, 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$. [[IMAGE0]] 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 $6 - 3 - 1 = 2$. [[IMAGE1]] 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 $(\frac{4}{5},\frac{18}{5})$.