, 2012

Solution 1

First, we calculate the side lengths of $\triangle ABC$: $$\begin{aligned} AB & = \sqrt{(0-3)^2+(5-0)^2}=\sqrt{34} \\ BC & = \sqrt{(3-8)^2+(0-3)^2}=\sqrt{34} \\ AC & = \sqrt{(0-8)^2+(5-3)^2}=\sqrt{68}\end{aligned}$$ Since $AB=BC$ and $AC=\sqrt{2}AB=\sqrt{2}BC$, then $\triangle ABC$ is an isosceles right-angled triangle, with the right angle at $B$. Therefore, $\angle ACB = 45^\circ$. [[IMAGE0]] Solution 2 As in Solution 1, $AB = BC = \sqrt{34}$. Line segment $AB$ has slope $\frac{5-0}{0-3}=-\frac{5}{3}$. Line segment $BC$ has slope $\frac{0-3}{3-8}=\frac{3}{5}$. Since the product of these two slopes is $-1$, then $AB$ and $BC$ are perpendicular. Therefore, $\triangle ABC$ is right-angled at $B$. Since $AB=BC$, then $\triangle ABC$ is an isosceles right-angled triangle, so $\angle ACB = 45^\circ$. Solution 3 As in Solution 1, $AB=BC=\sqrt{34}$ and $AC=\sqrt{68}$. Using the cosine law, $$\begin{aligned} AB^2 & = AC^2 + BC^2 - 2(AC)(BC)\cos(\angle ACB) \\ 34 & = 68 + 34 - 2(\sqrt{68})(\sqrt{34})\cos(\angle ACB)\\ 0 & = 68 - 2(\sqrt{2}\sqrt{34})(\sqrt{34})\cos(\angle ACB)\\ 0 & = 68 - 68\sqrt{2}\cos(\angle ACB)\\ 68 \sqrt{2} \cos(\angle ACB) & = 68 \\ \cos(\angle ACB) & = \tfrac{1}{\sqrt{2}}\end{aligned}$$ Since $\cos(\angle ACB) = \frac{1}{\sqrt{2}}$ and $0^\circ < \angle ACB < 180^\circ$, then $\angle ACB = 45^\circ$. Draw perpendiculars from $P$ and $Q$ to $X$ and $Y$, respectively, on $SR$. [[IMAGE1]] Since $PQ$ is parallel to $SR$ (because $PQRS$ is a trapezoid) and $PX$ and $QY$ are perpendicular to $SR$, then $PQYX$ is a rectangle. Thus, $XY=PQ=7$ and $PX=QY$. Since $\triangle PXS$ and $\triangle QYR$ are right-angled with $PS = QR$ and $PX=QY$, then these triangles are congruent, and so $SX = YR$. Since $XY = 7$ and $SR=15$, then $SX+7+YR=15$ or $2\times SX = 8$ and so $SX = 4$. By the Pythagorean Theorem in $\triangle PXS$, $$PX^2 = PS^2 - SX^2 = 8^2-4^2 = 64-16 = 48$$ Now $PR$ is the hypotenuse of right-angled $\triangle PXR$. Since $PR>0$, then by the Pythagorean Theorem, $$PR = \sqrt{PX^2+XR^2}=\sqrt{48+(7+4)^2}=\sqrt{48+11^2}=\sqrt{48+121}=\sqrt{169}=13$$ Therefore, $PR=13$.