algebra-precalculuscalculusgeometrytrigonometry
The definition of a lattice point is a point that has integer coordinates.
Find the number of lattice points $(x,y)$, where $-100\le x\le 100$ and $-100\le y\le 100$ are on the graph of the parametric equations \begin{align*} x&=30-40\cos t, y=-50 + 30\cos t. \end{align*}
How would I begin? There could be infinite possibilities for $t$, so I am not sure how to begin this problem.
$(\cos t,\sin t)$ is the constant speed parametrization of the unit circle. Adding to the coordinates shifts its center, while multiplying these terms scales the circle and hence changes its radius. So as you already recognized, $(1+\cos t,â3+\sin t)$ is a circle or radius $1$ with center $(1,-3)$. The other, $(2+4\cos s,kâ4\sin s)$, is a circle with center $(2,k)$ and radius $4$.
The fact that you have $-4\sin s$ instead of $+4\sin s$ means that the parametrization is the other way round from the conventional one: you start at the righternmost point, but then go down (negative $y$ direction) instead of up. Doesn't change which circles this describes, only what parameter $s$ corresponds to which point on that circle.
So you have two circles, and you want to know how to choose the $y$ coordinate for the center of the second in such a way that the two circles have exactly one point in common, i.e. they touch. There are two ways the circles might touch: they might touch on the outside, or the smaller one might touch the larger one from the inside. In the former case, the distance between the centers has to be the sum of the radii, in the latter case it's the difference of the radii. So you have touching circles for
$$ \sqrt{(2-1)^2+(k+3)^2}\in\{4+1,4-1\}=\{5,3\} $$
To solve this, square the equation and you get
\begin{align*} (2-1)^2+(k+3)^2 &= 5^2 & (2-1)^2+(k+3)^2 &= 3^2 \\ k^2 + 6k + 9 &= 25 - 1 & k^2 + 6k + 9 &= 9-1 \\ k^2 + 6k - 15 &= 0 & k^2 + 6k + 1 &= 0 \\ k_{1,2} = \frac{-6\pm\sqrt{36+60}}{2} &= -3\pm2\sqrt6 & k_{3,4} = \frac{-6\pm\sqrt{36-4}}{2} &= -3\pm2\sqrt2 & \end{align*}
if you look at the parametric graph on the intervals $0 \le t \le \pi/4$ it does an arc staring at $(1,0)$ to $(-1,-1)$ on the next $\pi/4 \le t \le \pi/2$ it retraces that arc back to $(1,0)$ for $\pi/2 \le t \le 3\pi/4$ it does an arc that is the reflection of the old arc traversed for $0 \le t \le \pi/4 $ on the $x$-axis. and finally for $3\pi/4 \le t \le \pi$ it is back to $(1,0)$ so the period is $\pi.$
Best Answer
You can basically ignore the cosine element - just think of it as a variable that can take any value between $-1$ and $1$ inclusive. The equations then define a line section between $(70,-80)$ and $(-10, -20)$:
$$\left .\begin{align}x&=30-40v\\ y&=-50 + 30v \end{align}\right \rbrace v\in [-1,1]$$
Then you just need to find values of $v$ in that range that make $40v$ and $30v$ both integers.
Clearly for $30v$ to be an integer, you need $v=\frac{k}{30}$, which gives 61 possibilities in range. Simultaneously you need $v=\frac{m}{40}$. This means that $30m=40k \implies 3m=4k$. Since $3$ and $4$ are co-prime, this will only happen when $3 \mid k$ and $4\mid m$, which means that the only solutions are for $v=\frac{n}{10}, n=\{-10,\ldots 10\}$
This gives a total of $21$ solutions.