Problem (Number Theory):
For some positive integers $k$, the parabola with equation $y=\frac{x^{2}}{k}-5$ intersects the circle with equation $x^{2}+y^{2}=25$ at exactly three distinct points $A, B$ and $C$. Determine all such positive integers $k$ for which the area of $\triangle A B C$ is an integer.

Solution:
First, we note that since $k$ is a positive integer, then $k \geq 1$.

Next, we note that the given parabola passes through the point $(0,-5)$ as does the given circle. (This is because if $x=0$, then $y=\frac{0^{2}}{k}-5=-5$ and if $(x, y)=(0,-5)$, then $x^{2}+y^{2}=0^{2}+(-5)^{2}=25$, so $(0,-5)$ satisfies each of the equations.)

Therefore, for every positive integer $k$, the two graphs intersect in at least one point.

If $y=-5$, then $x^{2}+(-5)^{2}=25$ and so $x^{2}=0$ or $x=0$. In other words, there is one point on both parabola and circle with $y=-5$, namely $(0,-5)$.

Now, the given circle with equation $x^{2}+y^{2}=25=5^{2}$ has centre $(0,0)$ and radius 5 .

This means that the $y$-coordinates of points on this circle satisfy $-5 \leq y \leq 5$.

To find the other points of intersection, we re-write $y=\frac{x^{2}}{k}-5$ as $k y=x^{2}-5 k$ or $x^{2}=k y+5 k$ and substitute into $x^{2}+y^{2}=25$ to obtain

$$
\begin{aligned}
(k y+5 k)+y^{2} & =25 \\
y^{2}+k y+(5 k-25) & =0 \\
(y+5)(y+(k-5)) & =0
\end{aligned}
$$

and so $y=-5$ or $y=5-k$.

(We note that since the two graphs intersect at $y=-5$, then $(y+5)$ was going to be a factor of the quadratic equation $y^{2}+k y+(5 k-25)=0$. If we had not seen this, we could have used the quadratic formula.)

Therefore, for $y=5-k$ to give points on the circle, we need $-5 \leq 5-k$ and $5-k \leq 5$.

This gives $k \leq 10$ and $k \geq 0$.

Since $k$ is a positive integer, the possible values of $k$ to this point are $k=1,2,3,4,5,6,7,8,9,10$.

If $k=1$, then $y=5-1=4$. In this case, $x^{2}+4^{2}=25$ or $x^{2}=9$ and so $x= \pm 3$.

This gives the two points $(3,4)$ and $(-3,4)$ which lie on the parabola and circle.

Consider the three points $A(3,4), B(-3,4)$ and $C(0,-5)$.

Now $A B$ is horizontal with $A B=3-(-3)=6$. (This is the difference in $x$-coordinates.) The vertical distance from $A B$ to $C$ is $4-(-5)=9$. (This is the difference in $y$ coordinates.)

Therefore, the area of $\triangle A B C$ is $\frac{1}{2}(6)(9)=27$, which is a positive integer.

We now repeat these calculations for each of the other values of $k$ by making a table:

| $k$ | $y$ | $x= \pm \sqrt{25-y^{2}}$ | Base | Height | Area of triangle |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 4 | $\pm 3$ | $3-(-3)=6$ | $4-(-5)=9$ | 27 |
| 2 | 3 | $\pm 4$ | $4-(-4)=8$ | $3-(-5)=8$ | 32 |
| 3 | 2 | $\pm \sqrt{21}$ | $2 \sqrt{21}$ | 7 | $7 \sqrt{21}$ |
| 4 | 1 | $\pm \sqrt{24}$ | $2 \sqrt{24}$ | 6 | $6 \sqrt{24}$ |
| 5 | 0 | $\pm 5$ | 10 | 5 | 25 |
| 6 | -1 | $\pm \sqrt{24}$ | $2 \sqrt{24}$ | 4 | $4 \sqrt{24}$ |
| 7 | -2 | $\pm \sqrt{21}$ | $2 \sqrt{21}$ | 3 | $3 \sqrt{21}$ |
| 8 | -3 | $\pm 4$ | 8 | 2 | 8 |
| 9 | -4 | $\pm 3$ | 6 | 1 | 3 |
| 10 | -5 | 0 |  |  |  |

When $k=10$, we have $y=5-k=-5$ and $x=0$ only, so there is only one point of intersection.

Finally, the values of $k$ for which there are three points of intersection and for which the area of the resulting triangle is a positive integer are $k=1,2,5,8,9$.

Final answer: 1,2,5,8,9