Answer: 15.0

Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18$

Solution:
Let's use stars and bars .
Let the donuts be represented by $O$ s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us $4$ in all. The four donuts we want can be represented as $OOOO$ . Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, $O|OO|O$ represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in $\binom{6}{2}=15$ ways. Our answer is hence $\boxed{15}$ . Notice that this can be generalized to get the stars and bars (balls and urns) identity.

Solution:
Simple casework works here as well:
Set up the following ratios: \[4:0:0\] \[3:1:0\] \[2:2:0\] \[2:1:1\]
In three of these cases we see that there are two of the same ratios (so like two boxes would have $0$ ), and so if we swapped those two donuts, we would have the same case. Thus we get $\frac{3!}{2!1!}$ for those $3$ (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal $\binom{4}{3}=6.$ Thus, our answer is $3 \cdot 3+6 = \boxed{15}$


Answer: 803.0

Patio blocks that are hexagons $1$ unit on a side are used to outline a garden by placing the blocks edge to edge with $n$ on each side. The diagram indicates the path of blocks around the garden when $n=5$
AIME 2002 II Problem 4.gif
If $n=202$ , then the area of the garden enclosed by the path, not including the path itself, is $m\left(\sqrt3/2\right)$ square units, where $m$ is a positive integer. Find the remainder when $m$ is divided by $1000$

Solution:
When $n>1$ , the path of blocks has $6(n-1)$ blocks total in it. When $n=1$ , there is just one lonely block. Thus, the area of the garden enclosed by the path when $n=202$ is
\[(1+6+12+18+\cdots +1200)A=(1+6(1+2+3...+200))A\]
where $A$ is the area of one block. Then, because $n(n+1)/2$ is equal to the sum of the first $n$ integers:
\[(1+6(1+2+3...+200))=(1+6((200)(201)/2))A=120601A\]
Since $A=\dfrac{3\sqrt{3}}{2}$ , the area of the garden is
\[120601\cdot \dfrac{3\sqrt{3}}{2}=\dfrac{361803\sqrt{3}}{2}\]
$m=361803$ $\dfrac{m}{1000}=361$ Remainder $\boxed{803}$


Answer: 1.15

Patty has $20$ coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have $70$ cents more. How much are her coins worth?
$\textbf{(A)}\ \textdollar 1.15\qquad\textbf{(B)}\ \textdollar 1.20\qquad\textbf{(C)}\ \textdollar 1.25\qquad\textbf{(D)}\ \textdollar 1.30\qquad\textbf{(E)}\ \textdollar 1.35$

Solution: She has $n$ nickels and $d=20-n$ dimes. Their total cost is $5n+10d=5n+10(20-n)=200-5n$ cents. If the dimes were nickels and vice versa, she would have $10n+5d=10n+5(20-n)=100+5n$ cents. This value should be $70$ cents more than the previous one. We get $200-5n+70=100+5n$ , which solves to $n=17$ . Her coins are worth $200-5n=\boxed{1.15}$

Solution: Changing a nickel into a dime increases the sum by $5$ cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by $70$ cents, there are $70/5=14$ more nickels than dimes. As the total count is $20$ , this means that there are $17$ nickels and $3$ dimes, which is equal to $\boxed{1.15}$


Answer: 5.0

Paul owes Paula $35$ cents and has a pocket full of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution: The fewest amount of coins that can be used is $2$ (a quarter and a dime). The greatest amount is $7$ , if he only uses nickels. Therefore we have $7-2=\boxed{5}$


Answer: 48.0

Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60$

Solution:
Let Paula work at a rate of $p$ , the two helpers work at a combined rate of $h$ , and the time it takes to eat lunch be $L$ , where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:
\[(8-L)(p+h)=50\]
\[(6.2-L)h=24\]
\[(11.2-L)p=26\]
With three equations and three variables, we need to find the value of $L$ .
Adding the second and third equations together gives us $6.2h+11.2p-L(p+h)=50$ . Subtracting the first equation from this new one gives us $-1.8h+3.2p=0$ , so we get $h=\frac{16}{9}p$ .
Plugging into the second equation:
\[(6.2-L)\frac{16}{9}p=24\] \[(6.2-L)p=\frac{27}{2}\]
We can then subtract this from the third equation:
\[5p=26-\frac{27}{2}\] \[p=\frac{5}{2}\] Plugging $p$ into our third equation gives: \[L=\frac{4}{5}\]
Converting $L$ from hours to minutes gives us $L=48$ minutes, which is $\boxed{48}$

Solution: Because Paula worked from \[8:00 \text{A.M.}\] to \[7:12 \text{P.M.}\] , she worked for 11 hours and 12 minutes = 672 minutes. Since there is $100-50-24=26$ % of the house left, we get the equation $26a=672$ . Because $672$ is $22$ mod $26$ , looking at our answer choices, the only answer that is $22$ $\text{mod}$ $26$ is $48$ . So the answer is $\boxed{48}$


Answer: 48.0

Solution:
Let Paula work at a rate of $p$ , the two helpers work at a combined rate of $h$ , and the time it takes to eat lunch be $L$ , where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:
\[(8-L)(p+h)=50\]
\[(6.2-L)h=24\]
\[(11.2-L)p=26\]
With three equations and three variables, we need to find the value of $L$ .
Adding the second and third equations together gives us $6.2h+11.2p-L(p+h)=50$ . Subtracting the first equation from this new one gives us $-1.8h+3.2p=0$ , so we get $h=\frac{16}{9}p$ .
Plugging into the second equation:
\[(6.2-L)\frac{16}{9}p=24\] \[(6.2-L)p=\frac{27}{2}\]
We can then subtract this from the third equation:
\[5p=26-\frac{27}{2}\] \[p=\frac{5}{2}\] Plugging $p$ into our third equation gives: \[L=\frac{4}{5}\]
Converting $L$ from hours to minutes gives us $L=48$ minutes, which is $\boxed{48}$

Solution: Because Paula worked from \[8:00 \text{A.M.}\] to \[7:12 \text{P.M.}\] , she worked for 11 hours and 12 minutes = 672 minutes. Since there is $100-50-24=26$ % of the house left, we get the equation $26a=672$ . Because $672$ is $22$ mod $26$ , looking at our answer choices, the only answer that is $22$ $\text{mod}$ $26$ is $48$ . So the answer is $\boxed{48}$


Answer: 15.0

Paula the painter had just enough paint for 30 identically sized rooms.  Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms.  How many cans of paint did she use for the 25 rooms?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 15\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 25$

Solution: Losing three cans of paint corresponds to being able to paint five fewer rooms. So $\frac 35 \cdot 25 = \boxed{15}$


Answer: 15.0

Solution: Losing three cans of paint corresponds to being able to paint five fewer rooms. So $\frac 35 \cdot 25 = \boxed{15}$


Answer: 7.0

Pauline Bunyan can shovel snow at the rate of $20$ cubic yards for the first hour, $19$ cubic yards for the second, $18$ for the third, etc., always shoveling one cubic yard less per hour than the previous hour.  If her driveway is $4$ yards wide, $10$ yards long, and covered with snow $3$ yards deep, then the number of hours it will take her to shovel it clean is closest to
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 12$

Solution: Her driveway has $(4)(10)(3)=120$ cubic yards of snow. After the first hour she would have $120-20=100$ cubic yards, then $100-19=81$ $81-18=63$ $63-17=46$ $46-16=30$ $30-15=15$ , and $15-14=1$ cubic yard after the seventh hour. It will take her a little more than seven hours to shovel it clean, which is closest to $\boxed{7}$


Answer: 6.0

Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is
[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7));[/asy]
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \  } 5 \qquad \mathrm{(D) \  } 5.5 \qquad \mathrm{(E) \  }6$

Solution: We draw in the rectangle bounding the given quadrilateral and label the points as shown. The area of rectangle $ABCD$ is $(3)(4) = 12$ , while the areas of the triangles $AEH$ $EBF$ $FCG$ , and $GDH$ are, respectively, \begin{align*}&\text{area of } \triangle AEH = \frac{1}{2}(1)(2) = 1, \\ &\text{area of } \triangle EBF = \frac{1}{2}(3)(2) = 3, \\ &\text{area of } \triangle FCG = \frac{1}{2}(1)(1) = \frac{1}{2}, \text{ and} \\ &\text{area of } \triangle GDH = \frac{1}{2}(3)(1) = \frac{3}{2}.\end{align*} Hence the area of the given quadrilateral $EFGH$ is $12-\left(1+3+\frac{1}{2}+\frac{3}{2}\right) = \boxed{6}$

Solution: The number of lattice points (i.e. pegs on the board) strictly inside the quadrilateral is $5$ and the number of lattice points on its boundary is $4$ . Therefore, by Pick's theorem , its area is $5+\frac{4}{2}-1 = \boxed{6}$


Answer: 64.0

Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, all nickels,dimes, and quarters, whose total value is $8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?
$\text {(A) } 0 \qquad \text {(B) } 13 \qquad \text {(C) } 37 \qquad \text {(D) } 64 \qquad \text {(E) } 83$

Solution:
Where $a,b,c$ is the number of nickels, dimes, and quarters, respectively, we can set up two equations:
\[(1)\ 5a+10b+25c=835\ \ \ \ (2)\ a+b+c=100\]
Eliminate $a$ by subtracting $(2)$ from $(1)/5$ to get $b+4c=67$ . Of the integer solutions $(b,c)$ to this equation, the number of dimes $b$ is least in $(3,16)$ and greatest in $(67,0)$ , yielding a difference of $67-3=\boxed{64}$


Answer: 1.0

Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?
$\textbf{(A) }0\quad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution: Given $n$ games, there must be a total of $n$ wins and $n$ losses. Hence, $4 + 3 + K = 2 + 3 + 3$ where $K$ is Kyler's wins. $K = 1$ , so our final answer is $\boxed{1}.$ ~CHECKMATE2021


Answer: 79.0

Pick two consecutive positive integers whose sum is less than $100$ . Square both
of those integers and then find the difference of the squares. Which of the
following could be the difference?
$\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131$

Solution:
Let the smaller of the two numbers be $x$ . Then, the problem states that $(x+1)+x<100$ $(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$ $2x+1$ is obviously odd, so only answer choices C and E need to be considered.
$2x+1=131$ contradicts the fact that $2x+1<100$ , so the answer is $\boxed{79}$

Solution: Since for two consecutive numbers $a$ and $b$ , the difference between their squares are $a^2-b^2=(a+b)(a-b)$ , which equals to $a+b$ , because $a$ and $b$ are consecutive. And because they are consecutive, one number must be even, and the other odd. Since the sum of an even and an odd number is always odd, and that the sum of $a$ and $b$ is less than 100, you can eliminate all answers expect for $\boxed{79}$


Answer: 6.0

Placing no more than one $\text{X}$ in each small square , what is the greatest number of $\text{X}$ 's that can be put on the grid shown without getting three $\text{X}$ 's in a row vertically, horizontally, or diagonally?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
[asy] for(int a=0; a<4; ++a)  {   draw((a,0)--(a,3));  } for(int b=0; b<4; ++b)  {   draw((0,b)--(3,b));  } [/asy]

Solution:
By the Pigeonhole Principle , if there are at least $7$ $\text{X}$ 's, then there will be some row with $3$ $\text{X}$ 's.  We can put in $6$ by leaving out the three boxes in one of the main diagonals.
$\rightarrow \boxed{6}$


Answer: 58.0

Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$ . Find the degree measure of $\angle BAG$
[asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label("$A$",A,dir(0)); label("$B$",B,dir(75)); label("$C$",C,dir(90)); label("$D$",D,dir(105)); label("$E$",E,dir(180)); label("$F$",F,dir(225)); label("$G$",G,dir(260)); label("$H$",H,dir(280)); label("$I$",I,dir(315)); [/asy]

Solution:
Let $O$ be the center of the circle with $ABCDE$ on it.
Let $x$ be the degree measurement of $\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ in circle $O$
and $y$ be the degree measurement of $\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$ in circle $C$
$\angle ECA$ is, therefore, $5y$ by way of circle $C$ and \[\frac{360-4x}{2}=180-2x\] by way of circle $O$ $\angle ABD$ is $180 - \frac{3x}{2}$ by way of circle $O$ , and \[\angle AHG = 180 - \frac{3y}{2}\] by way of circle $C$
This means that:
\[180-\frac{3x}{2}=180-\frac{3y}{2}+12\]
which when simplified yields \[\frac{3x}{2}+12=\frac{3y}{2}\] or \[x+8=y\] Since: \[5y=180-2x\] and \[5x+40=180-2x\] So: \[7x=140\Longleftrightarrow x=20\] \[y=28\] $\angle BAG$ is equal to $\angle BAE$ $\angle EAG$ , which equates to $\frac{3x}{2} + y$ .
Plugging in yields $30+28$ , or $\boxed{058}$

Solution:
Let $m$ be the degree measurement of $\angle GCH$ . Since $G,H$ lie on a circle with center $C$ $\angle GHC=\frac{180-m}{2}=90-\frac{m}{2}$
Since $\angle ACH=2 \angle GCH=2m$ $\angle AHC=\frac{180-2m}{2}=90-m$ . Adding $\angle GHC$ and $\angle AHC$ gives $\angle AHG=180-\frac{3m}{2}$ , and $\angle ABD=\angle AHG+12=192-\frac{3m}{2}$ . Since $AE$ is parallel to $BD$ $\angle DBA=180-\angle ABD=\frac{3m}{2}-12=$ $\overarc{BE}$
We are given that $A,B,C,D,E$ are evenly distributed on a circle. Hence,
$\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ $=\frac{\angle DBA}{3}=\frac{m}{2}-4$
Here comes the key: Draw a line through $C$ parallel to $AE$ , and select a point $X$ to the right of point $C$
$\angle ACX$ $\overarc{AB}$ $\overarc{BC}$ $m-8$
Let the midpoint of $\overline{HG}$ be $Y$ , then $\angle YCX=\angle ACX+\angle ACY=(m-8)+\frac{5m}{2}=90$ . Solving gives $m=28$
The rest of the solution proceeds as in solution 1, which gives $\boxed{058}$

Solution:
Let $\angle GAH = \varphi \implies \overset{\Large\frown} {GH} = 2\varphi \implies$ \[\overset{\Large\frown} {EF} = \overset{\Large\frown} {FG} = \overset{\Large\frown} {HI} = \overset{\Large\frown} {IA} = 2\varphi \implies\] \[\angle AGH = 2\varphi, \angle ACE = 10 \varphi.\]
\[BD||GH \implies \angle AJB = \angle AGH = 2 \varphi.\] \[\triangle AHG: \hspace{10mm}  \angle AHG = \beta = 180^\circ – 3 \varphi.\] $\hspace{10mm} \triangle ABJ:  \hspace{10mm} \angle BAG + \angle ABD = \alpha + \gamma = 180^\circ + 2 \varphi.$
Let arc $\overset{\Large\frown} {AB} = 2\psi \implies$
$\angle ACE = \frac {360^\circ – 8 \psi}{2}= 180^\circ – 4 \psi,  \angle ABD = \gamma =\frac {360^\circ – 6 \psi}{2} =180^\circ – 3 \psi.$ $\gamma –  \beta = 3(\varphi – \psi) = 12^\circ \implies \psi = \varphi – 4^\circ \implies 10 \varphi = 180^\circ – 4(\varphi – 4^\circ) \implies 14 \varphi = 196^\circ \implies \varphi = 14^\circ.$
Therefore $\gamma = 180^\circ – 3 \cdot (14^\circ – 4^\circ) = 150^\circ \implies  \alpha = 180^\circ + 2 \cdot 14^\circ – 150^\circ = \boxed{058}.$


Answer: 42.0

Point $B$ is in the exterior of the regular $n$ -sided polygon $A_1A_2\cdots A_n$ , and $A_1A_2B$ is an equilateral triangle . What is the largest value of $n$ for which $A_1$ $A_n$ , and $B$ are consecutive vertices of a regular polygon?

Solution:
1997 AIME-6.png
Let the other regular polygon have $m$ sides. Using the interior angle of a regular polygon formula, we have $\angle A_2A_1A_n = \frac{(n-2)180}{n}$ $\angle A_nA_1B = \frac{(m-2)180}{m}$ , and $\angle A_2A_1B = 60^{\circ}$ . Since those three angles add up to $360^{\circ}$
\begin{eqnarray*} \frac{(n-2)180}{n} + \frac{(m-2)180}{m} &=& 300\\ m(n-2)180 + n(m-2)180 &=& 300mn\\ 360mn - 360m - 360n &=& 300mn\\ mn - 6m - 6n &=& 0 \end{eqnarray*} Using SFFT
\begin{eqnarray*} (m-6)(n-6) &=& 36 \end{eqnarray*} Clearly $n$ is maximized when $m = 7, n = \boxed{042}$

Solution:
As above, find that $mn - 6m - 6n = 0$ using the formula for the interior angle of a polygon.
Solve for $n$ to find that $n = \frac{6m}{m-6}$ .  Clearly, $m>6$ for $n$ to be positive.
With this restriction of $m>6$ , the larger $m$ gets, the smaller the fraction $\frac{6m}{m-6}$ becomes.  This can be proven either by calculus, by noting that $n = \frac{6m}{m-6}$ is a  transformed hyperbola, or by dividing out the rational function to get $n = 6 + \frac{36}{m - 6}.$
Either way, minimizng $m$ will maximize $n$ , and the smallest integer $m$ such that $n$ is positive is $m=7$ , giving $n = \boxed{042}$

Solution: From the formula for the measure for an individual angle of a regular n-gon, $180 - \frac{360}{n}$ , the measure of $\angle A_2A_1A_n = 180 - \frac{360}{n}$ . Together with the fact that an equilateral triangle has angles measuring 60 degrees, the measure of $\angle A_nA_1B = 120 + \frac{360}{n}$ (Notice that this value decreases as $n$ increases; hence, we are looking for the least possible value of $\angle A_nA_1B$ ). For $A_n, A_1, B$ to be vertices of a regular polygon, $\angle A_nA_1B$ must be of the form $180 - \frac{360}{n}$ , where $n$ is a natural number greater than or equal to 3. It is obvious that $\angle A_nA_1B > 120$ . The least angle satisfying this condition is $180 - \frac{360}{7}$ . Equating this with $120 + \frac{360}{n}$ and solving yields $n = \boxed{042}$


Answer: 380.0

Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21.$ Point $D$ is not on $\overline{AC}$ so that $AD = CD,$ and $AD$ and $BD$ are integers . Let $s$ be the sum of all possible perimeters of $\triangle ACD$ . Find $s.$

Solution:
Denote the height of $\triangle ACD$ as $h$ $x = AD = CD$ , and $y = BD$ . Using the Pythagorean theorem , we find that $h^2 = y^2 - 6^2$ and $h^2 = x^2 - 15^2$ . Thus, $y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189$ . The LHS is difference of squares , so $(x + y)(x - y) = 189$ . As both $x,\ y$ are integers, $x+y,\ x-y$ must be integral divisors of $189$
The pairs of divisors of $189$ are $(1,189)\ (3,63)\ (7,27)\ (9,21)$ . This yields the four potential sets for $(x,y)$ as $(95,94)\ (33,30)\ (17,10)\ (15,6)$ . The last is not a possibility since it simply degenerates into a line . The sum of the three possible perimeters of $\triangle ACD$ is equal to $3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{380}$


Answer: 507.0

Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ . Find $x^2$

Solution:
Let $A$ be the origin, so $B=(16,0)$ and $C=(20,0).$ Using equilateral triangle properties tells us that $D=(8,8\sqrt3)$ and $E=(18,2\sqrt3)$ as well. Therefore, $M=(9,\sqrt3)$ and $N=(14,4\sqrt3).$ Applying the Shoelace Theorem to triangle $BMN$ gives
\[x=\dfrac 1 2 |16\sqrt3+36\sqrt3+0-(0+14\sqrt3+64\sqrt3)| =13\sqrt3,\]
so $x^2=\boxed{507}.$

Solution:
Note that $AB=DB=16$ and $BE=BC=4$ . Also, $\angle ABE = \angle DBC = 120^{\circ}$ . Thus, $\triangle ABE \cong \triangle DBC$ by SAS.
From this, it is clear that a $60^{\circ}$ rotation about $B$ will map $\triangle ABE$ to $\triangle DBC$ .
This rotation also maps $M$ to $N$ . Thus, $BM=BN$ and $\angle MBN=60^{\circ}$ . Thus, $\triangle BMN$ is equilateral.
Using the Law of Cosines on $\triangle ABE$ \[AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)\] \[AE = 4\sqrt{21}\] Thus, $AM=ME=2\sqrt{21}$
Using Stewart's Theorem on $\triangle ABE$ \[AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME\] \[BM = 2\sqrt{13}\]
Calculating the area of $\triangle BMN$ \[[BMN] = \frac{\sqrt{3}}{4} BM^2\] \[[BMN] = 13\sqrt{3}\] Thus, $x=13\sqrt{3}$ , so $x^2 = 507$ . Our final answer is $\boxed{507}$

Solution:
$AB = BD, BE = BC, \angle ABE = \angle CBD \implies  \triangle ABE \cong \triangle DBC$
Medians are equal, so $MB = MN, \angle ABM = \angle DBN \implies$ $\angle MBN = \angle ABD - \angle ABM + \angle DBN = 60^\circ \implies$
$\triangle MNB$ is  equilateral triangle.
The height of $\triangle BCE$ is $2 \sqrt{3},$ distance from $A$ to midpoint $BC$ is $16 + 2 = 18 \implies \frac {AE^2}{4} =\frac{ (16 + 2)^2 +2^2 \cdot 3}{4} = 81 + 3 = 84.$
$BM$ is the median of $\triangle ABE \implies$ $BM^2 = \frac {AB^2}{2} + \frac {BE^2}{2} - \frac {AE^2}{4}=16 \cdot 8 + 4 \cdot 2 - 84 = 52.$
The area of $\triangle BMN$
\[[BMN] = \frac{\sqrt{3}}{4} BM^2 =13 \sqrt{3} \implies \boxed{507}.\]


Answer: 2.0

Point $D$ is on side $CB$ of triangle $ABC$ . If $\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6$ ,
then the length of $AD$ is
$\textbf{(A)} \ 2 \qquad  \textbf{(B)} \ 2.5 \qquad  \textbf{(C)} \ 3 \qquad  \textbf{(D)} \ 3.5 \qquad  \textbf{(E)} \ 4$

Solution: Let $AD = y$ . Since $AD$ bisects $\angle{BAC}$ , the Angle Bisector Theorem gives $\frac{DB}{CD} = \frac{AB}{AC} = 2$ , so let $CD = x$ and $DB = 2x$ . Applying the Law of Cosines to $\triangle CAD$ gives $x^2 = 3^2 + y^2 - 3y$ , and to $\triangle DAB$ gives $(2x)^2 = 6^2 + y^2 - 6y$ . Subtracting $4$ times the first equation from the second equation therefore yields $0 = 6y - 3y^2 \Rightarrow y(y-2) = 0$ , so $y$ is $0$ or $2$ . But since $y \neq 0$ $y$ is the length of a side of a triangle), $y$ must be $2$ , so the answer is $\boxed{2}$


Answer: 36.0

Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

Solution: Points are defined as shown. It is pretty easy to show that $\triangle AFE \sim \triangle AGH$ by spiral similarity at $A$ by some short angle chasing. Now, note that $AD$ is the altitude of $\triangle AFE$ , as the altitude of $AGH$ . We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that $AD/2 = \frac{\sqrt{18}}{2}$ , the altitude of $\triangle AFE$ . Similarly, the altitude of $\triangle AGH$ is the altitude of $\triangle ABC$ , or $\frac{3\sqrt{7}}{2}$ . However, it's not too hard to see that $GB = HC = 1$ , and therefore $[AGH] = [ABC]$ . From here, we get that the area of $\triangle ABC$ is $\frac{15\sqrt{7}}{14} \implies \boxed{036}$ , by similarity. ~awang11

Solution:
Let $M_A$ $M_B$ $M_C$ be the midpoints of arcs $BC$ $CA$ $AB$ . By Fact 5, we know that $M_AB = M_AC = M_AI$ , and so by Ptolmey's theorem, we deduce that \[AB\cdot M_AC + AC\cdot M_AB = BC\cdot M_AA \implies M_AA = 2M_AI.\] In particular, we have $AI = IM_A$  Now the key claim is that:
Claim: $\triangle DEF$ and $\triangle M_AM_BM_C$ are homothetic at $I$ with ratio $2$
Proof. First, we show that $D$ is the midpoint of $M_AI$ . Indeed, we have \[\frac{ID}{DM_A} = \frac{BI}{BM_A}\cdot \frac{\sin\angle IBC}{\sin \angle CBM_A} = \frac{BI}{AI}\cdot\frac{\sin \angle B/2}{\sin \angle A/2} = 1\] by Ratio lemma and Law of Sines.
Now observe that:
Combining these facts gives that $\overline{EF}$ is a midline in $\triangle IM_BM_C$ , which proves the claim. $\blacksquare$
To finish, we compute $[M_AM_BM_C]$ , noting that $[AEF] = [DEF] = \tfrac{1}{4}[M_AM_BM_C]$
By Heron's, we can calculate the circumradius $R = 8/\sqrt{7}$ , and by Law of Cosines, we get \begin{align*}\cos A &= \frac{9}{16}\implies \cos A/2 = \frac{5}{\sqrt{32}} \\ \cos B &= \frac{1}{8} \implies \cos B/2 = \frac{3}{4} \\ \cos C &= \frac{3}{4} \implies \cos C/2 = \sqrt{\frac{7}{8}}.\end{align*} Then using $[XYZ] = 2R^2\sin X\sin Y\sin Z$ , we can compute \[[M_AM_BM_C] = 2\cdot \frac{64}{7}\cdot \frac{5}{\sqrt{32}}\cdot \frac{3}{4}\cdot \frac{\sqrt{7}}{\sqrt{8}} = \frac{30\sqrt{7}}{7}.\] Thus $[AEF] = 15\sqrt{7}/14$ , which gives a final answer of $\boxed{036}$

Solution:
Let $\overline{BC}$ lie on the x-axis and $B$ be the origin. $C$ is $(5,0)$ . Use Heron's formula to compute the area of triangle $ABC$ . We have $s=\frac{15}{2}$ . and $[ABC]=\sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{2^4}}=\frac{15\sqrt{7}}{4}$ . We now find the altitude, which is $\frac{\frac{15\sqrt{7}}{2}}{5}=\frac{3\sqrt{7}}{2}$ , which is the y-coordinate of $A$ . We now find the x-coordinate of $A$ , which satisfies $x^2 + (\frac{3\sqrt{7}}{2})^{2}=16$ , which gives $x=\frac{1}{2}$ since the triangle is acute. Now using the Angle Bisector Theorem, we have $\frac{4}{6}=\frac{BD}{CD}$ and $BD+CD=5$ to get $BD=2$ . The coordinates of D are $(2,0)$ .
Since we want the area of triangle $AEF$ , we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is $(\frac{5}{4}, \frac{3\sqrt{7}}{4})$ and the slope of AD is $-\sqrt{7}$ . The slope of the perpendicular bisector is $\frac{1}{\sqrt{7}}$ . The equation is(in point slope form) $y-\frac{3\sqrt{7}}{4}=\frac{1}{\sqrt{7}}(x-\frac{5}{4})$ .
The slope of AB, or in trig words, the tangent of $\angle ABC$ is $3\sqrt{7}$ .
Finding $\sin{\angle ABC}=\frac{\frac{3\sqrt{7}}{2}}{4}=\frac{3\sqrt{7}}{8}$ and $\cos{\angle ABC}=\frac{\frac{1}{2}}{4}=\frac{1}{8}$ . Plugging this in to half angle tangent, it gives $\frac{\frac{3\sqrt{7}}{8}}{1+\frac{1}{8}}=\frac{\sqrt{7}}{3}$ as the slope of the angle bisector, since it passes through $B$ , the equation is $y=\frac{\sqrt{7}}{3}x$ .
Similarly, the equation for the angle bisector of $C$ will be $y=-\frac{1}{\sqrt{7}}(x-5)$ .
For $E$ use the B-angle bisector and the perpendicular bisector of AD equations to intersect at $(3,\sqrt{7})$ .
For $F$ use the C-angle bisector and the perpendicular bisector of AD equations to intersect at $(\frac{1}{2}, \frac{9}{2\sqrt{7}})$ .
The area of AEF is equal to $\frac{EF \cdot \frac{AD}{2}}{2}$ since AD is the altitude of that triangle with EF as the base, with $\frac{AD}{2}$ being the height. $EF=\frac{5\sqrt{2}}{\sqrt{7}}$ and $AD=3\sqrt{2}$ , so $[AEF]=\frac{15}{2\sqrt{7}}=\frac{15\sqrt{7}}{14}$ which gives $\boxed{036}$ . NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo

Solution:
Let $B=(0,0)$ and $BC$ be the line $y=0$ .
We compute that $\cos{\angle{ABC}}=\frac{1}{8}$ , so $\tan{\angle{ABC}}=3\sqrt{7}$ .
Thus, $A$ lies on the line $y=3x\sqrt{7}$ . The length of $AB$ at a point $x$ is $8x$ , so $x=\frac{1}{2}$
We now have the coordinates $A=\left(\frac{1}{2},\frac{3\sqrt{7}}{2}\right)$ $B=(0,0)$ and $C=(5,0)$ .
We also have $D=(2,0)$ by the angle-bisector theorem and $M=\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)$ by taking the midpoint.
We have that because $\cos{\angle{ABC}}=\frac{1}{8}$ $\cos{\frac{\angle{ABC}}{2}}=\frac{3}{4}$ by half angle formula.
We also compute $\cos{\angle{ACB}}=\frac{3}{4}$ , so $\cos{\frac{\angle{ACB}}{2}}=\frac{\sqrt{14}}{4}$
Now, $AD$ has slope $-\frac{\frac{3\sqrt{7}}{2}}{2-\frac{1}{2}}=-\sqrt{7}$ , so it's perpendicular bisector has slope $\frac{\sqrt{7}}{7}$ and goes through $\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)$
We find that this line has equation $y=\frac{\sqrt{7}}{7}x+\frac{4\sqrt{7}}{7}$
As $\cos{\angle{CBI}}=\frac{3}{4}$ , we have that line $BI$ has form $y=\frac{\sqrt{7}}{3}x$ .
Solving for the intersection point of these two lines, we get $x=3$ and thus $E=\left(3, \sqrt{7}\right)$
We also have that because $\cos{\angle{ICB}}=\frac{\sqrt{14}}{4}$ $CI$ has form $y=-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}$
Intersecting the line $CI$ and the perpendicular bisector of $AD$ yields $-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}=\frac{x\sqrt{7}}{7}+\frac{4\sqrt{7}}{7}$
Solving this, we get $x=\frac{1}{2}$ and so $F=\left(\frac{1}{2},\frac{9\sqrt{7}}{14}\right)$
We now compute $EF=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{5\sqrt{7}}{14}\right)^2}=\frac{5\sqrt{14}}{7}$ .
We also have $MA=\sqrt{\left(\frac{3}{4}\right)^2+\left(\frac{3\sqrt{7}}{4}\right)^2}=\frac{3\sqrt{2}}{2}$
As ${MA}\perp{EF}$ , we have $[\triangle{AEF}]=\frac{1}{2}\left(\frac{3\sqrt{2}}{2}\times\frac{5\sqrt{14}}{7}\right)=\frac{15\sqrt{7}}{14}$
The desired answer is $15+7+14=\boxed{036}$ ~Imayormaynotknowcalculus

Solution:
As usual, we will use homogenized barycentric coordinates.
We have that $AD$ will have form $3z=2y$ . Similarly, $CF$ has form $5y=6x$ and $BE$ has form $5z=4x$ .
Since $A=(1,0,0)$ and $D=\left(0,\frac{3}{5},\frac{2}{5}\right)$ , we also have $M=\left(\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$ .
It remains to determine the equation of the line formed by the perpendicular bisector of $AD$
This can be found using EFFT. Let a point $T$ on $EF$ have coordinates $(x, y, z)$ .
We then have that the displacement vector $\overrightarrow{AD}=\left(-1, \frac{3}{5}, \frac{2}{5}\right)$ and that the displacement vector $\overrightarrow{TM}$ has form $\left(x-\frac{1}{2},y-\frac{3}{10},z-\frac{1}{5}\right)$ .
Now, by EFFT, we have $5^2\left(\frac{3}{5}\times\left(z-\frac{1}{5}\right)+\frac{2}{5}\times\left(y-\frac{3}{10}\right)\right)+6^2\left(-1\times\left(z-\frac{1}{5}\right)+\frac{2}{5}\times\left(x-\frac{1}{2}\right)\right)+4^2\left(-1\times\left(y-\frac{3}{10}\right)+\frac{3}{5}\times\left(x-\frac{1}{2}\right)\right)=0$ .
This equates to $8x-2y-7z=2$
Now, intersecting this with $BE$ , we have $5z=4x$ $8x-2y-7z=2$ , and $x+y+z=1$ .
This yields $x=\frac{2}{3}$ $y=-\frac{1}{5}$ , and $z=\frac{8}{15}$ , or $E=\left(\frac{2}{3},-\frac{1}{5},\frac{8}{15}\right)$
Similarly, intersecting this with $CF$ , we have $5y=6x$ $8x-2y-7z=2$ , and $x+y+z=1$ .
Solving this, we obtain $x=\frac{3}{7}$ $y=\frac{18}{35}$ , and $z=\frac{2}{35}$ , or $F=\left(\frac{3}{7},\frac{18}{35},\frac{2}{35}\right)$
We finish by invoking the Barycentric Distance Formula twice; our first displacement vector being $\overrightarrow{FE}=\left(\frac{5}{21},-\frac{5}{7},\frac{10}{21}\right)$ .
We then have $FE^2=-25\left(-\frac{5}{7}\cdot\frac{10}{21}\right)-36\left(\frac{5}{21}\cdot\frac{10}{21}\right)-16\left(\frac{5}{21}\cdot-\frac{5}{7}\right)=\frac{50}{7}$ , thus $FE=\frac{5\sqrt{14}}{7}$
Our second displacement vector is $\overrightarrow{AM}=\left(-\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$ .
As a result, $AM^2=-25\left(\frac{3}{10}\cdot\frac{1}{5}\right)-36\left(-\frac{1}{2}\cdot\frac{1}{5}\right)-16\left(-\frac{1}{2}\cdot\frac{3}{10}\right)=\frac{9}{2}$ , so $AM=\frac{3\sqrt{2}}{2}$
As ${AM}\perp{EF}$ , the desired area is $\frac{\frac{5\sqrt{14}}{7}\times\frac{3\sqrt{2}}{2}}{2}={\frac{15\sqrt{7}}{14}}\implies{m+n+p=\boxed{036}$ . ~Imayormaynotknowcalculus

Solution:
To get the area of $\triangle AEF$ , we try to find $AM$ and $\angle EAF$
Since $AD$ is the angle bisector, we can get that $BD=2$ and $CD=3$ . By applying Stewart's Theorem, we can get that $AD=3\sqrt{2}$ . Therefore $AM=\frac{3\sqrt{2}}{2}$
Since $EF$ is the perpendicular bisector of $AD$ , we know that $AE = DE$ . Since $BE$ is the angle bisector of $\angle BAC$ ,
we know that $\angle ABE = \angle DBE$ . By applying the Law of Sines to $\triangle ABE$ and $\triangle DBE$ , we know that $\sin \angle BAE = \sin \angle BDE$ .  Since $BD$ is not equal to $AB$ and therefore these two triangles are not congruent, we know that $\angle BAE$ and $\angle BDE$ are supplementary.  Then we know that $\angle ABD$ and $\angle AED$ are also supplementary. Given that $AE=DE$ , we can get that $\angle DAE$ is half of $\angle ABC$ .  Similarly, we have $\angle DAF$ is half of $\angle ACB$
By applying the Law of Cosines, we get $\cos \angle ABC = \frac{1}{8}$ , and then $\sin \angle ABC = \frac{3\sqrt{7}}{8}$ .  Similarly, we can get $\cos \angle ACB = \frac{3}{4}$ and $\sin \angle ACB = \frac{\sqrt{7}}{4}$ . Based on some trig identities, we can compute that $\tan \angle DAE = \frac{\sin \angle ABC}{1 + \cos \angle ABC} = \frac{\sqrt{7}}{3}$ , and $\tan \angle DAF = \frac{\sqrt{7}}{7}$
Finally, the area of $\triangle AEF$ equals $\frac{1}{2}AM^2(\tan \angle DAE + \tan \angle DAF)=\frac{15\sqrt{7}}{14}$ . Therefore, the final answer is $15+7+14=\boxed{036}$ . ~xamydad

Solution: First and foremost $\big[\triangle{AEF}\big]=\big[\triangle{DEF}\big]$ as $EF$ is the perpendicular bisector of $AD$ . Now   note that quadrilateral $ABDF$ is cyclic, because $\angle{ABF}=\angle{FBD}$ and $FA=FD$ . Similarly quadrilateral $AEDC$ is cyclic, \[\implies \angle{EDA}=\dfrac{C}{2}, \quad \angle{FDA}=\dfrac{B}{2}\] Let $A'$ $B'$ $C'$ be the $A$ $B$ , and $C$ excenters of $\triangle{ABC}$ respectively. Then it follows that $\triangle{DEF} \sim \triangle{A'C'B'}$ . By angle bisector theorem we have $BD=2 \implies \dfrac{ID}{IA}=\dfrac{BD}{BA}=\dfrac{1}{2}$ . Now let the feet of the perpendiculars from $I$ and $A'$ to $BC$ be $X$ and $Y$ resptively. Then by tangents we have \[BX=s-AC=\dfrac{3}{2} \implies XD=2-\dfrac{3}{2}=\dfrac{1}{2}\] \[CY=s-AC \implies YD=3-\dfrac{3}{2}=\dfrac{3}{2} \implies \dfrac{ID}{DA'}=\dfrac{XD}{YD}=\dfrac{1}{3} \implies \big[\triangle{DEF}\big]=\dfrac{1}{16}\big[\triangle{A'C'B'}\big]\] From the previous ratios, $AI:ID:DA'=2:1:3 \implies AD=DA' \implies \big[\triangle{ABC}\big]=\big[\triangle{A'BC}\big]$ Similarly we can find that $\big[\triangle{B'AC}\big]=2\big[\triangle{ABC}\big]$ and $\big[\triangle{C'AB}\big]=\dfrac{4}{7}\big[\triangle{ABC}\big]$ and thus \[\big[\triangle{A'B'C'}\big]=\bigg(1+1+2+\dfrac{4}{7}\bigg)\big[\triangle{ABC}\big]=\dfrac{32}{7}\big[\triangle{ABC}\big] \implies \big[\triangle{DEF}\big]=\dfrac{2}{7}\big[\triangle{ABC}\big]=\dfrac{15\sqrt{7}}{14} \implies m+n+p = \boxed{036}\] -tkhalid

Solution:
Trig values we use here:
$\cos A = \frac{9}{16}$
$\cos \frac{A}{2} = \frac{5}{4\sqrt2}$
$\sin \frac{A}{2} = \frac{\sqrt7}{4\sqrt2}$
$\cos \frac{B}{2} = \frac{3}{4}$
$\cos \frac{C}{2} = \frac{\sqrt7}{2\sqrt2}$
First let the incenter be $I$ . Let $M$ be the midpoint of minor arc $BC$ on $(ABC)$ and let $K$ be the foot of $M$ to $BC$
We can find $AD$ using Stewart's Theorem: from Angle Bisector Theorem $BD = 2$ and $CD = 3$ . Then it is easy to find that $AD = 3\sqrt3$
Now we trig bash for $DI = MI - MD$ . Notice that $MI = MB$ from the Incenter Excenter Lemma. We obtain that $MB = \frac{BK}{\cos \frac{A}{2}} = \frac{\frac{5}{2}}{\frac{5}{4\sqrt2}}=2\sqrt2$ . To get $MD$ we angle chase to get $\angle KDM = \frac{A}{2}+C$ . Then \[\cos(\frac{A}{2}+C) = \cos\frac{A}{2}\cos C - \sin\frac{A}{2}\sin C = \frac{1}{2\sqrt2} = \frac{\frac{1}{2}}{MD}\] gives $MD = \sqrt{2}$ . This means $DI = \sqrt2$
Now let $AI \cap EF = G$ . It is easy to angle chase $\angle GIE = 90- \frac{B}{2}$ and $\angle GIF = 90- \frac{C}{2}$ . Since $GI = GD - ID = \frac{3\sqrt2}{2}-\sqrt2=\frac{\sqrt2}{2}$ , we compute that \[EF = EG + FG = \frac{\sqrt2}{2}(\cot B/2 + \cot C/2) = \frac{\sqrt2}{2}(\frac{3}{\sqrt7}+\sqrt7) = \frac{5\sqrt{14}}{7}\] which implies \[[AEF] = AG*EF/2 = \frac{3\sqrt2}{2} * \frac{5\sqrt{14}}{7} / 2 = \frac{15\sqrt7}{14}\] which gives an answer of $\boxed{36}$ . ~Leonard_my_dude~

Solution:
Firstly, it is easy to find $BD=2,CD=3$ with angle bisector theorem.
Using LOC and some trig formulas we get all those values: $cos\angle{B}=\frac{1}{8},sin\angle{B}=\frac{3\sqrt{7}}{8},cos\angle{\frac{B}{2}}=\frac{3}{4},cos\angle{\frac{{ACB}}{2}}=\frac{\sqrt{14}}{4}$ Now we find the coordinates of points $A,E,F$ and we apply shoelace theorem later. Point A's coordinates is $(4*\frac{1}{8},4*\frac{3\sqrt{7}}{8})=(\frac{1}{2},\frac{3\sqrt{7}}{2})$ , Let $AJ$ is perpendicular to $BC$ $tan\angle{JAD}=\frac{2-\frac{1}{2}}{\frac{3\sqrt{7}}{2}}=\frac{\sqrt{7}}{7}$ , which means the slope of $FE$ is $\frac{\sqrt{7}}{7}$ . Find the coordinate of $M$ , it is easy, $(\frac{5}{4},\frac{3\sqrt{7}}{4})$ , the function $EF$ is $y=\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}$ . Now find the intersection of $EF,EB$ $\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}=\frac{\sqrt{7}x}{3}$ , getting that $x=3,E(3,\sqrt{7})$ Now we look at line segment $CF$ , since $cos\angle{\frac{ACB}{2}}=\frac{\sqrt{14}}{4},tan\angle{\frac{ACB}{2}}=\frac{\sqrt{7}}{7}$ . Since the line passes $(5,0)$ , we can set the equation to get the $CF:y=-\frac{\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}$ , find the intersection of $CF,EF$ $-\frac{\sqrt{7}x}{7}+\frac{5\sqrt{7}}{7}=\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}$ , getting that $F:(\frac{1}{2},\frac{9\sqrt{7}}{14})$ and in the end we use shoelace theorem with coordinates of $A,F,E$ getting the area $\frac{15\sqrt{7}}{14}$ leads to the final answer $\boxed{36}$


Answer: 3.0

Point $E$ is on side $AB$ of square $ABCD$ . If $EB$ has length one and $EC$ has length two, then the area of the square is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5$

Solution: Note that $\triangle BCE$ is a right triangle. Thus, we do Pythagorean theorem to find that side $BC=\sqrt{3}$ . Since this is the side length of the square, the area of $ABCD$ is $\boxed{3}$


Answer: 108.0

Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); label("$B$",(6,6),NE); label("$C$",(6,0),SE); label("$D$",(0,0),SW); label("$E$",(3,0),S); label("$F$",(4,2),E); [/asy]
$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$

Solution:
Let the area of $\triangle CEF$ be $x$ . Thus, the area of triangle $\triangle ACD$ is $45+x$ and the area of the square is $2(45+x) = 90+2x$
By AA similarity, $\triangle CEF \sim \triangle ABF$ with a 1:2 ratio, so the area of triangle $\triangle ABF$ is $4x$ . Now, consider trapezoid $ABED$ . Its area is $45+4x$ , which is three-fourths the area of the square. We set up an equation in $x$
\[45+4x = \frac{3}{4}\left(90+2x\right)\] Solving, we get $x = 9$ . The area of square $ABCD$ is $90+2x = 90 + 2 \cdot 9 = \boxed{108}$

Solution:
We can use analytic geometry for this problem.
Let us start by giving $D$ the coordinate $(0,0)$ $A$ the coordinate $(0,1)$ , and so forth. $\overline{AC}$ and $\overline{EB}$ can be represented by the equations $y=-x+1$ and $y=2x-1$ , respectively. Solving for their intersection gives point $F$ coordinates $\left(\frac{2}{3},\frac{1}{3}\right)$
Now, $\triangle$ $EFC$ ’s area is simply $\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}$ or $\frac{1}{12}$ . This means that pentagon $ABCEF$ ’s area is $\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$ of the entire square, and it follows that quadrilateral $AFED$ ’s area is $\frac{5}{12}$ of the square.
The area of the square is then $\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{108}$

Solution: $\triangle ABC$ has half the area of the square. $\triangle FEC$ has base equal to half the square side length, and by AA Similarity with $\triangle FBA$ , it has 1/(1+2)= 1/3 the height, so has $\dfrac1{12}$ th area of square. Thus, the area of the quadrilateral is $1-1/2-1/12=5/12$ th the area of the square. The area of the square is then $45\cdot\dfrac{12}{5}=\boxed{108}$

Solution:
Extend $\overline{AD}$ and $\overline{BE}$ to meet at $X$ . Drop an altitude from $F$ to $\overline{CE}$ and call it $h$ . Also, call $\overline{CE}$ $x$ . As stated before, we have $\triangle ABF \sim \triangle CEF$ , so the ratio of their heights is in a $1:2$ ratio, making the altitude from $F$ to $\overline{AB}$ $2h$ . Note that this means that the side of the square is $3h$ . In addition, $\triangle XDE \sim \triangle XAB$ by AA Similarity in a $1:2$ ratio. This means that the side length of the square is $2x$ , making $3h=2x$
Now, note that $[ADEF]=[XAB]-[XDE]-[ABF]$ . We have $[\triangle XAB]=(4x)(2x)/2=4x^2,$ $[\triangle XDE]=(x)(2x)/2=x^2,$ and $[\triangle ABF]=(2x)(2h)/2=(2x)(4x/3)/2=4x^2/3.$ Subtracting makes $[ADEF]=4x^2-x^2-4x^2/3=5x^2/3.$ We are given that $[ADEF]=45,$ so $5x^2/3=45 \Rightarrow x^2=27.$ Therefore, $x= 3 \sqrt{3},$ so our answer is $(2x)^2=4x^2=4(27)=\boxed{108}.$


Answer: 108.0

Solution:
Solution with Cartesian and Barycentric Coordinates:
We start with the following:
Claim: Given a square $ABCD$ , let $E$ be the midpoint of $\overline{DC}$ and let $BE\cap AC = F$ . Then $\frac {AF}{FC}=2$
Proof: We use Cartesian coordinates. Let $D$ be the origin, $A=(0,1),C=(0,1),B=(1,1)$ . We have that $\overline{AC}$ and $\overline{EB}$ are governed by the equations $y=-x+1$ and $y=2x-1$ , respectively. Solving, $F=\left(\frac{2}{3},\frac{1}{3}\right)$ . The result follows. $\square$
Now, we apply Barycentric Coordinates w.r.t. $\triangle ACD$ . We let $A=(1,0,0),D=(0,1,0),C=(0,0,1)$ . Then $E=(0,\tfrac 12,\tfrac 12),F=(\tfrac 13,0,\tfrac 23)$
In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\ x_{2} &y_{2}  &z_{2} \\   x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that \[\frac{[FEC]}{[ACD]}=\begin{vmatrix} 0&0&1\\ 0&\tfrac 12&\tfrac 12\\ \tfrac 13&0&\tfrac 23 \end{vmatrix}=\frac16.\] Let $[FEC]=x$ so that $[ACD]=45+x$ . Then, we have $\frac{x}{x+45}=\frac 16 \Rightarrow x=9$ , so the answer is $2(45+9)=\boxed{108}$


Answer: 12.0

Point $F$ is taken in side $AD$ of square $ABCD$ . At $C$ a perpendicular is drawn to $CF$ , meeting $AB$ extended at $E$ .
The area of $ABCD$ is $256$ square inches and the area of $\triangle CEF$ is $200$ square inches. Then the number of inches in $BE$ is:
[asy] size(6cm); pair A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1), E = (1.3, 0), F = (0, 0.7); draw(A--B--C--D--cycle); draw(F--C--E--B); label("$A$", A, SW); label("$B$", B, S); label("$C$", C, N); label("$D$", D, NW); label("$E$", E, SE); label("$F$", F, W); //Credit to MSTang for the asymptote[/asy]
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$

Solution:
Because $ABCD$ is a square $DC = CB = 16$ $DC \perp DA$ , and $CB \perp BA$ .  Also, because $\angle DCF + \angle FCB = \angle FCB + \angle BCE = 90^\circ$ $\angle DCF = \angle BCE$ .  Thus, by ASA Congruency, $\triangle DCF \cong \triangle BCF$
From the congruency, $CF = CE$ .  Using the area formula for a triangle, $CE = 20$ .  Finally, by the Pythagorean Theorem $BE = \sqrt{20^2 - 16^2} = 12$ , which is answer choice $\boxed{12}$


Answer: 16.0

Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$ $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$ .
If $EF = 32$ and $GF = 24$ , then $BE$ equals:
[asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E = intersectionpoints(A--C, B--F)[0]; draw(A--D--C--B--cycle); draw(A--C); draw(D--F--B); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$F$", F, N); label("$G$", G, NE); label("$E$", E, SE); //Credit to MSTang for the asymptote[/asy]
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 16$

Solution:
Let $BE = x$ and $BC = y$ .  Since $AF \parallel BC$ , by AA Similarity, $\triangle AFE \sim \triangle CBE$ .  That means $\frac{AF}{CB} = \frac{FE}{BE}$ .  Substituting in values results in \[\frac{AF}{y} = \frac{32}{x}\] Thus, $AF = \frac{32y}{x}$ , so $FD = \frac{32y - xy}{x}$
In addition, $DC \parallel AB$ , so by AA Similarity, $\triangle FDG = \triangle FAB$ .  That means \[\frac{\frac{32y-xy}{x}}{\frac{32y}{x}} = \frac{24}{x+32}\] Cross multiply to get \[\frac{y(32-x)}{x} (x+32) = \frac{32y}{x} \cdot 24\] Since $x \ne 0$ and $y \ne 0$ \[(32-x)(32+x) = 32 \cdot 24\] \[32^2 - x^2 = 32 \cdot 24\] \[32 \cdot 8 = x^2\] Thus, $x = 16$ , which is answer choice $\boxed{16}$


Answer: 12.0

Point $P$ is $9$ units from the center of a circle of radius $15$ . How many different chords of the circle contain $P$ and have integer lengths?
(A) 11  (B) 12  (C) 13  (D) 14  (E) 29

Solution:
Let $O$ be the center of the circle, and let the chord passing through $P$ that is perpendicular to $OP$ intersect the circle at $Q$ and $R$ . Then $OP = 9$ and $OQ = 15$ , so by the Pythagorean Theorem, $PQ = 12$ . By symmetry, $PR = 12$
  Let $AB$ be the diameter passing through $P$.    Then the longest chord of the circle that passes through $P$ is $AB$, which has length 30, and the shortest chord is $QR$, which has length 24. If we rotate the red chord (while ensuring it passes through $P$), we can create all possible lengths between 24 and 30. Indeed, we see that for each positive integer $n=25,26,27,28,29$, there are two chords of length $n$ passing through $P$, as seen in this picture:
Therefore, there are $2 + 2(29 - 25 + 1) = \boxed{12}$ chords of integer length passing through $P$


Answer: 108.0

Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Solution:
Let $[RST]$ be the area of polygon $RST$ .  We'll make use of the following fact: if $P$ is a point in the interior of triangle $XYZ$ , and line $XP$ intersects line $YZ$ at point $L$ , then $\dfrac{XP}{PL} = \frac{[XPY] + [ZPX]}{[YPZ]}.$
This is true because triangles $XPY$ and $YPL$ have their areas in ratio $XP:PL$ (as they share a common height from $Y$ ), and the same is true of triangles $ZPY$ and $LPZ$
We'll also use the related fact that $\dfrac{[XPY]}{[ZPX]} = \dfrac{YL}{LZ}$ .  This is slightly more well known, as it is used in the standard proof of Ceva's theorem
Now we'll apply these results to the problem at hand.
Since $AP = PD = 6$ , this means that $[APB] + [APC] = [BPC]$ ; thus $\triangle BPC$ has half the area of $\triangle ABC$ .  And since $PE = 3 = \dfrac{1}{3}BP$ , we can conclude that $\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$ , and thus $\dfrac{1}{4}$ of the area of $\triangle ABC$ .  This means that $\triangle APB$ is left with $\dfrac{1}{4}$ of the area of triangle $ABC$ \[[BPC]: [APC]: [APB] = 2:1:1.\] Since $[APC] = [APB]$ , and since $\dfrac{[APC]}{[APB]} = \dfrac{CD}{DB}$ , this means that $D$ is the midpoint of $BC$
Furthermore, we know that $\dfrac{CP}{PF} = \dfrac{[APC] + [BPC]}{[APB]} = 3$ , so $CP = \dfrac{3}{4} \cdot CF = 15$
We now apply Stewart's theorem to segment $PD$ in $\triangle BPC$ —or rather, the simplified version for a median.  This tells us that \[2 BD^2 + 2 PD^2 = BP^2+ CP^2.\] Plugging in we know, we learn that \begin{align*} 2 BD^2 + 2 \cdot 36 &= 81 + 225 = 306, \\ BD^2 &= 117. \end{align*} Happily, $BP^2 + PD^2 = 81 + 36$ is also equal to 117.  Therefore $\triangle BPD$ is a right triangle with a right angle at $B$ ; its area is thus $\dfrac{1}{2} \cdot 9 \cdot 6 = 27$ .  As $PD$ is a median of $\triangle BPC$ , the area of $BPC$ is twice this, or 54.  And we already know that $\triangle BPC$ has half the area of $\triangle ABC$ , which must therefore be $\boxed{108}$

Solution:
Because we're given three concurrent cevians and their lengths, it seems very tempting to apply Mass points . We immediately see that $w_E = 3$ $w_B = 1$ , and $w_A = w_D = 2$ . Now, we recall that the masses on the three sides of the triangle must be balanced out, so $w_C = 1$ and $w_F = 3$ . Thus, $CP = 15$ and $PF = 5$
Recalling that $w_C = w_B = 1$ , we see that $DC = DB$ and $DP$ is a median to $BC$ in $\triangle BCP$ . Applying Stewart's Theorem $BC^2 + 12^2 = 2(15^2 + 9^2)$ , and $BC = 6\sqrt {13}$ . Now notice that $2[BCP] = [ABC]$ , because both triangles share the same base and the $h_{\triangle ABC} = 2h_{\triangle BCP}$ . Applying Heron's formula on triangle $BCP$ with sides $15$ $9$ , and $6\sqrt{13}$ $[BCP] = 54$ and $[ABC] = \boxed{108}$

Solution:
Using a different form of Ceva's Theorem , we have $\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}$
Solving $4y = x + y$ and $x + y = 20$ , we obtain $x = CP = 15$ and $y = FP = 5$
Let $Q$ be the point on $AB$ such that $FC \parallel QD$ .
Since $AP = PD$ and $FP\parallel QD$ $QD = 2FP = 10$ . (Stewart's Theorem)
Also, since $FC\parallel QD$ and $QD = \frac{FC}{2}$ , we see that $FQ = QB$ $BD = DC$ , etc. ( Stewart's Theorem )
Similarly, we have $PR = RB$ $= \frac12PB = 7.5$ ) and thus $RD = \frac12PC = 4.5$
$PDR$ is a $3-4-5$ right triangle , so $\angle PDR$ $\angle ADQ$ ) is $90^\circ$ .
Therefore, the area of $\triangle ADQ = \frac12\cdot 12\cdot 6 = 36$ .
Using area ratio, $\triangle ABC = \triangle ADB\times 2 = \left(\triangle ADQ\times \frac32\right)\times 2 = 36\cdot 3 = \boxed{108}$

Solution: First, let $[AEP]=a, [AFP]=b,$ and $[ECP]=c.$ Thus, we can easily find that $\frac{[AEP]}{[BPD]}=\frac{3}{9}=\frac{1}{3} \Leftrightarrow [BPD]=3[AEP]=3a.$ Now, $\frac{[ABP]}{[BPD]}=\frac{6}{6}=1\Leftrightarrow [ABP]=3a.$ In the same manner, we find that $[CPD]=a+c.$ Now, we can find that $\frac{[BPC]}{[PEC]}=\frac{9}{3}=3 \Leftrightarrow \frac{(3a)+(a+c)}{c}=3 \Leftrightarrow c=2a.$ We can now use this to find that $\frac{[APC]}{[AFP]}=\frac{[BPC]}{[BFP]}=\frac{PC}{FP} \Leftrightarrow \frac{3a}{b}=\frac{6a}{3a-b} \Leftrightarrow a=b.$ Plugging this value in, we find that $\frac{FC}{FP}=3 \Leftrightarrow PC=15, FP=5.$ Now, since $\frac{[AEP]}{[PEC]}=\frac{a}{2a}=\frac{1}{2},$ we can find that $2AE=EC.$ Setting $AC=b,$ we can apply Stewart's Theorem on triangle $APC$ to find that $(15)(15)(\frac{b}{3})+(6)(6)(\frac{2b}{3})=(\frac{2b}{3})(\frac{b}{3})(b)+(b)(3)(3).$ Solving, we find that $b=\sqrt{405} \Leftrightarrow AE=\frac{b}{3}=\sqrt{45}.$ But, $3^2+6^2=45,$ meaning that $\angle{APE}=90 \Leftrightarrow [APE]=\frac{(6)(3)}{2}=9=a.$ Since $[ABC]=a+a+2a+2a+3a+3a=12a=(12)(9)=108,$ we conclude that the answer is $\boxed{108}$

Solution:
Firstly, since they all meet at one single point, denoting the mass of them separately. Assuming $M(A)=6;M(D)=6;M(B)=3;M(E)=9$ ; we can get that $M(P)=12;M(F)=9;M(C)=3$ ; which leads to the ratio between segments, \[\frac{CE}{AE}=2;\frac{BF}{AF}=2;\frac{BD}{CD}=1.\] Denoting that $CE=2x;AE=x; AF=y; BF=2y; CD=z; DB=z.$
Now we know three cevians' length, Applying Stewart theorem to them, getting three different equations: \begin{align} (3x)^2 \cdot 2y+(2z)^2 \cdot y&=(3y)(2y^2+400), \\ (3y)^2 \cdot z+(3x)^2 \cdot z&=(2z)(z^2+144), \\ (2z)^2 \cdot x+(3y)^2 \cdot x&=(3x)(2x^2+144). \end{align} After solving the system of equation, we get that $x=3\sqrt{5};y=\sqrt{13};z=3\sqrt{13}$
pulling $x,y,z$ back to get the length of $AC=9\sqrt{5};AB=3\sqrt{13};BC=6\sqrt{13}$ ; now we can apply Heron's formula here, which is \[\sqrt\frac{(9\sqrt{5}+9\sqrt{13})(9\sqrt{13}-9\sqrt{5})(9\sqrt{5}+3\sqrt{13})(9\sqrt{5}-3\sqrt{13})}{16}=108.\]
Our answer is $\boxed{108}$

Solution: In Solution 5, instead of finding all of $x, y, z$ , we only need $y, z$ . This is because after we solve for $y, z$ , we can notice that $\triangle BAD$ is isosceles with $AB = BD$ . Because $P$ is the midpoint of the base, $BP$ is an altitude of $\triangle BAD$ . Therefore, $[BAD] = \frac{(AD)(BP)}{2} = \frac{12 \cdot 9}{2} = 54$ . Using the same altitude property, we can find that $[ABC] = 2[BAD] = 2 \cdot 54 = \boxed{108}$

Solution: Set $AF=x,$ and use mass points to find that $PF=5$ and $BF=2x.$ Using Stewart's Theorem on $APB,$ we find that $AB=3\sqrt{13}.$ Then we notice that $APB$ is right, which means the area of $APB$ is $27.$ Because $CF=4\cdot PF,$ the area of $ABC$ is $4$ times the area of $APB,$ which means the area of $ABC=4\cdot 27=\boxed{108}.$

Solution:
We try to solve this using only elementary concepts. Let the areas of triangles $BCP$ $ACP$ and $ABP$ be $X$ $Y$ and $Z$ respectively. Then $\frac{X}{Y+Z}=\frac{6}{6}=1$ and $\frac{Y}{X+Z}=\frac{3}{9}=\frac{1}{3}$ . Hence $\frac{X}{2}=Y=Z$ . Similarly $\frac{FP}{PC}=\frac{Z}{X+Y}=\frac{1}{3}$ and since $CF=20$ we then have $FP=5$ . Additionally we now see that triangles $FPE$ and $CPB$ are similar, so $FE \parallel BC$ and $\frac{FE}{BC} = \frac{1}{3}$ . Hence $\frac{AF}{FB}=\frac{1}{2}$ . Now construct a point $K$ on segment $BP$ such that $BK=6$ and $KP=3$ , we will have $FK \parallel AP$ , and hence $\frac{FK}{AP} = \frac{FK}{6} = \frac{2}{3}$ , giving $FK=4$ . Triangle $FKP$ is therefore a 3-4-5 triangle! So $FK \perp BE$ and so $AP \perp BE$ . Then it is easy to calculate that $Z = \frac{1}{2} \times 6 \times 9 = 27$ and the area of triangle $ABC = X+Y+Z = 4Z = 4 \times 27 = \boxed{108}$ .
~Leole

Solution: We start with mass points as in Solution 2, and receive $BF:AF = 2$ $BD:CD = 1$ $CE:AE = 2$ Law of Cosines on triangles $ADB$ and $ADC$ with $\theta = \angle ADB$ and $BD=DC=x$ gives \[36+x^2-12x\cos \theta = 81\] \[36+x^2-12x\cos (180-\theta) = 36+x^2+12x\cos \theta = 225\] Adding them: $72+2x^2=306 \implies x=3\sqrt{13}$ , so $BC = 6\sqrt{13}$ . Similarly, $AB = 3\sqrt{13}$ and $AC = 9\sqrt{5}$ . Using Heron's, \[[ \triangle ABC ]= \sqrt{\left(\dfrac{9\sqrt{13}+9\sqrt{5}}{2}\right)\left(\dfrac{9\sqrt{13}09\sqrt{5}}{2}\right)\left(\dfrac{3\sqrt{13}+9\sqrt{5}}{2}\right)\left(\dfrac{-3\sqrt{13}+9\sqrt{5}}{2}\right)} = \boxed{108}.\]


Answer: 2.0

Point $P$ is outside circle $C$ on the plane. At most how many points on $C$ are $3$ cm from $P$
$\textbf{(A)} \ 1 \qquad  \textbf{(B)} \ 2 \qquad  \textbf{(C)} \ 3 \qquad  \textbf{(D)} \ 4 \qquad  \textbf{(E)} \ 8$

Solution: The points $3$ cm away from $P$ can be represented as a circle centered at $P$ with radius $3$ cm. The maximum number of intersection points of two circles is $\boxed{2}$


Answer: 96.0

Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$

Solution:
This is a combination of Solutions 1 and 2.
First, draw $O_1P,O_2P,BP,DP$ . Then, observe that $\angle BAP=45$ implies that $\angle BO_1P=90$ . So, $\triangle BO_1P$ is a $45-90-45$ triangle. Similarly, observe that $DO_2P$ is too. So, a rotation of $\angle O_1PO_2$ to $\angle BPO_2$ adds $45$ degrees. Then, moving to $BPD$ subtracts $45$ degrees. Hence, $\angle BPD=120$ . Let the intersection of $BD$ and $AC$ be $Q$ . Then $BQP$ is a $30-90-60$ triangle, hence $QP=\frac{6\sqrt{2}}{\sqrt{3}}$ (We know that $BQ$ is $6\sqrt{2}$ ), or $QP=2\sqrt{6}$ Finally, $AP=QP+AQ=2\sqrt{6}+6\sqrt{2}=\sqrt{24}+\sqrt{72} \Rightarrow \boxed{096}$

Solution:
Use vectors. In an $xy$ plane, let $(-s,0)$ be $A$ $(0,s)$ be $B$ $(s,0)$ be $C$ $(0,-s)$ be $D$ , and $(p,0)$ be P, where $s=|AB|/\sqrt{2}=6\sqrt{2}$ . It remains to find $p$
The line $y=-x$ is the perpendicular bisector of $AB$ and $CD$ , so $O_1$ and $O_2$ lies on the line. Now compute the perpendicular bisector of $AP$ . The center has coordinate $(\frac{p-s}{2},0)$ , and the segment is part of the $x$ -axis, so the perpendicular bisector has equation $x=\frac{p-s}{2}$ . Since $O_1$ is the circumcenter of triangle $ABP$ , it lies on the perpendicular bisector of both $AB$ and $AP$ , so \[O_1=(\frac{p-s}{2},-\frac{p-s}{2})\] Similarly, \[O_2=(\frac{p+s}{2},-\frac{p+s}{2})\] The relation $\angle O_1PO_2=120^\circ$ can now be written using dot product as \[\vec{PO_1}\cdot\vec{PO_2}=|\vec{PO_1}|\cdot|\vec{PO_2}|\cos 120^\circ=-\frac{1}{2}|\vec{PO_1}|\cdot|\vec{PO_2}|\] Computation of both sides yields \[\frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2}\] Solve for $p$ gives $p=s/\sqrt{3}=2\sqrt{6}$ , so $AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}$ . The answer is 72+24 $\Rightarrow\boxed{096}$

Solution:
Translate $\triangle{ABP}$ so that the image of $AB$ coincides $DC$ . Let the image of $P$ be $P’$
$\angle{DPC}=\angle{CPB}$ by symmetry, and $\angle{APB}=\angle{DP’C}$ because translation preserves angles. Thus $\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ$ . Therefore, quadrilateral $CPDP’$ is cyclic. Thus the image of $O_1$ coincides with $O_2$
$O_1P$ is parallel to $O_2P’$ so $\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ$ , so $\angle{PDP’}=60^\circ$ and $\angle{PDC}=15^\circ$ , thus $\angle{ADP}=75^{\circ}$
Let $M$ be the foot of the perpendicular from $D$ to $AC$ . Then $\triangle{AMD}$ is a 45-45-90 triangle and $\triangle{DMP}$ is a 30-60-90 triangle. Thus
$AM=6\sqrt{2}$ and $MP=\frac{6\sqrt{2}}{\sqrt{3}}$
This gives us $AP=AM+MP=\sqrt{72}+\sqrt{24}$ , and the answer is $72+24=\boxed{096}.$

Solution:
Reflect $O_1$ across $AP$ to $O_1'$ . By symmetry $O_1’$ is the circumcenter of $\triangle{ADP}$
$\angle{DO_1’P}$ $2*\angle{DAP} = 90^\circ$ , so $\angle{O_1’PD}=45^\circ$
similarly $\angle{DO_2P}$ $2*\angle{DCP} = 90^\circ$ , so $\angle{O_2PD}=45^\circ$
Therefore $\angle{O_1’PO_2}=90^\circ$ , so that $\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ$
By symmetry, $\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ$
Therefore, since $O_1’$ is the circumcenter of $\triangle{ADP}$ $\angle{ADP}$ $0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ$
Therefore $\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ$
Using sine rule in $\triangle{ADP}$ $AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}$ , and the answer is $72+24=\boxed{096}.$

Solution:
Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set $A=(0, 12)$ $B=(12,12)$ $C=(12, 0)$ $D=(0, 0)$ . Let this $P=(a, 12-a)$ for some $a$
We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that $O_{1}$ is the intersection of $x=6$ and, knowing the midpoint of $AP$ to be $(\frac{a}{2}, \frac{12-a}{2})$ and thus the equation to be $y=x+(12-a)$ , we get $(6, 18-a)$ . Likewise for $O_{2}$ it's $(6, 6-a)$ . Now what do we see? $O_{1}P=O_{2}P$ (just look at the coordinates)! So both of those distances are $4\sqrt{3}$ . Solving for $a$ we get it to be $6+2\sqrt{3}$ , since $AP>CP$ . Multiply by $\sqrt{2}$ because we are looking for $AP$ to get the answer of $\boxed{096}$

Solution:
Let $\angle APD = \theta$ . Then $\angle ADP = 180^{\circ}-45^{\circ}-\theta=135^{\circ}-\theta \implies \angle PDC=\theta-45^{\circ}$ .
Realize that because $O_1$ is a circumcenter, $\angle DO_1P=\angle DCP=45^{\circ} \implies \angle DPO_1=\frac{180^{\circ}-\angle DO_1P}{2}=45^{\circ}$ . Then $\angle O_2PA=75^{\circ}-\theta \implies \angle AOP = 180^{\circ}-2\angle O_2PA = 2\theta+30^{\circ} \implies \angle ABP=\theta + 15^{\circ} \implies \angle PBC=75-\theta$ .
Now, because $P$ lies on diagonal $AC$ $\triangle PDC \cong \triangle PBC \implies \angle PDC = \angle PBC \implies \theta-45^{\circ}=75^{\circ}-\theta \implies \theta = 60^{\circ}$ .
To finish, we look at $\triangle ADP$ . Drop a perpendicular from $D$ to $AP$ at $E$ . Then $\triangle ADE$ is a $45-45-90$ and $\triangle PDE$ is a $30-60-90$ . Therefore, $DE=EA=6\sqrt{2}, EP=2\sqrt{6}$ , so $AP=AE+EP=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24} \implies \boxed{096}$ $\blacksquare$ ~msc

Solution:
Both $O_1$ and $O_2$ lie on the perpendicular bisector of $AB$
Claim: $O_1O_2=12$ and $O_1P=O_2P$
Proof. Translate $O_1$ and $P$ $12$ units down, and let their images be $O_1'$ and $P'$ , respectively. Note that $\triangle ABP\cong\triangle DCP'$ . Additionally, \[\angle CP'D = \angle BPA = 180^{\circ} - \angle BPC = 180^{\circ} - \angle CPD,\] so $CPDP'$ is cyclic. This means $O_1'$ and $O_2$ coincide, so $O_1O_2=12$ . This also means the circumradii of both triangles are equal, so $O_1P=O_2P$ $\blacksquare$
Let the perpendicular from $P$ intersect $O_1O_2$ at $X$ and $AD$ at $Y$ . Since $\triangle O_1XP$ is 30-60-90, $XP=\frac{6}{\sqrt{3}} = 2\sqrt3$ . Since $YX=6$ $PY=6+2\sqrt3$ , so $AP=6\sqrt2+2\sqrt6 = \sqrt{72}+\sqrt{24} \implies\boxed{96}$

Solution:
\[BP = DP,    \angle PAB =\angle PCD  = 45^\circ \implies  O_{1}P =O_{2}P\] by the Law of Sines.
\[AB= CD, O_{1}A = O_{2}D = O_{1}B = O_{2}C \implies \triangle AO_{1}B = \triangle DO_{2}C.\] They have translational symmetry. The translation vector is \[\vec {AD} \implies  O_{1}O_{2} = AD.\]
\[\triangle PO_{1}O_{2} = \triangle O_{1}AB \implies \angle AO_{1}B = 120^\circ \implies\] \[\angle APB = 60^\circ \implies  \angle ABP = 180^\circ – 60^\circ – 45^\circ = 75^\circ.\] By the Law of Sines \[AP = \frac {AB \cdot \sin 75^\circ}{\sin 60^\circ} = \frac {AB \cdot \sin (30^\circ + 45^\circ)} {\sin 60^\circ}\] \[AP  = AB \cdot (\sin 45^\circ + \cos 45^\circ \cdot \tan 30^\circ),\] \[AP = \frac {AB}{\sqrt{2}} (1 + \frac {1}{\sqrt{3}}) = 6 \sqrt {2} + 2 \sqrt {6} \implies \boxed{096}\] vladimir.shelomovskii@gmail.com, vvsss


Answer: 463.0

Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent.  The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers.  Find $m+n.$

Solution:
Drop perpendiculars from $P$ to the three sides of $\triangle ABC$ and let them meet $\overline{AB}, \overline{BC},$ and $\overline{CA}$ at $D, E,$ and $F$ respectively.
Let $BE = x, CF = y,$ and $AD = z$ .  We have that \begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\ FP&=y\tan\theta\end{align*} We can then use the tool of calculating area in two ways \begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\\ &=\frac{1}{2}(13)(z\tan\theta)+\frac{1}{2}(14)(x\tan\theta)+\frac{1}{2}(15)(y\tan\theta)\\ &=\frac{1}{2}\tan\theta(13z+14x+15y)\end{align*} On the other hand, \begin{align*}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{21\cdot6\cdot7\cdot8}\\ &=84\end{align*} We still need $13z+14x+15y$ though. We have all these right triangles and we haven't even touched Pythagoras . So we give it a shot: \begin{align}x^2+x^2\tan^2\theta&=z^2\tan^2\theta+(13-z)^2\\ z^2+z^2\tan^2\theta&=y^2\tan^2\theta+(15-y)^2\\ y^2+y^2\tan^2\theta&=x^2\tan^2\theta+(14-x)^2\end{align} Adding $(1) + (2) + (3)$ gives \begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\ \Rightarrow13z+14x+15y&=295\end{align*} Recall that we found that $[ABC]=\frac{1}{2}\tan\theta(13z+14x+15y)=84$ . Plugging in $13z+14x+15y=295$ , we get $\tan\theta=\frac{168}{295}$ , giving us $\boxed{463}$ for an answer.

Solution:
Let $AB=c$ $BC=a$ $AC=b$ $PA=x$ $PB=y$ , and $PC=z$
So by the Law of Cosines , we have: \begin{align*}x^2 &= z^2 + b^2 - 2bz\cos{\theta}\\ y^2 &= x^2 + c^2 - 2cx\cos{\theta}\\ z^2 &= y^2 + a^2 - 2ay\cos{\theta}\end{align*} Adding these equations and rearranging, we have: \[a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}\qquad(1)\] Now $[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84$ , by Heron's formula
Now the area of a triangle, $[A] = \frac {mn\sin{\beta}}{2}$ , where $m$ and $n$ are sides on either side of an angle, $\beta$ . So, \begin{align*}[CAP] &= \frac {bz\sin{\theta}}{2}\\ [ABP] &= \frac {cx\sin{\theta}}{2}\\ [BCP] &= \frac {ay\sin{\theta}}{2}\end{align*} Adding these equations yields: \begin{align*}[ABC]= 84 &= \frac {(bz + cx + ay)\sin{\theta}}{2}\\ \Rightarrow 168&= (bz + cx + ay)\sin{\theta}\qquad (2)\end{align*} Dividing $(2)$ by $(1)$ , we have: \begin{align*}\frac {168}{a^2 + b^2 + c^2} &= \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}\\ \Rightarrow \tan{\theta} = \frac {336}{a^2 + b^2 + c^2} &= \frac {336}{14^2 + 15^2 + 13^2} = \frac {336}{590} = \frac {168}{295}\end{align*} Thus, $m + n = 168 + 295 = \boxed{463}$

Solution: Let $\angle{PAB} = \angle{PBC} = \angle{PCA} = x.$ Then, using Law of Cosines on the three triangles containing vertex $P,$ we have \begin{align*} b^2 &= a^2 + 169 - 26a \cos x \\ c^2 &= b^2 + 196 - 28b \cos x \\ a^2 &= c^2 + 225 - 30c \cos x. \end{align*} Add the three equations up and rearrange to obtain \[(13a + 14b + 15c) \cos x = 295.\] Also, using $[ABC] = \frac{1}{2}ab \sin \angle C$ we have \[[ABC] = [APB] + [BPC] + [CPA] = \dfrac{\sin x}{2}(13a + 14b + 15c) = 84 \iff (13a + 14b + 15c) \sin x = 168.\] Divide the two equations to obtain $\tan x = \frac{168}{295} \iff \boxed{463}.~\square$

Solution:
Firstly, denote angles $ABC$ $BCA$ , and $CAB$ as $B$ $A$ , and $C$ respectively. Let $\angle{PAB}=x$ .
Notice that by angle chasing that $\angle{BPC}=180-C$ and $\angle{BPA}=180-B$ .
Using the nice properties of the 13-14-15 triangle, we have $\sin B = \frac{12}{13}$ and $\sin C = \frac{4}{5}$ $\cos C$ is easily computed, so we have $\cos C=\frac{3}{5}$
Using Law of Sines, \begin{align*} \frac{BP}{\sin x} &= \frac{13}{\sin (180 - B)} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin (180 - C)} \end{align*} hence, \begin{align*} \frac{BP}{\sin x} &= \frac{13}{\sin B} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin C} \end{align*} Now, computation carries the rest. \begin{align*} \frac{13 \sin x}{\sin B} &= \frac{14 \sin (C-x)}{\sin C}  \\ \frac{169 \sin x}{12} &= \frac{210 \sin (C-x)}{12} \\ 169 \sin x &= 210 (\sin C \cos x - \cos C \sin x) \\ 169 \sin x &= 210 (\frac{4}{5} \cos x - \frac{3}{5} \sin x) \\ 169 \sin x &= 168 \cos x - 126 \sin x \\ 295 \sin x &= 168 \cos x \\ \tan x &= \frac{168}{295} \end{align*} Extracting yields $168 + 295 = \boxed{463}$


Answer: 2.0

Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$

Solution:
Since points on the graph make the equation true, substitute $\sqrt{\pi}$ in to the equation and then solve to find $a$ and $b$
$y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1$
$y^2 + \pi^2 = 2\pi y + 1$
$y^2 - 2\pi y + \pi^2 = 1$
$(y-\pi)^2 = 1$
$y-\pi = \pm 1$
$y = \pi + 1$
$y = \pi - 1$
There are only two solutions to the equation $(y-\pi)^2 = 1$ , so one of them is the value of $a$ and the other is $b$ . The order does not matter because of the absolute value sign.
$| (\pi + 1) - (\pi - 1) | = 2$
The answer is $\boxed{2}$

Solution:
This solution is very closely related to Solution #1 and just simplifies the problem earlier to make it easier.
$y^2 + x^4 = 2x^2 y + 1$ can be written as $x^4-2x^2y+y^2=1$ . Recognizing that this is a binomial square, simplify this to $(x^2-y)^2=1$ . This gives us two equations:
$x^2-y=1$ and $x^2-y=-1$
One of these $y$ 's is $a$ and one is $b$ . Substituting $\sqrt{\pi}$ for $x$ , we get $a=\pi+1$ and $b=\pi-1$
So, $|a-b|=|(\pi+1)-(\pi-1)|=2$
The answer is $\boxed{2}$

Solution:
This solution is similar to Solution #1 but uses a different way to find $y$ at the end.
Just like Solution #1, we arrive at the conclusion that $y^2 - 2\pi y + \pi^2 = 1$
Simplifying we get:
$y^2 - 2\pi y + \pi^2 -1 = 0$
We now can factor this quadratic. We must find two terms that multiply to $\pi^2 -1$ and add to $2\pi$
These terms are $\pi+1$ and $\pi-1$
Subtracting one from the other, we get $2$
Thus, the answer is $\boxed{2}$


Answer: 10.0

Points $A = (3,9)$ $B = (1,1)$ $C = (5,3)$ , and $D=(a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ABCD$ . The quadrilateral formed by joining the midpoints of $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ , and $\overline{DA}$ is a square. What is the sum of the coordinates of point $D$
$\text{(A) }7 \qquad \text{(B) }9 \qquad \text{(C) }10 \qquad \text{(D) }12 \qquad \text{(E) }16$

Solution:
We already know two vertices of the square: $(A+B)/2 = (2,5)$ and $(B+C)/2 = (3,2)$
There are only two possibilities for the other vertices of the square: either they are $(6,3)$ and $(5,6)$ , or they are $(0,1)$ and $(-1,4)$ . The second case would give us $D$ outside the first quadrant, hence the first case is the correct one. As $(6,3)$ is the midpoint of $CD$ , we can compute $D=(7,3)$ , and $7+3=\boxed{10}$


Answer: 18.0

Solution: This quadrilateral is a trapezoid, because $AB\parallel CD$ but $BC$ is not parallel to $AD$ . The area of a trapezoid is the product of its height and its median, where the median is the average of the side lengths of the bases. The two bases are $AB$ and $CD$ , which have lengths $2$ and $4$ , respectively, so the length of the median is $\frac{2+4}{2}=3$ $CB$ is perpendicular to the bases, so it is the height, and has length $6$ . Therefore, the area of the trapezoid is $(3)(6)=18, \boxed{18}$

Solution:
Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a rectangle.  Since the areas do not overlap, the area of the entire trapezoid is the sum of the area of the triangle and the area of the rectangle.
$A_{trap} = A_{tri} + A_{rect}$
$A_{trap} = \frac{1}{2}bh + lw$
$A_{trap} = \frac{1}{2}\cdot 6 \cdot 2 + 6\cdot 2$
$A_{trap}  = 6 + 12 = 18 \rightarrow \boxed{18}$


Answer: 800.0

Points $A$ $B$ , and $C$ lie in that order along a straight path where the distance from $A$ to $C$ is $1800$ meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at $A$ and running toward $C$ , Paul starting at $B$ and running toward $C$ , and Eve starting at $C$ and running toward $A$ . When Paul meets Eve, he turns around and runs toward $A$ . Paul and Ina both arrive at $B$ at the same time. Find the number of meters from $A$ to $B$

Solution:
We know that in the same amount of time given, Ina will run twice the distance of Eve, and Paul would run quadruple the distance of Eve. Let's consider the time it takes for Paul to meet Eve: Paul would've run 4 times the distance of Eve, which we can denote as $d$ . Thus, the distance between $B$ and $C$ is $4d+d=5d$ . In that given time, Ina would've run twice the distance $d$ , or $2d$ units.
Now, when Paul starts running back towards $A$ , the same amount of time would pass since he will meet Ina at his starting point. Thus, we know that he travels another $4d$ units and Ina travels another $2d$ units.
Therefore, drawing out the diagram, we find that $2d+2d+4d+d=9d=1800 \implies d=200$ , and distance between $A$ and $B$ is the distance Ina traveled, or $4d=4(200)=\boxed{800}$

Solution:
Let $x$ be the distance from $A$ to $B$ . Then the distance from $B$ to $C$ is $1800-x$ . Since Eve is the slowest, we can call her speed $v$ , so that Ina's speed is $2v$ and Paul's speed is $4v$
For Paul and Eve to meet, they must cover a total distance of $1800-x$ which takes them a time of $\frac{1800-x}{4v+v}$ . Paul must run the same distance back to $B$ , so his total time is $\frac{2(1800-x)}{5v}$
For Ina to reach $B$ , she must run a distance of $x$ at a speed of $2v$ , taking her a time of $\frac{x}{2v}$
Since Paul and Ina reach $B$ at the same time, we know that $\frac{2(1800-x)}{5v} = \frac{x}{2v}$ (notice that $v$ cancels out on both sides). Solving the equation gives $x = \boxed{800}$


Answer: 3.0

Points $A$ and $B$ are 10 units apart.  Points $B$ and $C$ are 4 units apart.  Points $C$ and $D$ are 3 units apart.  If $A$ and $D$ are as close as possible, then the number of units between them is
$\text{(A)}\ 0 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 11 \qquad \text{(E)}\ 17$

Solution:
If $AB = 10$ and $BC=4$ , then $(10 - 4) \le AC \le (10 + 4)$ by the triangle inequality .  In the triangle inequality, the equality is only reached when the "triangle" $ABC$ is really a degenerate triangle, and $ABC$ are collinear.
Simplifying, this means the smallest value $AC$ can be is $6$
Applying the triangle inequality on $ACD$ with $AC = 6$ and $CD = 3$ , we know that $6 - 3 \le AD \le 6 + 3$ when $AC$ is minimized.  If $AC$ were larger, then $AD$ could be larger, but we want the smallest $AD$ possible, and not the largest.  Thus, $AD$ must be at least $3$ , but cannot be smaller than $3$ .  Therefore, $\boxed{3}$ is the answer.


Answer: 12.0

Points $A$ and $C$ lie on a circle centered at $O$ , each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$ . What is $\frac{BD}{BO}$
$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {\sqrt3}{3} \qquad \text{(D) } \frac {\sqrt2}{2} \qquad \text{(E) } \frac {\sqrt3}{2}$

Solution:
As $\triangle ABC$ is equilateral, we have $\angle BAC = \angle BCA = 60^\circ$ , hence $\angle OAC = \angle OCA = 30^\circ$ . Then $\angle AOC = 120^\circ$ , and from symmetry we have $\angle AOB = \angle COB = 60^\circ$ . Thus, this gives us $\angle ABO = \angle CBO = 30^\circ$
We know that $DO = AO$ , as $D$ lies on the circle. From $\triangle ABO$ we also have $AO = BO \sin 30^\circ = \frac{BO}2$ , Hence $DO = \frac{BO}2$ , therefore $BD = BO - DO = \frac{BO}2$ , and $\frac{BD}{BO} = \boxed{12}$


Answer: 12.0

Solution:
As in the previous solution, we find out that $\angle AOB = \angle COB = 60^\circ$ . Hence $\triangle AOD$ and $\triangle COD$ are both equilateral.
We then have $\angle SCD = \angle SAD = 30^\circ$ , hence $D$ is the incenter of $\triangle ABC$ , and as $\triangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 \cdot SD = BD$ , and as $SD = SO$ , we have $2\cdot SD = SD + SO = OD$ , therefore $BD=OD$ , and as before we conclude that $\frac{BD}{BO} = \boxed{12}$


Answer: 49.0

Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$

Solution: Since $AB = AC$ , then $\triangle ABC$ is isosceles, so $BD = CD$ . Therefore, the coordinates of $C$ are $(-1 - 3, 3 + 6) = \boxed{4,9}$

Solution: Calculating the equation of the line running between points $B$ and $D$ $y = -2x + 1$ . The only coordinate of $C$ that is also on this line is $\boxed{4,9}$

Solution: Similar to the first solution, because the triangle is isosceles, then the line drawn in the middle separates the triangle into two smaller congruent triangles. To get from $B$ to $D$ , we go to the right $3$ and up $6$ . Then to get to point $C$ from point $D$ , we go to the right $3$ and up $6$ , getting us the coordinates $\boxed{4,9}$

Solution:
As stated in solution 1, the triangle is isosceles.

This means that $D(-1, 3)$ is the midpoint of $B(2, -3)$ and $C(x,y)$ . So $\frac{x+2}{2}$ $= -1$ and so $x = -4$ . Similarly for $y$ , we have $\frac{y-3}{2}$ $=3$ and so $y=9$ . So our final answer is $\boxed{4,9}$


Answer: 30.0

Points $A, B, C$ and $D$ are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy]
$\textbf{(A)}\ 15 \qquad  \textbf{(B)}\ 20 \qquad  \textbf{(C)}\ 24 \qquad  \textbf{(D)}\ 30 \qquad  \textbf{(E)}\ 40$

Solution: Drawing segments $AC$ and $BD$ , the number of triangles outside square $ABCD$ is the same as the number of triangles inside the square. Thus areas must be equal so the area of $ABCD$ is half the area of the larger square which is $\frac{60}{2}=\boxed{30}$

Solution:
If the side length of the larger square is $x$ , the side length of the smaller square is $\frac{\sqrt{2} \cdot x}{2}$ . Therefore the area of the smaller square is $\frac{x^2}{2}$ , half of the larger square's area, $x^2$
Thus, the area of the smaller square in the picture is $\frac{60}{2}=\boxed{30}$


Answer: 9.0

Points $A,B,C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$ . Point $E$ is not on the line, and $BE = CE = 10$ . The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$ . Find $AB$
$\text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 19/2$

Solution: Let $M$ be the foot of the altitude from $E$ to $BC.$ Then $MB=MC=6$ because $\triangle BEC$ is isosceles. By the Pythagorean triple $(6,8,10)$ the altitude is $8.$ Since $(8,15,17)$ is the only primitive Pythagorean triple with leg $8,$ we test $AE=DE=17,AM=DM=15.$ Since $2(10+10+12)=(17+17+2\cdot 15)$ this works, giving us $AB=15-6=\boxed{9}.$


Answer: 6.0

Solution:
Since $\triangle FGH$ is fairly small relative to the rest of the diagram, we can make an underestimate by using the current diagram.  All triangles are right-isosceles triangles.
$CD = \frac {CG}{2} = 3, DE = \frac{CD}{2} = \frac{3}{2}, EF = \frac{DE}{2} = \frac{3}{4}$
$CB = CD = 3, DK = DE = \frac{3}{2}, EL = EF = \frac{3}{4}$
$[CBD] = \frac{1}{2}3^2 = \frac{9}{2}$
$[DKE] = \frac{1}{2}(\frac{3}{2})^2 = \frac{9}{8}$
$[ELF] = \frac{1}{2}(\frac{3}{4})^2 = \frac{9}{32}$
The sum of the shaded regions is $\frac{9}{2} + \frac{9}{8} + \frac{9}{32} = \frac{189}{32} \approx 5.9$
$5.9$ is an underestimate, as some portion (but not all) of $\triangle FGH$ will be shaded in future iterations.
If you shade all of $\triangle FGH$ , this will add an additional $\frac{9}{32}$ to the area, giving $\frac{198}{32} \approx 6.2$ , which is an overestimate.
Thus, $6 \rightarrow \boxed{6}$ is the only answer that is both over the underestimate and under the overestimate.

Solution:
In iteration $1$ , congruent triangles $\triangle ABJ,  \triangle BDJ,$ and $\triangle BDC$ are created, with one of them being shaded.
In iteration $2$ , three more congruent triangles are created, with one of them being shaded.
As the process continues indefnitely, in each row, $\frac{1}{3}$ of each triplet of new congruent triangles will be shaded.  The "fourth triangle" at the top ( $\triangle FGH$ in the diagram) will gradually shrink,
leaving about $\frac{1}{3}$ of the area shaded.  This means $\frac{1}{3}\left(\frac{1}{2}6\cdot 6\right) = 6$ square units will be shaded when the process goes on indefinitely, giving $\boxed{6}$

Solution:
Using Solution 1 as a template, note that the sum of the areas forms a geometric series
$\frac{9}{2} + \frac{9}{8} + \frac{9}{32} + \frac{9}{128} + ...$
This is the sum of a geometric series with first term $a_1 = \frac{9}{2}$ and common ratio $r = \frac{1}{4}$ This is the easiest way to do this problem.
The sum of an infinite geometric series with $|r|<1$ is shown by the formula. $S_{\infty} = \frac{a_1}{1 - r}$ Insert the values to get $\frac{\frac{9}{2}}{1 - \frac{1}{4}} = \frac{9}{2}\cdot\frac{4}{3} = 6$ , giving an answer of $\boxed{6}$

Solution: Find the area of the bottom triangle, which is 4.5. Notice that the area of the triangles is divided by 4 every time. 4.5*5/4≈5.7, and 5.7+(1/4)≈5.9. We can clearly see that the sum is approaching answer choice $\boxed{6}$

Solution: Obviously, the area of $DBC$ is $\frac{1}{3}$ of $AJDC$ , the area of $DEA$ is $\frac{1}{3}$ of $JIED$ , if the progress is going to infinity, the shaded triangles will be $\frac{1}{3}$ of the triangle $ACG$ . However, 100 times is much enough. The answer is $\frac{1}{3}\times 6\times 6=\boxed{6}$


Answer: 2.0

Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$
$\mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3}$

Solution: Since triangle $BEF$ is equilateral, $EA=FC$ , and $EAB$ and $FCB$ are $SAS$ congruent. Thus, triangle $DEF$ is an isosceles right triangle. So we let $DE=x$ . Thus $EF=EB=FB=x\sqrt{2}$ . If we go angle chasing, we find out that $\angle AEB=75^{\circ}$ , thus $\angle ABE=15^{\circ}$ $\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}$ . Thus $\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}$ , or $AE=\frac{x(\sqrt{3}-1)}{2}$ . Thus $AB=\frac{x(\sqrt{3}+1)}{2}$ , and $[ABE]=\frac{x^2}{4}$ , and $[DEF]=\frac{x^2}{2}$ . Thus the ratio of the areas is $\boxed{2}$

Solution: WLOG, let the side length of $ABCD$ be 1. Let $DE = x$ . It suffices that $AE = 1 - x$ . Then triangles $ABE$ and $CBF$ are congruent by HL, so $CF = AE$ and $DE = DF$ . We find that $BE = EF = x \sqrt{2}$ , and so, by the Pythagorean Theorem, we have $(1 - x)^2 + 1 = 2x^2.$ This yields $x^2 + 2x = 2$ , so $x^2 = 2 - 2x$ . Thus, the desired ratio of areas is \[\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{2}.\]

Solution: Assume $AB=1$ . Then, $FC$ is $x$ and $ED$ is $1-x$ . We see that using $HL$ $FCB$ is congruent to EAB. Using Pythagoras of triangles $FCB$ and $FDE$ we get $2{(1-x)}^2=x^2+1$ . Expanding, we get $2x^2-4x+2=x^2+1$ . Simplifying gives $x^2-4x+1=0$ solving using completing the square (or other methods) gives 2 answers: $2-\sqrt{3}$ and $2+\sqrt{3}$ . Because $x < 1$ $x=2-\sqrt{3}$ . Using the areas, the answer is $\boxed{2}$

Solution: First, since $\bigtriangleup BEF$ is equilateral and $ABCD$ is a square, by the Hypothenuse Leg Theorem, $\bigtriangleup ABE$ is congruent to $\bigtriangleup CBF$ . Then, assume length $AB = BC = x$ and length $DE = DF = y$ , then $AE = FC = x - y$ $\bigtriangleup BEF$ is equilateral, so $EF = EB$ and $EB^2 = EF^2$ , it is given that $ABCD$ is a square and $\bigtriangleup DEF$ and $\bigtriangleup ABE$ are right triangles. Then we use the Pythagorean theorem to prove that $AB^2 + AE^2 = EB^2$ and since we know that $EB^2 = EF^2$ and $EF^2 = DE^2 + DF^2$ , which means $AB^2 + AE^2 = DE^2 + DF^2$ . Now we plug in the variables and the equation becomes $x^2 + (x+y)^2 = 2y^2$ , expand and simplify and you get $2x^2 - 2xy = y^2$ . We want the ratio of area of $\bigtriangleup DEF$ to $\bigtriangleup ABE$ . Expressed in our variables, the ratio of the area is $\frac{y^2}{x^2 - xy}$ and we know $2x^2 - 2xy = y^2$ , so the ratio must be $2$ . So, the answer is $\boxed{2}$


Answer: 1.0

Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB$ . As $P$ moves along a line that is parallel to side $AB$ , how many of the four quantities listed below change?
(a) the length of the segment $MN$
(b) the perimeter of $\triangle PAB$
(c) the area of $\triangle PAB$
(d) the area of trapezoid $ABNM$
[asy] draw((2,0)--(8,0)--(6,4)--cycle); draw((4,2)--(7,2)); draw((1,4)--(9,4),Arrows); label("$A$",(2,0),SW); label("$B$",(8,0),SE); label("$M$",(4,2),W); label("$N$",(7,2),E); label("$P$",(6,4),N); [/asy]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution:
(a) Triangles $ABP$ and $MNP$ are similar, and since $PM=\frac{1}{2}AP$ $MN=\frac{1}{2}AB$
(b) We see the perimeter changes. For example, imagine if P was extremely far to the left.
(c) The area clearly doesn't change, as both the base $AB$ and its corresponding height remain the same.
(d) The bases $AB$ and $MN$ do not change, and neither does the height, so the area of the trapezoid remains the same.
Only $1$ quantity changes, so the correct answer is $\boxed{1}$


Answer: 70.0

Points $P$ and $Q$ are both in the line segment $AB$ and on the same side of its midpoint. $P$ divides $AB$ in the ratio $2:3$ ,
and $Q$ divides $AB$ in the ratio $3:4$ . If $PQ=2$ , then the length of $AB$ is:
$\textbf{(A)}\ 60\qquad \textbf{(B)}\ 70\qquad \textbf{(C)}\ 75\qquad \textbf{(D)}\ 80\qquad \textbf{(E)}\ 85$

Solution:
Draw diagram as shown, where $P$ and $Q$ are on the same side.  Let $AP = x$ and $QB = y$
Since $P$ divides $AB$ in the ratio $2:3$ $\frac{x}{y+2} = \frac{2}{3}$ .  Since $Q$ divides $AB$ in the ratio $3:4$ $\frac{x+2}{y} = \frac{3}{4}$ .  Cross multiply to get a system of equations \[3x = 2y+4\] \[4x+8=3y\] Solve the system to get $x = 28$ and $y = 40$ .  Thus, $AB = 40 + 28 + 2 = 70$ , so the answer is $\boxed{70}$


Answer: 8.0

Points $P$ and $Q$ lie in a plane with $PQ=8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ $Q$ , and $R$ is a right triangle with area $12$ square units?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\  6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$

Solution:
Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$
We construct a circle with diameter $\overline{PQ}.$ All such locations for $R$ are shown below:

We apply casework to the right angle of $\triangle PQR:$
Together, there are $\boxed{8}$ such locations for $R.$

Solution:
Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$
Without the loss of generality, let $P=(-4,0)$ and $Q=(4,0).$ We conclude that the $y$ -coordinate of $R$ must be $\pm3.$
We apply casework to the right angle of $\triangle PQR:$
Together, there are $2+2+4=\boxed{8}$ such locations for $R.$


Answer: 4.0

Points $R$ $S$ and $T$ are vertices of an equilateral triangle, and points $X$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?
[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

Solution:
There are $6$ points in the figure, and $3$ of them are needed to form a triangle, so there are ${6\choose{3}} =20$ possible triplets of the $6$ points. However, some of these created congruent triangles, and some don't even make triangles at all.
Case 1: Triangles congruent to $\triangle RST$ There is obviously only $1$ of these: $\triangle RST$ itself.
Case 2: Triangles congruent to $\triangle SYZ$ There are $4$ of these: $\triangle SYZ, \triangle RXY, \triangle TXZ,$ and $\triangle XYZ$
Case 3: Triangles congruent to $\triangle RSX$ There are $6$ of these: $\triangle RSX, \triangle TSX, \triangle STY, \triangle RTY, \triangle RSZ,$ and $\triangle RTZ$
Case 4: Triangles congruent to $\triangle SYX$ There are again $6$ of these: $\triangle SYX, \triangle SZX, \triangle TYZ, \triangle TYX, \triangle RXZ,$ and $\triangle RYZ$
However, if we add these up, we accounted for only $1+4+6+6=17$ of the $20$ possible triplets. We see that the remaining triplets don't even form triangles; they are $SYR, RXT,$ and $TZS$ . Adding these $3$ into the total yields for all of the possible triplets, so we see that there are only $4$ possible non-congruent, non-degenerate triangles, $\boxed{4}$

Solution: We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is $3+1 = \boxed{4}$

Solution: Notice that 20 is obviously too high (There are only 20 ways to choose 3 of the points to form a triangle or a line in total!) and you can count 4 distinct triangles quickly: $\triangle RYX$ $\triangle RYT$ $\triangle RYZ$ $\triangle RST$ . So the answer is $\boxed{4}$


Answer: 163.0

Polyhedron $ABCDEFG$ has six faces.  Face $ABCD$ is a square with $AB = 12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF = AG = 8,$ and $GF = 6;$ and face $CDE$ has $CE = DE = 14.$ The other three faces are $ADEG, BCEF,$ and $EFG.$ The distance from $E$ to face $ABCD$ is 12.  Given that $EG^2 = p - q\sqrt {r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p + q + r.$

Solution:
We let $A$ be the origin, or $(0,0,0)$ $B = (0,0,12)$ , and $D = (12,0,0)$ . Draw the perpendiculars from F and G to AB, and let their intersections be X and Y, respectively. By symmetry, $FX = GY = \frac{12-6}2 = 3$ , so $G = (a,b,3)$ , where a and b are variables.
We can now calculate the coordinates of E. Drawing the perpendicular from E to CD and letting the intersection be Z, we have $CZ = DZ = 6$ and $EZ = 4\sqrt{10}$ . Therefore, the x coordinate of $E$ is $12-\sqrt{(4\sqrt{10})^2-12^2}=12-\sqrt{16}=12-4=8$ , so $E = (8,12,6)$
We also know that $A,D,E,$ and $G$ are coplanar, so they all lie on the plane $z = Ax+By+C$ . Since $(0,0,0)$ is on it, then $C = 0$ . Also, since $(12,0,0)$ is contained, then $A = 0$ . Finally, since $(8,12,6)$ is on the plane, then $B = \frac 12$ . Therefore, $b = 6$ . Since $GA = 8$ , then $a^2+6^2+3^2=8^2$ , or $a = \pm \sqrt{19}$ . Therefore, the two permissible values of $EG^2$ are $(8 \pm \sqrt{19})^2+6^2+3^2 = 128 \pm 16\sqrt{19}$ . The only one that satisfies the conditions of the problem is $128 - 16\sqrt{19}$ , from which the answer is $128+16+19=\boxed{163}$


Answer: 1950.0

Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$

Solution: The following system of equations can be formed with $P$ representing the number of students in Portia's high school and $L$ representing the number of students in Lara's high school: \begin{align*} P&=3L, \\ P+L&=2600. \end{align*} Substituting $P=3L$ gives $4L=2600.$ Solving for $L$ gives $L=650.$ Since we need to find $P,$ we multiply $650$ by $3$ to get $P=\boxed{1950}.$

Solution: Suppose Lara's high school has $x$ students, so Portia's high school has $3x$ students. We have $x+3x=2600,$ or $4x=2600.$ The answer is \[3x=2600\cdot\frac 34=650\cdot3=\boxed{1950}.\]

Solution: Clearly, $2600$ is $4$ times the number of students in Lara's high school. Therefore, Lara's high school has $2600\div4=650$ students, and Portia's high school has $650\cdot3=\boxed{1950}$ students.

Solution: The number of students in Portia's high school must be a multiple of $3.$ This eliminates $\textbf{(B)},\textbf{(D)},$ and $\textbf{(E)}.$ Since $\textbf{(A)}$ is too small (as it is clear that $600+\frac{600}{3}<2600$ ), we are left with $\boxed{1950}.$


Answer: 501.0

Positive integers $a$ $b$ $c$ , and $d$ satisfy $a > b > c > d$ $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$

Solution: Using the difference of squares $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010$ , where equality must hold so $b = a - 1$ and $d = c - 1$ . Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $\boxed{501}$

Solution: Since $a+b$ must be greater than $1005$ , it follows that the only possible value for $a-b$ is $1$ (otherwise the quantity $a^2 - b^2$ would be greater than $2010$ ). Therefore the only possible ordered pairs for $(a,b)$ are $(504, 503)$ $(505, 504)$ , ... , $(1004, 1003)$ , so $a$ has $\boxed{501}$ possible values.


Answer: 15.0

Positive integers $a$ and $b$ are each less than $6$ . What is the smallest possible value for $2 \cdot a - a \cdot b$
$\textbf{(A)}\ -20\qquad\textbf{{(B)}}\ -15\qquad\textbf{{(C)}}\ -10\qquad\textbf{{(D)}}\ 0\qquad\textbf{{(E)}}\ 2$

Solution: Factoring the equation gives $a(2 - b)$ . From this we can see that to obtain the least possible value, $2 - b$ should be negative, and should be as small as possible. To do so, $b$ should be maximized. Because $2 - b$ is negative, we should maximize the positive value of $a$ as well. The maximum values of both $a$ and $b$ are $5$ , so the answer is $5(2 - 5) = \boxed{15}$


Answer: 8.0

Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$ -axis at the same point. What is the sum of all possible $x$ -coordinates of these points of intersection?
$\textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8}$

Solution: Note that when $y=0$ , the $x$ values of the equations should be equal by the problem statement. We have that \[0 = ax + 5 \implies x = -\dfrac{5}{a}\] \[0 = 3x+b \implies x= -\dfrac{b}{3}\] Which means that \[-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15\] The only possible pairs $(a,b)$ then are $(a,b) = (1,15), (3,5), (5,3), (15, 1)$ . These pairs give respective $x$ -values of $-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}$ which have a sum of $\boxed{8}$

Solution: Going off of Solution 1, for the first equation, notice that the value of $x$ cannot be less than $-5$ . We also know for the first equation that the values of $x$ have to be $5$ divided by something. Also, for the second equation, the values of $x$ can only be $-\frac13,-\frac23,-\frac33, \dots$ . Therefore, we see that, the only values common between the two sequences are $-1, -5, -\frac13,-\frac53$ , and adding them up, we get for our answer, $\boxed{8}$


Answer: 881.0

Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$

Solution: To simplify, we write this logarithmic expression as an exponential one. Just looking at  the first log, it has a base of 2 and an argument of the expression in parenthesis. Therefore, we can make 2 the base, 0 the exponent, and the argument the result. That means $\log_{2^a}(\log_{2^b}(2^{1000}))=1$ (because $2^0=1$ ). Doing this again, we get $\log_{2^b}(2^{1000})=2^a$ . Doing the process one more time, we finally eliminate all of the logs, getting ${(2^{b})}^{(2^a)}=2^{1000}$ . Using the property that ${(a^x)^{y}}=a^{xy}$ , we simplify to $2^{b\cdot2^{a}}=2^{1000}$ . Eliminating equal bases leaves $b\cdot2^a=1000$ . The largest $a$ such that $2^a$ divides $1000$ is $3$ , so we only need to check $1$ $2$ , and $3$ . When $a=1$ $b=500$ ; when $a=2$ $b=250$ ; when $a=3$ $b=125$ . Summing all the $a$ 's and $b$ 's gives the answer of $\boxed{881}$

Solution:
We proceed as in Solution 1, raising $2$ to both sides to achieve $\log_{2^a}(\log_{2^b}(2^{1000})) = 1.$ We raise $2^a$ to both sides to get $\log_{2^b}(2^{1000})=2^a$ , then simplify to get $\dfrac{1000}b=2^a$
At this point, we want both $a$ and $b$ to be integers. Thus, $2^a$ can only be a power of $2$ . To help us see the next step, we factorize $1000$ $\dfrac{2^35^3}b=2^a.$ It should be clear that $a$ must be from $1$ to $3$ ; when $a=1$ $b=500$ ; when $a=2$ $b=250$ ; and finally, when $a=3$ $b=125.$ We sum all the pairs to get $\boxed{881}.$


Answer: 1002.0

Positive integers $a,b,$ and $c$ are chosen so that $a<b<c$ , and the system of equations
has exactly one solution. What is the minimum value of $c$
$\mathrm{(A)}\ 668 \qquad\mathrm{(B)}\ 669 \qquad\mathrm{(C)}\ 1002 \qquad\mathrm{(D)}\ 2003 \qquad\mathrm{(E)}\ 2004$

Solution:
Consider the graph of $f(x)=|x-a|+|x-b|+|x-c|$
When $x<a$ , the slope is $-3$
When $a<x<b$ , the slope is $-1$
When $b<x<c$ , the slope is $1$
When $c<x$ , the slope is $3$
Setting $x=b$ gives $y=|b-a|+|b-b|+|b-c|=c-a$ , so $(b,c-a)$ is a point on $f(x)$ . In fact, it is the minimum of $f(x)$ considering the slope of lines to the left and right of $(b,c-a)$ . Thus, graphing this will produce a figure that looks like a cup:  From the graph, it is clear that $f(x)$ and $2x+y=2003$ have one intersection point if and only if they intersect at $x=a$ . Since the line where $a<x<b$ has slope $-1$ , the positive difference in $y$ -coordinates from $x=a$ to $x=b$ must be $b-a$ . Together with the fact that $(b,c-a)$ is on $f(x)$ , we see that $P=(a,c-a+b-a)$ . Since this point is on $x=a$ , the only intersection point with $2x+y=2003$ , we have $2 \cdot a+(b+c-2a)=2003 \implies b+c=2003$ . As $c>b$ , the smallest possible value of $c$ occurs when $b=1001$ and $c=1002$ . This is indeed a solution as $a=1000$ puts $P$ on $y=2003-2x$ , and thus the answer is $\boxed{1002}$

Solution:
We just showed that whenever $c<1002$ , the system has at least two different solutions: one with $x\leq a$ and one with $x>a$
We also showed that for $c=1002$ there are some $a,b$ for which the system has exactly one solution.
Hence the optimal value of $c$ is $\boxed{1002}$