Answer: 129.0

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression , the last three terms form a geometric progression , and the first and fourth terms differ by $30$ . Find the sum of the four terms.

Solution:
Denote the first term as $a$ , and the common difference between the first three terms as $d$ . The four numbers thus are in the form $a,\ a+d,\ a+2d,\ \frac{(a + 2d)^2}{a + d}$
Since the first and fourth terms differ by $30$ , we have that $\frac{(a + 2d)^2}{a + d} - a = 30$ . Multiplying out by the denominator, \[(a^2 + 4ad + 4d^2) - a(a + d) = 30(a + d).\] This simplifies to $3ad + 4d^2 = 30a + 30d$ , which upon rearranging yields $2d(2d - 15) = 3a(10 - d)$
Both $a$ and $d$ are positive integers, so $2d - 15$ and $10 - d$ must have the same sign. Try if they are both positive (notice if they are both negative , then $d > 10$ and $d < \frac{15}{2}$ , which is a contradiction). Then, $d = 8, 9$ . Directly substituting and testing shows that $d \neq 8$ , but that if $d = 9$ then $a = 18$ . Alternatively, note that $3|2d$ or $3|2d-15$ implies that $3|d$ , so only $9$ may work. Hence, the four terms are $18,\ 27,\ 36,\ 48$ , which indeed fits the given conditions. Their sum is $\boxed{129}$

Solution: The sequence is of the form $a-d,$ $a,$ $a+d,$ $\frac{(a+d)^2}{a}$ . Since the first and last terms differ by 30, we have \[\frac{(a+d)^2}{a}-a+d=30\] \[d^2+3ad=30a\] \[d^2+3ad-30a=0\] \[d=\frac{-3a + \sqrt{9a^2+120a}}{2}.\] Let $9a^2+120a=x^2$ , where $x$ is an integer. This yields the following: \[9a^2+120a-x^2=0\] \[a=\frac{-120 + \sqrt{14400+36x^2}}{18}\] \[a=\frac{-20 + \sqrt{400+x^2}}{3}.\] We then set $400+x^2=y^2$ , where $y$ is an integer. Factoring using difference of squares, we have \[400=2^4 \cdot 5^2=(y+x)(y-x).\] Then, noticing that $y+x > y-x$ , we set up several systems of equations involving the factors of $400$ . The second system we set up in this manner, \[y+x=2^3 \cdot 5^2\] \[y-x=2,\] yields the solution $y=101, x=99$ . Plugging back in, we get that $a=27 \implies d=9$ , so the sequence is $18,$ $27,$ $36,$ $48,$ and the answer is $\boxed{129}.$

Solution: We represent the values as $a-d$ $a$ $a+d$ , and $\frac{(a+d)^2}{a}$ Take the difference between the first and last values \[\frac{(a+d)^2}{a}-a+d=30\] Manipulating the values by expanding and then long division we see \[\frac{a^2+2ad+d^2}{a}-a+d=30\] \[\frac{(a+2d)a+d^2}{a}-a+d=30\] \[a+2d+\frac{d^2}{a}-a+d=30\] Combining like terms we get \[3d+\frac{d^2}{a}=30\] And looking at the values we know that since the second term must be positive (since both a and d are positive), $d$ must be a maximum of 9, which offers 9 possible values (since $d$ must be a positive integer.) We can resort to guesswork by this time, and thus receive the result of of $d=9$ at which $a=27$ . The solution is thus $\boxed{129}.$


Answer: 18.0

In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$

Solution:
Call the trapezoid $ABCD$ with $AB$ as the smaller base and $CD$ as the longer. Let the point where an altitude intersects the larger base be $E$ , where $E$ is closer to $D$
Subtract the two bases and divide to find that $ED$ is $\log 8$ . The altitude can be expressed as $\frac{4}{3} \log 8$ . Therefore, the two legs are $\frac{5}{3} \log 8$ , or $\log 32$
The perimeter is thus $\log 32 + \log 32 + \log 192 + \log 3$ which is $\log 2^{16} 3^2$ . So $p + q = \boxed{018}$

Solution:
Set the base of the log as 2. Then call the trapezoid $ABCD$ with $CD$ as the longer base. Then have the two feet of the altitudes be $E$ and $F$ , with $E$ and $F$ in position from left to right respectively. Then, $CF$ and $ED$ are $\log 192 - \log 3 = \log 64$ (from the log subtraction identity. Then $CF=EF=3$ (isosceles trapezoid and $\log 64$ being 6. Then the 2 legs of the trapezoid is $\sqrt{3^2+4^2}=5=\log 32$
And we have the answer:
$\log 192 + \log 32 + \log 32 + \log 3 = \log(192 \cdot 32 \cdot 32 \cdot 3) = \log(2^6 \cdot 3 \cdot 2^5 \cdot 2^5 \cdot 3) = \log(2^{16} \cdot 3^2) \Rightarrow 16+2 = \boxed{18}$

Solution:
Let $ABCD$ be the trapezoid, where $\overline{AB} || \overline{CD}$ and $AB = \log 3$ and $CD = \log 192$ . Draw altitudes from $A$ and $B$ to $\overline{CD}$ with feet at $E$ and $F$ , respectively. $AB = \log 3$ , so $EF = \log 3$ . Now, we attempt to find $DE + FC$ , or what's left of $CD$ after we take out $EF$ . We make use of the two logarithmic rules:
\[\log(xy) = \log x + \log y\]
\[\log(x^a) = a\log(x)\]
\[CD = \log 192 = \log (3 \cdot 2^6) = \log 3 + \log(2^6) = \log 3 + 6\log 2\]
Thus, since $CD = DE + EF + FC = \log 3 + 6\log 2$ $CD - EF = \log 3 + 6\log 2 - \log 3 = 6\log 2 = DE + FC$
Now, why was finding $DE + FC$ important? Absolutely no reason! Just kidding, lol 🤣 Now, we essentially "glue" triangles $\triangle DAE$ and $\triangle BFC$ together to get $\triangle XC'D'$ , where $X$ is the point where $A$ and $B$ became one. Note we can do this because $\triangle DAE$ and $\triangle BFC$ are both right triangles with a common leg length (the altitude of trapezoid $ABCD$ ).
Triangle $XC'D'$ has a base of $C'D'$ , which is just equal to $DE + FC = 6\log 2$ . It is equal to $DE + FC$ because when we brought triangles $\triangle DAE$ and $\triangle BFC$ together, the length of $CD$ was not changed except for taking out $EF$
$XC' = XD'$ since $AD = BC$ because the problem tells us we have an isosceles trapezoid. Drop and altitude from $X$ to $C'D'$ The altitude has length $\log 16 = 4\log 2$ . The altitude also bisects $C'D'$ since $\triangle XC'D'$ is isosceles. Let the foot of the altitude be $M$ . Then $MD' = 3\log 2$ (Remember that C'D' was $6\log 2$ , and then it got bisected by the altitude). Thus, the hypotenuse, $XD'$ must be $5\log 2$ from the Pythagorean Theorem or by noticing that you have a 3-4-5 right triangle with a similarity ratio of $\log 2$ . Since $XD' = XC' = BC = AD$ $BC = AD = 5\log 2 = \log 2^5$
Now, we have $CD = \log (3 \cdot 2^6)$ $AB = \log 3$ , and $BC = AD = \log 2^5$ . Thus, their sum is
\[\log (3 \cdot 2^6) + \log 3 + \log 2^5 + \log 2^5 = \log (2^16 \cdot 3^2)\]
Thus, $p + q = 16 + 2 = \boxed{18}$ . ~Extremelysupercooldude


Answer: 704.0

In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss delivers them in the order $1, 2, 3, 4, 5, 6, 7, 8, 9$
While leaving for lunch, the secretary tells a colleague that letter $8$ has already been typed but says nothing else about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based on the above information, how many such after-lunch typing orders are possible? (That there are no letters left to be typed is one of the possibilities.)

Solution:
Re-stating the problem for clarity, let $S$ be a set arranged in increasing order. At any time an element can be appended to the end of $S$ , or the last element of $S$ can be removed. The question asks for the number of different orders in which all of the remaining elements of $S$ can be removed, given that $8$ had been removed already.
Since $8$ had already been added to the pile, the numbers $1 \ldots 7$ had already been added at some time to the pile; $9$ might or might not have been added yet. So currently $S$ is a subset of $\{1, 2, \ldots 7\}$ , possibly with $9$ at the end. Given that $S$ has $k$ elements, there are $k+1$ intervals for $9$ to be inserted, or $9$ might have already been placed, giving $k+2$ different possibilities.
Thus, the answer is $\sum_{k=0}^{7} {7 \choose k}(k+2)$ $= 1 \cdot 2 + 7 \cdot 3 + 21 \cdot 4 + 35 \cdot 5 + 35 \cdot 6 + 21 \cdot 7 + 7 \cdot 8 + 1 \cdot 9$ $= \boxed{704}$


Answer: 13.0

In base $10$ , the number $2013$ ends in the digit $3$ .  In base $9$ , on the other hand, the same number is written as $(2676)_{9}$ and ends in the digit $6$ .  For how many positive integers $b$ does the base- $b$ -representation of $2013$ end in the digit $3$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18$

Solution: We want the integers $b$ such that $2013\equiv 3\pmod{b} \Rightarrow b$ is a factor of $2010$ . Since $2010=2 \cdot 3 \cdot 5 \cdot 67$ , it has $(1+1)(1+1)(1+1)(1+1)=16$ factors. Since $b$ cannot equal $1, 2,$ or $3$ , as these cannot have the digit $3$ in their base representations, our answer is $16-3=\boxed{13}$


Answer: 11.0

In checking the petty cash a clerk counts $q$ quarters, $d$ dimes, $n$ nickels, and $c$ cents. Later he discovers that $x$ of the nickels were counted as quarters and $x$ of the dimes were counted as cents. To correct the total obtained the clerk must:
$\textbf{(A)}\ \text{make no correction}\qquad\textbf{(B)}\ \text{subtract 11 cents}\qquad\textbf{(C)}\ \text{subtract 11}x\text{ cents}\\ \textbf{(D)}\ \text{add 11}x\text{ cents}\qquad\textbf{(E)}\ \text{add }x\text{ cents}$

Solution:
If the clerk mistook $x$ nickels as quarters, then every mistake inflates the total by $20$ cents. In order to correct this, we have to subtract $20$ cents $x$ times, for a total of $20x$ cents. We can do the same for the $x$ dimes that were turned into pennies (or cents). This exchange would increase the total value of the coins by $9x$
As a total, you get $-20x + 9x = -11x$ . In order to correct the amount, the clerk should $\boxed{11}$


Answer: 46.0

In convex hexagon $ABCDEF$ , all six sides are congruent, $\angle A$ and $\angle D$ are right angles , and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent . The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$

Solution:
Let the side length be called $x$ , so $x=AB=BC=CD=DE=EF=AF$
2006 II AIME-1.png
The diagonal $BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}$ . Then the areas of the triangles AFB and CDE in total are $\frac{x^2}{2}\cdot 2$ ,
and the area of the rectangle BCEF equals $x\cdot x\sqrt{2}=x^2\sqrt{2}$
Then we have to solve the equation
Therefore, $AB$ is $\boxed{046}$

Solution: Because $\angle B$ $\angle C$ $\angle E$ , and $\angle F$ are congruent, the degree-measure of each of them is ${{720-2\cdot90}\over4}= 135$ . Lines $BF$ and $CE$ divide the hexagonal region into two right triangles and a rectangle. Let $AB=x$ . Then $BF=x\sqrt2$ . Thus \begin{align*} 2116(\sqrt2+1)&=[ABCDEF]\\ &=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), \end{align*} so $x^2=2116$ , and $x=\boxed{046}$


Answer: 777.0

In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$ . Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$

Solution:
By the Law of Cosines on $\triangle ABD$ at angle $A$ and on $\triangle BCD$ at angle $C$ (note $\angle C = \angle A$ ),
\[180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A\] \[(AD^2 - BC^2) = 360(AD - BC) \cos A\] \[(AD - BC)(AD + BC) = 360(AD - BC) \cos A\] \[(AD + BC) = 360 \cos A\] We know that $AD + BC = 640 - 360 = 280$ $\cos A = \dfrac{280}{360} =  \dfrac{7}{9} = 0.777 \ldots$
$\lfloor 1000 \cos  A \rfloor = \boxed{777}$

Solution:
Notice that $AB = CD$ , and $BD = DB$ , and $\angle{DAB} \cong \angle{BCD}$ , so we have side-side-angle matching on triangles $ABD$ and $CDB$ .  Since the problem does not allow $\triangle{ABD} \cong \triangle{CDB}$ , we know that $\angle{ADB}$ is not a right angle, and there is a unique other triangle with the matching side-side-angle.
Extend $AD$ to $C'$ so that $\triangle{ABC'}$ is isosceles with $AB = C'B$ .  Then notice that $\triangle{DC'B}$ has matching side-side-angle, and yet $\triangle{ADB} \not\cong \triangle{C'DB}$ because $\angle{ADB}$ is not right.  Therefore $\triangle{C'DB}$ is the unique triangle mentioned above, so $\triangle{CDB}$ is congruent, in some order of vertices, to $\triangle{C'DB}$ .  Since $\triangle{CDB} \cong \triangle{C'DB}$ would imply $\triangle{CDB} = \triangle{C'DB}$ , making quadrilateral $ABCD$ degenerate, we must have $\triangle{CDB} \cong \triangle{C'BD}$
Since the perimeter of $ABCD$ is $640$ $AD + BC = 640 - 180 - 180 = 280$ .  Hence $280 = AD + BC = AD + DC'$ .  Drop the altitude of $\triangle{ABC'}$ from $B$ and call the foot $P$ .  Then right triangle trigonometry on $\triangle{APB}$ shows that $\cos{A} = AP/AB = 140/180 = 7/9$ , so $\lfloor 1000 \cos A \rfloor = \boxed{777}$


Answer: 90.0

In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$

Solution:
Let $\angle MKN=\alpha$ and $\angle LNK=\beta$ . Let $P$ be the project of $L$ onto line $NK$ . Note $\angle KLP=\beta$
Then, $KP=28\sin\beta=8\cos\alpha$ .
Furthermore, $KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha$
Dividing the equations gives \[\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}\]
Thus, $MK=\frac{MN}{\tan\alpha}=98$ , so $MO=MK-KO=\boxed{090}$

Solution: Because $\angle KLN = \angle KMN = 90^{\circ}$ $KLMN$ is a cyclic quadrilateral. Hence, by Power of Point, \[KO\cdot KM = KL^2 \implies KM=\dfrac{28^2}{8}=98 \implies MO=98-8=\boxed{090}\] as desired.

Solution: First, let $P$ be the intersection of $LO$ and $KN$ as shown above. Note that $m\angle KPL = 90^{\circ}$ as given in the problem. Since $\angle KPL \cong \angle KLN$ and $\angle PKL \cong \angle LKN$ $\triangle PKL \sim \triangle LKN$ by AA similarity. Similarly, $\triangle KMN \sim \triangle KPO$ . Using these similarities we see that \[\frac{KP}{KL} = \frac{KL}{KN}\] \[KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}\] and \[\frac{KP}{KO} = \frac{KM}{KN}\] \[KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}\] Combining the two equations, we get \[\frac{8\cdot KM}{KN} = \frac{784}{KN}\] \[8 \cdot KM = 28^2\] \[KM = 98\] Since $KM = KO + MO$ , we get $MO = 98 -8 = \boxed{090}$

Solution:
Extend $KL$ and $NM$ past $L$ and $M$ respectively to meet at $P$ . Let $H$ be the intersection of diagonals $KM$ and $LN$ (this is the orthocenter of $\triangle KNP$ ).
As $\triangle KOL \sim \triangle KHP$ (as $LO \parallel PH$ , using the fact that $H$ is the orthocenter), we may let $OH = 8k$ and $LP = 28k$
Then using similarity with triangles $\triangle KLH$ and $\triangle KMP$ we have
\[\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}\]
Cross-multiplying and dividing by $4+4k$ gives $2(8+8k+HM) = 28 \cdot 7 = 196$ so $MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}$ . (Solution by scrabbler94)

Solution: First, let $P$ be the intersection of $LO$ and $KN$ . We can use the right triangles in the problem to create equations. Let $a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,$ and $g=NC.$ We are trying to find $d.$ We can find $7$ equations. They are \[4225+d^2=c^2,\] \[4225+d^2+16d+64=a^2+2ab+b^2,\] \[a^2+e^2=c^2,\] \[b^2+e^2=64,\] \[b^2+e^2+2ef+f^2=784,\] \[a^2+e^2+2ef+f^2=g^2,\] and \[g^2+784=a^2+2ab+b^2.\] We can subtract the fifth equation from the sixth equation to get $a^2-b^2=g^2-784.$ We can subtract the fourth equation from the third equation to get $a^2-b^2=c^2-64.$ Combining these equations gives $c^2-64=g^2-784$ so $g^2=c^2+720.$ Substituting this into the seventh equation gives $c^2+1504=a^2+2ab+b^2.$ Substituting this into the second equation gives $4225+d^2+16d+64=c^2+1504$ . Subtracting the first equation from this gives $16d+64=1504.$ Solving this equation, we find that $d=\boxed{090}.$ (Solution by DottedCaculator)

Solution: Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$ . Furthermore, $(XLN)$ is tangent to $\overline{KL}$ . Then, \[KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,\] and $MO=KM-KO=\boxed{090}$

Solution:
(Diagram by vedadehhc)
Call the base of the altitude from $L$ to $NK$ point $P$ . Let $PO=x$ . Now, we have that $KP=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$ . Using Power of a Point, we have
\[(KO)(OM)=(LO)(OQ)\] $Q$ is the intersection of $OL$ with the circle $\neq L$
\[8(MO)=(\sqrt{720+x^2}-x)(\sqrt{720+x^2}+x)\]
\[8(MO)=720\]
\[MO=\boxed{090}\]

Solution: Note that since $\angle KLN = \angle KMN$ , quadrilateral $KLMN$ is cyclic. Therefore, we have \[\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,\] so $\triangle KLO \sim \triangle KML$ , giving \[\frac{KM}{28} = \frac{28}{8} \implies KM = 98.\] Therefore, $OM = 98-8 = \boxed{90}$

Solution:
By Pythagorean Theorem, $KM^2+65^2 = KN^2 = 28^2 + LN^2$ . Thus, $LN^2 = KM^2 + 65^2 - 28^2$
By Pythagorean Theorem, $KP^2 + LP^2 = 28^2$ , and $PN^2 + LP^2 = LN^2$
\[PN^2 = (KN - KP)^2 = (\sqrt{KM^2 + 65^2} - KP)^2\]
It follows that \[(\sqrt{KM^2 + 65^2} - KP)^2 + LP^2 = KM^2 + 65^2 - 28^2\]
\[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + KP^2 + LP^2 = KM^2 + 65^2 - 28^2\]
Since $KP^2 + LP^2 = 28^2$ \[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + 28^2 = KM^2 + 65^2 - 28^2\]
\[-2\sqrt{KM^2 + 65^2}(KP) = -2 \times 28^2\]
\[KP = \frac{28^2}{\sqrt{KM^2 + 65^2}}\]
$\angle OKP = \angle NKM$ (it's the same angle) and $\angle OPK = \angle KMN = 90^{\circ}$ . Thus, $\triangle KOP \sim \triangle KNM$
Thus,
\[\frac{KO}{KN} = \frac{KP}{KM}\]
\[\frac{8}{\sqrt{KM^2 + 65^2}} = \frac{\frac{28^2}{\sqrt{KM^2 + 65^2}}}{KM}\]
Multiplying both sides by $\sqrt{KM^2 + 65^2}$
\[8 = \frac{28^2}{KM}\]
\[KM = 98\]
Therefore, $OM = 98-8 = \boxed{90}$


Answer: 210.0

In counting $n$ colored balls, some red and some black, it was found that $49$ of the first $50$ counted were red.
Thereafter, $7$ out of every $8$ counted were red. If, in all, $90$ % or more of the balls counted were red, the maximum value of $n$ is:
$\textbf{(A)}\ 225 \qquad \textbf{(B)}\ 210 \qquad \textbf{(C)}\ 200 \qquad \textbf{(D)}\ 180 \qquad \textbf{(E)}\ 175$

Solution:
The desired percentage of red balls is more than $90$ percent, so write an inequality
\[\frac{49+7x}{50+8x} \ge 0.9\]
Since $x >0$ , the sign does not need to be swapped after multiplying both sides by $50+8x$
\[49+7x \ge 45+7.2x\] \[4 \ge 0.2x\] \[20 \ge x\]
Thus, up to $20$ batches of balls can be used, so a total of $20 \cdot 8 + 50 = 210$ balls can be counted while satisfying the requirements.  The answer is $\boxed{210}$


Answer: 23.0

In equiangular octagon $CAROLINE$ $CA = RO = LI = NE =$ $\sqrt{2}$ and $AR = OL = IN = EC = 1$ . The self-intersecting octagon $CORNELIA$ encloses six non-overlapping triangular regions. Let $K$ be the area enclosed by $CORNELIA$ , that is, the total area of the six triangular regions. Then $K =$ $\dfrac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a + b$

Solution:
We can draw $CORNELIA$ and introduce some points.
2018 AIME II Problem 4.png
The diagram is essentially a 3x3 grid where each of the 9 squares making up the grid have a side length of 1.
In order to find the area of $CORNELIA$ , we need to find 4 times the area of $\bigtriangleup$ $ACY$ and 2 times the area of $\bigtriangleup$ $YZW$
Using similar triangles $\bigtriangleup$ $ARW$ and $\bigtriangleup$ $YZW$ (We look at their heights), $YZ$ $=$ $\frac{1}{3}$ . Therefore, the area of $\bigtriangleup$ $YZW$ is $\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}$ $=$ $\frac{1}{12}$
Since $YZ$ $=$ $\frac{1}{3}$ and $XY = ZQ$ $XY$ $=$ $\frac{1}{3}$ and $CY$ $=$ $\frac{4}{3}$
Therefore, the area of $\bigtriangleup$ $ACY$ is $\frac{4}{3}\cdot$ $1$ $\cdot$ $\frac{1}{2}$ $=$ $\frac{2}{3}$
Our final answer is $\frac{1}{12}$ $\cdot$ $2$ $+$ $\frac{2}{3}$ $\cdot$ $4$ $=$ $\frac{17}{6}$
$17 + 6 =$ $\boxed{023}$


Answer: 20.0

In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$

Solution:
Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier.
Let $M$ be the midpoint of $\overline{DE}$ . Then $\Delta MCA$ is a 30-60-90 triangle with $MC = \dfrac{3}{2}$ $AC = 3$ and $AM = \dfrac{3\sqrt{3}}{2}$ . Since the triangle $\Delta AME$ is right, then we can find the length of $\overline{AE}$ by pythagorean theorem, $AE = \sqrt{7}$ . Therefore, since $\Delta AME$ is a right triangle, we can easily find $\sin(\angle EAM) = \dfrac{1}{2\sqrt{7}}$ and $\cos(\angle EAM) = \sqrt{1-\sin(\angle EAM)^2}=\dfrac{3\sqrt{3}}{2\sqrt{7}}$ . So we can use the double angle formula for sine, $\sin(\angle EAD) = 2\sin(\angle EAM)\cos(\angle EAM) = \dfrac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$

Solution: We find that, as before, $AE = \sqrt{7}$ , and also the area of $\Delta DAE$ is 1/3 the area of $\Delta ABC$ . Thus, using the area formula, $1/2 \cdot 7 \cdot \sin(\angle EAD) = 3\sqrt{3}/4$ , and $\sin(\angle EAD) = \dfrac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}.$

Solution: Let A be the origin of the complex plane, B be $1+i\sqrt{3}$ , and C be $2$ . Also, WLOG, let D have a greater imaginary part than E. Then, D is $\frac{4}{3}+\frac{2i\sqrt{3}}{3}$ and E is $\frac{5}{3}+\frac{i\sqrt{3}}{3}$ . Then, $\sin(\angle DAE) = Im\left(\dfrac{\frac{4}{3}+\frac{2i\sqrt{3}}{3}}{ \frac{5}{3}+\frac{i\sqrt{3}}{3}}\right) = Im\left(\frac{26+6i\sqrt{3}}{28}\right) =  \frac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$

Solution: Without loss of generality, say that the side length of triangle ABC is 3. EC is 1 and by the law of cosines, $AE^2=1+3^2-2(1)(3)\cos(\angle DAE)$ or $AE=\sqrt7$ The same goes for AD. DE equals 1 because AD and AE trisect BC. By the law of cosines, $\cos(\angle DAE)=(1-7-7)/-2(\sqrt7)(\sqrt7)=13/14$ Since $\sin^2(\angle DAE)=1-cos^2(\angle DAE)$ Then $\sin^2(\angle DAE)= 1-\frac{169}{196}=\frac{27}{196}$ So $\sin(\angle DAE)=\frac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$

Solution: Setting up a convinient coordinate system, we let $A$ be at point $(0, 0)$ $B$ be at point $(3, 3\sqrt3)$ , and $C$ be at point $(6, 0)$ . Then $D$ and $E$ will be at points $(4, 2\sqrt3)$ and $(5, \sqrt3)$ . Then $\cos(\angle DAE) = \frac{\vec{AD}\cdot\vec{AE}}{\|\vec{AD}\| \|\vec{AE}\|} = \frac{4\cdot5 + 2\sqrt{3}\cdot\sqrt{3}}{28}=\frac{13}{14}$ . From here, we see that $\sin(\angle DAE) = \sqrt{1-\cos^2(\angle DAE)} = \frac{3\sqrt3}{14}\Longrightarrow\boxed{020}$


Answer: 0.0

In how many ways can $10001$ be written as the sum of two primes?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution:
For the sum of two numbers to be odd, one must be odd and the other must be even, because all odd numbers are of the form $2n+1$ where n is an integer, and all even numbers are of the form $2m$ where m is an integer. \[2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1\] and $m+n$ is an integer because $m$ and $n$ are both integers.
The only even prime number is $2,$ so our only combination could be $2$ and $9999.$ However, $9999$ is clearly divisible by $3$ ,
so the number of ways $10001$ can be written as the sum of two primes is $\boxed{0}$

Solution: First, we noticed that 10001 is equal to 5000+5001, if you subtract n to 5000 and add n to 5001, you always get an even number, even number is never a prime number except 2. We also see that whenever an addend is an odd number, the other addend will be even, so having an odd number as an addend is not possible, other than 9999 and 2, because 2 is a prime. We try 2 and 9999 but we can see 9999 is divisible by 3 and 9 clearly. So the answer is $\boxed{0}$

Solution: It is helpful to know and understand the Goldbach Conjecture - that every even number can be written as the sum of $2$ primes - and also, that the ${\textbf{odd numbers}\ }$ that are the sum of two primes are exactly two more than a prime. This is because to make the sum of two numbers odd, you must have one even and one odd. There is only one even prime, which is two, so the sum will be of the form $2+p$ . Hence, the odd numbers that are the sum of two primes are exactly $2$ more than a prime. Relating to the problem, $10001$ is not $2$ more than a prime, because $10001-2=9999$ and we can easily see that $9999$ is divisible by $3$ . Therefore, $10001$ cannot be written as the sum of two primes, and the answer is $\boxed{0}$ ~mk


Answer: 7.0

In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution:
Factor $345=3\cdot 5\cdot 23$
Suppose we take an odd number $k$ of consecutive integers, with the median as $m$ . Then $mk=345$ with $\tfrac12k<m$ .
Looking at the factors of $345$ , the possible values of $k$ are $3,5,15,23$ with medians as $115,69,23,15$ respectively.
Suppose instead we take an even number $2k$ of consecutive integers, with median being the average of $m$ and $m+1$ . Then $k(2m+1)=345$ with $k\le m$ .
Looking again at the factors of $345$ , the possible values of $k$ are $1,3,5$ with medians $(172,173),(57,58),(34,35)$ respectively.
Thus the answer is $\boxed{7}$

Solution:
We need to find consecutive numbers (an arithmetic sequence that increases by $1$ ) that sums to $345$ . This calls for the sum of an arithmetic sequence given that the first term is $k$ , the last term is $g$ and with $n$ elements, which is: $\frac {n \cdot (k+g)}{2}$
We look for sequences of $n$ consecutive numbers starting at $k$ and ending at $k+n-1$ . We can now substitute $g$ with $k+n-1$ . Now we substitute our new value of $g$ into $\frac {n \cdot (k+g)}{2}$ to get that the sum is $\frac {n \cdot (k+k+n-1)}{2} = 345$
This simplifies to $\frac {n \cdot (2k+n-1)}{2} = 345$ . This gives a nice equation. We multiply out the 2 to get that $n \cdot (2k+n-1)=690$ . This leaves us with 2 integers that multiply to $690$ which leads us to think of factors of $690$ . We know the factors of $690$ are: $1,2,3,5,6,10,15,23,30,46,69,115,138,230,345,690$ . So through inspection (checking), we see that only $2,3,5,6,10,15$ and $23$ work. This gives us the answer of $\boxed{7}$ ways.

Solution: At the very end of Solution 2, where we find the factors of 690, instead of inspection, notice that all numbers will work until you get to $30$ , and that is because $\frac{345}{30}=11.5$ , which means $11$ and $12$ must be the middle 2 numbers; however, a sequence of length $30$ with middle numbers $11$ and $12$ that consists only of integers would go into the negatives, so any number from 30 onwards wouldn't work, and since $1$ is a trivial, non-counted solution, we get $\boxed{7}$ -ColtsFan10

Solution:
The median of the sequence $m$ is either an integer or a half integer. Let $m=\frac{i}{2}, i \in N$ , then $P=i\cdot n=2\cdot 3 \cdot 5 \cdot 23$
On the other hand we have two constraints:
1) $m \geq \frac{n+1}{2} \iff i>n$ because the integers in the sequence are all positive, and $n>1$
2) If $n$ is odd then $m$ is an integer, $i$ is even; if $n$ is even then $m$ is a half integer, $i$ is odd. Therefore, $n$ and $i$ have opposite parity.
Now $P$ has $16$ factors and it is not a perfect square. There are $8-1=7$ choices for $1 < n < \sqrt{P}$ . Also since $2|P, 4\nmid P$ , we know $n$ and $\frac{P}{n}$ must have opposite parity. Therefore the answer is $\boxed{7}$


Answer: 7.0

Solution:
We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.
For the first case, we can cleverly choose the convenient form of our sequence to be \[a-n,\cdots, a-1, a, a+1, \cdots, a+n\]
because then our sum will just be $(2n+1)a$ . We now have \[(2n+1)a = 345\] and $a$ will have a solution when $\frac{345}{2n+1}$ is an integer, namely when $2n+1$ is a divisor of 345. We check that \[2n+1 = 3, 5, 15, 23\] work, and no more, because $2n+1=1$ does not satisfy the requirements of two or more consecutive integers, and when $2n+1$ equals the next biggest factor, $69$ , there must be negative integers in the sequence. Our solutions are $\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}$
For the even cases, we choose our sequence to be of the form: \[a-(n-1), \cdots, a, a+1, \cdots, a+n\] so the sum is $\frac{(2n)(2a+1)}{2} = n(2a+1)$ . In this case, we find our solutions to be $\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}$
We have found all 7 solutions and our answer is $\boxed{7}$

Solution:
The sum from $a$ to $b$ where $a$ and $b$ are integers and $a>b$ is
$S=\dfrac{(a-b+1)(a+b)}{2}$
$345=\dfrac{(a-b+1)(a+b)}{2}$
$2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)$
Let $c=a-b+1$ and $d=a+b$
$2\cdot 3\cdot 5\cdot 23=c\cdot d$
If we factor $690$ into all of its factor groups $(\text{exg}~ (10,69) ~\text{or} ~(15,46))$ we will have several ordered pairs $(c,d)$ where $c<d$
The number of possible values for $c$ is half the number of factors of $690$ which is $\frac{1}{2}\cdot2\cdot2\cdot2\cdot2=8$
However, we have one extraneous case of $(1,690)$ because here, $a=b$ and we have the sum of one consecutive number which is not allowed by the question.
Thus the answer is $8-1=7$
$\boxed{7}$

Solution:
The consecutive sums can be written as $345=kn+\sum_{i=1}^{k-1}{i}$ where $k$ is the number of terms in a sequence, and $n$ is the first term. Then, $\{k,n\}\in \mathbb{N}$ and $k\geq2$ . Evaluating the sum and rearranging yields $n=\frac{345}{k}-\frac{k-1}{2}$
The prime factorization of $345$ is $1\cdot3\cdot5\cdot23$ . Then, $3\cdot5$ $3\cdot23$ , and $5\cdot23$ are also divisors.  As odd divisors of $345$ , note that they all produce integer solutions to $n$ as $k$ . Only $k!=1$ is not valid, as $k\geq2$ . Similarly, quickly notice that $k=2$ is a solution. Multiplying an odd divisor by $2$ always yields an integer solution (see below). As such, the even integer solutions are $k=2, 2\cdot3, 2\cdot5, 2\cdot15 ... 2\cdot345$
Note that the function is decreasing for increasing values of $k$ and that $\frac{345}{k}-\frac{k-1}{2}=4$ for $k=23$ . Thus, when $n$ is negative, $k$ is only slightly more than $k=23$ . Recall $\{k,n\}\in \mathbb{N}$ . Since the next highest solution, $k=2\cdot23=46$ is twice $k=23$ $k\leq23$ . Thus, the remaining solutions are when $k=2,6,10,3,5,15,23$ $\implies \boxed{7}$

Solution:
We're dealing with an increasing arithmetic progression of common difference 1. Let $x$ be the number of terms in a summation. Let $y$ be the first term in a summation. The sum of an arithmetic progression is the average of the first term and the last term multiplied by the number of terms. The problem tells us that the sum must be 345.
\begin{align*} x \cdot \frac{y+y+[(x-1)1]}{2}&=345 \\ 2xy+x^2-x&=690 \end{align*}
In order to satisfy the constraints of the problem, x and y must be positive integers. Maybe we can make this into a Diophantine thing! In fact, if we just factor out that $x$ ... voilà!
\[(x)(x+2y-1)=690\]
There are 16 possible factor pairs to try (for brevity, I will not enumerate them here). Notice that the expression in the right parenthesis is $2y-1$ more than the expression in the parenthesis on the left. $y$ is at least 1. Thus, the expression in the right parenthesis will always be greater than the expression on the left. This eliminates 8 factor pairs. The problem also says the "increasing sequence" has to have "two or more" terms, so $x \geqslant 2$ . This eliminates the factor pair $1 \cdot 690$ . With brief testing, we find that the the other 7 factor pairs produce 7 viable ordered pairs. This means we have found $\boxed{7}$ ways to write 345 in the silly way outlined by the problem.

Solution:
The sum of all integers from $x$ to $y$ inclusive is equal to
\[\frac{y(y+1)}{2}-\left(\frac{x(x+1)}{2}\right)+x.\]
In words, $x$ added to sum of the first $x$ positive inegers subtracted from the sum of the first positive $y$ integers.
Setting this equal to $345$ and multiplying by $2$ to clear fractions, we can see that \[y(y+1)-x(x+1)+2x\] \[=y^2-x^2+y+x\] \[=(y+x)(y-x+1)=690.\] Now we know that, $y+x$ and $y-x+1$ multiply together to $690$ , and they're both integers. Now, we can set $y+x$ and $y-x+1$ equal to factors of $690$ . Clearly, $x>\frac{1}{2},$ so $y+x$ will take the larger of the two factors.
We can factorize $690$ as $1\cdot2\cdot3\cdot5\cdot23$
We also know that when solving for $y$ , we can add the two systems in $y+x$ and $y-x+1$ to get a new equation in terms of $2y+1.$ In order for $y$ to be an integer, the sum of the two factors must be odd.
We also know that there is only one factor of $2$ , so either $y+x$ or $y-x+1$ will be even. The number of odd factors multiplied together can either be $1, 2$ or $3$ (0 doesn't work, since $(690, 1)$ doesn't). There are a total of $3$ odd numbers present in our new prime factorization, now excluding $1$
Therefore, the number of combinations that yield integral values for both $x$ and $y$ are \[{3\choose1}+{3\choose2}+{3\choose3}= \boxed{7}\]


Answer: 7.0

Solution:
By the sum of an arithmetic sequence... this ultimately comes to $n+n+1+n+2....+n+p=345=(2n+p)(p+1)=690=23\cdot3\cdot5\cdot2$
Quick testing (would take you roughly a minute)
We see that the first 7 values of $p$ that work are
$p=1,2,4,5,9,14,22$
We see that each one of them works.
Hence, the answer is $\boxed{7}$


Answer: 0.0

In how many ways can $47$ be written as the sum of two primes
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{more than 3}$

Solution:
For $47$ to be written as the sum of two integers , one must be odd and the other must be even.  There is only one even prime, namely $2$ , so one of the numbers must be $2$ , making the other $45$
However, $45$ is not prime, so there are no ways to write $47$ as the sum of two primes $\rightarrow \boxed{0}$


Answer: 24.0

In how many ways can the letters in $\textbf{BEEKEEPER}$ be rearranged so that two or more $\textbf{E}$ s do not appear together?
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120$

Solution: All valid arrangements of the letters must be of the form \[\textbf{E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E}.\] The problem is equivalent to counting the arrangements of $\textbf{B},\textbf{K},\textbf{P},$ and $\textbf{R}$ into the four blanks, in which there are $4!=\boxed{24}$ ways.


Answer: 32.0

In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$

Solution:
We write out the $5!=120$ cases, then filter out the valid ones:
$13254,14253,14352,15243,15342,21435,21534,23154,24153,24351,25143,25341,\linebreak  31425,31524,32415,32514,34152,34251,35142,35241,41325,41523,42315,42513,\linebreak 43512,45132,45231,51324,51423,52314,52413,53412.$
We count these out and get $\boxed{32}$ permutations that work.

Solution:
By symmetry with respect to $3,$ note that $(x_1,x_2,x_3,x_4,x_5)$ is a valid sequence if and only if $(6-x_1,6-x_2,6-x_3,6-x_4,6-x_5)$ is a valid sequence. We enumerate the valid sequences that start with $1,2,31,$ or $32,$ as shown below:

There are $16$ valid sequences that start with $1,2,31,$ or $32.$ By symmetry, there are $16$ valid sequences that start with $5,4,35,$ or $34.$ So, the answer is $16+16=\boxed{32}.$

Solution:
Reading the terms from left to right, we have two cases for the consecutive digits, where $+$ means increase and $-$ means decrease:
$\textbf{Case \#1: }\boldsymbol{+,-,+,-}$
$\textbf{Case \#2: }\boldsymbol{-,+,-,+}$
For $\text{Case \#1},$ note that for the second and fourth terms, one term must be $5,$ and the other term must be either $3$ or $4.$ We have four subcases:
$(1) \ \underline{\hspace{3mm}}3\underline{\hspace{3mm}}5\underline{\hspace{3mm}}$
$(2) \ \underline{\hspace{3mm}}5\underline{\hspace{3mm}}3\underline{\hspace{3mm}}$
$(3) \ \underline{\hspace{3mm}}4\underline{\hspace{3mm}}5\underline{\hspace{3mm}}$
$(4) \ \underline{\hspace{3mm}}5\underline{\hspace{3mm}}4\underline{\hspace{3mm}}$
For $(1),$ the first two blanks must be $1$ and $2$ in some order, and the last blank must be $4.$ So, we get $2$ possibilities. Similarly, $(2)$ also has $2$ possibilities.
For $(3),$ there are no restrictions for the numbers $1, 2,$ and $3.$ So, we get $3!=6$ possibilities. Similarly, $(4)$ also has $6$ possibilities.
Together, $\text{Case \#1}$ has $2+2+6+6=16$ possibilities. By symmetry, $\text{Case \#2}$ also has $16$ possibilities.
Finally, the answer is $16+16=\boxed{32}.$

Solution:
Like Solution 3, we have two cases. Due to symmetry, we just need to count one of the cases. For the purpose of this solution, we will be doing $-,+,-,+$ . Instead of starting with 5, we start with 1.
There are two ways to place it:
_1_ _ _
_ _ _1_
Now we place 2, it can either be next to 1 and on the outside, or is place in where 1 would go in the other case. So now we have another two "sub case":
_1_2_(case 1)
21_ _ _(case 2)
There are 3! ways to arrange the rest for case 1, since there is no restriction.
For case 2, we need to consider how many ways to arrange 3,4,5 in a a>b<c fashion. It should seem pretty obvious that b has to be 3, so there will be 2! way to put 4 and 5.
Now we find our result, times 2 for symmetry, times 2 for placement of 1 and times (3!+2!) for the two different cases for placement of 2. This give us $2*2*(3!+2!)=4*(6+2)=\boxed{32}$

Solution:
We only need to find the # of rearrangements when 5 is the 4th digit and 5th digit. Find the total, and multiply by 2. Then we can get the answer by adding the case when 5 is the third digit.
Case $1$ 5 is the 5th digit. __ __ __ __ 5
Then $4$ can only be either 1st digit or the 3rd digit.
4 __ __ __ 5, then the only way is that $3$ is the 3rd digit, so it can be either $231$ or $132$ , give us $2$ results.
__ __ 4 __ 5, then the 1st digit must be $2$ or $3$ $2$ gives us $1$ way, and $3$ gives us $2$ ways. (Can't be $1$ because the first digit would increasing). Therefore, $4$ in the middle and $5$ in the last would result in $3$ ways.
Case $2$ $5$ is the fourth digit. __ __ __ 5 __
Then the last digit can be all of the 4 numbers $1$ $2$ $3$ , and $4$ . Let's say if the last digit is $4$ , then the 2nd digit would be the largest for the remaining digits to prevent increasing order or decreasing order. Then the remaining two are interchangeable, give us $2!$ ways. All of the $4$ can work, so case $2$ would result in $2!+2!+2!+2!=8$ ways.
Case $3$ $5$ is in the middle. __ __ 5 __ __
Then there are only two cases:
1. $42513$ , then 4 and 3 are interchangeable, which results in $2!*2!$ . Or it can be $43512$ , then 4 and 2 are interchangeable, but it can not be $23514$ , so there can only be 2 possible ways: $43512$ $21534$
Therefore, case 3 would result in $4+2=6$ ways.
$8+3+2=13$ , so the total ways for case 1 and case 2 with both increasing and decreasing would be $13*2=26.$
Finally, we have $26+6=\boxed{32}.$

Solution:
First, we list the triples that are invalid:
543, 542, 541, 532, 531, 521, 432, 431, 321
By symmetry, there are the same amount of increasing triplets as there are decreasing ones. This yields 18 invalid 3 digit permutations in total.
Suppose the triplet is ABC and the other 2 digits are X and Y. We then have 3 ways to arrange a triplet with 2 other digits.
ABCXY, XABCY, XYABC
X and Y can be arranged 2 ways.
XY, YX
This produces 18*3*2=108 permutations of invalid results. We have 5!
ways to arrange 5 numbers so 120-108=12.
Now, we must account for overcounting. For example, when 543 is counted,
it only registers as one invalid permutation but in fact, it is 3 whole
invalid permutations. We then complete this for the rest of the list:
54321 has 543, 432, and 321
54213 has 542 and 421
54123 has 541 and 123
53214 has 532 and 321
53124 has 531 and 124
52134 has 521 and 134
43215 has 432 and 321
43125 has 431 and 125
32145 has 321 and 145
This produces 19 values that we have overcounted but this value itself is also overcounted. We already counted 9 of the terms. This brings the final value of overcounted terms down to 10 for the decreasing triplets. By symmetry, 10 increasing triplets were overcounted.
This gives us $120-108+20=\boxed{32}.$

Solution:
It is easier to consider the complement of the desired cases, so try to find the cases that DO have three integers in increasing order. First, write down the sets of three numbers that feature the numbers in increasing order. They are 123, 124, 125, 134, 135, 145, 234, 245, 345. Each of these can be in three positions: the three are in the front with two more numbers in the back, the three are in the middle with two on each side, and two in the front and the set of three in the back. Now, count the number of combinations of 5 numbers that each of the set of three can form that have no been previously accounted for. Also, if the set features both 3 increasing and 3 decreasing, then do not count it because we will separately count them.
\[\begin{tabular}[t]{|l|c|c|c|} \hline Three Increasing Numbers&Front&Middle&Back\\\hline 123&2&2&2\\\hline 124&2&2&2\\\hline 125&1&2&2\\\hline 134&2&2&2\\\hline 135&1&2&2\\\hline 145&1&2&2\\\hline 234&2&1&1\\\hline 245&1&1&1\\\hline 345&1&0&1\\\hline \end{tabular}\]
This gives us a total of 41 possibilities. These account for the possibilities with ONLY increasing numbers. Mirror over to the other side to get the set of combinations with either at least 1 set of 3 increasing numbers in a row or only at least 1 set of 3 decreasing numbers, but not both. We can count the ones with both separately. 14532, 12543, 13542, 54123, 53124, 52134. Total of 6.
$41*2+6=82+6=88$ . This is the compliment. There are a total of $5!=120$ $120-88=32$
Thus, the answer is $\boxed{32}$

Solution:
First, note that there is a symmetry from $+,-,+,-$ to $-,+,-,+$ as follows: $a, b, c, d, e \leftrightarrow 6-a, 6-b, 6-c, 6-d, 6-e$ . Now, consider the placement of 5 in the $+,-,+,-$ case. Clearly, 5 is the maximum value, so it must be placed in the 2nd position or the 4th position, but we also have symmetry $a, b, c, d, e \leftrightarrow e, d, c, b, a$ . We will find the number of sequences with 5 in the second position by using casework on the position of the 4.
Case 1:
4 5 __ __ __: 3 is the maximum value, so it must go into the 4th position, leaving us 2 ways to fill in 1 and 2.
Case 2:
__ 5 __ 4 __: 1, 2, and 3 are all less than 4 or 5, so they can be filled in any manner: 3! = 6.
Overall, we have 8 ways to fill in the sequences. By our two types of symmetry, there are $8 \cdot 2 \cdot 2 = \boxed{32}$


Answer: 100.0

In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?
$\text{(A)}\ 50\text{ yrs.} \qquad \text{(B)}\ 75\text{ yrs.} \qquad \text{(C)}\ 100\text{ yrs.} \qquad \text{(D)}\ 125\text{ yrs.} \qquad \text{(E)}\ 150\text{ yrs.}$

Solution:
We can divide the total area by how much will be occupied per person:
$\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}$ can stay on the island at its maximum capacity.
We can divide 16600 by the current population in $1998$ which is 200 to see by what factor the population increases:
$\frac{16600}{200}=83$ -fold increase in population.
Thus, the population increases by a factor $83$ .  This is very close to $3 \times 3 \times 3 \times 3 = 81$ , and so there are about $4$ triplings of the island's population.
It takes $4\times25=100=\boxed{100}$ years to triple the island's population four times in succession.

Solution:
We can continue the pattern, and because the pattern increases numbers rapidly, it won't be hard. $\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}$ can live on the island at its maximum capacity.
$1998: 200 \text{ people}$ $2023: 600 \text{ people}$ $2048: 1800 \text{ people}$ $2073: 5400 \text{ people}$ $2098: 16200 \text{ people}$ After the year $2098$ , it will not be possible for the next increase to occur, because when $16200$ is tripled, it is way more than the maximum capacity. Thus, the answer is $2098-1998 = 100$ , or $\boxed{100}$


Answer: 242.0

In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$

Solution:
We have the following diagram:
Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$ , and $YZ$ be the height of $\triangle AZB$ . As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$ . We can apply this logic to triangles $BCQ$ and $XCQ$ as well, giving us $BC = CX = 333$ . Since $CD = 650$ $XW = DW + CX - CD = 16$
Additionally, we can see that $\triangle XZW$ is similar to $\triangle PQZ$ and $\triangle AZB$ . We know that $\frac{XW}{AB} = \frac{16}{500}$ . So, we can say that the height of the triangle $AZB$ is $500u$ while the height of the triangle $XZW$ is $16u$ . After that, we can figure out the distance from $Y$ to $PQ: \frac{500+16}{2} = 258u$ and the height of triangle $PZQ: 500-258 = 242u$
Finally, since the ratio between the height of $PZQ$ to the height of $AZB$ is $242:500$ and $AB$ is $500$ $PQ = \boxed{242}.$

Solution:
Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$ . The diagram looks like this:  Because the trapezoid is isosceles, by symmetry $PQ$ is parallel to $AB$ and $CD$ . Therefore, $\angle PAB \cong \angle APP'$ by interior angles and $\angle PAB \cong \angle PAD$ by the problem statement. Thus, $\triangle P'AP$ is isosceles with $P'P = P'A$ . By symmetry, $P'DP$ is also isosceles, and thus $P'A = \frac{AD}{2}$ . Similarly, the same thing is happening on the right side of the trapezoid, and thus $P'Q'$ is the midline of the trapezoid. Then, $PQ = P'Q' - (P'P + Q'Q)$
Since $P'P = P'A = \frac{AD}{2}, Q'Q = Q'B = \frac{BC}{2}$ and $AD = BC = 333$ , we have $P'P + Q'Q = \frac{333}{2} + \frac{333}{2} = 333$ . The length of the midline of a trapezoid is the average of their bases, so $P'Q' = \frac{500+650}{2} = 575$ . Finally, $PQ = 575 - 333 = \boxed{242}$

Solution:
We have the following diagram:  Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$ , respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$ , respectively.
Claim: quadrilaterals $AZWD$ and $BYXC$ are rhombuses.
Proof: Since $\angle DAB + \angle ADC = 180^{\circ}$ $\angle ADP + \angle PAD = 90^{\circ}$ . Therefore, triangles $APD$ $APZ$ $DPW$ and $PZW$ are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, $\triangle PZW$ is congruent to the other three. Therefore, $AD = DW = WZ = AZ$ , so $AZWD$ is a rhombus. By symmetry, $BYXC$ is also a rhombus.
Extend line $PQ$ to meet $\overline{AD}$ and $\overline{BC}$ at $R$ and $S$ , respectively. Because of rhombus properties, $RP = QS = \frac{333}{2}$ . Also, by rhombus properties, $R$ and $S$ are the midpoints of segments $AD$ and $BC$ , respectively; therefore, by trapezoid properties, $RS = \frac{AB + CD}{2} = 575$ . Finally, $PQ = RS - RP - QS = \boxed{242}$

Solution:
Let $X$ and $Y$ be the feet of the altitudes from $P$ and $Q$ , respectively, to $AB$ , and let $Z$ and $W$ be the feet of the altitudes from $P$ and $Q$ , respectively, to $CD$ . Side $AB$ is parallel to side $CD$ , so $XYWZ$ is a rectangle with width $PQ$ . Furthermore, because $CD - AB = 650-500 = 150$ and trapezoid $ABCD$ is isosceles, $WC - YB = ZD - XA = 75$
Also because $ABCD$ is isosceles, $\angle ABC + \angle BCD$ is half the total sum of angles in $ABCD$ , or $180^{\circ}$ . Since $BQ$ and $CQ$ bisect $\angle ABC$ and $\angle BCD$ , respectively, we have $\angle QBC + \angle QCB = 90^{\circ}$ , so $\angle BQC = 90^{\circ}$
Letting $BQ = 333k$ , applying Pythagoras to $\triangle BQC$ yields $QC = 333\sqrt{1-k^2}$ . We then proceed using similar triangles: $\angle BYQ = \angle BQC = 90^{\circ}$ and $\angle YBQ = \angle QBC$ , so by AA similarity $YB = 333k^2$ . Likewise, $\angle CWQ = \angle BQC = 90^{\circ}$ and $\angle WCQ = \angle QCB$ , so by AA similarity $WC = 333(1 - k^2)$ . Thus $WC + YB = 333$
Adding our two equations for $WC$ and $YB$ gives $2WC = 75 + 333 = 408$ . Therefore, the answer is $PQ = ZW = CD - 2WC = 650 - 408 = \boxed{242}$

Solution:
This will be my first solution on AoPS. My apologies in advance for any errors.
Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that $P$ is equidistant from $AB, AD,$ and $CD$ and $Q$ is equidistant from $AB, BC,$ and $CD.$ If we let the feet of the altitudes from $P$ to $AB, AD,$ and $CD$ be called $E, F,$ and $G$ respectively, we can say that $PE = PF = PG.$ Analogously, we let the feet of the altitudes from $Q$ to $AB, BC,$ and $CD$ be $H, I,$ and $J$ respectively. Thus, $QH = QI = QJ.$ Because $ABCD$ is an isosceles trapezoid, we can say that all of the altitudes are equal to each other.
By SA as well as SS congruence for right triangles, we find that triangles $AEP, AFP, BHQ,$ and $BIQ$ are congruent. Similarly, $DFP, DGP, CJQ,$ and $CIQ$ by the same reasoning. Additionally, $EH = GJ = PQ$ since $EHQP$ and $GJQP$ are congruent rectangles.
If we then let $x = AE = AF = BH = BI,$ let $y = CI = CJ = DG = DF,$ and let $z = EH = GJ = PQ,$ we can create the following system of equations with the given side length information: \begin{align*} 2x + z &= 500, \\ 2y + z &= 650, \\ x + y &= 333. \end{align*} Adding the first two equations, subtracting by twice the second, and dividing by $2$ yields $z = PQ = \boxed{242}.$

Solution: Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$ . The diagram looks like this:  Since $\angle A + \angle D=\angle B + \angle C = 180^{\circ}$ , it follows that $\angle P'AP+\angle P'DP = \angle Q'BQ + \angle Q'CQ = 90^{\circ}$ . Thus, $\angle APD = \angle BQC = 90^{\circ}$ , implying that $\triangle APD$ and $\triangle BQC$ are right triangles. Since $P'P$ and $Q'Q$ are medians, $P'P+Q'Q=\frac{333\times2}{2}=333$ . Since $P'Q'=\frac{500+650}{2}=575$ , we have $PQ+P'P+Q'Q=575$ , or $PQ=575-333=\boxed{242}$

Solution: Let $PQ = x$ . Note that since $AP$ bisects $\angle{A}$ and $DP$ bisects $\angle{D}$ , we have \[\angle{APD} = 180^{\circ}-\tfrac12 \angle{A}-\tfrac12 \angle{D}=90^{\circ}.\] Let $\angle{ADP}=\theta$ . We have that $\angle{ADC} = 2\theta.$ Now, drop an altitude from $A$ to $CD$ at $E$ . Notice that $DE=\tfrac{650-500}{2}=75$ . By the definition of cosine, we have \[\cos{2\theta}=1-2\cos^2{\theta}=\tfrac{75}{333}=\tfrac{25}{111} \implies \cos{\theta}=\tfrac{2\sqrt{1887}}{111}.\] Notice, however, that we can also apply this to $\triangle{APD}$ ; we have \[\cos{\theta}=\tfrac{DP}{333} \implies DP=6\sqrt{1887}.\] By the Pythagorean Theorem, we get \[AP=\sqrt{333^2-(6\sqrt{1887})^2}=3\sqrt{4773}.\] Then, drop an altitude from $P$ to $AB$ at $F$ ; if $AF=y$ , then $PQ=x=500-2y$ . Because $AP$ is an angle bisector, we see that $\angle{BAP}=\angle{DAP}=90^{\circ}-\theta$ . Again, by the definition of cosine, we have \[\cos{(90^{\circ}-\theta)}=\sin{\theta}=\tfrac{\sqrt{4773}}{111}=\tfrac{y}{3\sqrt{4773}} \implies y=129.\] Finally, $PQ=500-2y=\boxed{242}$

Solution:
As in solution 4, $\angle APD = 90^{\circ}$ . Set $k = AX$ and $x = DP$
We know that $DZ = AX + \frac{DC-AB}{2}$ , so $DZ = k + \frac{650-500}{2} = k + 75$
$\triangle DPZ \sim \triangle APD$ by AA, so we have $\frac{PD}{AD} = \frac{ZD}{PD}$ , resulting in
\[\frac{x}{333} = \frac{k+75}{x} \text{ (1)}\]
$\triangle APX \sim \triangle ADP$ by AA, so we have $\frac{AP}{AD} = \frac{AX}{AP}$ , resulting in
\[\frac{\sqrt{333^2-x^2}}{333} = \frac{k}{\sqrt{333^2-k^2}} \text{ (2)}\]
From $\text{(1)}$ , we have $x^2 = 333k + 333(75) = 333k + 24975$ . From $\text{(2)}$ , we have $333^2 - x^2 = 333k$ , or $x^2 = 333^2 - 333k$ . Thus, $333k + 24975 = 333^2 - 333k$ . Solving for $k$ yields $k = 129$
By symmetry, $YB = AX = 129$ . Thus, $PQ = XY = AB - 2AX = 500 - 2(129) = \boxed{242}$


Answer: 875.0

In isosceles triangle $\triangle ABC$ $A$ is located at the origin and $B$ is located at $(20,0)$ .  Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$ .  If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$ , where $p,q,r,s$ are integers.  Find $\frac{p-q+r-s}2$

Solution:
Redefine the points in the same manner as the last time ( $\triangle AB'C'$ , intersect at $D$ $E$ , and $F$ ). This time, notice that $[ADEF] = [\triangle AB'C'] - ([\triangle ADC'] + [\triangle EFB'])$
The area of $[\triangle AB'C'] = [\triangle ABC]$ . The altitude of $\triangle ABC$ is clearly $10 \tan 75 = 10 \tan (30 + 45)$ . The tangent addition rule yields $10(2 + \sqrt{3})$ (see above). Thus, $[\triangle ABC] = \frac{1}{2} 20 \cdot (20 + 10\sqrt{3}) = 200 + 100\sqrt{3}$
The area of $[\triangle ADC']$ (with a side on the y-axis) can be found by splitting it into two triangles, $30-60-90$ and $15-75-90$ right triangles $AC' = AC = \frac{10}{\sin 15}$ . The sine subtraction rule shows that $\frac{10}{\sin 15} = \frac{10}{\frac{\sqrt{6} - \sqrt{2}}4} = \frac{40}{\sqrt{6} - \sqrt{2}} = 10(\sqrt{6} + \sqrt{2})$ $AC'$ , in terms of the height of $\triangle ADC'$ , is equal to $h(\sqrt{3} + \tan 75) = h(\sqrt{3} + 2 + \sqrt{3})$ \begin{align*} [ADC'] &= \frac 12 AC' \cdot h \\ &= \frac 12 (10\sqrt{6} + 10\sqrt{2})\left(\frac{10\sqrt{6} + 10\sqrt{2}}{2\sqrt{3} + 2}\right) \\ &= \frac{(800 + 400\sqrt{3})}{(2 + \sqrt{3})}\cdot\frac{2 - \sqrt{3}}{2-\sqrt{3}} \\ &= \frac{400\sqrt{3} + 400}8 = 50\sqrt{3} + 50 \end{align*}
The area of $[\triangle EFB']$ was found in the previous solution to be $- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750$
Therefore, $[ADEF]$ $= (200 + 100\sqrt{3}) - \left((50 + 50\sqrt{3}) + (-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750)\right)$ $= 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$ , and our answer is $\boxed{875}$

Solution:
Call the points of the intersections of the triangles $D$ $E$ , and $F$ as noted in the diagram (the points are different from those in the diagram for solution 1). $\overline{AD}$ bisects $\angle EDE'$
Through HL congruency, we can find that $\triangle AED$ is congruent to $\triangle AE'D$ . This divides the region $AEDF$ (which we are trying to solve for) into two congruent triangles and an isosceles right triangle
Since $FE' = AE' = AE$ , we find that $[AE'F] = \frac 12 (5\sqrt{6} + 5\sqrt{2})^2 = 100 + 50\sqrt{3}$
Now, we need to find $[AED] = [AE'D]$ . The acute angles of the triangles are $\frac{15}{2}$ and $90 - \frac{15}{2}$ . By repeated application of the half-angle formula , we can find that $\tan \frac{15}{2} = \sqrt{2} - \sqrt{3} + \sqrt{6} - 2$
The area of $[AED] = \frac 12 \left(20 \cos 15\right)^2 \left(\tan \frac{15}{2}\right)$ . Thus, $[AED] + [AE'D] = 2\left(\frac 12((5\sqrt{6} + 5\sqrt{2})^2 \cdot (\sqrt{2} - \sqrt{3} + \sqrt{6} - 2))\right)$ , which eventually simplifies to $500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$
Adding them together, we find that the solution is $[AEDF] = [AE'F] + [AED] + [AE'D]$ $= 100 + 50\sqrt{3} + 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600=$ $= 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$ , and the answer is $\boxed{875}$

Solution:
From the given information, calculate the coordinates of all the points (by finding the equations of the lines, equating). Then, use the shoelace method to calculate the area of the intersection.
Now, we can equate the equations to find the intersections of all the points.
We take these points and tie them together by shoelace, and the answer should come out to be $\boxed{875}$


Answer: 224.0

In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161$ . What is the correct value of the product of $a$ and $b$
$\textbf{(A)}\ 116 \qquad\textbf{(B)}\  161 \qquad\textbf{(C)}\  204 \qquad\textbf{(D)}\ 214 \qquad\textbf{(E)}\ 224$

Solution: We have $161 = 7 \cdot 23.$ Since $a$ has two digits, the factors must be $23$ and $7,$ so $a = 32$ and $b = 7.$ Then, $ab = 7 \times 32 = \boxed{224}.$


Answer: 224.0

In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161.$ What is the correct value of the product of $a$ and $b$
$\textbf{(A)}\ 116 \qquad \textbf{(B)}\ 161 \qquad \textbf{(C)}\ 204 \qquad \textbf{(D)}\ 214 \qquad \textbf{(E)}\ 224$

Solution:
Taking the prime factorization of $161$ reveals that it is equal to $23*7.$ Therefore, the only ways to represent $161$ as a product of two positive integers is $161*1$ and $23*7.$ Because neither $161$ nor $1$ is a two-digit number, we know that $a$ and $b$ are $23$ and $7.$ Because $23$ is a two-digit number, we know that a, with its two digits reversed, gives $23.$ Therefore, $a = 32$ and $b = 7.$ Multiplying our two correct values of $a$ and $b$ yields
\[a*b = 32*7 =\]
\[= \boxed{224}\]


Answer: 400.0

In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
$\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000$

Solution:
The length $L$ of the rectangle is $\frac{1000}{25}=40$ meters. The perimeter $P$ is $\frac{1000}{10}=100$ meters. Since $P_{rect} = 2L + 2W$ , we plug values in to get:
$100 = 2\cdot 40 + 2W$
$100 = 80 + 2W$
$2W = 20$
$W = 10$ meters
Since $A_{rect} = LW$ , the area is $40\cdot 10=400$ square meters or $\boxed{400}$


Answer: 766.0

In order to complete a large job, $1000$ workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then $100$ workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional $100$ workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the $800$ workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

Solution: A train is traveling at $1000$ miles per hour and has one hour to reach its destination $1000$ miles away. After $15$ minutes and $250$ miles it slows to $900$ mph, and thus takes $\frac{250}{900}(60)=\frac{50}{3}$ minutes to travel the next $250$ miles. Then it slows to $800$ mph, so the next quarter takes $\frac{250}{800}(60)=\frac{150}{8}$ . The train then has $60-15-\frac{50}{3}-\frac{150}{8}=230/24$ minutes left to travel 250 miles, and doing the arithmetic shows that during this last stretch it must travel more than $1565$ mph to make it on time. Therefore the company needs to add $1566-800 = \boxed{766}$ more workers.

Solution:
Let each worker's speed be $w$ , the entire time be $t$ , and the total work be $1$
From the initial problem statement, we have $1000w\cdot\frac{1}{4}t=\frac{1}{4}$
We need to find the time the next quarter takes to complete the same amount of work, which is $\frac{1000}{900}\cdot\frac{1}{4}\cdot t=\frac{5}{18}t$
Similarly, we find that the time the third quarter takes is $\frac{1000}{800}\cdot\frac{1}{4}\cdot t=\frac{5}{16}t$
Finally, the time the last quarter takes is $t\left[1-\left(\frac{1}{4}+\frac{5}{18}+\frac{5}{16}\right)\right]=\frac{23}{144}t$
We then let the number of workers needed be $x$ , so we have the equation $\left(800+x\right)w\cdot\frac{23}{144}t=\frac{1}{4}$ . Dividing by the first equation, we have
\[\frac{800+x}{1000}\cdot\frac{\frac{23}{144}}{\frac{1}{4}}=1\] \[\frac{800+x}{1000}\cdot\frac{23}{36}=1\] \[800+x=\frac{36000}{23}\] \[x=\frac{17600}{23}.\]
We can't have a part of a worker, so we take the ceiling of $x$ , which we find to be $\boxed{766}$

Solution: Suppose $1000$ workers can complete one quarter of the job in one day. After the first day, there were $900$ workers remaining so the second quarter was completed in $\frac{10}{9}$ days. Now there are only $800$ workers remaining so the third quarter can be completed in $\frac{10}{8}$ days. It has been $1+\frac{10}{9}+\frac{10}{8}=\frac{121}{36}$ days since the job started, and we still have $\frac{23}{36}$ days to complete the last quarter of the job, so we need $\frac{36}{23}$ of $1000$ workers, or $1565.2173913$ . Since we can’t hire fractional workers, $1566-800=\boxed{766}$ must be the answer.


Answer: 5.0

In order to pass $B$ going $40$ mph on a two-lane highway, $A$ , going $50$ mph, must gain $30$ feet.
Meantime, $C, 210$ feet from $A$ , is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds,
then, in order to pass safely, $A$ must increase his speed by:
$\textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph}$

Solution: Let $V_A, V_B, V_C$ be $A, B, C$ 's velocity, respectively. We want to pass $B$ before we collide with $C$ . Since $A$ and $B$ are going in the same direction and $V_A>V_B$ $A$ will pass $B$ in $\frac{30\mathrm{ft}}{V_A-V_B}$ time. Since $A$ and $C$ are going in opposite directions, their relative velocity is $V_A+V_C$ , so the amount of time before $A$ will collide with $C$ is given by $\frac{210\mathrm{ft}}{V_A+V_C}$ . We want to pass $B$ before we collide with $C$ , so $V_A$ must satisfy the inequality $\frac{30\mathrm{ft}}{V_A-V_B}<\frac{210\mathrm{ft}}{V_A+V_C}$ . We can eliminate all the units, simplifying the inequality to $\frac{1}{V_A-V_B} < \frac{7}{V_A+V_C}$ . Solving this and substituting our known values of $V_B$ and $V_C$ yields $330<6V_A$ , so $A$ must increase his speed by $\boxed{5}$ miles per hour.


Answer: 110.0

In our number system the base is ten. If the base were changed to four you would count as follows: $1,2,3,10,11,12,13,20,21,22,23,30,\ldots$ The twentieth number would be:
$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 38 \qquad\textbf{(C)}\ 44 \qquad\textbf{(D)}\ 104 \qquad\textbf{(E)}\ 110$

Solution:
The $20^{\text{th}}$ number will be the value of $20_{10}$ in base $4$ . Thus, we see \[20_{10} = (1) \cdot 4^2 + (1) \cdot 4^1 + 0 \cdot 4^0\] \[= 110_{4}\]
$\boxed{110}$


Answer: 777.0

In parallelogram $ABCD$ , let $O$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}$ . Angles $CAB$ and $DBC$ are each twice as large as angle $DBA$ , and angle $ACB$ is $r$ times as large as angle $AOB$ . Find $\lfloor 1000r \rfloor$

Solution:
Let $\theta = \angle DBA$ . Then $\angle CAB = \angle DBC = 2\theta$ $\angle AOB = 180 - 3\theta$ , and $\angle ACB = 180 - 5\theta$ . Since $ABCD$ is a parallelogram, it follows that $OA = OC$ . By the Law of Sines on $\triangle ABO,\, \triangle BCO$
Dividing the two equalities yields
\[\frac{\sin 2\theta}{\sin \theta} = \frac{\sin (180 - 5\theta)}{\sin 2\theta} \Longrightarrow \sin^2 2\theta = \sin 5\theta \sin \theta.\]
Pythagorean and product-to-sum identities yield
\[1 - \cos^2 2 \theta = \frac{\cos 4\theta - \cos 6 \theta}{2},\]
and the double and triple angle ( $\cos 3x = 4\cos^3 x - 3\cos x$ ) formulas further simplify this to
\[4\cos^3 2\theta - 4\cos^2 2\theta - 3\cos 2\theta + 3 = (4\cos^2 2\theta - 3)(\cos 2\theta - 1) = 0\]
The only value of $\theta$ that fits in this context comes from $4 \cos^2 2\theta - 3 = 0 \Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}$ . The answer is $\lfloor 1000r \rfloor = \left\lfloor 1000 \cdot \frac{180 - 5\theta}{180 - 3\theta} \right\rfloor = \left \lfloor \frac{7000}{9} \right \rfloor = \boxed{777}$


Answer: 177.0

In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$

Solution:
One of the ways to solve this problem is to make this parallelogram a straight line.
So the whole length of the line is $APC$ $AMC$ or $ANC$ ), and $ABC$ is $1000x+2009x=3009x.$
$AP$ $AM$ or $AN$ ) is $17x.$
So the answer is $3009x/17x = \boxed{177}$

Solution: Draw a diagram with all the given points and lines involved. Construct parallel lines $\overline{DF_2F_1}$ and $\overline{BB_1B_2}$ to $\overline{MN}$ , where for the lines the endpoints are on $\overline{AM}$ and $\overline{AN}$ , respectively, and each point refers to an intersection. Also, draw the median of quadrilateral $BB_2DF_1$ $\overline{E_1E_2E_3}$ where the points are in order from top to bottom. Clearly, by similar triangles, $BB_2 = \frac {1000}{17}MN$ and $DF_1 = \frac {2009}{17}MN$ . It is not difficult to see that $E_2$ is the center of quadrilateral $ABCD$ and thus the midpoint of $\overline{AC}$ as well as the midpoint of $\overline{B_1}{F_2}$ (all of this is easily proven with symmetry). From more triangle similarity, $E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP$ $= \boxed{177}AP$

Solution: Assume, for the ease of computation, that $AM=AN=17$ $AB=1000$ , and $AD=2009$ .  Now, let line $MN$ intersect line $CD$ at point $X$ and let $Y$ be a point such that $XY\parallel AD$ and $AY\parallel DX$ .  As a result, $ADXY$ is a parallelogram.  By construction, $\triangle MAN\sim \triangle MYX$ so \[\frac{MY}{MA}=\frac{YX}{AN}=\frac{AD}{AN}=\frac{2009}{17}\implies MY=2009\] and $AY=DX=2009-17$ .  Also, because $AM\parallel XC$ , we have $\triangle PAM\sim \triangle PCX$ so \[\frac{PC}{PA}=\frac{CX}{AM}=\frac{DX+CD}{AM}=\frac{2009-17+1000}{17}=176.\] Hence, $\frac{AC}{AP}=\frac{PC}{PA}+1=\boxed{177}.$

Solution:
Assign $A = (0,0)$ . Since there are no constraints in the problem against this, assume $ABCD$ to be a rectangle with dimensions $1000 \times 2009.$ Now, we can assign \[A=(0,0)\] \[B=(1000, 0)\] \[C=(1000,-2009)\] \[D=(0,-2009).\]
Then, since $\frac{AM}{AB} = \frac{17}{1000}$ and $AB = 1000$ , we can place $M$ at $(17, 0).$ Similarly, place $AN$ at $(0, 17).$ Then, the equation of line $MN$ is $y=x-17,$ and the equation of $AC$ is $y=\frac{-2009}{1000}x.$ Solve to find point $P$ at $\left( \frac{1000}{177}, \frac{-2009}{177} \right)$
We can calculate vectors to represent the distances: \[\overrightarrow{AC}= <1000, -2009>\] \[\overrightarrow{AP}= \frac{1}{177}<1000, -2009>.\] In this way, we can see that \[AC:AP = 177:1,\] and our answer is $\boxed{177}.$


Answer: 13.0

In quadrilateral $ABCD$ $AB = 5$ $BC = 17$ $CD = 5$ $DA = 9$ , and $BD$ is an integer. What is $BD$
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

Solution:
By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$ , and also $BD + CD > BC$ , hence $BD > BC - CD = 17 - 5 = 12$
We get that $12 < BD < 14$ , and as we know that $BD$ is an integer, we must have $BD=\boxed{13}$


Answer: 13.0

Solution:
By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$ , and also $BD + CD > BC$ , hence $BD > BC - CD = 17 - 5 = 12$
We get that $12 < BD < 14$ , and as we know that $BD$ is an integer, we must have $BD=\boxed{13}$


Answer: 84.0

In quadrilateral $ABCD$ $\angle B$ is a right angle diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$ $AB=18$ $BC=21$ , and $CD=14$ . Find the perimeter of $ABCD$

Solution:
From the problem statement, we construct the following diagram:
Using the Pythagorean Theorem
Substituting $(AB)^2 + (BC)^2$ for $(AC)^2$
Plugging in the given information:
So the perimeter is $18+21+14+31=84$ , and the answer is $\boxed{084}$


Answer: 69.0

In quadrilateral $ABCD$ $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ $AB = 8$ $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
Extend $\overline{AD}$ and $\overline{BC}$ to meet at $E$ . Then, since $\angle BAD = \angle ADC$ and $\angle ABD = \angle DCE$ , we know that $\triangle ABD \sim \triangle DCE$ . Hence $\angle ADB = \angle DEC$ , and $\triangle BDE$ is isosceles . Then $BD = BE = 10$
Using the similarity, we have:
\[\frac{AB}{BD} = \frac 8{10} = \frac{CD}{CE} = \frac{CD}{16} \Longrightarrow CD = \frac{64}5\]
The answer is $m+n = \boxed{069}$

Solution: Draw a line from $B$ , parallel to $\overline{AD}$ , and let it meet $\overline{CD}$ at $M$ . Note that $\triangle{DAB}$ is similar to $\triangle{BMC}$ by AA similarity, since $\angle{ABD}=\angle{MCB}$ and since $BM$ is parallel to $CD$ then $\angle{BMC}=\angle{ADM}=\angle{DAB}$ . Now since $ADMB$ is an isosceles trapezoid, $MD=8$ . By the similarity, we have $MC=AB\cdot \frac{BC}{BD}=8\cdot \frac{6}{10}=\frac{24}{5}$ , hence $CD=MC+MD=\frac{24}{5}+8=\frac{64}{5}\implies 64+5=\boxed{069}$

Solution: Since $\angle{BAD}=\angle{ADM}$ , if we extend AB and DC, they must meet at one point to form a isosceles triangle $\triangle{ADM}$ .Now, since the problem told that $\angle{ABD}=\angle{BCD}$ , we can imply that $\angle{DBM}=\angle{BCM}$ Since $\angle{M}=\angle{M}$ , so $\triangle{CBM}\sim\triangle{BDM}$ . Assume the length of $BM=x$ ;Since $\frac{BC}{MB}=\frac{DB}{MD}$ we can get $\frac{6}{x}=\frac{10}{8+x}$ , we get that $x=12$ .So $AM=DM=20$ similarly, we use the same pair of similar triangle we get $\frac{CM}{BM}=\frac{BM}{DM}$ , we get that $CM=\frac{36}{5}$ . Finally, $CD=MD-MC=\frac{64}{5}\implies 64+5=69=\boxed{069}$ ~bluesoul

Solution: Denote $\angle{BAD}=\angle{CDA}=x$ , and $\angle{ABD}=\angle{BCD}=y$ . Note that $\angle{ADB}=180^\circ-x-y$ , and $\angle{DBC}=360^\circ-2x-2y$ . This motivates us to draw the angle bisector of $\angle{DBC}$ because $\angle{DBC} = 2 \angle{ADB}$ , so we do so and consider the intersection with $CD$ as $E$ . By the angle bisector theorem, we have $\frac{CE}{DE} = \frac{BC}{BD} = \frac{3}{5}$ , so we write $CE=3z$ and $DE=5z$ . We also know that $\angle{EBC}=\angle{ADB}$ and $\angle{BCE}=\angle{DBA}$ , so $\triangle{ADB} \sim \triangle{EBC}$ . Hence, $\frac{CE}{BC}=\frac{AB}{BD}$ , so we have $3z=\frac{24}{5}$ . As $CD=8z$ , it must be that $CD=\frac{64}{5}$ , so the final answer is $\boxed{069}$


Answer: 17.0

In quadrilateral $ABCD$ , sides $\overline{AB}$ and $\overline{BC}$ both have length 10, sides $\overline{CD}$ and $\overline{DA}$ both have length 17, and the measure of angle $ADC$ is $60^\circ$ . What is the length of diagonal $\overline{AC}$
[asy]draw((0,0)--(17,0)); draw(rotate(301, (17,0))*(0,0)--(17,0)); picture p; draw(p, (0,0)--(0,10)); draw(p, rotate(115, (0,10))*(0,0)--(0,10)); add(rotate(3)*p);  draw((0,0)--(8.25,14.5), linetype("8 8"));  label("$A$", (8.25, 14.5), N); label("$B$", (-0.25, 10), W); label("$C$", (0,0), SW); label("$D$", (17, 0), E);[/asy]
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5$

Solution: Because $\overline{AD} = \overline{CD}$ $\triangle ADC$ is an isosceles triangle with $\angle DAC = \angle DCA$ . Angles in a triangle add up to $180^\circ$ , and since $\angle ADC=60^\circ$ , the other two angles are also $60^\circ$ , and $\triangle ADC$ is an equilateral triangle. Therefore $\overline{AC}=\overline{DA}=\boxed{17}$

Solution: We can divide $\overline{CD}$ in half and connect this point to A, dividing $\triangle ADC$ in half. This means the base will be $\frac{17}{2}$ and the hypotenuse will be 17. By using the Pythagorean's Theorem, we see that if the base and height are shared, the hypotenuse should be the same. This tells us that the length of $\overline{AC} = \boxed{17}$

Solution: Take an equilateral triangle with side length $17$ $\triangle ADC$ is congruent to this by $SAS$ , hence it is equilateral. The answer is $\boxed{17}$


Answer: 166.0

In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ , intersecting at $O$ $BO=4$ $OD = 6$ $AO=8$ $OC=3$ , and $AB=6$ . The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

Solution: After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of $AB, BD(BD = BO + OD)$ , but we want to find the value of $AD$ . We can apply stewart's theorem now, letting $m = 4, n = 6, AD = X, AB = 6$ , and we have $10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6$ , and we see that $x = \boxed{166}$

Solution:
 (Diagram not to scale)
Since $AO \cdot OC = BO \cdot OD$ $ABCD$ is cyclic through power of a point. From the given information, we see that $\triangle{AOB}\sim \triangle{DOC}$ and $\triangle{BOC} \sim \triangle{AOD}$ . Hence, we can find $CD=\frac{9}{2}$ and $AD=2 \cdot BC$ . Letting $BC$ be $x$ , we can use Ptolemy's to get \[6 \cdot \frac{9}{2} + 2x^2=10 \cdot 11 \implies x=\sqrt{\frac{83}{2}}\] Since we are solving for $AD=2x=2\cdot\sqrt{\frac{83}{2}}=\sqrt{4\cdot\frac{83}{2}} = \boxed{166}$

Solution: The solution says it all. Since $\angle AOD$ is supplementary to $\angle AOB$ $cos(\angle AOD) = cos(180^{\circ} - \angle AOB)=-cos(\angle AOB)$ . The law of cosines on $\triangle AOB$ gives us $cos(\angle AOB)=\frac {8^2+4^2-6^2}{(2)(8)(4)}=\frac {11}{16}$ . Again, we can use the law of cosines on $\triangle AOD$ , which gives us \[AD=\sqrt {8^2+6^2-2(8)(6)cos(\angle AOD)}\] \[=\sqrt {8^2+6^2-(2)(8)(6)cos(\angle 180^{\circ} - AOB)}\] \[=\sqrt {8^2+6^2+(2)(8)(6)cos(\angle AOB)}\] \[=\sqrt {8^2+6^2+(2)(8)(6)(\frac {11}{16})}\] \[=\sqrt{166}\] which gives us $\boxed{166}$


Answer: 150.0

In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers , find $p+q.$

Solution: Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$ . Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$ , and $BC=8$ and $EC=2$ . We are given that $DC=12$ . Since $\angle CED = 120^{\circ}$ , using Law of Cosines on $\bigtriangleup CED$ gives \[12^2=DE^2+4-2(2)(DE)(\cos 120^{\circ})\] which gives \[144-4=DE^2+2DE\] . Adding $1$ to both sides gives $141=(DE+1)^2$ , so $DE=\sqrt{141}-1$ $\bigtriangleup DAP$ and $\bigtriangleup EBQ$ are both $30-60-90$ , so $AP=5$ and $BQ=5$ $PQ=DE$ , and therefore $AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141} \rightarrow (p,q)=(9,141) \rightarrow \boxed{150}$

Solution: Draw the perpendiculars from $C$ and $D$ to $AB$ , labeling the intersection points as $E$ and $F$ . This forms 2 $30-60-90$ right triangles , so $AE = 5$ and $BF = 4$ . Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$ , we find another right triangle $\triangle DGC$ $DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$ . The Pythagorean Theorem yields that $GC^2 = 12^2 - \sqrt{3}^2 = 141$ , so $EF = GC = \sqrt{141}$ . Therefore, $AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{141}$ , and $p + q = \boxed{150}$


Answer: 36.0

In rectangle $ABCD$ $AB = 12$ and $BC = 10$ .  Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9$ $DF = 8$ $\overline{BE} \parallel \overline{DF}$ $\overline{EF} \parallel \overline{AB}$ , and line $BE$ intersects segment $\overline{AD}$ .  The length $EF$ can be expressed in the form $m \sqrt{n} - p$ , where $m$ $n$ , and $p$ are positive integers and $n$ is not divisible by the square of any prime.  Find $m + n + p$

Solution:
Let us call the point where $\overline{EF}$ intersects $\overline{AD}$ point $G$ , and the point where $\overline{EF}$ intersects $\overline{BC}$ point $H$ . Since angles $FHB$ and $EGA$ are both right angles, and angles $BEF$ and $DFE$ are congruent due to parallelism, right triangles $BHE$ and $DGF$ are similar. This implies that $\frac{BH}{GD} = \frac{9}{8}$ . Since $BC=10$ $BH+GD=BH+HC=BC=10$ . ( $HC$ is the same as $GD$ because they are opposite sides of a rectangle.) Now, we have a system:
$\frac{BH}{GD}=\frac{9}8$
$BH+GD=10$
Solving this system (easiest by substitution), we get that:
$BH=\frac{90}{17}$
$GD=\frac{80}{17}$
Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles:
$\sqrt{9^2-\left(\frac{90}{17}\right)^2}$ and $\sqrt{8^2-\left(\frac{80}{17}\right)^2}$
Notice that adding these two sides would give us twelve plus the overlap $EF$ . This means that:
$EF= \sqrt{9^2-\left(\frac{90}{17}\right)^2}+\sqrt{8^2-\left(\frac{80}{17}\right)^2}-12=3\sqrt{21}-12$
Since $21$ isn't divisible by any perfect square, our answer is:
$3+21+12=\boxed{36}$

Solution:
Extend lines $BE$ and $CD$ to meet at point $G$ .
Draw the altitude $GH$ from point $G$ to line $BA$ extended.
$GE=DF=8,$ $GB=17$
In right $\bigtriangleup GHB$ $GH=10$ $GB=17$ , thus by Pythagoras Theorem we have: $HB=\sqrt{17^2-10^2}=3\sqrt{21}$
$HA=EF=3\sqrt{21}-12$
Thus our answer is: $3+21+12=\boxed{36}$

Solution: We notice that since $\overline{BE}||\overline{DF}$ $\overline{EF}$ is the diagonal of a rectangle. Now, let us extend the width lines to intersect with $\overline{BE}$ and $\overline{DF}$ , respectively, to form this rectangle. Let us call the length of $\overline{EF}$ $d$ , the perpendicular distance between $\overline{EF}$ and $\overline{AD}$ $k$ , and the perpendicular distance between $\overline{EF}$ and $\overline{CD}$ $x$ . We now can begin the similar triangles. When drawing the diagram (where $E$ is closer to $\overline{AD}$ than $F$ ), we put the similar right triangles in the same position so that we can begin solving for our variables and finding ratios. Since all of these triangles are similar, we find that $\frac{dx}{8}=\frac{10d-dx}{9}$ , which solves for $x=\frac{80}{17}$ . Completing the same thing for $k$ , we see $\frac{12d-dk}{9}=\frac{dk+d^2}{8}$ , which solves for $k=\frac{96-9d}{17}$ . We are now ready to find both the length and the width of the rectangle. We find these to be $\frac{10d}{17}$ and $\frac{12d+d^2}{17}$ . Now let us use the Pythagorean to solve for $d$ . We square the length and the width and multiply both sides by 289 (the denominator of the LHS) to reach the equation $(12+d^2)^2+100d^2=289d^2$ . Expanding and simplifying, we find that $d^4+24d^3=45d^2$ . Dividing both sides by $d^2$ and moving everything to the LHS, we see that $d^2+24d-45=0$ . Applying the quadratic formula, $d=\frac{-24\pm \sqrt{756}}{2}=-12\pm \sqrt{189}$ . Since $d>0$ $d=\sqrt{189}-12=3\sqrt{21}-12$ . Finally, $m+n+p=3+21+12=\boxed{036}$ . -Gideontz


Answer: 16.0

In rectangle $ABCD$ $AB=1$ $BC=2$ , and points $E$ $F$ , and $G$ are midpoints of $\overline{BC}$ $\overline{CD}$ , and $\overline{AD}$ , respectively. Point $H$ is the midpoint of $\overline{GE}$ . What is the area of the shaded region?
[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1);  // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3);  draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey);  label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,E); label("$F$",F,S); label("$G$",G,W); label("$H$",H,N);  label("$\frac12$",(0.25,0),S); label("$\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); [/asy]
$\textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16$

Solution: Denote $D=(0,0)$ . Then $A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)$ . Let the intersection of $AF$ and $DH$ be $X$ , and the intersection of $BF$ and $CH$ be $Y$ . Then we want to find the coordinates of $X$ so we can find $XY$ . From our points, the slope of $AF$ is $\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4$ , and its $y$ -intercept is just $2$ . Thus the equation for $AF$ is $y = -4x + 2$ . We can also quickly find that the equation of $DH$ is $y = 2x$ . Setting the equations equal, we have $2x = -4x +2 \implies x = \frac13$ . Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\frac13$ , so $XY = 1 - 2 \cdot \frac13 = \frac13$ . Now the area of the kite is simply the product of the two diagonals over $2$ . Since the length $HF = 1$ , our answer is $\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{16}$

Solution: From the diagram in Solution 1, let $e$ be the height of $XHY$ and $f$ be the height of $XFY$ . It is clear that their sum is $1$ as they are parallel to $GD$ . Let $k$ be the ratio of the sides of the similar triangles $XFY$ and $AFB$ , which are similar because $XY$ is parallel to $AB$ and the triangles share angle $F$ . Then $k = f/2$ , as 2 is the height of $AFB$ . Since $XHY$ and $DHC$ are similar for the same reasons as $XFY$ and $AFB$ , the height of $XHY$ will be equal to the base, like in $DHC$ , making $XY = e$ . However, $XY$ is also the base of $XFY$ , so $k = e / AB$ where $AB = 1$ so $k = e$ . Subbing into $k = f/2$ gives a system of linear equations, $e + f = 1$ and $e = f/2$ . Solving yields $e = XY = 1/3$ and $f = \frac{2}{3}$ , and since the area of the kite is simply the product of the two diagonals over $2$ and $HF = 1$ , our answer is $\frac{\frac{1}{3} \cdot 1}{2} = \boxed{16}$


Answer: 141.0

In rectangle $ABCD$ $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$

Solution: From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$ , and $ABE$ are also right triangles. By $AA$ $\triangle FBA \sim \triangle BCA$ , and $\triangle FBA \sim \triangle ABE$ , so $\triangle ABE \sim \triangle BCA$ . This gives $\frac {AE}{AB}= \frac {AB}{BC}$ $AE=\frac{AD}{2}$ and $BC=AD$ , so $\frac {AD}{2AB}= \frac {AB}{AD}$ , or $(AD)^2=2(AB)^2$ , so $AD=AB \sqrt{2}$ , or $100 \sqrt{2}$ , so the answer is $\boxed{141}$

Solution: Let $x$ be the ratio of $BC$ to $AB$ . On the coordinate plane, plot $A=(0,0)$ $B=(100,0)$ $C=(100,100x)$ , and $D=(0,100x)$ . Then $E=(0,50x)$ . Furthermore, the slope of $\overline{AC}$ is $x$ and the slope of $\overline{BE}$ is $-x/2$ . They are perpendicular, so they multiply to $-1$ , that is, \[x\cdot-\frac{x}{2}=-1,\] which implies that $-x^2=-2$ or $x=\sqrt 2$ . Therefore $AD=100\sqrt 2\approx 141.42$ so $\lfloor AD\rfloor=\boxed{141}$

Solution:
Draw $CX$ and $EX$ to form a parallelogram $AEXC$ . Since $EX \parallel AC$ $\angle BEX=90^\circ$ by the problem statement, so $\triangle BEX$ is right.
Letting $AE=y$ , we have $BE=\sqrt{100^2+y^2}$ and $AC=EX=\sqrt{100^2+(2y)^2}$ . Since $CX=EA$ $\sqrt{100^2+y^2}^2+\sqrt{100^2+(2y)^2}=(3y)^2$ . Solving this, we have \[100^2+ 100^2 + y^2 + 4y^2 = 9y^2\] \[2\cdot 100^2 = 4y^2\] \[\frac{100^2}{2}=y^2\] \[\frac{100}{\sqrt{2}}=y\] \[\frac{100\sqrt{2}}{2}=y\] \[100\sqrt{2}=2y=AD\] , so the answer is $\boxed{141}$


Answer: 12.0

Solution: Using the triangle area formula for triangles: $A = \frac{bh}{2},$ where $A$ is the area, $b$ is the base, and $h$ is the height. This equation gives us $A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\boxed{12}$

Solution: A triangle with the same height and base as a rectangle is half of the rectangle's area.  This means that a triangle with half of the base of the rectangle and also the same height means its area is one quarter of the rectangle's area.  Therefore, we get $\frac{48}{4} =\boxed{12}$

Solution: We can find the area of the entire rectangle, DCBA to be $8 \cdot 6=48$ and find DCM area to be $\frac{6 \cdot 4}{2} = 12$ and BCA to be $\frac{6 \cdot 8}{2}=24$ $48-12-24=$ $\boxed{12}$

Solution: A triangle is half of a rectangle. So since M is the midpoint of DA, we can see that triangle DAC is half of the whole rectangle. And it is also true that triangle MAC is half of triangle DAC, so triangle MAC would just be 1/4 of the whole rectangle. Since the rectangle's area is 48, 1/4 of 48 would be 12. Which gives us the answer $\boxed{12}$

Solution: We can subtract the total areas of triangles DCM and ABC from the rectangle ABCD. For triangle DCM, the base is 4 and the height is 6, so we multiply 4 and 6, then divide by 2 to get 12. For triangle ABC, the base is 4 and the height is 8, so we multiply 4 and 8, then divide by 2 to get 24. We add 24 and 12 to get 36. Then, we calculate the area of rectangle ABCD, which is 48. We subtract 36 from 48, resulting in $\boxed{12}$


Answer: 20.0

In rectangle $ABCD$ $\overline{AB}=20$ and $\overline{BC}=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $\overline{AE}$
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$

Solution: Note that $\tan 15^\circ=2-\sqrt{3}=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3$ . (It is important to memorize the sin, cos, and tan values of $15^\circ$ and $75^\circ$ .) Therefore, we have $DE=10\sqrt 3$ . Since $ADE$ is a $30-60-90$ triangle, $AE=2 \cdot AD=2 \cdot 10=\boxed{20}$

Solution: Let $F$ be a point on line $\overline{CD}$ such that points $C$ and $F$ are distinct and that $\angle EBF = 15^\circ$ . By the angle bisector theorem, $\frac{BC}{BF} = \frac{CE}{EF}$ . Since $\triangle BFC$ is a $30-60-90$ right triangle, $CF = \frac{10\sqrt{3}}{3}$ and $BF = \frac{20\sqrt{3}}{3}$ . Additionally, \[CE + EF = CF = \frac{10\sqrt{3}}{3}\] Now, substituting in the obtained values, we get $\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{CE}{EF} \Rightarrow \frac{2\sqrt{3}}{3}CE = EF$ and $CE + EF = \frac{10\sqrt{3}}{3}$ . Substituting the first equation into the second yields $\frac{2\sqrt{3}}{3}CE + CE = \frac{10\sqrt{3}}{3} \Rightarrow CE = 20 - 10\sqrt{3}$ , so $DE = 10\sqrt{3}$ . Because $\triangle ADE$ is a $30-60-90$ triangle, $AE = \boxed{20}$

Solution: Reflect $\triangle{ECB}$ over line segment $\overline{CD}$ . Let the point $F$ be the point where the right angle is of our newly reflected triangle. By subtracting $90 - (15+15) = 60$ to find $\angle ABF$ , we see that $\triangle{ABC}$ is a $30-60-90$ right triangle. By using complementary angles once more, we can see that $\angle{EAD}$ is a $60^\circ$ angle, and we've found that $\triangle{EAD}$ is a $30-60-90$ right triangle. From here, we can use the $1-2-\sqrt{3}$ properties of a $30-60-90$ right triangle to see that $\overline{AE}=\boxed{20}.$

Solution:
Square.PNG
We are going to use pure Euclidian geometry to prove $AE=AB$
Reflect rectangle $ABCD$ along line $CD$ . Let the square be $ABFG$ as shown. Construct equilateral triangle $\triangle EFH$
Because $HF=EF$ $GF=BF=20$ , and $\angle GFH=\angle BFE=15^{\circ}$ $\triangle GFH\cong \triangle BFE$ by $SAS$
So, $GH=BE$ $GH=HE=HF$
Because $GH=HE=HF$ $\angle GHF= \angle BEF=75^{\circ} + 75^{\circ} = 150^{\circ}$ $\angle GHE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}$ $\angle GHE=\angle GHF$
$\triangle GHE \cong \triangle GHF$ by $SAS$
So, $GF=GE$ . By the reflection, $AE=GE=GF=AB$ $AE=AB=\boxed{20}$

Solution:
Rectangle.PNG
We are going to use pure Euclidian geometry to prove $AE=AB$
Construct equilateral triangle $\triangle BEF$ , and let $GF$ be the height of $\triangle ABF$
$\angle GBF=90^{\circ}-15^{\circ}-60^{\circ}=15^{\circ}$ $\angle GBF=\angle CBE$ $\angle BGF=\angle BCE=90^{\circ}$ $BF=BE$
$\triangle BGF \cong \triangle BCE$ by $AAS$
$BG=BC=10, AG=20-10=10$ $AG=BG$ $GF=GF$ , by $HL$ $\triangle AGF \cong \triangle BGF$
So, $AF=BF=EF$
$\angle AFB=75^{\circ}+75^{\circ}=150^{\circ}$ $\angle AFE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}$ $\angle AFB=\angle AFE$ $AF=AF$ $BF=EF$
$\triangle AFB \cong \triangle AFE$ by $SAS$
So, $AE=AB=\boxed{20}$

Solution: All trigonometric functions in this solution are in degrees. We know \[\sin\left(a+b\right)=\sin\left(a\right)\cos\left(b\right)+\sin\left(b\right)\cos\left(a\right)\] so \[\sin\left(15\right)=\sin\left(45-30\right)=\sin\left(45\right)\cos\left(-30\right)+\sin\left(-30\right)\cos\left(45\right)\] \[=\frac{\sqrt{2}}{2}\cdot\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{-\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{2}-\sqrt{6}}{4}\] \[=\frac{\sqrt{2}-\sqrt{6}}{4}\] Let $EC=x$ , then $BE=\sqrt{x^{2}+100}$ . By the definition of sine, \[\frac{x}{\sqrt{x^{2}+100}}=\frac{\sqrt{2}-\sqrt{6}}{4}\] Squaring both sides, \[\frac{x^{2}}{x^{2}+100}=\frac{\left(\sqrt{2}-\sqrt{6}\right)^{2}}{16}=\frac{2-2\sqrt{12}+6}{16}=\frac{8-4\sqrt{3}}{16}=\frac{2-\sqrt{3}}{4}\] Cross-multiplying, \[4x^{2}=\left(2-\sqrt{3}\right)\left(x^{2}+100\right)=2x^{2}+200-\sqrt{3}x^{2}-100\sqrt{3}\] Simplifying, \[\left(2+\sqrt{3}\right)x^{2}=200-100\sqrt{3}\] \[x^{2}=\frac{200-100\sqrt{3}}{2+\sqrt{3}}=\frac{100\left(2-\sqrt{3}\right)}{2+\sqrt{3}}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}\] Let $\frac{2-\sqrt{3}}{2+\sqrt{3}}=p$ . Notice that $\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^{2}-\sqrt{3}^{2}=1$ so $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$ $p$ is then \[\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\frac{1}{2+\sqrt{3}}}{2+\sqrt{3}}=\frac{1}{\left(2+\sqrt{3}\right)^{2}}\] Recall that \[x^{2}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}\] which we now know is \[100\cdot\frac{1}{\left(2+\sqrt{3}\right)^{2}}=\frac{100}{\left(2+\sqrt{3}\right)^{2}}=\left(\frac{10}{2+\sqrt{3}}\right)^{2}\] Therefore \[x=\frac{10}{2+\sqrt{3}}\] Rationalizing the denominator, \[\frac{10}{2+\sqrt{3}}\cdot\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{20-10\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\] Which by difference of squares reduces to \[20-10\sqrt{3}\] so $EC=20-10\sqrt{3}$ $ED$ is then $20-\left(20-10\sqrt{3}\right)=10\sqrt{3}$ and since we know $AD=10$ , by the Pythagorean theorem, $AE = 20$ . The answer is $\boxed{20}$

Solution:
The ratio between the side lengths of a 15, 75, 90 triangle are $\sqrt{3}-1, \sqrt{3}+1, 2\sqrt{2}$ . Therefore, $\cos 75^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}}$ . Using triangular ratios, $\frac{\sqrt{3}+1}{10}=\frac{2\sqrt{2}}{BE}$ $BE=10\sqrt{2}(\sqrt{3}-1)$
From the law of cosines, $(AE)^2=(AB)^2+(BE)^2-2(AB)(BE)\cos 75^\circ$
So, $(AE)^2=20^2+(10\sqrt{2}(\sqrt{3}-1))^2-2(20)(10\sqrt{2}(\sqrt{3}-1))(\frac{\sqrt{3}-1}{2\sqrt{2}})$
At this stage, it is easier if we do not expand the numbers yet. Further simplifying, we have $(AE)^2=20^2+200(\sqrt{3}-1)^2-40(10\sqrt{2}(\sqrt{3}-1))(\frac{(\sqrt{3}-1)(\sqrt{2})}{4})$
Now, we have $(AE)^2=20^2+200(\sqrt{3}-1)^2-10(10\sqrt{2}(\sqrt{3}-1))(\sqrt{3}-1)(\sqrt{2})$
Combing like terms, $(AE)^2=20^2+200(\sqrt{3}-1)^2-(100(\sqrt{2})^2)(\sqrt{3}-1)^2)$
This equals, $(AE)^2=20^2+200(\sqrt{3}-1)^2-200(\sqrt{3}-1)^2$
The $200(\sqrt{3}-1)^2)$ cancel out each other. We are left with $(AE)^2=20^2$ . So, $AE=\boxed{20}$ .
~hwan


Answer: 3.0

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

Solution:
Since $\angle C = 90^\circ$ , each of the three smaller angles is $30^\circ$ , and $\triangle BEC$ and $\triangle CDF$ are both $30-60-90$ triangles.

Defining the variables as illustrated above, we have $x = 6\sqrt{3}$ from $\triangle BEC$
Then $x-2 = 6\sqrt{3} - 2$ , and $y = \sqrt{3} (6 \sqrt{3} - 2) = 18 - 2\sqrt{3}$
The area of the rectangle is thus $xy = 6\sqrt{3}(18 - 2\sqrt{3}) = 108\sqrt{3} - 36$
Using the approximation $\sqrt{3} \approx 1.7$ , we get an area of just under $147.6$ , which is closest to answer $\boxed{3} > 1.7$ ).


Answer: 150.0

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


Answer: 3.0

In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$

Solution:
$[CEF] = \frac{[ABCD]}{4} = \frac{35}{2}$
$\triangle CEH \sim \triangle AFH$ $\frac{HE}{HF} = \frac{CE}{AF} = \frac{3}{2}$ $\frac{HE}{EF} = \frac{3}{5}$
$[CEH] = \frac{HE}{EF} \cdot [CEF] = \frac{3}{5} \cdot \frac{35}{2} = \frac{21}{2}$
$\triangle CEH \sim \triangle AFH$ $\frac{AH}{HC} = \frac{AF}{CE} = \frac{2}{3}$ $\frac{AH}{AC} = \frac{2}{5}$ $\frac{CH}{AC} = \frac{3}{5}$
$\triangle CEJ \sim \triangle AGJ$ $\frac{AJ}{JC} = \frac{AG}{CE} = \frac{4}{3}$ $\frac{AJ}{AC} = \frac{4}{7}$
$\frac{HJ}{AC} = \frac{AJ}{AC} - \frac{AH}{AC} = \frac{4}{7} - \frac{2}{5} = \frac{6}{35}$
$\frac{HJ}{CH} = \frac{HJ}{AC} \cdot \frac{AC}{CH} = \frac{6}{35} \cdot \frac{5}{3} = \frac{2}{7}$
$[EHJ] = \frac{HJ}{CH} \cdot [CEH] = \frac{2}{7} \cdot \frac{21}{2} = \boxed{3}$

Solution:
We need one more pair of ratios to fully define our mass point system.  Let's use $\triangle AFH \sim \triangle CEH\implies EH:HF = 3:2$ and now do mass points on $\triangle AEG$

Now it's just a standard "Area Reduction by Ratios"™ problem going from:
\[[ABCD]\xrightarrow[]{\frac{1}{2}}[AEB]\xrightarrow[]{\frac{2}{3}}[AEG]\xrightarrow[]{\frac{3}{7}}[AEJ]\xrightarrow[]{\frac{3}{10}}[HEJ]\]
or,
\[70 \cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{7}\cdot \frac{3}{10} = \boxed{3}\]


Answer: 3.0

Solution:
 Note that the triangles $AFH$ and $CEH$ are similar, as they have the same angles. Hence $\frac {AH}{HC} = \frac{AF}{EC} = \frac 23$
Also, triangles $AGJ$ and $CEJ$ are similar, hence $\frac {AJ}{JC} = \frac {AG}{EC} = \frac 43$
We can now compute $[EHJ]$ as $[ACD]-[AHD]-[DEH]-[EJC]$ . We have:
Therefore $[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-\frac {21}2-\frac{15}2 = \boxed{3}$

Solution:
As in the previous solution, we note the similar triangles and prove that $H$ is in $2/5$ and $J$ in $4/7$ of $AC$
We can then compute that $HJ = AC \cdot \left( \frac 47 - \frac 25 \right) = AC \cdot \frac{6}{35}$
As $E$ is the midpoint of $CD$ , the height from $E$ onto $AC$ is $1/2$ of the height from $D$ onto $AC$ . Therefore we have $[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}$

Solution: Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are $7$ and $10$ (or any other two positive real numbers that multiply to 70). We can find $H$ and $J$ by intersecting lines, and then we calculate the area of $EHJ$ using shoelace formula. This yields $\boxed{3}$

Solution: Note that triangle $AFH$ is similar to triangle $CEH$ with ratio $\frac{2}{3}$ . Similarly, triangle $AGJ$ is similar to triangle $ECJ$ with ratio $\frac{4}{3}$ . Thus, if $AC = a$ then we know that $AH = \frac{2}{5}a$ and $JC = \frac{3}{7}a$ meaning $HJ = \frac{6}{35}a$ and thus the ratio of $HJ$ to $JC$ is $\frac{\frac{6}{35}}{\frac{3}{7}} = \frac{2}{5}$ which equals the ratio of the areas of $HJE$ to $JEC$ . If $y = AD, x = DC$ , then we know that $JEC = \text{(altitude from J to EC)} \cdot EC = \frac{3}{7}y \cdot \frac{1}{2}x \cdot \frac{1}{2}$ and since $xy = 70$ and we want to find $\frac{2}{5}$ of this, we get our answer is $\frac{2}{5} \cdot \frac{3}{7} \cdot \frac{1}{2} \cdot 70 \cdot \frac{1}{2} = \boxed{3}$ . -SuperJJ


Answer: 40400.0

In rectangle $ABCD$ , we have $A=(6,-22)$ $B=(2006,178)$ $D=(8,y)$ , for some integer $y$ . What is the area of rectangle $ABCD$
$\mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400$

Solution:
This solution is the same as Solution 1 up to the point where we find that $y=-42$
We build right triangles so we can use the Pythagorean Theorem.  The triangle with hypotenuse $AB$ has legs $200$ and $2000$ , while the triangle with hypotenuse $AD$ has legs $2$ and $20$ .  Aha!  The two triangles are similar by SAS, with one triangle having side lengths $100$ times the other!
Let $AD=x$ .  Then from our reasoning above, we have $AB=100x$ .  Finally, the area of the rectangle is $100x(x)=100x^2=100(20^2+2^2)=100(400+4)=100(404)=\boxed{40400}$

Solution: We do not need to solve for y. We form a right triangle with $AB$ as the hypotenuse and two adjacent sides lengths 200 and 2000, respectively. We form another right triangle with $AD$ as the hypotenuse and 2 is one of the lengths of the adjacent sides. Those two triangles are similar because $AD$ and $AB$ are perpendicular. $\frac{AB}{AD} = \frac{200}{2}$ , so the area $AB \cdot AD = \frac {AB^2}{100} = \frac {2000^2 + 200^2}{100} = \boxed{40400}$

Solution:
In order to find the area of rectangle $ABCD$ , we need to find $AB$ first. Using the distance formula , we can derive:
$AB = \sqrt{(2006-6)^2 + (178-(-22))^2} = \sqrt{(2000)^2 + (200)^2 } = 200\sqrt{101}$
Now we can look at the answer choices. Because all of them are integers, then we know that $AD$ has to also contain $\sqrt{101}$ to ensure that $AB*AD$ is an integer. Once you know that, you can guarantee that the answer must be a multiple of 101. (If you're wondering why there can't be a fraction like $1/101$ , there can't because $y$ is an integer the number under the square root in $AD$ , is an integer.) Having that information narrows down the answer choices to just $(B)$ $4040$ , and $(E)$ $40400$
Looking back at what we found for $AB$ , which is $200\sqrt{101}$ , and what we know about $AD$ , that it has to contain $\sqrt{101}$ . we know that the answer is at least $200\sqrt{101}*\sqrt{101}=20200$ , which is already greater than answer choice $(B)$ , which is $4040$ . From this, we can conclude that the answer is $\boxed{40400}$