Number of points $C=(x,y)$ on the circle $x^2+y^2=16$ such that the area of the triangle whose vertices are $A,B$ and $C$ is a positive integer

Question

Let $A(-4,0)$ and $B(4,0)$.Number of points $C=(x,y)$ on the circle $x^2+y^2=16$ such that the area of the triangle whose vertices are $A,B$ and $C$ is a positive integer,is...

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I found the area of triangle $ABC=4|y|$ but i cannot count in how many cases the area will be a positive integer.The answer given is $62$.Is there a method by which i can solve this.Please help me.Thanks.

Answer 1

What you missed

You are absolutely right. The area is $4|y|$, and for that to be an integer, $y$ can take $64$ (Divide the circle into units of $\frac14$) values because $0 \le |y| \le 4$. However, $2$ of these values are points $A$ and $B$ itself (so they will not help form a triangle), hence the answer is $\color{blue}{62}$

Answer 2

Hint: $-4 \leq y \leq 4$, so the possible $y$ values are $$\frac{1}{4}, -\frac{1}{4}, \frac{2}{4}, -\frac{2}{4}, \frac{3}{4}, - \frac{3}{4}, \dots, \frac{16}{4}, -\frac{16}{4}$$

Then, how many times is each $y$ value realized on the circle?