, 2024

Since $\triangle ABD$ is
right-angled at $B$ and has $\angle ADB = 45\degree$, then $\angle BAD = 45\degree$.

Similarly, $\triangle CPD$ is
right-angled and isosceles with $\$\angle$ PCD =
$45\degree\$.$

Further, $\triangle APN$ and $\triangle CBN$ are also both right-angled
and isosceles.

Since $CB = 6$ and $NB = CB$, then $NB = 6$.

Since $AB = 12$ and $NB = 6$, then $AN = AB - NB = 6$.

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Since $\triangle APN$ is
right-angled and isosceles, then its sides are in the ratio $1:1:\sqrt{2}$.

Thus, $AP = PN = $\frac{1}{\sqrt{2}}$ AN =
$\frac{6}{\sqrt{2}} = 3\sqrt{2}\$.$

Alternatively, if $AP = PN = x$, then
the Pythagorean Theorem gives $AN^2 = AP^2 +
PN^2$and so$6^2 = 2x^2$
which gives $AP^2 = x^2 = 18$.

Thus, the area of $\triangle APN$ is
$\$\frac{1}{2} \cdot AP \cdot$ PN = $\frac{1}{2} \cdot 3\sqrt{2} \cdot 3\sqrt{2}$ = 9$.
The line with equation $y = -3x +
6$has$y$-intercept$6$, which means that$OB = 6$.

To find the $x$-intercept of this
line, we set $y = 0$ and obtain the
equation $-3x + 6 = 0$ which gives
$3x = 6$ or $x = 2$. This means that $OA = 2$.

Since $\triangle ABO$ is right-angled
at $O$, its area is $\$\frac{1}{2} \cdot OB \cdot$ OA = $\frac{1}{2} \cdot 6 \cdot$ 2 = 6$.

Since the area of $\triangle ACD$ is
$\frac{1}{2}$ of the area of $\triangle ABO$, then the area of $\triangle ACD$ is $3$.

Next, we note that the line with equation $y
= mx + 1$has$y$-intercept
$1$; thus, $OD = 1$.

This means that the area of $\$\triangle$
ADO$is$$\frac{1}{2} \cdot OD \cdot$
OA = $\frac{1}{2} \cdot 1 \cdot$ 2 = 1$.

We can determine the area of $\$\triangle$
BCD$by subtracting the areas of$$\triangle$ ACD$and$$\triangle$ ADO$from that of$$\triangle$ ABO$, which tells us that the
area of $\triangle BCD$ is $6 - 3 - 1 = 2$.

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Now, we can consider $BD$, which
has length $6 - 1 = 5$, as the base
of $\triangle BCD$; the corresponding
height of $\triangle BCD$ is the
distance from $C$ to the $y$-axis, which we call $h$.

Thus, $\$\frac{1}{2} \cdot 5 \cdot$ h =
2$and so$h =
$\frac{4}{5}\$.$

This means that $C$ has $x$-coordinate $\frac{4}{5}$.

Since $C$ is on the line with
equation $y = -3x + 6$, we have $y = -3 $\cdot \frac{4}{5}$ + 6 =
$\frac{18}{5}\$.$

Therefore, the coordinates of $C$ are
$(\frac{4}{5},\frac{18}{5})$.