mat 2005 Q4

mat · Uk Proof
4. An $n \times n$ square array contains 0 s and 1 s. Such a square is given below with $n = 3$.
001
100
110

Two types of operation $C$ and $R$ may be performed on such an array.
  • The first operation $C$ takes the first and second columns (on the left) and replaces them with a single column by comparing the two elements in each row as follows: if the two elements are the same then $C$ replaces them with a 1 , and if they differ $C$ replaces them with a 0 .
  • The second operation $R$ takes the first and second rows (from the top) and replaces them with a single row by comparing the two elements in each column as follows: if the two elements are the same then $R$ replaces them with a 1 , and if they differ $R$ replaces them with a 0 .

By way of example, the effects of performing $R$ then $C$ on the square above are given below.
001
100
110
$\xrightarrow { R }$
010
110
$\xrightarrow { C }$
00
10

(a) If $R$ then $C$ are performed (in that order) on a $2 \times 2$ array then only a single number (0 or 1 ) remains.
(i) Write down, in the grids on the next page, the eight $2 \times 2$ arrays which, when $R$ then $C$ are performed, produce a 1.
(ii) By grouping your answers accordingly, show that if
$a$$b$
$c$$d$

is amongst your answers to part (i) then so is
$a$$c$
$b$$d$

Explain why this means that doing $R$ then $C$ on a $2 \times 2$ array produces the same answer as doing $C$ first then $R$.
(b) Consider now an $n \times n$ square array containing 0 s and 1 s , and the effects of performing $R$ then $C$ or $C$ then $R$ on the square.
(i) Explain why the effect on the right $n - 2$ columns is the same whether the order is $R$ then $C$ or $C$ then $R$. [This then also applies to the bottom $n - 2$ rows.]
(ii) Deduce that performing $R$ then $C$ on an $n \times n$ square produces the same result as performing $C$ then $R$.
l

  1. (a) Three points $P , A , B$ lie on a circle which has centre $O$. The point $C$ is where $P O$ extends to meet $A B$ as shown in the diagram below. [Figure]

Show that $\measuredangle A O C = 2 \measuredangle A P C$ and $\measuredangle B O C = 2 \measuredangle B P C$. Why does this mean that $\angle A P B$ is independent of the choice of the point $P$ ?
(b) Four points $K , L , M , N$ lie on a circle and the lines $L K$ and $M N$ meet outside the circle at a point $S$, as shown in the diagram below. [Figure]
Using part (a) and the Sine Rule show that
$$\frac { K S } { N S } = \frac { S M } { S L }$$
[You may also assume that part (b) holds true in the special case when $M = N$ in which case the line $S M$ is the tangent to the circle at $M$.]
(c) A tower has height $h$. Assuming the earth to be a perfect sphere of radius $r$, determine the greatest distance $x$ from the top of the tower at which an observer can still see it. [Figure]
4. An $n \times n$ square array contains 0 s and 1 s. Such a square is given below with $n = 3$.

\begin{center}
\begin{tabular}{ | l | l | l | }
\hline
0 & 0 & 1 \\
\hline
1 & 0 & 0 \\
\hline
1 & 1 & 0 \\
\hline
\end{tabular}
\end{center}

Two types of operation $C$ and $R$ may be performed on such an array.

\begin{itemize}
  \item The first operation $C$ takes the first and second columns (on the left) and replaces them with a single column by comparing the two elements in each row as follows: if the two elements are the same then $C$ replaces them with a 1 , and if they differ $C$ replaces them with a 0 .
  \item The second operation $R$ takes the first and second rows (from the top) and replaces them with a single row by comparing the two elements in each column as follows: if the two elements are the same then $R$ replaces them with a 1 , and if they differ $R$ replaces them with a 0 .
\end{itemize}

By way of example, the effects of performing $R$ then $C$ on the square above are given below.

\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
0 & 0 & 1 \\
\hline
1 & 0 & 0 \\
\hline
1 & 1 & 0 \\
\hline
\end{tabular}
\end{center}$\xrightarrow { R }$\begin{tabular}{ | l | l | l | }
\hline
0 & 1 & 0 \\
\hline
1 & 1 & 0 \\
\hline
\end{tabular}$\xrightarrow { C }$\begin{tabular}{ | l | l | }
\hline
0 & 0 \\
\hline
1 & 0 \\
\hline
\end{tabular}

(a) If $R$ then $C$ are performed (in that order) on a $2 \times 2$ array then only a single number (0 or 1 ) remains.\\
(i) Write down, in the grids on the next page, the eight $2 \times 2$ arrays which, when $R$ then $C$ are performed, produce a 1.\\
(ii) By grouping your answers accordingly, show that if

\begin{center}
\begin{tabular}{ | l | l | }
\hline
$a$ & $b$ \\
\hline
$c$ & $d$ \\
\hline
\end{tabular}
\end{center}

is amongst your answers to part (i) then so is

\begin{center}
\begin{tabular}{ | l | l | }
\hline
$a$ & $c$ \\
\hline
$b$ & $d$ \\
\hline
\end{tabular}
\end{center}

Explain why this means that doing $R$ then $C$ on a $2 \times 2$ array produces the same answer as doing $C$ first then $R$.\\
(b) Consider now an $n \times n$ square array containing 0 s and 1 s , and the effects of performing $R$ then $C$ or $C$ then $R$ on the square.\\
(i) Explain why the effect on the right $n - 2$ columns is the same whether the order is $R$ then $C$ or $C$ then $R$. [This then also applies to the bottom $n - 2$ rows.]\\
(ii) Deduce that performing $R$ then $C$ on an $n \times n$ square produces the same result as performing $C$ then $R$.

\begin{center}
\begin{tabular}{ l }
\hline
l \\
\hline
\end{tabular}
\end{center}

\begin{enumerate}
  \setcounter{enumi}{4}
  \item (a) Three points $P , A , B$ lie on a circle which has centre $O$. The point $C$ is where $P O$ extends to meet $A B$ as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{72d6ed00-a6cf-4a23-b714-2f5a1bbd9967-14_401_406_461_833}
\end{enumerate}

Show that $\measuredangle A O C = 2 \measuredangle A P C$ and $\measuredangle B O C = 2 \measuredangle B P C$. Why does this mean that $\angle A P B$ is independent of the choice of the point $P$ ?\\
(b) Four points $K , L , M , N$ lie on a circle and the lines $L K$ and $M N$ meet outside the circle at a point $S$, as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{72d6ed00-a6cf-4a23-b714-2f5a1bbd9967-14_412_704_1158_683}

Using part (a) and the Sine Rule show that

$$\frac { K S } { N S } = \frac { S M } { S L }$$

[You may also assume that part (b) holds true in the special case when $M = N$ in which case the line $S M$ is the tangent to the circle at $M$.]\\
(c) A tower has height $h$. Assuming the earth to be a perfect sphere of radius $r$, determine the greatest distance $x$ from the top of the tower at which an observer can still see it.\\
\includegraphics[max width=\textwidth, alt={}, center]{72d6ed00-a6cf-4a23-b714-2f5a1bbd9967-14_387_332_2087_868}
Paper Questions