ . Since lengths cannot be negative, $x=3$ . Since $x=3$ , that means the altitude $BF=2\cdot3=6$ , or $[ABC]=60$ . Thus $[ABCD]=[ACD]+[ABC]=300+60=\boxed{360}$

Solution: Let the points be $A(-10,0)$ $\:B(x,y)$ $\:C(10,0)$ $\:D(10,30)$ ,and $\:E(-5,0)$ , respectively. Since $B$ lies on line $DE$ , we know that $y=2x+10$ . Furthermore, since $\angle{ABC}=90^\circ$ $B$ lies on the circle with diameter $AC$ , so $x^2+y^2=100$ . Solving for $x$ and $y$ with these equations, we get the solutions $(0,10)$ and $(-8,-6)$ . We immediately discard the $(0,10)$ solution as $y$ should be negative. Thus, we conclude that $[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{360}$

Solution: Let $\angle C = \angle{ACB}$ and $\angle{B} = \angle{CBE}.$ Using Law of Sines on $\triangle{BCE}$ we get \[\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}\] and LoS on $\triangle{ABE}$ yields \[\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.\] Divide the two to get $\tan{B} = 3 \tan{C}.$ Now, \[\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}\] and solve the quadratic, taking the positive solution (C is acute) to get $\tan{C} = \frac{1}{3}.$ So if $AB = a,$ then $BC = 3a$ and $[ABC] = \frac{3a^2}{2}.$ By Pythagorean Theorem, $10a^2 = 400 \iff \frac{3a^2}{2} = 60$ and the answer is $300 + 60 \iff \boxed{360}.$

Solution:
Denote $EB$ as $x$ . By the Law of Cosines: \[AB^2 = 25 + x^2 - 10x\cos(\angle DEC)\] \[BC^2 = 225 + x^2 + 30x\cos(\angle DEC)\]
Adding these up yields: \[400 = 250 + 2x^2 + 20x\cos(\angle DEC) \Longrightarrow x^2 + \frac{10x}{\sqrt{5}} - 75 = 0\] By the quadratic formula, $x = 3\sqrt5$
Observe: \[[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DEC) = (3)(20) = 60\]
Thus the desired area is $\frac{1}{2}(30)(20) + 60 = \boxed{360}$

Solution:
Let $C = (0, 0)$ and $D = (0, 30)$ . Then $E = (-15, 0), A = (-20, 0),$ and $B$ lies on the line $y=2x+30.$ So the coordinates of $B$ are \[(x, 2x+30).\]
We can make this a vector problem. $\overrightarrow{\mathbf{B}} = \begin{pmatrix} x \\ 2x+30 \end{pmatrix}.$ We notice that point $B$ forms a right angle, meaning vectors $\overrightarrow{\mathbf{BC}}$ and $\overrightarrow{\mathbf{BA}}$ are orthogonal, and their dot-product is $0$
We determine $\overrightarrow{\mathbf{BC}}$ and $\overrightarrow{\mathbf{BA}}$ to be $\begin{pmatrix} -x \\ -2x-30 \end{pmatrix}$ and $\begin{pmatrix} -20-x \\ -2x-30 \end{pmatrix}$ , respectively. (To get this, we use the fact that $\overrightarrow{\mathbf{BC}} = \overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}}$ and similarly, $\overrightarrow{\mathbf{BA}} = \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}}.$
Equating the cross-product to $0$ gets us the quadratic $-x(-20-x)+(-2x-30)(-2x-30)=0.$ The solutions are $x=-18, -10.$ Since $B$ clearly has a more negative x-coordinate than $E$ , we take $x=-18$ . So $B = (-18, -6).$
From here, there are multiple ways to get the area of $\Delta{ABC}$ to be $60$ , and since the area of $\Delta{ACD}$ is $300$ , we get our final answer to be \[60 + 300 = \boxed{360}.\]

Solution:
Let $F$ be the midpoint of $AC$ , and draw $FG // CD$ where $G$ is on $BD$ . We have $EF=5,FC=10$
$\Delta EFG \sim \Delta ECD \implies FG=10=FA=FC$ . Therefore $ABCG$ is a cyclic quadrilateral.
Notice that $\angle EFG=90^\circ, EG=\sqrt{5^2+10^2}=5\sqrt{5} \implies BE=\frac{AE\cdot EC}{EG}=\frac{5\cdot 15}{5\sqrt{5}}=3\sqrt{5}$ via Power of a Point.
The altitude from $B$ to $AC$ is then equal to $GF\cdot \frac{BE}{GE}=10\cdot \frac{3\sqrt 5}{5 \sqrt 5}=6$
Finally, the total area of $ABCD$ is equal to $\frac 12 \cdot 20 \left(30+6 \right) =\boxed{360}.$

Solution:
Let $AB = x$ $BC = y$
Looking at the diagram we have $x^2 + y^2 = 20^2$ $DE = \sqrt{30^2+15^2} = 15\sqrt{5}$ $[ACD] = \frac{1}{2} \cdot 20 \cdot 30 = 300$
Because $\triangle CEF \sim \triangle CAB$ $EF = AB \cdot \frac{CE}{CA} = x \cdot \frac{15}{20} = \frac{3x}{4}$
$BF = BC - CF = BC - BC \cdot \frac{CE}{CA} = \frac{1}{4} \cdot BC = \frac{y}{4}$
$BE = \sqrt{ \left( \frac{3x}{4} \right) ^2 + \left( \frac{y}{4} \right) ^2 } = \frac{ \sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$ ,  we get $BE = \frac{ \sqrt{8x^2 + 400} }{4} = \frac{ \sqrt{2x^2 + 100} }{2}$
$[ABC] = \frac{1}{2} \cdot x \cdot y$
Because $\triangle ABC$ and $\triangle ACD$ share the same base, $\frac{[ABC]}{[ACD]} = \frac{BE}{DE}$
$[ABC] = [ACD] \cdot \frac{BE}{DE} = 300 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ 15 \sqrt{5} }$
$\frac{1}{2} \cdot x \cdot y = 20 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ \sqrt{5} }$
$xy = 4 \sqrt{10x^2 + 500}$
By $x^2 + y^2 = 400$ $y = \sqrt{400 - x^2}$ . So, $x \cdot  \sqrt{400 - x^2} = 4 \sqrt{10x^2 + 500}$
$x^2 (400 - x^2) = 16 (10x^2 + 500)$
Let $x^2 = a$ $a (400 - a) = 16 (10a + 500)$ $400a - a^2 = 160a + 8000$ $a^2 - 240a + 8000 = 0$ $(a-200)(a-40) = 0$
Because $x < 20$ $a$ can only equal 40. $a = 40$ $x = 2 \sqrt{10}$ $y = 6 \sqrt{10}$
$[ABC] = \frac{1}{2} \cdot 2 \sqrt{10} \cdot 6 \sqrt{10} = 60$
$[ABCD] = [ABC] + [ACD] = 60 + 300 = \boxed{360}$

Solution:
Drop perpendiculars $\overline{AF}$ and $\overline{CG}$ to $\overline{BD}.$ Notice that since $\angle AEF=\angle CEG$ (since they are vertical angles) and $\angle AFE=\angle CGE=90^\circ,$ triangles $AEF$ and $CEG$ are similar. Therefore, we have
\[x/EF=CE/AE=15/5=3,\]
where $EG=x.$ Therefore, $EF=x/3.$
Additionally, angle chasing shows that triangles $CEG$ and $DCG$ are also similar. This gives $CG/x=DC/CE=30/15=2,$ so $CG=2x.$ Thus, applying the Pythagorean Theorem to triangle $CEG$ gives
\[x^2+(2x)^2=15^2,\]
so $EG=x=3\sqrt 5.$ Our pairs of similar triangles then allow us to fill in the following lengths (in this order):
\[EF=x/3=\sqrt 5, CG=2x=6\sqrt 5, AF=CG/3=2\sqrt 5, DG=2\cdot CG=12\sqrt 5.\]
Now, let $BF=y.$ Angle chasing shows that triangle $ABF$ and $BCG$ are similar, so $BG/AF=CG/BF.$ Plugging in known lengths gives
\[\dfrac{y+4\sqrt 5}{2\sqrt 5}=\dfrac{6\sqrt 5}{y}.\]
This gives $y=2\sqrt 5.$ Now we know all the lengths that make up $BD,$ which allows us to find
\[BD=2\sqrt 5+\sqrt 5+3\sqrt 5+12\sqrt 5=18\sqrt 5.\]
Therefore,
\begin{align*} [ABCD] &= [ABD]+[CBD] \\ &= (BD)(AF)/2+(BD)(CG)/2 \\ &= (18\sqrt 5)(2\sqrt 5)/2+(18\sqrt 5)(6\sqrt 5)/2 \\ &= \boxed{360}


Answer: 1565.0

Quadrilateral $ABCD$ with side lengths $AB=7, BC=24, CD=20, DA=15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{a\pi-b}{c},$ where $a,b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a+b+c?$
$\textbf{(A) } 260 \qquad \textbf{(B) } 855 \qquad \textbf{(C) } 1235 \qquad \textbf{(D) } 1565 \qquad \textbf{(E) } 1997$

Solution:
Opposite angles of every cyclic quadrilateral are supplementary, so \[\angle B + \angle D = 180^{\circ}.\] We claim that $AC=25.$ We can prove it by contradiction:
By the Inscribed Angle Theorem, we conclude that $\overline{AC}$ is the diameter of the circle. So, the radius of the circle is $r=\frac{AC}{2}=\frac{25}{2}.$
The area of the requested region is \[\pi r^2 - \frac12\cdot AB\cdot BC - \frac12\cdot AD\cdot DC = \frac{625\pi}{4}-\frac{168}{2}-\frac{300}{2}=\frac{625\pi-936}{4}.\] Therefore, the answer is $a+b+c=\boxed{1565}.$

Solution:
When we look at the side lengths of the quadrilateral we see $7$ and $24,$ which screams out $25$ because of Pythagorean triplets. As a result, we can draw a line through points $A$ and $C$ to make a diameter of $25.$ See Solution 1 for a rigorous proof.
This can also be shown using the Law of Cosines: Since $7^2+24^2-2\cdot7\cdot24\cdot\cos B=15^2+20^2-2\cdot15\cdot20\cdot\cos D$ and $\cos B + \cos D = 0,$ it follows that $\cos B = \cos D = 0.$
Since the diameter is $25,$ we can see the area of the circle is just $\frac{625\pi}{4}$ from the formula of the area of the circle with just a diameter.
Then we can use Brahmagupta Formula $\sqrt{(s - a)(s - b)(s - c)(s - d)}$ where $a,b,c,d$ are side lengths, and $s$ is semi-perimeter to find the area of the quadrilateral.
If we just plug the values in, we get $\sqrt{54756}=234.$ So now the area of the region we are trying to find is $\frac{625\pi}{4} - 234 = \frac{625\pi-936}{4}.$
Therefore, the answer is $a+b+c=\boxed{1565}.$

Solution:
We can guess that this quadrilateral is actually made of two right triangles: $\triangle CDA$ has a $3 \text{-} 4 \text{-} 5$ ratio in the side lengths, and $\triangle ABC$ is a $7 \text{-} 24 \text{-} 25$ triangle.
(See Solution 1 for a proof.)
Next, we can choose one of these triangles and use the circumradius formula to find the radius. Let's choose the $15-20-25$ triangle. The area of the triangle is equal to the product of the side lengths divided by $4$ times the circumradius. Therefore, $150 = \frac{15\cdot20\cdot25}{4r}$ . Solving this simple algebraic equation gives us $r = \frac{25}{2}$
Plugging in the values, we have $\frac{25}{2}^2\cdot\pi - \left(\frac{15\cdot20}{2}+\frac{7\cdot24}{2}\right) = \frac{625\cdot\pi}{4} - 234$ . Rewriting this gives us $\frac{625\pi-936}{4}$
Therefore, adding these values gets us $\boxed{1565}.$


Answer: 204.0

Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?
$\textbf{(A)} \ 96 \qquad  \textbf{(B)} \ 108 \qquad  \textbf{(C)} \ 156 \qquad  \textbf{(D)} \ 204 \qquad  \textbf{(E)} \ 372$

Solution:
We tackle the problem by sorting it by how many stores are involved in the transaction.
1) 2 stores are involved. There are $\binom{4}{2} = 6$ ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. $6 \cdot 2 = 12$ total arrangements.
2) 3 stores are involved. There are $\binom{4}{3} = 4$ ways to choose which of the stores are involved. We then break the problem down to into two subsections - when the parents and grouped together or sold separately.
Separately: All children must be in one store. There are $3!$ ways to arrange this. $6$ ways in total.
Together: Both parents are in one store and the 3 children are split between the other two. There are $\binom{3}{2}$ ways to split the children and $3!$ ways to choose to which store each group will be sold. $3! \cdot \binom{3}{2} = 18$
$(6 + 18) \cdot 4 = 96$ total arrangements.
3) All 4 stores are involved. We break down the problem as previously shown.
Separately: All children must be split between two stores. There are $\binom{3}{2} = 3$ ways to arrange this. We can then arrange which group is sold to which store in $4!$ ways. $4! \cdot 3 = 72$
Together: Both parents are in one store and the 3 children are each in another store. There are $4! = 24$ ways to arrange this.
$24 + 72 = 96$ total arrangements.
Final Answer: $12 + 96 + 96 = \boxed{204}$


Answer: 9.0

Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighbourhood picnic?
$\textbf{(A)}\hspace{.05in}6 \qquad \textbf{(B)}\hspace{.05in}6\dfrac23 \qquad \textbf{(C)}\hspace{.05in}7\dfrac12 \qquad \textbf{(D)}\hspace{.05in}8 \qquad \textbf{(E)}\hspace{.05in}9$

Solution: Since Rachelle uses $3$ pounds of meat to make $8$ hamburgers, she uses $\frac{3}{8}$ pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or $\frac{3}{8} \cdot 24 = \boxed{9}$


Answer: 7.0

Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution:
So, let there be $x$ pairs of $$1$ socks, $y$ pairs of $$3$ socks, and $z$ pairs of $$4$ socks.
We have $x+y+z=12$ $x+3y+4z=24$ , and $x,y,z \ge 1$
Now, we subtract to find $2y+3z=12$ , and $y,z \ge 1$ .
It follows that $2y$ is a multiple of $3$ and $3z$ is a multiple of $3$ . Since sum of 2 multiples of 3 = multiple of 3, so we must have $2y=6$
Therefore, $y=3$ , and it follows that $z=2$ . Now, $x=12-y-z=12-3-2=\boxed{7}$ , as desired.

Solution:
Since the total cost of the socks was $$24$ and Ralph bought $12$ pairs, the average cost of each pair of socks is $\frac{$24}{12} = $2$
There are two ways to make packages of socks that average to $$2$ . You can have:
$\bullet$ Two $$1$ pairs and one $$4$ pair (package adds up to $$6$
$\bullet$ One $$1$ pair and one $$3$ pair (package adds up to $$4$
Now, we need to solve \[6a+4b=24,\] where $a$ is the number of $$6$ packages and $b$ is the number of $$4$ packages. We see our only solution (that has at least one of each pair of sock) is $a=2, b=3$ , which yields the answer of $2\times2+3\times1 = \boxed{7}$

Solution: Since there are 12 pairs of socks, and Ralph bought at least one pair of each, there are $12-3=9$ pairs of socks left. Also, the sum of the three pairs of socks is $1+3+4=8$ . This means that there are $24-8=16$ dollars left. If there are only $1$ dollar socks left, then we would have $9\cdot1=9$ dollars wasted, which leaves $7$ more dollars. If we replace one pair with a $3$ dollar pair, then we would waste an additional $2$ dollars. If we replace one pair with a $4$ dollar pair, then we would waste an additional $3$ dollars. The only way $7$ can be represented as a sum of $2$ s and $3$ s is $2+2+3$ . If we change $3$ pairs, we would have $6$ pairs left. Adding the one pair from previously, we have $\boxed{7}$ pairs.

Solution: Let the amount of $1$ dollar socks be $a$ $3$ dollar socks be $b$ , and $4$ dollar socks be $c$ . We then know that $a+b+c=12$ and $a+3b+4c=24$ . We can make $a+b+c=12$ into $a=12-b-c$ and then plug that into the other equation, producing $12-b-c+3b+4c=24$ which simplifies to $2b+3c=12$ . It's not hard to see $b=3$ and $c=2$ . Now that we know $b$ and $c$ , we know that $a=7$ , meaning the number of $1$ dollar socks Ralph bought is $\boxed{7}$

Solution:
If Ralph bought one sock of each kind, he already used $$8$ , so there are $$16$ left and 9 socks. If we split the $$16$ into four $$4$ sections, (as it is the smallest possible number that 1, 3, 4, can make in different ways that in all use at least each of the numbers once,) if Ralph bought a $$3$ pair, he would need to buy a $$1$ pair in order for it to add up to a multiple of four. Similarly, if Ralph bought a $$1$ pair, he would either need to buy three $$1$ pairs or a $$3$ pair. If Ralph bought a $$4$ pair, it would already make a group. Now, the problem is just how we can split 9 into 4 groups of 1, 2, or 4. We clearly see that $1 + 2 + 2 + 4 = 9$ , or a $$4$ pair, two $$3$ pairs, and six $$1$ pairs. Because we subtracted the necessary one of each kind, there are two $$4$ pairs, three $$3$ pairs, and seven $$1$ pairs. Therefore, the number of $$1$ pairs Ralph bought is $\boxed{7}$ .
~strongstephen


Answer: 4.0

Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given $2$ cards out of a set of $10$ cards numbered $1,2,3, \dots,10.$ The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon-- $11,$ Oscar-- $4,$ Aditi-- $7,$ Tyrone-- $16,$ Kim-- $17.$ Which of the following statements is true?
$\textbf{(A) }\text{Ravon was given card 3.}$
$\textbf{(B) }\text{Aditi was given card 3.}$
$\textbf{(C) }\text{Ravon was given card 4.}$
$\textbf{(D) }\text{Aditi was given card 4.}$
$\textbf{(E) }\text{Tyrone was given card 7.}$

Solution: By logical deduction, we consider the scores from lowest to highest: \begin{align*} \text{Oscar's score is 4.} &\implies \text{Oscar is given cards 1 and 3.} \\ &\implies \text{Aditi is given cards 2 and 5.} \\ &\implies \text{Ravon is given cards 4 and 7.} && (\bigstar) \\ &\implies \text{Tyrone is given cards 6 and 10.} \\ &\implies \text{Kim is given cards 8 and 9.} \end{align*} Therefore, the answer is $\boxed{4.}$


Answer: 16.0

Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$

Solution: Let Ray and Tom drive 40 miles. Ray's car would require $\frac{40}{40}=1$ gallon of gas and Tom's car would require $\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles, on $1+4=5$ gallons of gas, for a combined rate of $\frac{80}{5}=$ $\boxed{16}$

Solution: Taking the harmonic mean of the two rates, we get \[\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{16}.\]

Solution: Let the number of miles that Ray and Tom each drive be denoted by $m$ . Thus the number of gallons Ray's car uses can be represented by $\frac{m}{40}$ and the number of gallons that Tom's car uses can likewise be expressed as $\frac{m}{10}$ . Thus the total amount of gallons used by both cars can be expressed as $\frac{m}{40} + \frac{m}{10} = \frac{m}{8}$ . The total distance that both drive is equal to $2m$ so the total miles per gallon can be expressed as $\frac{2m}{\frac{m}{8}} = \boxed{16}$


Answer: 16.0

Solution: Let Ray and Tom drive 40 miles. Ray's car would require $\frac{40}{40}=1$ gallon of gas and Tom's car would require $\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles, on $1+4=5$ gallons of gas, for a combined rate of $\frac{80}{5}=$ $\boxed{16}$

Solution: Taking the harmonic mean of the two rates, we get \[\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{16}.\]

Solution: Let the number of miles that Ray and Tom each drive be denoted by $m$ . Thus the number of gallons Ray's car uses can be represented by $\frac{m}{40}$ and the number of gallons that Tom's car uses can likewise be expressed as $\frac{m}{10}$ . Thus the total amount of gallons used by both cars can be expressed as $\frac{m}{40} + \frac{m}{10} = \frac{m}{8}$ . The total distance that both drive is equal to $2m$ so the total miles per gallon can be expressed as $\frac{2m}{\frac{m}{8}} = \boxed{16}$


Answer: 420.0

Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$

Solution:
Because the coefficient of $x^2$ in both $p(x)$ and $q(x)$ is 0, the remaining root of $p(x)$ is $-(r+s)$ , and the remaining root of $q(x)$ is $-(r+s+1)$ . The coefficients of $x$ in $p(x)$ and $q(x)$ are both equal to $a$ , and equating the two coefficients gives \[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2\] from which $s = \tfrac 12 (5r+13)$ . The product of the roots of $p(x)$ differs from that of $q(x)$ by $240$ , so \[(r+4)\cdot \tfrac 12 (5r+7)\cdot \tfrac 12(7r+15)- r\cdot \tfrac 12 (5r+13)\cdot \tfrac 12(7r+13)=240\] from which $r^2+4r-5 =0$ , with roots $r=1$ and $r=-5$
If $r = 1$ , then the roots of $p(x)$ are $r=1$ $s=9$ , and $-(r+s)=-10$ , and $b=-rst=90$
If $r = -5$ , then the roots of $p(x)$ are $r=-5$ $s=-6$ , and $-(r+s)=11$ , and $b=-rst=-330$
Therefore the requested sum is $|- 330| + |90| = \boxed{420}$

Solution:
Let $r$ $s$ , and $-r-s$ be the roots of $p(x)$ (per Vieta's). Then $r^3 + ar + b = 0$ and similarly for $s$ . Also, \[q(r+4) = (r+4)^3 + a(r+4) + b  + 240 = 12r^2 + 48r + 304 + 4a = 0\]
Set up a similar equation for $s$
\[q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.\]
Simplifying and adding the equations gives \begin{align}\tag{*} r^2 - s^2 + 4r + 3s + 49 &= 0 \end{align} Now, let's deal with the $ax$ terms. Plugging the roots $r$ $s$ , and $-r-s$ into $p(x)$ yields a long polynomial, and plugging the roots $r+4$ $s-3$ , and $-1-r-s$ into $q(x)$ yields another long polynomial. Equating the coefficients of $x$ in both polynomials, we get: \[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),\] which eventually simplifies to \[s = \frac{13 + 5r}{2}.\] Substitution into (*) should give $r = -5$ and $r = 1$ , corresponding to $s = -6$ and $s = 9$ , and $|b| = 330, 90$ , for an answer of $\boxed{420}$

Solution:
The roots of $p(x)$ are $r$ $s$ , and $-r-s$ since they sum to $0$ by Vieta's Formula (co-efficient of $x^2$ term is $0$ ).
Similarly, the roots of $q(x)$ are $r + 4$ $s - 3$ , and $-r-s-1$ , as they too sum to $0$
Then:
$a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2$ and $-b = rs(-r-s)$ from $p(x)$ and
$a=(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2$ and $-(b+240)=(r+4)(s-3)(-r-s-1)$ from $q(x)$
From these equations, we can write that \[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 = a\] and simplifying gives \[2s-5r-13=0 \Rightarrow s = \frac{5r+13}{2}.\]
We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get \[rs(r+s) = b\] \[(r+4)(s-3)(r+s+1)=b + 240.\] Subtracting the first equation from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$
Expanding, simplifying, substituting $s = \frac{5r+13}{2}$ , and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$ , so $r = -5, 1$ . Then $s = -6, 9$
Finally, we substitute back into $b=rs(r+s)$ to get $b = (-5)(-6)(-5-6) = -330$ , or $b = (1)(9)(1 + 9) = 90$
The answer is $|-330|+|90| = \boxed{420}$

Solution:
By Vieta's, we know that the sum of roots of $p(x)$ is $0$ . Therefore,
the roots of $p$ are $r, s, -r-s$ . By similar reasoning, the roots of $q(x)$ are $r + 4, s - 3, -r - s - 1$ . Thus, $p(x) = (x - r)(x - s)(x + r + s)$ and $q(x) = (x - r - 4)(x - s + 3)(x + r + s + 1)$
Since $p(x)$ and $q(x)$ have the same coefficient for $x$ , we can go ahead
and match those up to get \begin{align*}     rs - r(r + s) - s(r + s) &= (r + 4)(s - 3) - (r + 4)(r + s + 1) - (s - 3)(r + s + 1) \\     0 &= -13 - 5r + 2s \\     s &= \frac{5r + 13}{2} \end{align*}
At this point, we can go ahead and compare the constant term in $p(x)$ and $q(x)$ . Doing so is certainly valid, but we can actually do this another way. Notice that $p(s) = 0$ . Therefore, $q(s) = 240$ . If we plug that into
our expression, we get that \begin{align*}     q(s)  &= 3(s - r - 4)(r + 2s + 1) \\     240 &= 3(s - r - 4)(r + 2s + 1) \\     240 &= 3\left( \frac{3r + 5}{2} \right)(6r + 14) \\     80 &= (3r + 5)(3r + 7) \\     0 &= r^2 + 4r - 5 \end{align*} This tells us that $(r, s) = (1, 9)$ or $(-5, -6)$ . Since $-b$ is the product of the roots, we have that the two possibilities are $1 \cdot 9 \cdot (-10) = -90$ and $(-5)(-6)11 = 330$ . Adding the absolute values of these gives us $\boxed{420}$


Answer: 10.0

Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution:
Since $x,y,z$ are all reals located in $[0, n]$ , the number of choices for each one is continuous so we use geometric probability.
WLOG( Without loss of generality ), assume that $n\geq x \geq y \geq z \geq 0$ . Then the set of points $(x,y,z)$ is a tetrahedron, or a triangular pyramid. The point $(x,y,z)$ distributes uniformly in this region. If this is not easy to understand, read Solution II.
The altitude of the tetrahedron is $n$ and the base is an isosceles right triangle with a leg length $n$ . The volume is $V_1=\dfrac{n^3}{6}$ , as shown in the first figure in red.

Now we will find the region with points satisfying $|x-y|\geq1$ $|y-z|\geq1$ $|z-x|\geq1$
Since $n\geq x \geq y \geq z \geq 0$ , we have $x-y\geq1$ $y-z\geq1$
The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too.

The volume of this region is $V_2=\dfrac{(n-2)^3}{6}$
So the probability is $p=\dfrac{V_2}{V_1}=\dfrac{(n-2)^3}{n^3}$
Substituting $n$ with the values in the choices, we find that when $n=10$ $p=\frac{512}{1000}>\frac{1}{2}$ , when $n=9$ $p=\frac{343}{729}<\frac{1}{2}$ . So $n\geq 10$
So the answer is $\boxed{10}$

Solution:
Because $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ , which means that $x$ $y$ , and $z$ distributes uniformly and independently in the interval $[0,n]$ . So the point $(x, y, z)$ distributes uniformly in the cubic $0\leqslant x, y, z \leqslant n$ , as shown in the figure below. The volume of this cubic is $V_0=n^3$
Cubic.png
As we want to find the probablity of the incident $A=\big\{ |x-y| \geq 1, |y-z| \geq1, |z-x| \geq 1 \big\}$ ,
we should find the volume of the region of points such that $|x-y|\geq 1$ $|y-z|\geq 1$ $|z-x|\geq 1$ and $0\leq x, y, z \leq n$
Now we will find the region $\big\{ (x,y,z)\ | \ 0\leq x, y, z \leq n, |x-y|\geq 1, |y-z|\geq 1, |z-x|\geq 1 \big\}$
The region can be generated by cutting off 3 slices corresponding to $|x-y|< 1$ $|y-z|< 1$ , and $|z-x|< 1$ , respectively, from the cubic.
After cutting off a slice corresponding to $|x-y|< 1$ , we get two triangular prisms, as shown in the figure.
2.png
In order to observe the object clearly, we rotate the object by the $z$ axis, as shown.
3.png
We can draw the slice corresponding to $|y-z|< 1$ on the object.
4B.png
After cutting off the slice corresponding to $|y-z|< 1$ , we have 4 pieces left.
5.png
After cutting off the slice corresponding to $|z-x|< 1$ , we have 6 congruent triangular prisms.
6B.png
Here we draw all the pictures in colors in order to explain the solution clearly. That does not mean that the students should do it in the examination. They can draw a figure with lines only, as shown below.
7.png
Every triangular pyramid has an altitude $n-2$ and a base of isoceless right triangle with leg length $n-2$ , so the volume is $(n-2)^3/6$ .
Then the volume of the region $\big\{ (x,y,z)\ | \ 0\leqslant x, y, z \leqslant n, |x-y|\geqslant 1, |y-z|\geqslant 1, |z-x|\geqslant 1 \big\}$ is $V_A=6\times(n-2)^3/6$ $(n-2)^3$
So the probability of the incident $A$ is $P(A)=\dfrac{V_A}{V_0}$ $\dfrac{(n-2)^3}{n^3}$
Then we can get the answer the same way as Solution I.
The answer is $\boxed{10}$

Solution: If $x$ $y$ , and $z$ are separated by at least one, then by subtracting the minimum space between the three variables (which is $2$ ), $x$ $y$ , and $z$ can be chosen randomly in the interval $[0,n-2]$ . The probability is hence $\dfrac{(n-2)^3}{n^3} > \dfrac{1}{2}.$ $\boxed{10}$ is the minimum value in the answer choices that satisfies this inequality.


Answer: 10.0

Solution: Imagine Points $x$ $y$ $z$ as the "starting points" of three "blocks" of real numbers that have length $1$ . We are just trying to find the probability that those three "blocks" do not overlap. To do this we can set each unit of $1$ into $\mu$ equal little increments, and take the limit of the probability as $\mu$ approaches $\infty$ . (This is because there are indefinitely many real numbers in a given interval.) We see that the total number of arrangements for the three blocks of $1$ , without considering the rule that no two blocks shall overlap is $(n\mu + 1)^3$ . We see that the number of ways to arrange the three blocks such that no three of them are overlapping is simply $\dbinom{(n-2)\mu + 3}{3}$ . Taking the limit as $\mu$ approaches infinity, we obtain our closed form to be $\lim_{\mu \to \infty} \frac{\dbinom{(n-2)\mu + 3}{3}}{ (n\mu + 1)^3  }.$ Dividing leading coefficients, we obtain $\lim_{\mu \to \infty} \frac{\dbinom{(n-2)\mu + 3}{3}}{ (n\mu + 1)^3  } =   \frac{\mu(n-2)^3}{\mu n^3} = \frac{(n-2)^3}{n^3}$ . Solving the inequality $\frac{(n-2)^3}{n^3} > \frac{1}{2}$ , we get the least value of $n$ as $n=\boxed{10}$


Answer: 440.0

Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \cdot y = -2$ . What is the value of
\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?\]
$\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480$

Solution:
\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}\]
Continuing to combine \[\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}\] From the givens, it can be concluded that $x^2y^2=4$ . Also, \[(x+y)^2=x^2+2xy+y^2=16\] This means that $x^2+y^2=20$ . Substituting this information into $\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}$ , we have $\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{440}$ . ~PCChess

Solution: As above, we need to calculate $\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}$ . Note that $x,y,$ are the roots of $x^2-4x-2$ and so $x^3=4x^2+2x$ and $y^3=4y^2+2y$ . Thus $x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88$ where $x^2+y^2=20$ and $x^2y^2=4$ as in the previous solution. Thus the answer is $\frac{(20)(88)}{4}=\boxed{440}$ . Note( $x^2+y^2=(x+y)^2-2xy=20$ , and $x^2y^2 = (xy)^2 = 4$

Solution: Note that $( x^3 + y^3 ) ( \frac{1}{y^2} + \frac{1}{x^2} ) = x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y.$ Now, we only need to find the values of $x^3 + y^3$ and $\frac{1}{y^2} + \frac{1}{x^2}.$ Recall that $x^3 + y^3 = (x + y) (x^2 - xy + y^2),$ and that $x^2 - xy + y^2 = (x + y)^2 - 3xy.$ We are able to solve the second equation, and doing so gets us $4^2 - 3(-2) = 22.$ Plugging this into the first equation, we get $x^3 + y^3 = 4(22) = 88.$ In order to find the value of $\frac{1}{y^2} + \frac{1}{x^2},$ we find a common denominator so that we can add them together. This gets us $\frac{x^2}{x^2 y^2} + \frac{y^2}{x^2 y^2} = \frac{x^2 + y^2}{(xy)^2}.$ Recalling that $x^2 + y^2 = (x+y)^2 - 2xy$ and solving this equation, we get $4^2 - 2(-2) = 20.$ Plugging this into the first equation, we get $\frac{1}{y^2} + \frac{1}{x^2} = \frac{20}{(-2)^2} = 5.$ Solving the original equation, we get $x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{440}.$ emerald_block

Solution:
This is basically bashing using Vieta's formulas to find $x$ and $y$ (which I highly do not recommend, I only wrote this solution for fun).
We use Vieta's to find a quadratic relating $x$ and $y$ . We set $x$ and $y$ to be the roots of the quadratic $Q ( n ) = n^2 - 4n - 2$ (because $x + y = 4$ , and $xy = -2$ ). We can solve the quadratic to get the roots $2 + \sqrt{6}$ and $2 - \sqrt{6}$ $x$ and $y$ are "interchangeable", meaning that it doesn't matter which solution $x$ or $y$ is, because it'll return the same result when plugged in. So we plug in $2 + \sqrt{6}$ for $x$ and $2 - \sqrt{6}$ and get $\boxed{440}$ as our answer.

Solution:
This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.
We first change the original expression to $4 + \frac{x^5 + y^5}{x^2 y^2}$ , because $x + y = 4$ . This is equal to $4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8$ . We can factor and reduce $x^4 + y^4$ to $(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392$ . Now our expression is just $400 - (x^3 y + x y^3)$ . We factor $x^3 y + x y^3$ to get $(xy)(x^2 + y^2) = -40$ . So the answer would be $400 - (-40)  = \boxed{440}$

Solution:
We first simplify the expression to \[x + y + \frac{x^5 + y^5}{x^2y^2}.\] Then, we can solve for $x$ and $y$ given the system of equations in the problem.
Since $xy = -2,$ we can substitute $\frac{-2}{x}$ for $y$ .
Thus, this becomes the equation \[x - \frac{2}{x} = 4.\] Multiplying both sides by $x$ , we obtain $x^2 - 2 = 4x,$ or \[x^2 - 4x - 2 = 0.\] By the quadratic formula we obtain $x = 2 \pm \sqrt{6}$ .
We also easily find that given $x = 2 \pm \sqrt{6}$ $y$ equals the conjugate of $x$ .
Thus, plugging our values in for $x$ and $y$ , our expression equals \[4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}\] By the binomial theorem, we observe that every second terms of the expansions $x^5$ and $y^5$ will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of $x^5 + y^5$ .
Thus, our expression equals \[4 + \frac{2(2^5 + \tbinom{5}{2}2^3 \times 6 + \tbinom{5}{4}2 \times 36)}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}.\] which equals \[4 + \frac{2(872)}{4}\] which equals $\boxed{440}$

Solution:
As before, simplify the expression to \[x + y + \frac{x^5 + y^5}{x^2y^2}.\] Since $x + y = 4$ and $x^2y^2 = 4$ , we substitute that in to obtain \[4 + \frac{x^5 + y^5}{4}.\] Now, we must solve for $x^5 + y^5$ . Start by squaring $x + y$ , to obtain \[x^2 + 2xy + y^2 = 16\] Simplifying, $x^2 + y^2 = 20$ . Squaring once more, we obtain \[x^4 + y^4 + 2x^2y^2 = 400\] Once again simplifying, $x^4 + y^4 = 392$ . Now, to obtain the fifth powers of $x$ and $y$ , we multiply both sides by $x + y$ .
We now have \[x^5 + x^4y + xy^4 + y^5 = 1568\] ,                                                                                or \[x^5 + y^5 + xy(x^3 + y^3) = 1568\] We now solve for $x^3 + y^3$ $(x + y)^3=x^3 + y^3 + 3xy(x + y) = 64$ , so $x^3 + y^3 = 88$ .
Plugging this back into $x^5 + x^4y + xy^4 + y^5 = 1568$ , we find that $x^5 + y^5 = 1744$ , so we have \[4 + \frac{1744}{4}.\] . This equals 440, so our answer is $\boxed{440}$

Solution:
We can use Newton Sums to solve this problem.
We start by noticing that we can rewrite the equation as $\frac{x^3}{y^2} + \frac{y^3}{x^2} + x + y.$ Then, we know that $x + y = 4,$ so we have $\frac{x^3}{y^2} + \frac{y^3}{x^2} + 4.$ We can use the equation $x \cdot y = -2$ to write $x = \frac{-2}{y}$ and $y = \frac{-2}{x}.$ Next, we can plug in these values of $x$ and $y$ to get $\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5}{4} + \frac{y^5}{4},$ which is the same as \[\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5 + y^5}{4}.\] Then, we use Newton sums where $S_n$ is the elementary symmetric sum of the sequence and $P_n$ is the power sum ( $x^n + y^n$ ). Using this, we can make the following Newton sums: \[P_1 = S_1\] \[P_2 = P_1 S_1 - 2S_2\] \[P_3 = P_2 S_1 - P_1 S_2\] \[P_4 = P_3 S_1 - P_2 S_2\] \[P_5 = P_4 S_1 - P_3 S_2.\] We also know that $S_1$ is 4 because $x + y$ is four, and we know that $S_2$ is $-2$ because $x \cdot y$ is $-2$ as well.
Then, we can plug in values! We have \[P_1 = S_1 = 4\] \[P_2 = P_1 S_1 - 2S_2 = 16 - (-4) = 20\] \[P_3 = P_2 S_1 - P_1 S_2 = 80 - (-8) = 88\] \[P_4 = P_3 S_1 - P_2 S_2 = 88 \cdot 4 - (-40) = 392\] \[P_5 = P_4 S_1 - P_3 S_2 = 392 \cdot 4 - (-2) \cdot 88 = 1744.\] We earlier noted that $\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5 + y^5}{4},$ so we have that this equals $\frac{1744}{4},$ or $436.$ Then, plugging this back into the original equation, this is $436 + 4$ or $440,$ so our answer is $\boxed{440}.$

Solution:
As in the first solution, we get the expression to be $\frac{x^3+y^3}{x^2} + \frac{x^3+y^3}{y^2}.$
Then, since the numerators are the same, we can put the two fractions as a common denominator and multiply the numerator by $x^2y^2.$ This gets us $\frac{(x^2 + y^2)(x^3 + y^3)}{x^2y^2}.$
Now, since we know $x+y=4$ and $xy=-2,$ instead of solving for $x$ and $y,$ we will try to manipulate the above expression them into a manner that we can substitute the sum and product that we know. Also, another form of $x^3+y^3$ is $(x+y)(x^2-xy+y^2).$
Thus, we can convert the current expression to $\frac{(x^2 + y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}.$
Doing some algebraic multiplications, we get $\frac{((x+y)^2 - 2xy)(x+y)((x+y)^2 - 3xy)}{(xy)^2}.$
Since we know $x+y=4$ and $xy=-2,$ we have $\frac{(16-(-4))(4)(16-(-6))}{4} = \frac{20 \cdot 4 \cdot 22}{4} = 20 \cdot 22 = 440.$
Therefore the answer is $\boxed{440}.$

Solution: We give all of the terms in this expression a common denominator. $x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = \frac{x^3y^2 + x^5 + y^5 + x^2y^3}{x^2y^2}$ . We can find $x^2y^2 = (xy)^2 = (-2)^2 = 4$ and we can find $x^3y^2 + x^2y^3 = x^2y^2(x + y) = 16$ . Our expression $\frac{xy^2 + x^5 + y^5 + x^2y}{x^2y^2}$ is now $\frac{16 + x^5 + y^5}{4}$ . Now all we need to find is $x^5 + y^5$ . Using the binomial theorem, $(x + y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5 = 4^5 = 1024$ . The extra terms that we don't need is $5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 = 5xy(x^3 + y^3) + 10x^2y^2(x + y) = -10(x^3 + y^3) + 160$ . What's $x^3 + y^3$ ? We use the same method. Using the binomial theorem, $(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 = x^3 + y^3 + 3xy(x + y) = x^3 + y^3 - 24 = 4^3 = 64$ . Now we know that $x^3 + y^3 = 64 + 24 = 88$ , and plugging that into $-10(x^3 + y^3) + 160$ gives $-10(88) + 160 = -880 + 160 = -720$ . Now we see that those extra terms have a sum of $-720$ . Thus $(x + y)^5 = x^5 + y^5 - 720 = 1024$ so $x^5 + y^5 = 1024 + 720 = 1744$ . Remember our goal: we want to find $\frac{-8 + x^5 + y^5}{4}$ . Using $x^5 + y^5 = 1744$ $\frac{16 + x^5 + y^5}{4} = \frac{16 + 1744}{4} = \frac{1760}{4} = \boxed{440}$

Solution: Since we know $x+y=4$ , we can simplify to $4 + \frac{x^3}{y^2} + \frac{y^3}{x^2}$ . Then, when you add the fractions, you get $\frac{x^5 + y^5}{x^2y^2} = \frac{x^5 + y^5}{4}$ . To find $x^5 + y^5$ , we expand $(x+y)^5 = 4^5$ and then simplify. $x^5+y^5+5x^4y+5xy^4+10x^3y^2+10x^2y^3 = 1024$ . We can use the fact that $xy = -2$ to simplify to $x^5 + y^5-10x^3 - 10y^3 + 40x + 40y = 1024$ . We can factor to get $x^5 + y^5 -10(x^3 + y^3) +40(x+y)$ . To find $x^3 + y^3$ , we simplify $(x+y)^3 = 4^3$ $x^3+y^3+3x^2y+3y^2x = 64$ . We can use the same process as we did before and get $x^3 + y^3 -6(x+y) = 64$ (using the fact that $xy = -2$ ). Therefore, $x^3 + y^3 = 88$ . Now, plugging it back in to our other expansion, we get $x^5 + y^5 + 160 - 880 = 1024$ $x^5 +y^5 = 1744$ . Now to get the final answer, all we need to do is to plug it back into the original equation and get $\frac{1744}{4} + 4 = \boxed{440}$


Answer: 2.0

Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$ . What is $x+y$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution: If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$ . Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$ . We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$ . Notice this is a circle with radius $0$ , which only contains one point. So, the only point is $(5, -3)$ , so the sum is $5 + (-3) = 2 \implies \boxed{2}$ . ~ asdf334

Solution: If we move every term including $x$ or $y$ to the LHS, we get \[x^2 - 10x + y^2 + 6y = -34.\] We can complete the square to find that this equation becomes \[(x - 5)^2 + (y + 3)^2 = 0.\] Since the square of any real number is nonnegative, we know that the sum is greater than or equal to $0$ .  Equality holds when the value inside the parhentheses is equal to $0$ .  We find that \[(x,y) = (5,-3)\] and the sum we are looking for is \[5+(-3)=2 \implies \boxed{2}.\] - Honestly


Answer: 2.0

Solution: If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$ . Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$ . We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$ . Notice this is a circle with radius $0$ , which only contains one point. So, the only point is $(5, -3)$ , so the sum is $5 + (-3) = 2 \implies \boxed{2}$ . ~ asdf334

Solution: If we move every term including $x$ or $y$ to the LHS, we get \[x^2 - 10x + y^2 + 6y = -34.\] We can complete the square to find that this equation becomes \[(x - 5)^2 + (y + 3)^2 = 0.\] Since the square of any real number is nonnegative, we know that the sum is greater than or equal to $0$ .  Equality holds when the value inside the parhentheses is equal to $0$ .  We find that \[(x,y) = (5,-3)\] and the sum we are looking for is \[5+(-3)=2 \implies \boxed{2}.\] - Honestly


Answer: 208.0

Recall that the conjugate of the complex number $w = a + bi$ , where $a$ and $b$ are real numbers and $i = \sqrt{-1}$ , is the complex number $\overline{w} = a - bi$ . For any complex number $z$ , let $f(z) = 4i\hspace{1pt}\overline{z}$ . The polynomial \[P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1\] has four complex roots: $z_1$ $z_2$ $z_3$ , and $z_4$ . Let \[Q(z) = z^4 + Az^3 + Bz^2 + Cz + D\] be the polynomial whose roots are $f(z_1)$ $f(z_2)$ $f(z_3)$ , and $f(z_4)$ , where the coefficients $A,$ $B,$ $C,$ and $D$ are complex numbers. What is $B + D?$
$(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304$

Solution:
By Vieta's formulas, $z_1z_2+z_1z_3+\dots+z_3z_4=3$ , and $B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).$
Since $\overline{a}\cdot\overline{b}=\overline{ab},$ \[B=(4i)^2\left(\overline{z_1z_2}+\overline{z_1z_3}+\overline{z_1z_4}+\overline{z_2z_3}+\overline{z_2z_4}+\overline{z_3z_4}\right).\] Since $\overline{a}+\overline{b}=\overline{a+b},$ \[B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48\]
Also, $z_1z_2z_3z_4=1,$ and \[D=(4i)^4\left(\overline{z}_1\,\overline{z}_2\,\overline{z}_3\,\overline{z}_4\right)=256\left(\overline{z_1z_2z_3z_4}\right)=256(\overline{1})=256.\]
Our answer is $B+D=256-48=\boxed{208}.$


Answer: 10.0

Rectangle $ABCD$ , pictured below, shares $50\%$ of its area with square $EFGH$ . Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$ . What is $\frac{AB}{AD}$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

Solution:
From the problem statement, we know that \[\frac{[ABCD]}{2} = \frac{[EFGH]}{5} \Rightarrow [ABCD]=\frac{2[EFGH]}{5}\]
If we let $a = EF$ and $b  = AD$ , we see \[[ABCD] = 2ab = \frac{2a^2}{5} \Rightarrow b = \frac{a}{5}\] . Hence, $\frac{AB}{AD} = \frac{2a}{b} = 2a(\frac{5}{a}) = \boxed{10}$


Answer: 69.0

Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$ , and segment $CD$ at distinct points $N$ $U$ , and $T$ , respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$ . Suppose that $AU = 84$ $AN = 126$ , and $UB = 168$ . Then $DA$ can be represented as $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$

Solution:
The center of the semicircle is also the midpoint of $AB$ . Let this point be O. Let $h$ be the length of $AD$
Rescale everything by 42, so $AU = 2, AN = 3, UB = 4$ . Then $AB = 6$ so $OA = OB = 3$
Since $ON$ is a radius of the semicircle, $ON = 3$ . Thus $OAN$ is an equilateral triangle.
Let $X$ $Y$ , and $Z$ be the areas of triangle $OUN$ , sector $ONB$ , and trapezoid $UBCT$ respectively.
$X = \frac {1}{2}(UO)(NO)\sin{O} = \frac {1}{2}(1)(3)\sin{60^\circ} = \frac {3}{4}\sqrt {3}$
$Y = \frac {1}{3}\pi(3)^2 = 3\pi$
To find $Z$ we have to find the length of $TC$ . Project $T$ and $N$ onto $AB$ to get points $T'$ and $N'$ . Notice that $UNN'$ and $TUT'$ are similar. Thus:
$\frac {TT'}{UT'} = \frac {UN'}{NN'} \implies \frac {TT'}{h} = \frac {1/2}{3\sqrt {3}/2} \implies TT' = \frac {\sqrt {3}}{9}h$
Then $TC = T'C - T'T = UB - TT' = 4 - \frac {\sqrt {3}}{9}h$ . So:
$Z = \frac {1}{2}(BU + TC)(CB) = \frac {1}{2}\left(8 - \frac {\sqrt {3}}{9}h\right)h = 4h - \frac {\sqrt {3}}{18}h^2$
Let $L$ be the area of the side of line $l$ containing regions $X, Y, Z$ . Then
$L = X + Y + Z = \frac {3}{4}\sqrt {3} + 3\pi + 4h - \frac {\sqrt {3}}{18}h^2$
Obviously, the $L$ is greater than the area on the other side of line $l$ . This other area is equal to the total area minus $L$ . Thus:
$\frac {2}{1} = \frac {L}{6h + \frac {9}{2}{\pi} - L} \implies 12h + 9\pi = 3L$
Now just solve for $h$
\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\ 0 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\ h^2 & = \frac {9}{4}(6) \\ h & = \frac {3}{2}\sqrt {6} \end{align*}
Don't forget to un-rescale at the end to get $AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}$
Finally, the answer is $63 + 6 = \boxed{069}$

Solution:
Let $O$ be the center of the semicircle. It follows that $AU + UO = AN = NO = 126$ , so triangle $ANO$ is equilateral
Let $Y$ be the foot of the altitude from $N$ , such that $NY = 63\sqrt{3}$ and $NU = 21$
Finally, denote $DT = a$ , and $AD = x$ . Extend $U$ to point $Z$ so that $Z$ is on $CD$ and $UZ$ is perpendicular to $CD$ . It then follows that $ZT = a-84$ . Since $NYU$ and $UZT$ are similar
$\frac {x}{a-84} = \frac {63\sqrt{3}}{21} = 3\sqrt{3}$
Given that line $NT$ divides $R$ into a ratio of $1:2$ , we can also say that
$(x)(\frac{84+a}{2}) + \frac {126^2\pi}{6} - (63)(21)(\sqrt{3}) = (\frac{1}{3})(252x + \frac{126^2\pi}{2})$
where the first term is the area of trapezoid $AUTD$ , the second and third terms denote the areas of $\frac{1}{6}$ a full circle, and the area of $NUO$ , respectively, and the fourth term on the right side of the equation is equal to $R$ . Cancelling out the $\frac{126^2\pi}{6}$ on both sides, we obtain
$(x)(\frac{84+a}{2}) - \frac{252x}{3} = (63)(21)(\sqrt{3})$
By adding and collecting like terms, $\frac{3ax - 252x}{6} = (63)(21)(\sqrt{3})$
$\frac{(3x)(a-84)}{6} = (63)(21)(\sqrt{3})$
Since $a - 84 = \frac{x}{3\sqrt{3}}$
$\frac {(3x)(\frac{x}{3\sqrt{3}})}{6} = (63)(21)(\sqrt{3})$
$\frac {x^2}{\sqrt{3}} = (63)(126)(\sqrt{3})$
$x^2 = (63)(126)(3) = (2)(3^5)(7^2)$
$x = AD = (7)(3^2)(\sqrt{6}) = 63\sqrt{6}$ , so the answer is $\boxed{069}.$

Solution:
Note that the total area of $\mathcal{R}$ is $252DA + \frac {126^2 \pi}{2}$ and thus one of the regions has area $84DA + \frac {126^2 \pi}{6}$
As in the above solutions we discover that $\angle AON = 60^\circ$ , thus sector $ANO$ of the semicircle has $\frac{1}{3}$ of the semicircle's area.
Similarly, dropping the $N'T'$ perpendicular we observe that $[AN'T'D] = 84DA$ , which is $\frac{1}{3}$ of the total rectangle.
Denoting the region to the left of $\overline {NT}$ as $\alpha$ and to the right as $\beta$ , it becomes clear that if $[\triangle UT'T] = [\triangle NUO]$ then the regions will have the desired ratio.
Using the 30-60-90 triangle, the slope of $NT$ , is ${-3\sqrt{3}}$ , and thus $[\triangle UT'T] = \frac {DA^2}{6\sqrt{3}}$
$[NUO]$ is most easily found by $\frac{absin(c)}{2}$ $[\triangle NUO] = \frac {126*42 * \frac {\sqrt{3}}{2}}{2}$
Equating, $\frac {126*42 * \frac {\sqrt{3}}{2}}{2} = \frac {DA^2}{6\sqrt{3}}$
Solving, $63 * 21 * 3 * 6 = DA^2$
$DA = 63 \sqrt{6} \longrightarrow \boxed{069}$

Solution:
Like above solutions, note that $ANO$ is equilateral with side length $126,$ where $O$ is the midpoint of $AB.$ Then, if we let $DA=a$ and set origin at $D=(0,0),$ we get $N=(63,a+63\sqrt{3}), U=(84,a).$ Line $NU$ is then $y-a=\sqrt{27}(x-84),$ so it intersects $CA,$ the $x$ -axis, at $x=(a/\sqrt{27}+84),$ giving us point $T.$ Now the area of region $R$ is $252a+\pi(126)^2 / 2,$ so one third of that is $84a+\pi(126)^2 / 6.$
The area of the smaller piece of $R$ is $[AUTD] + [ANU] + [\text{lune} AN] = \frac{1}{2} \cdot a(84+\frac{a}{\sqrt{27}}+84)+\frac{1}{2} \cdot 84 \cdot 63 \sqrt{3}+ \frac{\pi (126)^2}{6}-\frac{1}{2} \cdot 126 \cdot 63 \sqrt{2}$ $=\frac{a^2}{2\sqrt{27}}+84a-21\cdot 63\sqrt{2} + \frac{\pi(126)^2}{6}.$ Setting this equal to $84a+\pi(126)^2 / 6$ and canceling the $84a + \pi(126)^2$ yields $\frac{a^2}{2\sqrt{27}}=21 \cdot 63 \sqrt{3},$ so $a = 63 \sqrt{6}$ and the anser is $\boxed{069}.$ ~ rzlng

Solution:
Once we establish that $\Delta ANO$ is equilateral, we have \[[{\rm Sector } BON] = 2[{\rm Sector } AON],  [BCT'U]=2[ADT'U]\] \[\Rightarrow [\overset{\large\frown}{NB} CT'UO]=2[\overset{\large\frown}{NA} DT'UO]\] On the other hand, \[[\overset{\large\frown}{NB} CT]=2[\overset{\large\frown}{NA} DT]\] Therefore, $[UT'T]=[NUO]$
Now, $UO=42, NU=21 \Rightarrow [UT'T]=[NUO]=2[NN'U]$ . Also $\Delta UT'T \sim \Delta NN'U$ . Therefore, \[DA=UT'=\sqrt{2} NN'=\sqrt{2} \left(\frac{\sqrt{3}}{2}\cdot 126\right)=63\sqrt{6}\longrightarrow \boxed{069}\]


Answer: 13.0

Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$
[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy]
$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$

Solution: The area of $\bigtriangleup CDE$ is $\frac{DC\cdot CE}{2}$ . The area of $ABCD$ is $AB\cdot AD=5\cdot 6=30$ , which also must be equal to the area of $\bigtriangleup CDE$ , which, since $DC=5$ , must in turn equal $\frac{5\cdot CE}{2}$ . Through transitivity, then, $\frac{5\cdot CE}{2}=30$ , and $CE=12$ . Then, using the Pythagorean Theorem, you should be able to figure out that $\bigtriangleup CDE$ is a $5-12-13$ triangle, so $DE=\boxed{13}$

Solution: The area of the rectangle is $5\times6=30.$ Since the parallel line pairs are identical, $DC=5$ . Let $CE$ be $x$ $\dfrac{5x}{2}=30$ is the area of the right triangle. Solving for $x$ , we get $x=12.$ According to the Pythagorean Theorem, we have a $5-12-13$ triangle. So, the hypotenuse $DE$ has to be $\boxed{13}$

Solution: This problem can be solved with the Pythagorean Theorem ( $a^2 + b^2 = c^2$ ).  We know $AB = DC$ , so $DC = 5$ $CE$ is twice the length of $AD$ , so $CE = 12$ $5^2 + 12^2 = c^2$ $5^2 = 25$ $12^2 = 144$ $25 + 144 = 169$ $169$ has a square root of $13$ , so the hypotenuse or $DE$ is $13$ .  The answer is $\boxed{13}$