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\newtheorem{theorem}{Theorem}
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\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
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\lhead{\color{blue}  AMATYC}
\chead{ \LARGE Spring 2013 - Solutions}
\rhead{ page \ \thepage}
\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  2013}
\rfoot{\small   Last revised:  June 14, 2013}
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\begin{document}


\begin{enumerate}
\item Ms. Pham writes $2$ final exams, each with $25$ problems. If the exams
have $12$ problems in common, how many problems does she write? 
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A. $\ 24$

B. $\ 26$

C. \ $37$

D. \ $38$ \ 
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E. $49$ \ 
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Solution: \ Suppose she writes the $12$ question first that are common to
the final exam. \ Then she will have to write $13$ more for the first exam,
to have $25$ questions. \ She will also have to write $13$ more for the
second exam, to have $25$ questions. \ Thus she writes $12+13+13=\allowbreak
38$ questions. \ \ The correct answer is D.

\item A triangle has two sides of length $8.1$ and $1.4$. If the length of
the third side is an even integer, its length must be 
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A. $\ 2$

B. $\ 4$

C. \ $6$

D. \ $8$%
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E. \ $10$ \ \ 
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Solution: \ Recall the triangle inequality. \ If $a$, $b$, and $c$ are sides
of a triangle, then $a+b>c$. \ Let us denote the missing sides by $x$. \ The
triangle inequality gives us that%
\begin{equation*}
8+1.4>x\text{ \ \ \ \ and \ \ \ \ \ }1.4+x>8
\end{equation*}%
We solve these inequalities and get $6.6<x<9.4$. \ The only even integer in
this interval is $8$. \ The correct answer is D.

\item If $(a,b)$ is the solution to the system of equations $\left\{ 
\begin{array}{c}
\pi x+\left( \pi +e\right) y=\pi +2e \\ 
\left( \pi +3e\right) x+\left( \pi +4e\right) y=\pi +5e%
\end{array}%
\right. $ find $b-a$. 
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A. $\ -3$

B. $\ -1$

C. \ $0$

D. \ $1$

E. \ $3$ 
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Solution: \ This is just a fairly intimidating, however linear system in two
variables. \ Keep in mind, the problem is asking for the value of $y-x$. \ \
We have two options: we can solve the system for $x$ and $y$ and then add,
or perhaps, there is a shortcut to the value of $y-x$ without actually
finding $x$ or $y$. \ We first investigate what would happen if we edded or
subtracted the equaltions in the system. \ It looks very promising if we
subtract the first equation from the second. \ (We will just add the
opposite.)%
\begin{eqnarray*}
-\pi x-\left( \pi +e\right) y &=&-\pi -2e \\
\left( \pi +3e\right) x+\left( \pi +4e\right) y &=&\pi +5e
\end{eqnarray*}%
\begin{eqnarray*}
3ex+3ey &=&3e\text{ \ \ \ \ \ \ \ divide by }3e \\
x+y &=&1
\end{eqnarray*}%
Too bad we are not asked for the value of $x+y$. \ This is still useful
information. \ Let us look at the first equation again.%
\begin{eqnarray*}
\pi x+\left( \pi +e\right) y &=&\pi +2e \\
\pi x+\pi y+ey &=&\pi +2e \\
\pi \left( x+y\right) +ey &=&\pi +2e\text{ \ \ \ \ \ \ we know that }x+y=1 \\
\pi +ey &=&\pi +2e\text{ \ \ \ \ \ } \\
ey &=&2e\text{ \ \ \ \ \ \ \ \ \ \ \ \ divide by }e \\
y &=&2
\end{eqnarray*}%
If $y=2$ and $x+y=1,$ then $x=-1$ and so $y-x=2-\left( -1\right) =3$. \ The
correct answer is E.

\item The year $2013$ has the property that when its distinct prime factors $%
3$, $11$, and $61$ are each reduced by $1$ and written in increasing order
(that is, $2$, $10$, $60$) each number is a factor of the next. Find the
next year with this property.%
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A. $\ 2014$

B. $\ 2015$

C. \ $2016$ 
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D. \ $2017$

E. \ $2018$ \ \ 
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Solution: \ Let us look at the prime factorization of the potential answers.
\ $2014=2\cdot 19\cdot 53$ and after subtracting $1$, we obtain $1$, $18$,
and $52$. \ Since $18$ is not a factor of $52$, this number is ruled out as
the correct answer. \ 

$2015=5\cdot 13\cdot 31$ and after subtracting $1$, we obtain $4$, $12$, and 
$52$. \ Since $12$ is not a factor of $52$, this number is ruled out as the
correct answer. \ 

$2016=2^{5}\cdot 3^{2}\cdot 7$ $\ $and the prime factors are $2$, $3$, and $%
7.$ \ so it can not be the product of three prime number. After subtracting $%
1$, we obtain $1$, $2$, and $6$. \ This is clearly the solution so we could
stop looking. \ The answer is C.

$2017$ is actually a prime number. \ (We just have to check for prime
divisors until $45$ since $\sqrt{2017}\approx \allowbreak 44.\,\allowbreak
911$)

$2018=2\cdot 1009$ \ and it does not have three distinct prime factors.

\item If the lines with equations $y=2x+b$ and $y=mx-6$ intersect at a point
on the $x-$axis, then 
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A. $mb=12$

B. $mb+12=0$ 
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C. $m=3b$

D. $m+3b=0$

E. $3m=b$ \ \ 
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Solution: \ When we solve the system $\left\{ 
\begin{array}{c}
y=2x+b \\ 
y=mx-6%
\end{array}%
\right. $ the solution is a point on the \ $x-$axis, and so $y=0$%
\begin{eqnarray*}
0 &=&2x+b\text{ \ \ and \ \ }0=mx-6 \\
-\dfrac{b}{2} &=&x\text{ \ \ \ \ }\Longrightarrow \text{ \ \ \ \ }0=m\left( -%
\dfrac{b}{2}\right) -6
\end{eqnarray*}%
\begin{eqnarray*}
0 &=&m\left( -\dfrac{b}{2}\right) -6 \\
6 &=&-\dfrac{mb}{2} \\
-12 &=&mb \\
0 &=&mb+12
\end{eqnarray*}%
So the correct answer is B.

\item Find the smallest positive integer value of $n$ for which $\dfrac{1}{a}%
+\dfrac{1}{b}=\dfrac{1}{n}$ has at least three solutions $\left( a,b\right) $
with integers $a\geq b>0$ 
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A. $3$

B. $4$ 
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C. $5$

D. $6$

E. $7$ \ \ 
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Solution: \ 
\begin{eqnarray*}
\dfrac{1}{a}+\dfrac{1}{b} &=&\dfrac{1}{n}\text{ \ \ \ \ \ \ multiply by }abn
\\
bn+an &=&ab\text{ \ \ \ \ \ \ \ subtract }\left( an+bn\right) \\
0 &=&ab-an-bn
\end{eqnarray*}%
We will factor the right-hand side. \ Based on the three terms, we try $%
\left( a-n\right) \left( b-n\right) =ab-an-bn+n^{2}$. \ So, we can factor it
if we just add $n^{2}$ to both sides.%
\begin{eqnarray*}
n^{2} &=&ab-an-bn+n^{2} \\
n^{2} &=&\left( a-n\right) \left( b-n\right)
\end{eqnarray*}%
Before we proceed, let us notice that $a\geq b$ implies that $a-n\geq b-n$

Because $n=1$ and $2$ are not listed among the solutions offered, we don't
need to consider them.

Case 1. $n=3$%
\begin{equation*}
9=\left( a-3\right) \left( b-3\right)
\end{equation*}%
\begin{equation*}
\begin{array}{ccccc}
9 & = & \left( a-3\right) & \left( b-3\right) &  \\ 
&  & 9 & 1 & ~~\Longrightarrow a=12,b=4 \\ 
&  & 3 & 3 & ~~\Longrightarrow a=6,b=6%
\end{array}%
\end{equation*}%
This only gives us two solutions.

Case 2. \ $n=4$%
\begin{equation*}
16=\left( a-4\right) \left( b-4\right)
\end{equation*}%
\begin{equation*}
\begin{array}{ccccc}
16 & = & \left( a-4\right) & \left( b-4\right) &  \\ 
&  & 16 & 1 & ~~\Longrightarrow a=20,b=5 \\ 
&  & 8 & 2 & ~~\Longrightarrow a=12,b=6 \\ 
&  & 4 & 4 & ~~\Longrightarrow a=8,b=8%
\end{array}%
\end{equation*}%
So this is the first number among the ones listed that has at least three
solutions. \ The correct answer is B.

\item The equation $a^{3}+b^{2}+c^{2}=2013$ has a solution in positive
integers for which $b$ is a multiple of $5$. \ Find $a+b+c$ for this
solution. 
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A. $55$

B. $57$ \ 
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C. $59$

D. $61$

E. $63$ \ \ 
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Solution: \ \ 
\begin{equation*}
a^{3}+b^{2}+c^{2}=2013\text{ \ \ and \ \ \ }b=5k
\end{equation*}%
First note that $\sqrt[3]{2013}\approx 12.\,\allowbreak 626\,45$ and so $%
a\leq 12$. \ (Indeed, $12^{3}=\allowbreak 1728$ and so $a$ could be $12$ but 
$13^{3}=2197$ that is too large.)%
\begin{equation*}
\begin{array}{ccccccc}
x\text{ } &  & 0 & 1 & 2 & 3 & 4 \\ 
x^{2} &  & 0 & 1 & 4 & 4 & 1 \\ 
x^{3} &  & 0 & 1 & 3 & 2 & 4 \\ 
&  &  &  &  &  & 
\end{array}%
\end{equation*}%
Re-write modulo $5$ (That means to replace every number by the remainder we
get after division by $5.$ \ \ For example, $7\equiv 2$ and $10\equiv 0$)%
\begin{eqnarray*}
a^{3}+0^{2}+c^{2} &\equiv &3 \\
a^{3}+c^{2} &\equiv &3
\end{eqnarray*}%
Case 1. \ $a^{3}\equiv 3$ and $c^{2}\equiv 0$ \ \ \ \ \ \ $\Longrightarrow $
\ \ \ \ $a\equiv 2$ and $c\equiv 0$

Case 2. \ $a^{3}\equiv 2$ and $c^{2}\equiv 1$ \ \ \ \ \ \ $\Longrightarrow $
\ \ \ \ \ \ \ \ $a\equiv 3$ and $c\equiv 1$ or $4$

Case 3. \ $a^{3}\equiv 4$ and $c^{2}\equiv 4$ \ \ \ \ \ \ $\Longrightarrow $
\ \ \ \ \ \ \ \ $a\equiv 4$ and $c\equiv 2$ or $3$

Case 1. \ $a\equiv 2$ and $b\equiv c\equiv 0$

$a=2,7,12$

$a=2$%
\begin{eqnarray*}
2^{3}+b^{2}+c^{2} &=&2013 \\
b^{2}+c^{2} &=&2005
\end{eqnarray*}%
This is impossible because if both $b$ and $c$ are divisible by $5$ then
their squares are divisible by $25,$ then and so is their sum. $2005$ is not
divisible by $25$.

$a=7$%
\begin{eqnarray*}
7^{3}+b^{2}+c^{2} &=&2013 \\
b^{2}+c^{2} &=&1670
\end{eqnarray*}%
This is impossible because if both $b$ and $c$ are divisible by $5$ then
their squares are divisible by $25,$ then and so is their sum. $\ $But $1670$
is not divisible by $25$.

$a=12$%
\begin{eqnarray*}
12^{3}+b^{2}+c^{2} &=&2013 \\
b^{2}+c^{2} &=&285
\end{eqnarray*}%
This is impossible because if both $b$ and $c$ are divisible by $5$ then
their squares are divisible by $25,$ and so is their sum. But $285$ is not
divisible by $25$. \ 

Case 2. \ $a\equiv 3$ and $c\equiv 1$ or $4$

$a\equiv 3$ means that $a=3$ or $8$

$a=3$%
\begin{eqnarray*}
3^{3}+b^{2}+c^{2} &=&2013 \\
b^{2}+c^{2} &=&1986
\end{eqnarray*}%
This is impossible for divisibility by $9$.%
\begin{equation*}
\begin{array}{cccccccccccc}
x &  &  & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 
x^{2} &  &  & 0 & 1 & 4 & 0 & 7 & 7 & 0 & 4 & 1%
\end{array}%
\end{equation*}%
$1986$ gives a remainder of $6$ and there are no two squares modulo $9$ that
add up to $6$.

$a=8$%
\begin{eqnarray*}
8^{3}+b^{2}+c^{2} &=&2013 \\
b^{2}+c^{2} &=&1501
\end{eqnarray*}%
Now $b$ could be $5,10,15,20,25,30,$ or $35$%
\begin{equation*}
\begin{array}{cccccc}
b=5 &  & c^{2}=1988 &  & \text{no solution} &  \\ 
b=10 &  & c^{2}=1913 &  & \text{no solution} &  \\ 
b=15 &  & c^{2}=1788 &  & \text{no solution} &  \\ 
b=20 &  & c^{2}=1613 &  & \text{no solution} &  \\ 
b=25 &  & c^{2}=1388 &  & \text{no solution} &  \\ 
b=30 &  & c^{2}=1113 &  & \text{no solution} & 
\end{array}%
\end{equation*}%
Case 3. \ $a\equiv 4$ and $c\equiv 2$ or $3$

$a\equiv 4$ means that $a=4$ or $9$

$a=4$ \ \ \ \ 
\begin{eqnarray*}
4^{3}+b^{2}+c^{2} &=&2013 \\
b^{2}+c^{2} &=&1949
\end{eqnarray*}%
Now $b$ could be $5,10,15,20,25,30,$ or $35$%
\begin{equation*}
\begin{array}{cccccc}
b=5 &  & c^{2}=1924 &  & \text{no solution} &  \\ 
b=10 &  & c^{2}=1849 &  & c=43 &  \\ 
b=15 &  & c^{2}=1724 &  & \text{no solution} &  \\ 
b=20 &  & c^{2}=1549 &  & \text{no solution} &  \\ 
b=25 &  & c^{2}=1324 &  & \text{no solution} &  \\ 
b=30 &  & c^{2}=1049 &  & \text{no solution} & 
\end{array}%
\end{equation*}%
Thus $a=4$ and $b=10$ and $c=43.$ \ We check: $4^{3}+10^{2}+43^{2}=%
\allowbreak 2013.$ \ So the answer is $4+10+43=57$ which is B.

There should be a better way. \ 

\item Each letter A through Z of the alphabet is assigned a unique integer
from $2$ to $27$.

If \ A$\cdot $M$\cdot $A$\cdot $T$\cdot $Y$\cdot $C $=3^{2}\cdot 5^{2}\cdot
7\cdot 11^{2}$, find M + T + Y + C.%
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A. $30$ \ 
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B. $34$

C. $36$

D. $38$

E. $42$ \ \ 
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Solution: \ \ There are $5$ variables, A, M, T, Y, and C. \ \ They are all
different, and based on the product, all odd. \ In fact, based on the
product, the possible values are for these variables are $3,5,7,9,11,15,21,$
and $25$.%
\begin{equation*}
\text{A}^{2}\cdot \text{M}\cdot \text{T}\cdot \text{Y}\cdot \text{C}%
=3^{2}\cdot 5^{2}\cdot 7\cdot 11^{2}
\end{equation*}%
Let us also notice that M, T, Y, and C are symmetrical. A is the only
variable with a different role than the others. \ 

First we will establish that only A $=11$ is possible.

Suppose that A is divisible by $3.$ \ Then A$^{2}$ has two $3-$factors in
its prime factorization so the rest, M, Y, T, C all must have none. \ This
leaves us only the values $5,7,11,$ and $25$ for these but then the product
of these numbers has one too many $5$-factors. \ So, A is not divisible by $%
3 $. \ 

Suppose now that $A$ is divisible by $5$. \ Then A$^{2}$ has two $5-$factors
in its prime factorization so the rest, M, Y, T, C all must have none. \
This leaves us only the values $3,7,9,11,$ and $21$ for these. \ $9$ can not
be one of them because if $9$ is one of M, Y, T, C then the other three can
not be divisible by $3$ and so we only have $7$ and $11$ for the other three
values. \ So M, Y, T, C must be $3,$ $7,$ $11,$ and $21$. \ This is
impossible too, because now the product has one too many $7$-factors. \ So,
A is not divisible by $5$.

A can also not be divisible by $7$ because then the product would be
divisible by $49$. \ 

This leaves only one value for A, that is $11.$ \ 

This leaves us $3,5,7,9,15,21,$ and $25$ for M, Y, T, C where 
\begin{equation*}
\text{M}\cdot \text{T}\cdot \text{Y}\cdot \text{C}=3^{2}\cdot 5^{2}\cdot 7
\end{equation*}%
Next we claim that none of these can be $9$ or $25$. \ Suppose one is $9$. \
Then the rest of the numbers can have no $3-$factor in them, leaving us only 
$5,7,$ and $25$ for the other three. \ But their product has one too many $5$%
-factor. \ The argument goes similarly for $25$. \ So now we have the values
of $3,5,7,15$ and $21$. \ The product of these numbers is $3^{3}\cdot
5^{2}\cdot 7^{2}$ \ We need to get the product $3^{2}\cdot 5^{2}\cdot 7$ by
removing one of the numbers. \ So we remove $21$. \ This gives us \ A $=11$
\ and\ \ M, T, Y, C $=3,5,7,15$ and so the sum is $3+5+7+15=30$. \ \ \ The
correct answer is A.

\item The third-degree polynomial $P(x)$ has only nonnegative integer
coefficients. If $P\left( 0\right) \cdot P\left( 3\right) =139$ and $P\left(
1\right) \cdot P\left( 2\right) =689$, find $P\left( -1\right) $. \ 
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A. $-2$ \ 

B. $-1$ \ 
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C. $0$

D. $1$

E. $2$ \ \ 
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Solution: \ On the interval $\left[ 0,\infty \right) $,$\ \ $the polynomial $%
P\left( x\right) $ is increasing and has only integer values $P\left(
n\right) $ for integer $n$. \ We first look at the prime-factorization of $%
139$ and $689$.

Fortunately, $139$ is a prime. \ Since $P\left( x\right) $ is increasing,
this means that $P\left( 0\right) =1$ and $P\left( 3\right) =139$. \ On the
other hand, $689=13\cdot 53$. \ This means that either $P\left( 1\right) =1$
and $P\left( 2\right) =689$ or $P\left( 1\right) =13$ and $P\left( 2\right)
=53$. \ Because $P\left( x\right) $ is increasing, $P\left( 1\right) =1$ is
impossible because $P\left( 0\right) =1$ already. So we have the following:%
\begin{equation*}
P\left( 0\right) =1\text{ \ \ \ \ \ }P\left( 1\right) =13\text{\ \ \ \ \ }%
P\left( 2\right) =53\text{\ \ and\ \ }P\left( 3\right) =139
\end{equation*}%
\begin{equation*}
P\left( x\right) =ax^{3}+bx^{2}+cx+d\text{ \ \ \ \ \ \ \ \ \ \ \ }a,b,c,d\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\text{ \ \ }a,b,c,d\geq 0
\end{equation*}%
\begin{eqnarray*}
P\left( 0\right) &=&1\text{~~~~~}\Longrightarrow ~~~~a\cdot 0^{3}+b\cdot
0^{2}+c\cdot 0+d=1\text{~~~~~}\Longrightarrow ~~~~d=1 \\
P\left( 1\right) &=&13\text{~~~~~}\Longrightarrow ~~~~a\cdot 1^{3}+b\cdot
1^{2}+c\cdot 1+d=13\text{~~~~~}\Longrightarrow ~~~~a+b+c+d=13 \\
P\left( 2\right) &=&53\text{~~~~~}\Longrightarrow ~~~~a\cdot 2^{3}+b\cdot
2^{2}+c\cdot 2+d=53\text{~~~~~}\Longrightarrow ~~~~8a+4b+2c+d=53 \\
P\left( 3\right) &=&139\text{~~~~~}\Longrightarrow ~~~~a\cdot 3^{3}+b\cdot
3^{2}+c\cdot 3+d=139\text{~~~~~}\Longrightarrow ~~~~27a+9b+3c+d=139
\end{eqnarray*}%
\begin{eqnarray*}
a+b+c+1 &=&13\text{~~~~~}\Longrightarrow ~~~~a+b+c=12 \\
8a+4b+2c+1 &=&53\text{~~~~~}\Longrightarrow ~~~~8a+4b+2c=52\text{~~~~~}%
\Longrightarrow ~~~~4a+2b+c=26 \\
27a+9b+3c+1 &=&139\text{~~~~~}\Longrightarrow ~~~~27a+9b+3c=138\text{~~~~~}%
\Longrightarrow ~~~~9a+3b+c=46
\end{eqnarray*}%
So we have the following system of equations: \ (we already know $d=1$) \ 
\begin{eqnarray*}
a+b+c &=&12 \\
4a+2b+c &=&26 \\
9a+3b+c &=&46
\end{eqnarray*}%
We eliminate $c$ by subtracting the first equation from the other two%
\begin{eqnarray*}
3a+b &=&14 \\
8a+2b &=&34\text{~~~~}\Longrightarrow ~~~4a+b=17
\end{eqnarray*}%
Now we just have%
\begin{eqnarray*}
3a+b &=&14 \\
4a+b &=&17
\end{eqnarray*}%
After subtraction, we have $a=3$. \ Then $4a+b=17$ and $a=3$ gives us $b=5$.
\ Then $a+b+c=12$ gives us $c=4$. \ Thus%
\begin{equation*}
a=3\text{~~~~~~~~}b=5\text{~~~~~~~~}c=4\text{~~~~~~~~}d=1
\end{equation*}%
We check: if $P\left( x\right) =3x^{3}+5x^{2}+4x+1$, then $P\left( 0\right)
=1$, \ $P\left( 1\right) =13$, $P\left( 2\right) =53$ and $P\left( 3\right)
=139$. \ And $P\left( -1\right) =3\left( -1\right) ^{3}+5\left( -1\right)
^{2}+4\left( -1\right) +1=\allowbreak -1$. \ The correct answer is B.

\item Find the smallest positive value of $t$ such that $\cos t$ is the same
whether $t$ is in radians or in degrees. Write your answer (rounded to 3
decimal places) in the corresponding blank on the answer sheet.

Solution: \ Let $x$ be our number. \ Then $x%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$ and $x_{\text{rad}}$ denotes the two angles with $\cos x%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
=\cos x_{\text{rad}}$. \ First, let us notice that $x_{\text{rad}}$ denotes
the larger angle.

There are two ways the cosince of two angles are the same: either one is
co-terminal with the other, or one is co-terminal with the opposite of the
other. \ The smallest positive co-terminal angle is obtained when we add $360%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$ (or $2\pi $) to an angle. \ If we are looking for the smallest positive
angle like that, the second case is more promising because then $x_{\text{rad%
}}$ is still less than $2\pi $ while in the first case, $x_{\text{rad}}$ is
greater than $2\pi $.

So we will state that $x_{\text{rad}}$ is $2\pi $ more than the opposite of $%
x%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$. \ We need to decide first in how to write the equation: in degrees or in
radians. \ It doesn't matter, the answer should be the same number. \ We
will go with degrees.%
\begin{eqnarray*}
x_{\text{rad}}\text{ \ in degrees } &=&-x%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
+360%
%TCIMACRO{\U{b0} }%
%BeginExpansion
{{}^\circ}
%EndExpansion
\\
x_{\text{rad}}\cdot \left( \dfrac{180%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
}{\pi }\right) &=&-x%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
+360%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
\end{eqnarray*}%
So our equation is 
\begin{eqnarray*}
x\cdot \dfrac{180}{\pi } &=&-x+360\text{ \ \ \ \ \ \ \ \ \ \ add }x \\
x\left( \dfrac{180}{\pi }\right) +x &=&360\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ factor our }x \\
x\left( \dfrac{180}{\pi }+1\right) &=&360\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ divide by }\dfrac{180}{\pi }+1 \\
x &=&\dfrac{360}{\dfrac{180}{\pi }+1}\approx 6.\,\allowbreak 175
\end{eqnarray*}

\item In quadrilateral ABCD, AB $=6$, BC $=6$, CD $=8$, AD $=10$, and $%
\angle $C $=90%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$. If the angle bisector of $\angle $A meets diagonal BD at point E, find
BE. 
%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
%BeginExpansion
\begin{multicols}{5}%
%EndExpansion

A. $\dfrac{15}{4}$ \ 
%TCIMACRO{\TeXButton{correct}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

B. $4$ \ 

C. $5$

D. $\ 6$

E. $\dfrac{25}{4}$ \ \ 
%TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

Solution: \ Consider the picture shown below. \FRAME{dtbpF}{2.1681in}{%
2.3177in}{0pt}{}{}{pic11a.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
2.1681in;height 2.3177in;depth 0pt;original-width 0.0718in;original-height
0.0796in;cropleft "0";croptop "1";cropright "1.0431";cropbottom "0";filename
'pic11a.bmp';file-properties "XNPEU";}}\ First, diagonal BD is $10$ units
long by the Pythagorean theorem. \ \ Second, recall a theorem: in any
triangle, an angle bisector splits the opposite sides into two parts in the
same ratio as the ratio of the two sides around the angle. \ If we know this
theorem, the solution is immediate: \ the other two sides are in a ratio of $%
3$ to $5$, so BE must be $\dfrac{3}{8}$ of BD (the other part is $\dfrac{5}{8%
}$) and so the answer is $\dfrac{3}{8}\cdot 10=\dfrac{15}{4}$. \ The correct
answer is A. \ 

In case someone does not know this theorem, we will sort of derive it here:
\ we will prove that $\dfrac{\text{BE}}{\text{DE}}=\dfrac{3}{5}$.

The trick is to compare the areas of triangles\ AEB and ADE. \ (Recall that
the area of a triangle can be computed as $A=\dfrac{1}{2}ab\sin \gamma $)%
\begin{equation*}
\dfrac{A_{\text{AEB}}}{A_{\text{ADE}}}=\dfrac{\dfrac{1}{2}\cdot 6\cdot \text{%
AE}\cdot \sin \dfrac{\alpha }{2}}{\dfrac{1}{2}\cdot 10\cdot \text{AE}\cdot
\sin \dfrac{\alpha }{2}}=\dfrac{3}{5}
\end{equation*}%
The area of the same triangles can also be computed as $\dfrac{1}{2}ah$ and
also notice that these triangles have the same height. \ So,%
\begin{equation*}
\dfrac{A_{\text{AEB}}}{A_{\text{ADE}}}=\dfrac{3}{5}=\dfrac{\dfrac{1}{2}\cdot 
\text{EB}\cdot h}{\dfrac{1}{2}\cdot \text{ED}\cdot h}=\dfrac{\text{EB}}{%
\text{ED}}
\end{equation*}%
So now we know that the ratio of EB to ED is $3$ to $5$ and they add up to $%
10.$ \ Then EB is $\dfrac{3}{8}$ of $10$ which is $\dfrac{15}{4}$. \ 

\item Line L has intercepts $2$ and $4$, while line M has intercepts $4$ and 
$6$. If L and M intersect at $(a,b)$, which of the following could NOT be $%
3a+b$?

%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
%BeginExpansion
\begin{multicols}{5}%
%EndExpansion

A. $0$ \ 
%TCIMACRO{\TeXButton{correct}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

B. $4$ \ 

C. $\ 8$

D. $\ 12$

E. $\ 32$ \ \ 
%TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

Solution: \ There are four cases to consider. \ 

Case 1. \ Intercepts of L are $\left( 2,0\right) $ and $\left( 0,4\right) $
\ \ \ \ $\Longrightarrow $\ \ \ \ $y=-2x+4$

\qquad \qquad Intercepts of M are $\left( 4,0\right) $ and $\left(
0,6\right) $ \ \ \ \ $\Longrightarrow $\ \ \ \ $y=-\dfrac{3}{2}x+6$%
\begin{eqnarray*}
-2x+4 &=&-\dfrac{3}{2}x+6 \\
-2x &=&-\dfrac{3}{2}x+2\text{ \ \ \ \ multiply by }2 \\
-4x &=&-3x+4 \\
-x &=&4 \\
x &=&-4~~~~~\Longrightarrow ~~~~y=12~~~~~~~\Longrightarrow ~~~3a+b=0
\end{eqnarray*}%
Case 2. \ Intercepts of L are $\left( 2,0\right) $ and $\left( 0,4\right) $

\qquad \qquad Intercepts of M are $\left( 6,0\right) $ and $\left(
0,4\right) .$ \ The two lines clearly intersect at $\left( 0,4\right) $ and
so $3a+b=4$.

Case 3. \ Intercepts of L are $\left( 4,0\right) $ and $\left( 0,2\right) $

\qquad \qquad Intercepts of M are $\left( 4,0\right) $ and $\left(
0,6\right) .$ \ The two lines clearly intersect at $\left( 4,0\right) $ and
so $3a+b=12$.

Case 4. \ Intercepts of L are $\left( 4,0\right) $ and $\left( 0,2\right) $
\ \ \ \ $\Longrightarrow $\ \ \ \ $y=-\dfrac{1}{2}x+2$

\qquad \qquad Intercepts of M are $\left( 6,0\right) $ and $\left(
0,4\right) $ \ \ \ \ $\Longrightarrow $\ \ \ \ $y=-\dfrac{2}{3}x+4$%
\begin{eqnarray*}
-\dfrac{1}{2}x+2 &=&-\dfrac{2}{3}x+4 \\
-\dfrac{1}{2}x &=&-\dfrac{2}{3}x+2\text{ \ \ \ \ multiply by }6 \\
-3x &=&-4x+12\text{ } \\
x &=&12~~~~~\Longrightarrow ~~~~y=-4~~~~~~~\Longrightarrow ~~~3a+b=32
\end{eqnarray*}%
The only number we didn't get is $8$. \ \ The correct answer is C.

\item Sue traveled continuously starting on 1/1/2012. Her first trip was
less than 3 months, and each successive trip was 2 days longer than the
previous trip. If her last trip ended on 12/31/2012, which of these was the
length in days of one of her trips? \ 
%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
%BeginExpansion
\begin{multicols}{5}%
%EndExpansion

A. $54$ \ 
%TCIMACRO{\TeXButton{correct}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

B. $58$ \ 

C. $\ 65$

D. $\ 72$

E. $\ 77$ \ \ 
%TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

Solution: \ $2012$ was a leap year with $365$ days. \ Recall that the sum of
the first $n$ terms in an arithmetic sequence with first element $a$ and
common difference $d$ is $\ s_{n}=\dfrac{2a+\left( n-1\right) d}{2}n$. \ In
this case, $d=2$ and $s_{n}=365$%
\begin{equation*}
365=\dfrac{2a+\left( n-1\right) 2}{2}n=\left( a+\left( n-1\right) \right)
n=\left( a+n-1\right) n
\end{equation*}%
It clear that $n\geq 2$ and so $a+n-1$ is greater than $n$. \ These two are
also divisors of $365$. \ The prime-factorization of $365$ is $365=5\cdot 73$%
\begin{equation*}
\begin{array}{cccccccccccc}
& 365 & = & \left( a+n-1\right) & n &  &  &  &  &  &  &  \\ 
&  &  & 365 & 1 &  & \text{impossible} & \text{since} & n\geq 2 &  &  &  \\ 
&  &  & 73 & 5 &  &  &  &  &  &  & 
\end{array}%
\end{equation*}%
If $n=5$ and $a+n-1=73$, then $a+4=73$ so $a=69$ and so the lengths of the
five trips are%
\begin{equation*}
69,71,73,75,77
\end{equation*}%
and so the correct answer is E.

Solution: \ $2012$ was a leap year with $365$ days. \ Recall that the sum of
the first $n$ terms in an arithmetic sequence with first element $a$ and
common difference $d$ is $\ s_{n}=\dfrac{2a+\left( n-1\right) d}{2}n$. \ In
this case, $d=2$ and $s_{n}=365+n-1$%
\begin{eqnarray*}
364+n &=&\dfrac{2a+\left( n-1\right) 2}{2}n=\left( a+n-1\right) n \\
364 &=&\left( a+n-1\right) n-n \\
364 &=&\left( a+n-2\right) n\text{ \ \ \ \ \ \ \ \ where }n\geq 2
\end{eqnarray*}%
The prime factorization of $364$ is $364=2^{2}\cdot 7\cdot 13$.dd \ $%
a+52-2=7 $, Solution is: $-43$ \ 
\begin{equation*}
\begin{array}{cccccccccccc}
& 364 & = & \left( a+n-2\right) & n &  &  &  &  &  &  &  \\ 
&  &  & 364 & 1 &  & \text{impossible} & \text{since} & n\geq 2 &  &  &  \\ 
&  &  & 182 & 2 & \Longrightarrow & n=2 & \Longrightarrow & a=182 &  &  & 
\\ 
&  &  & 91 & 4 & \Longrightarrow & n=4 & \Longrightarrow & a=89 &  &  &  \\ 
&  &  & 52 & 7 & \Longrightarrow & n=7 & \Longrightarrow & a=47 &  &  &  \\ 
&  &  & 7 & 52 & \Longrightarrow & n=52 & \Longrightarrow & a=-43 &  &  & 
\\ 
&  &  & 4 & 91 & \Longrightarrow & n=91 & \Longrightarrow &  &  &  &  \\ 
&  &  &  &  &  &  &  &  &  &  & 
\end{array}%
\end{equation*}%
The first three months are $31+29+31=\allowbreak 91$ days long and so $a=89$
and $a=47$ are both possible.

2 trips - first is too long

3 trips: \ $x+x+2+x+4=367$ \ \ \ \ \ $x$ is not an integer

4 trips: \ $x+x+2+x+4+x+6=368$, Solution is: $89$

\item A binary string is a sequence of 1's and 0's, such as 10011 or
11101010. How many different binary strings of length 6 are there such that
no two are reversals of each other or add up to 111111?%
%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
%BeginExpansion
\begin{multicols}{5}%
%EndExpansion

A. $22$ \ 
%TCIMACRO{\TeXButton{correct}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

B. $23$ \ 

C. $\ 24$

D. $\ 25$

E. $\ 26$ \ \ 
%TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

Solution: \ There are $2^{6}=64$ six-long binary strings. \ We will count
the compement and subtract it from $64$. \ If we interpret these numbers as
written in base $2$, they are the numbers from $0$ to $63$. \ We can then
pair them up so that the pairs add up to 111111 - which is the same as two
numbers adding up to $63$. \ So the pairs are $0$ with 63, $1$ with $62$, \
\ and so on, $30$ with $33$, and finally $31$ with $32$. \ So if we select
one from the pair into our collection, then we cannot select the other. \
How about the reversal? \ Each number has a unique number as their reversal.
\ This is another number UNLESS the number is symmetrical and is therefore
its own reversal. \ How many such symmetrical numbers are there? \ We claim $%
8$. \ This is because we have complete freedom to select the first three
digits - giving us $8$ choice but then there is no choice but to duplicate
the triple backwards to get a symmetrical string. \ For example 110 will
give us the string 110011$.$ \ So, out of the $64$ strings 8 are symmetrical
and so the other 56 can be paired into $28$ pairs where they are each
other's reversal. \ The question is: can we pick just one from each of the $%
28$ pairs so that no two add up to 63?

Solution: \ The blue lines connect two numbers that add to 111111 and the
red lines connect strings that are reversals of each other.\FRAME{dtbpF}{%
4.2601in}{3.3572in}{0pt}{}{}{pic14.bmp}{\special{language "Scientific
Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file
"F";width 4.2601in;height 3.3572in;depth 0pt;original-width
0.1315in;original-height 0.0977in;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'pic14.bmp';file-properties "XNPEU";}}If we
wanted to collect numbers in a set such that no two are connected, then 

- we can pick one from each pairs such as $0$-$63$ or $7$-$56$

- we can pick exactly two from each of the squares. \ For example, the first
square, containing $1$-$32$-$31$-$62,$ we can either select the pair 1 and
31 or the pair $32$ and $62$.

This means that a \ maximum of $32$ such numbers can be collected.

\item In quadrilateral PQRS, $\angle $P = $\angle $Q = $\angle $S = $45%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$, $\angle $QPR = $\angle $RPS, and PR $=8\sqrt{2}$. \ Find the area of
quadrilateral PQRS to the nearest integer.%
%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
%BeginExpansion
\begin{multicols}{5}%
%EndExpansion

A. $60$

B. $61$ \ 

C. $\ 62$

D. $\ 63$

E. $\ 64$ \ 
%TCIMACRO{\TeXButton{correct}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion
\ \ 
%TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

Solution: \ Triangles PQR and RSP are congruent because they have two angles
in common and they share side PR. \ Angle QRS is $360%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
-3\cdot 45%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
=225%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$. \ \ Angle QRP is half of that, $112.5%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$.\FRAME{dtbpF}{2.1958in}{2.5399in}{0pt}{}{}{pic15.bmp}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
"USEDEF";valid_file "F";width 2.1958in;height 2.5399in;depth
0pt;original-width 0.115in;original-height 0.1392in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'pic15.bmp';file-properties
"XNPEU";}}\ We compute the length of RS using the Law of Sines%
\begin{equation*}
\dfrac{\text{RS}}{8\sqrt{2}}=\dfrac{\sin 22.5^{\circ }}{\sin 45^{\circ }}%
~~~~~\Longrightarrow ~~~~\text{RS}=8\sqrt{2}\left( \dfrac{\sin 22.5^{\circ }%
}{\dfrac{1}{\sqrt{2}}}\right) =16\sin 22.5%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
\end{equation*}%
Recall that the area of triangle $ABC$ is $\dfrac{1}{2}ab\sin \gamma $.

So triangle RSP has area 
\begin{eqnarray*}
A &=&\dfrac{1}{2}\cdot \text{RS}\cdot 8\sqrt{2}\sin 112.5%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
=\dfrac{1}{2}\cdot \left( 16\sin 22.5%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
\right) \cdot 8\sqrt{2}\sin 112.5%
%TCIMACRO{\U{b0} }%
%BeginExpansion
{{}^\circ}
%EndExpansion
\\
&=&64\sqrt{2}\sin 22.5%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
\sin 112.5%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
\end{eqnarray*}%
The area of the quadrilateral is twice that, 
\begin{equation*}
A_{\text{PQRS}}=128\sqrt{2}\sin 22.5%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
\sin 112.5%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
\approx 64
\end{equation*}%
So the correct answer is E.

\item The numbers $2$ and $1$ are the smallest positive integers for which
the square of the first is $2$ more than twice the square of the second. If $%
a$ and $b$ are the smallest such pair with $a>10$, find $a-b$. \ 
%TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}%
%BeginExpansion
\begin{multicols}{5}%
%EndExpansion

A. $\ 13$

B. $\ 15$ \ 

C. $\ 17$ 
%TCIMACRO{\TeXButton{correct}{\color{black}}}%
%BeginExpansion
\color{black}%
%EndExpansion

D. $\ 19$

E. $\ 21$ \ \ \ 
%TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}%
%BeginExpansion
\end{multicols}%
%EndExpansion

Solution: \ \ We need to find integers $a$, $b$ with $a>10$ and $%
a^{2}=2b^{2}+2$.

Claim 1. \ $b$ must be an odd number.

Proof: \ This is because of rules of divisibility by $4$. \ If we square an
odd number, the result will give a remainder $1$ after divided by $4$. \ If
we square an even number, it will be divisible by $4.$ \ (In other words, if 
$x$ is odd, then $x^{2}\equiv 1\qquad \func{mod}4$ and if $x$ is even, then $%
x^{2}\equiv 0\qquad \func{mod}4$) \ Let $n$ be an integer.%
\begin{eqnarray*}
\left( 2n+1\right) ^{2} &=&4n^{2}+4n+1=4\left( n^{2}+n\right) +1\text{ \ \ \
has a remainder }1\text{ after division by }4 \\
\left( 2n\right) ^{2} &=&4n^{2}\text{ \ \ \ \ divisible by }4
\end{eqnarray*}%
So if $b$ is an even number, then clearly $2b^{2}$ is divisible by $4$ but
then $2b^{2}+2$ is even but NOT divisible by $4$. \ But then $a^{2}$ even
but not divisible by $4$ is impossible. \ \ There is a shorter way to
discuss all this: $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\begin{array}{cccccc}
\text{reminder of }b\text{ after division by }4 &  & 0 & 1 & 2 & 3 \\ 
b^{2} &  & 0 & 1 & 0 & 1 \\ 
2b^{2}+2 &  & 2 & 0 & 2 & 0 \\ 
&  &  &  &  &  \\ 
\text{reminder of }a\text{ after division by }4 &  & 0 & 1 & 2 & 3 \\ 
a^{2} &  & 0 & 1 & 0 & 1%
\end{array}%
$

The table above shows the same information. \ It is also pretty clear that $%
a $ must be even.

Let us now follow what the last digit would be of $b$, $b^{2},$ and $%
a^{2}=2b^{2}+2$.

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\begin{array}{ccccccccccc}
\text{last digit of }b &  & 1 & 3 & 5 & 7 & 9 &  &  &  &  \\ 
\text{last digit of }b^{2} &  & 1 & 9 & 5 & 9 & 1 &  &  &  &  \\ 
\text{last digit of }2b^{2} &  & 2 & 8 & 0 & 8 & 2 &  &  &  &  \\ 
\text{last digit of }2b^{2}+2 &  & 4 & 0 & 2 & 0 & 4 &  &  &  &  \\ 
&  &  &  &  &  &  &  &  &  &  \\ 
&  &  &  &  &  &  &  &  &  &  \\ 
\text{last digit of }a &  & 0 & 2 & 4 & 6 & 8 &  &  &  &  \\ 
\text{last digit of }a^{2} &  & 0 & 4 & 6 & 6 & 4 &  &  &  & 
\end{array}%
$

Only $0$ and $4$ are possible for the last digit of both $a^{2}$ and $%
2b^{2}+2$. \ This means that either

Case 1. \ The last digit of $b$ is $1$ or $9$ and the last digit of $a$ is $%
4 $ or $8$ or

Case 2. \ The last digit of $b$ is $3$ or $7$ and the last digit of $a$ is $%
0 $

So the last digit of $b$ is either $1$ or $3$ or $7$ or $9$

Let's see.

If $b=3$ then $2b^{2}+2=20$ - not a square

If $b=7$ then $2b^{2}+2=100=10^{2}$ a square but $a>10$ is not true

If $b=9$ then $2b^{2}+2=164$ - not a square

If $b=11$ then $2b^{2}+2=244$ - not a square

If $b=13$ then $2b^{2}+2=340$ - not a square

If $b=17$ then $2b^{2}+2=580$ - not a square

If $b=19$ then $2b^{2}+2=724$ - not a square

If $b=21$ then $2b^{2}+2=884$ - not a square \ \ \ \ 

If $b=23$ then $2b^{2}+2=1060$ - not a square \ 

If $b=27$ then $2b^{2}+2=1460$ - not a square

If $b=29$ then $2b^{2}+2=1684$ - not a square

If $b=31$ then $2b^{2}+2=1924$ - not a square

If $b=33$ then $2b^{2}+2=2180$ - not a square

If $b=37$ then $2b^{2}+2=2740$ - not a square

If $b=39$ then $2b^{2}+2=3044$ - not a square

If $b=41$ then $2b^{2}+2=3364=58^{2}$ - so $a=58$

Then $a-b=58-41=17$ and so the correct answer is C.

\item A number is chosen at random from among all $5$-digit numbers
containing exactly one each of the digits $1$, $2$, $3$, $4$, and $5$. Find
the probability that no two adjacent digits in the number are consecutive
integers. \ 
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A. $\ \dfrac{1}{10}$

B. $\ \dfrac{7}{60}$ \ 
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C. $\ $ $\dfrac{2}{15}$

D. $\ \dfrac{3}{20}$

E. $\ \ \dfrac{1}{6}$ \ \ \ 
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Solution: \ If all digits are to eppear only once, the total number of such $%
5$- digit numbers is $5!=120$. \ That is the denominator. \ For the
numerator, we need to count all permutations of $1$, $2$, $3$, $4$, and $5$
in which no two adjacent digits in the number are consecutive integers. \
Notice that $1$ and $5$ have only one "neighbor" while $2$, $3$, and $4$
have two.

We will count these numbers, preferably in an increasing order.

Case 1. \ Suppose the number starts with $1$. \ After $1$ we have 3 choices:
3,4, or 5

$13524$ after $3$ we can only pick $5$ and then we can only pick $2$ and so
on

$14253$

$15$ - now we are stuck with $3$ consecutive integers - so, no number

3) $\ 24135$

4) $\ 24153$

5) $\ 25314$

6) $\ 31425$

7) $\ 31524$

8) $\ 35142$

9) $\ 35241$

10) $\ 41352$

11) $\ 42513$

12) $\ 42531$

$51$ - nope

13) $\ 52413$

14) $\ 53142$

and so the probability is $\dfrac{14}{120}=\dfrac{7}{60}$ which is choice B.

\item The triangular region with vertices $(0,0)$, $(4,0)$, and $(0,3)$ is
rotated $90^{\circ }$ counterclockwise around the origin. Find the area of
the figure formed by this rotation to the nearest hundredth. \ 
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A. $\ 19.96$

B. $\ 20.04$ \ 
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C. $\ $ $20.12$

D. $\ 20.20$

E. $\ \ 20.28$ \ \ \ 
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Solution: \ \FRAME{dtbpFX}{1.7063in}{1.7063in}{0pt}{}{}{Plot}{\special%
{language "Scientific Word";type "MAPLEPLOT";width 1.7063in;height
1.7063in;depth 0pt;display "USEDEF";plot_snapshots TRUE;mustRecompute
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\TEXUX{$y$};function \TEXUX{$\left( \left( 0,0\right) ,\left( 4,0\right)
\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness
3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle
"Line";function \TEXUX{$\left( \left( 0,0\right) ,\left( 0,3\right) \right)
$};linecolor "black";linestyle 1;pointstyle "point";linethickness
3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle
"Line";function \TEXUX{$\left( \left( 4,0\right) ,\left( 0,3\right) \right)
$};linecolor "black";linestyle 1;pointstyle "point";linethickness
3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle
"Line";function \TEXUX{$\left( \left( -3,0\right) ,\left( 0,0\right) \right)
$};linecolor "green";linestyle 1;pointstyle "point";linethickness
3;lineAttributes "Solid";curveColor "[flat::RGB:0x00008000]";curveStyle
"Line";function \TEXUX{$\left( \left( -0.1,4\right) ,\left( -0.1,0\right)
\right) $};linecolor "green";linestyle 1;pointstyle "point";linethickness
3;lineAttributes "Solid";curveColor "[flat::RGB:0x00008000]";curveStyle
"Line";function \TEXUX{$\sqrt{16-x^{2}}$};linecolor "green";linestyle
1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range
"0,4";num-x-gridlines 100;curveColor "[flat::RGB:0x00008000]";curveStyle
"Line";rangeset"X";function \TEXUX{$\left( \left( -3,0\right) ,\left(
0,4\right) \right) $};linecolor "green";linestyle 1;pointstyle
"point";linethickness 3;lineAttributes "Solid";curveColor
"[flat::RGB:0x00008000]";curveStyle "Line";function
\TEXUX{$\sqrt{9-x^{2}}$};linecolor "black";linestyle 1;pointstyle
"point";linethickness 3;lineAttributes "Solid";var1range
"-3,0";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle
"Line";rangeset"X";valid_file "T";tempfilename
'MONOI503.wmf';tempfile-properties "XPR";}}The area in the first quadrant is
just a quarter of a circle with radius $4$ and so the area is $A_{1}=\dfrac{1%
}{4}\left( 4\right) ^{2}\pi =4\pi $. \ The area in the second quadrant,
denoted by $A_{2}$ is more complex: \ 

Let $A_{T}$ denote the area of the triangle defined by $\left( 0,0\right) $, 
$\left( 0,4\right) $, and $\left( -3,0\right) $. \ Let $A_{C}$ denote the
area of the quarter of the circle in the second quadrant and let $A_{\text{%
int}}$ denote the area of the intersection of these two. \ The area we are
looking for is%
\begin{equation*}
A_{2}=A_{T}+A_{C}-A_{\text{int}}
\end{equation*}%
Clearly $A_{T}=\dfrac{1}{2}\left( 3\right) \left( 4\right) =6$ and $A_{C}=%
\dfrac{1}{4}\left( 3\right) ^{2}\pi =\dfrac{9}{4}\pi $. \ The real challenge
is to compute the area of the intersection\FRAME{dtbpFX}{1.7063in}{1.7063in}{%
0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width
1.7063in;height 1.7063in;depth 0pt;display "USEDEF";plot_snapshots
TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin "-5";xmax "5";xviewmin
"-5";xviewmax "5";yviewmin "-5";yviewmax
"5";viewset"XY";rangeset"X";plottype 4;plotticks 1;num-x-ticks
11;num-y-ticks 11;numpoints 100;plotstyle "patch";axesstyle "normal";xis
\TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function
\TEXUX{$\left( \left( 0,0\right) ,\left( 0,3\right) \right) $};linecolor
"black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes
"Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";function
\TEXUX{$\left( \left( -3,0\right) ,\left( 0,0\right) \right) $};linecolor
"green";linestyle 1;pointstyle "point";linethickness 3;lineAttributes
"Solid";curveColor "[flat::RGB:0x00008000]";curveStyle "Line";function
\TEXUX{$\left( \left( -0.1,4\right) ,\left( -0.1,0\right) \right)
$};linecolor "green";linestyle 1;pointstyle "point";linethickness
3;lineAttributes "Solid";curveColor "[flat::RGB:0x00008000]";curveStyle
"Line";function \TEXUX{$\left( \left( -3,0\right) ,\left( 0,4\right) \right)
$};linecolor "green";linestyle 1;pointstyle "point";linethickness
3;lineAttributes "Solid";curveColor "[flat::RGB:0x00008000]";curveStyle
"Line";function \TEXUX{$\sqrt{9-x^{2}}$};linecolor "black";linestyle
1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range
"-3,0";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle
"Line";rangeset"X";valid_file "T";tempfilename
'MONGX701.wmf';tempfile-properties "XPR";}}First we will compute the
coordinates of the point where the circle and the line intersect each other. 
$\ $This means finding the other solution (other than $\left( -3,0\right) $)
of the system $\left\{ 
\begin{array}{c}
x^{2}+y^{2}=9 \\ 
y=\dfrac{4}{3}x+4%
\end{array}%
\right. $

We wil use substitution. \ Because we know one of the solution of this
system is $\left( -3,0\right) $, we will try to solve for $y$ at the end. \
Quadratic equations are easy to factor if one of the root is zero. \ \ So we
first solve for $x$ in terms of $y$ and substitute that into the quadratic
equation.%
\begin{equation*}
y=\dfrac{4}{3}x+4~~~~~\implies ~~~~~~x=\dfrac{3}{4}\left( y-4\right) =\dfrac{%
3}{4}y-3
\end{equation*}%
\begin{eqnarray*}
x^{2}+y^{2} &=&9 \\
\left( \dfrac{3}{4}y-3\right) ^{2}+y^{2} &=&9 \\
\dfrac{9}{16}y^{2}-\dfrac{9}{2}y+9+y^{2} &=&9 \\
\dfrac{9}{16}y^{2}-\dfrac{9}{2}y+9+y^{2} &=&9 \\
\dfrac{25}{16}y^{2}-\dfrac{9}{2}y &=&0 \\
\dfrac{25}{16}y\left( y-\dfrac{~~\dfrac{9}{2}~~}{\dfrac{25}{16}}\right) &=&0%
\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\dfrac{~~\dfrac{9}{2}~~}{\dfrac{25}{16}}=%
\dfrac{9}{2}\cdot \dfrac{16}{25}=\dfrac{72}{25}
\end{eqnarray*}%
\begin{equation*}
y_{1}=0\text{ \ \ \ and \ \ \ \ }y_{2}=\dfrac{72}{25}=2.\,88
\end{equation*}%
Then $x_{2}=\dfrac{3}{4}y_{2}-3=\dfrac{3}{4}\left( 2.\,88\right) -3=-0.84$.
\ So the other intersection point is $\left( -0.84,2.88\right) $.

We separate the two areas as shown below. \ The area of the triangle is $%
\dfrac{1}{2}\left( 3\right) \left( 2.88\right) =4.\,\allowbreak 32$ as the
height belonging to the $3$ unit long side is $2.88$. \ For the area of the
sector, we need to find first the central angle. \ \FRAME{dtbpFX}{2.5097in}{%
1.7063in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type
"MAPLEPLOT";width 2.5097in;height 1.7063in;depth 0pt;display
"USEDEF";plot_snapshots TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin
"-5";xmax "5";xviewmin "-5";xviewmax "5";yviewmin "-1";yviewmax
"5";viewset"XY";rangeset"X";plottype 4;plotticks 1;num-x-ticks
11;num-y-ticks 11;constrained TRUE;numpoints 100;plotstyle "patch";axesstyle
"normal";xis \TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name
\TEXUX{$y$};function \TEXUX{$\left( \left( -3,0\right) ,\left( 0,0\right)
\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness
3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle
"Line";function \TEXUX{$\left( \left( 0,4\right) ,\left( 0,0\right) \right)
$};linecolor "green";linestyle 1;pointstyle "point";linethickness
3;lineAttributes "Solid";curveColor "[flat::RGB:0x00008000]";curveStyle
"Line";function \TEXUX{$\left( \left( -0.84,2.88\right) ,\left( -3,0\right)
\right) $};linecolor "green";linestyle 1;pointstyle "point";linethickness
3;lineAttributes "Solid";curveColor "[flat::RGB:0x00008000]";curveStyle
"Line";function \TEXUX{$\sqrt{9-x^{2}}$};linecolor "black";linestyle
1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range
"-0.84,0";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle
"Line";rangeset"X";function \TEXUX{$\left( \left( 0,0\right) ,\left(
-0.84,2.88\right) \right) $};linecolor "green";linestyle 1;pointstyle
"point";linethickness 3;lineAttributes "Solid";curveColor
"[flat::RGB:0x00008000]";curveStyle "Line";function \TEXUX{$\left( \left(
-0.84,0\right) ,\left( -0.84,2.88\right) \right) $};linecolor
"black";linestyle 3;pointstyle "point";linethickness 1;lineAttributes
"Dots";curveColor "[flat::RGB:0000000000]";curveStyle "Line";valid_file
"T";tempfilename 'MONPV504.wmf';tempfile-properties "XPR";}}Using right
triangle trigonometry, we write%
\begin{equation*}
\tan \alpha =\dfrac{0.84}{2.88}=\dfrac{7}{24}~~\implies \alpha =\tan
^{-1}\left( \dfrac{7}{24}\right)
\end{equation*}%
So the area of the sector is 
\begin{equation*}
A_{\text{sector}}=\dfrac{r^{2}\theta }{2}=\dfrac{\left( 3\right) ^{2}}{2}%
\tan ^{-1}\left( \dfrac{7}{24}\right) =\dfrac{9}{2}\tan ^{-1}\left( \dfrac{7%
}{24}\right) \approx 3.\,\allowbreak 429\,66
\end{equation*}%
So the area of the intersection is 
\begin{equation*}
A_{\text{int}}=4.\,32+\dfrac{9}{2}\tan ^{-1}\left( \dfrac{7}{24}\right)
\end{equation*}%
We are ready to compute the area: \ 
\begin{equation*}
A=A_{1}+A_{2}=A_{1}+A_{T}+A_{C}-A_{\text{int}}=4\pi +6+\dfrac{9}{4}\pi
-\left( 4.\,32+\dfrac{9}{2}\tan ^{-1}\left( \dfrac{7}{24}\right) \right)
\approx 20.\,\allowbreak 037\,880\,6
\end{equation*}

After rounding to the nearest hundredth, we get $20.04$ which is choice B.

\item For how many pairs of positive integers $(n,m)$ with $n,m<100$ are
both of the polynomials $x^{2}+mx+n$ and $x^{2}+mx-n$ factorable over the
integers? \ 
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A. $\ 4$

B. $\ 5$ \ 
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C. $\ $ $6$

D. $\ 7$

E. $\ \ 8$ \ \ \ 
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Solution: \ Let $s_{1}$,$s_{2}$ and $t_{1},t_{2}$ be the integers in the
factored form, i.e. 
\begin{equation*}
x^{2}+mx+n=\left( x-s_{1}\right) \left( x-s_{2}\right) \text{ \ \ and \ }%
x^{2}+mx-n=\left( x-t_{1}\right) \left( x-t_{2}\right)
\end{equation*}%
Then 
\begin{equation*}
x^{2}+mx+n=x^{2}-\left( s_{1}+s_{2}\right) x+s_{1}s_{2}\text{ \ \ and \ }%
x^{2}+mx-n=x^{2}-\left( t_{1}+t_{2}\right) x+t_{1}t_{2}
\end{equation*}%
\begin{eqnarray*}
-\left( s_{1}+s_{2}\right) &=&m \\
s_{1}s_{2} &=&n \\
-\left( t_{1}+t_{2}\right) &=&m \\
t_{1}t_{2} &=&-n
\end{eqnarray*}

$m^{2}-4n$ a square and $m^{2}+4n$ a square

$m^{2}-4n=A^{2}$

$m^{2}+4n=B^{2}$

$B^{2}-A^{2}=8n$

$\left( B+A\right) \left( B-A\right) =8n$

\item Triangles ACD and BCD (AD = 14, BD = 40) are inscribed in a semicircle
with diameter CD = 50. If AB $>25$, find the area of their union.

\item A. 625 B. 637.5 C. 652.5 D. 673.5 E. 675
\end{enumerate}

\end{document}