[Number Theory]
Problem: Determine all possible values for the area of a right-angled triangle with one side length equal to 60 and with the property that its side lengths form an arithmetic sequence.

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, $3,5,7,9$ are the first four terms of an arithmetic sequence.)

Solution: Since the three side lengths of a right-angled triangle form an arithemetic sequence and must include 60 , then the three side lengths are $60,60+d, 60+2 d$ or $60-d, 60,60+d$ or $60-2 d, 60-d, 60$, for some $d \geq 0$.

For a triangle with sides of length $60,60+d, 60+2 d$ to be right-angled, by the Pythagorean Theorem, the following equivalent equations must be true:

$$
\begin{aligned}
60^{2}+(60+d)^{2} & =(60+2 d)^{2} \\
3600+3600+120 d+d^{2} & =3600+240 d+4 d^{2} \\
0 & =3 d^{2}+120 d-3600 \\
0 & =d^{2}+40 d-1200 \\
0 & =(d+60)(d-20)
\end{aligned}
$$

(Note that, since $d \geq 0$, then $60+2 d$ must be the hypotenuse of the triangle.) Since $d \geq 0$, then $d=20$, which gives the triangle with side lengths $60,80,100$.

The longest side length is the hypotenuse and the shorter two sides meet at right angles, giving an area of $\frac{1}{2}(60)(80)=2400$.



For a triangle with sides of length $60-d, 60,60+d$ to be right-angled, by the Pythagorean Theorem, the following equivalent equations must be true:

$$
\begin{aligned}
(60-d)^{2}+60^{2} & =(60+d)^{2} \\
3600-120 d+d^{2}+3600 & =3600+120 d+d^{2} \\
3600 & =240 d \\
d & =15
\end{aligned}
$$

Since $d \geq 0$, then $d=15$ is admissible, which gives the triangle with side lengths 45, 60,75. Using a similar analysis, the area of this triangle is $\frac{1}{2}(45)(60)=1350$.

For a triangle with sides of length $60-2 d, 60-d, 60$ to be right-angled, by the Pythagorean Theorem, the following equivalent equations must be true:

$$
\begin{aligned}
(60-2 d)^{2}+(60-d)^{2} & =60^{2} \\
3600-240 d+4 d^{2}+3600-120 d+d^{2} & =3600 \\
5 d^{2}-360 d+3600 & =0 \\
d^{2}-72 d+720 & =0 \\
(d-60)(d-12) & =0
\end{aligned}
$$

Since $d \geq 0$, then $d=60$ or $d=12$, which give possible side lengths of $-60,0,60$ (which do not form a triangle) and 36,48,60 (which do form a triangle).

Using a similar analysis, the area of this triangle is $\frac{1}{2}(36)(48)=864$.

Therefore, the possible values for the area of such a triangle are 2400, 1350, and 864.
Suppose that a triangle has side lengths in arithemetic sequence.

Then the side lengths can be written as $a-d, a, a+d$ for some $a>0$ and $d \geq 0$.

Note that $a-d \leq a \leq a+d$.

For such a triangle to be right-angled, by the Pythagorean Theorem, the following equivalent equations are true:

$$
\begin{aligned}
(a-d)^{2}+a^{2} & =(a+d)^{2} \\
a^{2}-2 a d+d^{2}+a^{2} & =a^{2}+2 a d+d^{2} \\
a^{2} & =4 a d
\end{aligned}
$$

Since $a>0$, then $a=4 d$, and so the side lengths of the triangle are $a-d=3 d, a=4 d$, and $a+d=5 d$ for some $d \geq 0$.

(Note that such triangles are all similar to the 3-4-5 triangle.)

If such a triangle has 60 as a side length, then there are three possibilities:

(i) $3 d=60$ : This gives $d=20$ and side lengths $60,80,100$.

Since the triangle is right-angled and its hypotenuse has length 100, then its area will equal $\frac{1}{2}(60)(80)=2400$.

(ii) $4 d=60$ : This gives $d=15$ and side lengths $45,60,75$.

In a similar way to case (i), its area will equal $\frac{1}{2}(45)(60)=1350$.

(iii) $5 d=60$ : This gives $d=12$ and side lengths $36,48,60$.

In a similar way to case (i), its area will equal $\frac{1}{2}(36)(48)=864$.

Therefore, the possible values for the area of such a triangle are 2400, 1350, and 864 .

Final answer: 2400, 1350, 864