SAT1000 - P887

Extend the reflections of the triangle mirror so that the path of the ray is represented as a straight line, and label the diagram as follows, including $M$ (the midpoint of $A$ and $B$ ) and $N$ (the centroid of $\triangle A'BC$ ):

Since the centroid is located $\frac{2}{3}$ of the distance from a vertex to its opposite midpoint, $\frac{A'N}{A'M} = \frac{2}{3}$ . Since $\triangle A'MS \sim A'NT$ by AA similarity, $\frac{A'T}{A'S} = \frac{A'N}{A'M}$ or $\frac{A'T}{2} = \frac{2}{3}$ , which solves to $A'T = \frac{4}{3}$ . That means the $x$ -coordinate of $N$ is $AB - A'T = 4 - \frac{4}{3} = \frac{8}{3}$ . By similarity, the $y$ -coordinate of $N$ is also $\frac{8}{3}$ , so the coordinates of $N$ are $N(\frac{8}{3}, \frac{8}{3})$ .

Let $k = AP = A'P"$ . Then the coordinates of $P$ are $P(k, 0)$ and the coordinates of $P'$ are $P'(4, k + 4)$ , and the equation of the line through $P$ and $P'$ is $y = -\frac{k + 4}{k - 4}(x - k)$ .

$N(\frac{8}{3}, \frac{8}{3})$ is on $y = -\frac{k + 4}{k - 4}(x - k)$ , so $\frac{8}{3} = -\frac{k + 4}{k - 4}(\frac{8}{3} - k)$ , which solves to $k = \frac{4}{3}$ .

Therefore, $|AP| = k = \frac{4}{3}$ and $\lfloor 1000|AP| \rfloor = \boxed{1333}$ .

$\triangle {PQR}$ is a right-angled triangle with right angle at $P$ . Let $\angle {APR}=α$ , position coordinates of $Q$ be $(h, k)$ , and $|\overline {AP}|=a$ .

Then, $k=(h+a)\tan α, h=a+k\tan α\implies$

$h=a\sec 2α,k=a\tan 2α$

$\implies a(\sec 2α+\tan 2α)=4$

$\implies a=\dfrac {4\cos 2α}{1+\sin 2α}$

Centeroid of the triangle is at $(\frac{4}{3},\frac{4}{3})$

Therefore $\frac{4}{3}=(\frac{4}{3}+a) \tan α$

$\implies 2\tan^3 α-5\tan^2 α+4\tan α-1=0$

$\implies \tan α=1,\frac{1}{2}$

For $\tan α=1(α=45\degree)$ , the ray from $P$ will be incident normally to the surface $BC$ , and is so discarded.

For $\tan α=0.5,a=\dfrac {4}{3}\approx 1.333333$ , and the required answer is $\boxed {1333}$ .

Extend the reflected beam with dashed line behind the mirror and the respective image points be label with a prime. Let $AP = a$ ; then $P(a,0)$ and its image $P'(4,4+a$ . Note the centroid $O \left(\frac 43, \frac 43 \right)$ and its image $O' \left(\frac 83, \frac 83 \right)$ .

Then the line $PP'$ is given by:

$\begin{aligned} \frac {y-\frac 83}{x-\frac 83} & = \frac {BP'}{BP} = \frac {4+a}{4-a} \\ \implies y & = \frac {4+a}{4-a}\left(x - \frac 83\right) + \frac 83 \end{aligned}$

Since at $P(a,0)$ , $y=0$ when $x=a$ , we have:

$\begin{aligned} \frac {4+a}{4-a}\left(a - \frac 83\right) + \frac 83 & = 0 \\ (4+a)\left(a - \frac 83\right) + \frac 83 (4-a) & = 0 \\ 4a + a^2 - \frac {16}3 a & = 0 \\ \implies a & = \frac {16}3 - 4 = \frac 43 \end{aligned}$

Therefore, $\lfloor 1000a\rfloor = \boxed {1333}$ .