Find the lengths of the three sides of a triangle that satisfy the following conditions.
(1) The side lengths are integers;
(2) The perimeter is an integer multiple of the area.
Solution: Let the three sides of the triangle be $a, b, c$, and the perimeter is $k$ (where $k$ is an integer) times the area. According to Heron's formula, we have
\[a+b+c=k \sqrt{\frac{a+b+c}{2} \cdot \frac{b+c-a}{2} \cdot \frac{c+a-b}{2} \cdot \frac{a+b-c}{2}}\]
Then, \(4 \cdot \frac{a+b+c}{2}=k^{2} \cdot \frac{b+c-a}{2} \cdot \frac{c+a-b}{2} \cdot \frac{a+b-c}{2}\)
Let \(m=b+c-a, n=c+a-b, p=a+b-c\). Then \(m+n+p=a+b+c\).
Thus, \(16(m+n+p)=k^{2} m n p\), and
\(m \geqslant 1, n \geqslant 1, p \geqslant 1\).
Therefore, \(\frac{k^{2}}{16}=\frac{1}{m n}+\frac{1}{m p}+\frac{1}{n p} \leqslant 3\).
Hence, \(k^{2} \leqslant 48<49\), which means \(k<7\).
Since \(k\) is an integer, \(k=1,2,3,4,5,6\).
From equation (1), when \(k=1,2,3,5,6\),
\(\frac{b+c-a}{2}, \frac{c+a-b}{2}, \frac{a+b-c}{2}\) are integers.
Let \(\frac{b+c-a}{2}=x, \frac{c+a-b}{2}=y, \frac{a+b-c}{2}=z\).
Substituting into equation (1) gives
\[4(x+y+z)=k^{2} x y z\],

or \(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=\frac{k^{2}}{4}\).
Assume \(x \geqslant y \geqslant z\). From equation (3), \(\frac{k^{2}}{4} \leqslant \frac{3}{z^{2}}\), which means
\[z^{2} \leqslant \frac{12}{k^{2}}\].
(i) When \(k=1\), from equation (4), \(z^{2} \leqslant 12, z \leqslant 2 \sqrt{3}<4\). Thus, \(z=1,2,3\).
Therefore, equation (2) becomes
\[4(x+y+z)=x y z\]
(a) When \(z=1\), from equation (2),
\[x y=4(x+y+1)=4 x+4 y+4\],

which means \((x-4)(y-4)=20=20 \times 1=10 \times 2=5 \times 4\).
Solving gives \((x, y)=(24,5),(14,6),(9,8)\).
By transformation, \(a=y+z, b=z+x, c=x+y\).
Thus, \((a, b, c)=(6,25,29),(7,15,20),(9,10,17)\).
(b) When \(z=2\), similarly solving gives
\((x, y, z)=(10,3,2),(6,4,2)\).
Thus, \((a, b, c)=(5,12,13),(6,8,10)\).
(c) When \(z=3\), from equation (5),
\[3 x y=4 x+4 y+12\],

which means \((3 x-4)(3 y-4)=52=52 \times 1=26 \times 2=13 \times 4\).
This equation has only one integer solution \(x=10, y=2\). But \(y=2<3=z\), so when \(z=3\), \((x, y, z)\) has no integer solution.
(ii) When \(k=2\), equation (2) becomes
\[x+y+z=x y z\].
From equation (4), \(z^{2} \leqslant \frac{12}{4}=3\), which means \(z \leqslant \sqrt{3}\). Thus, \(z=1\).
Substituting \(z=1\) into equation (6) gives \(x y-x-y=1\), which means
\[(x-1)(y-1)=2 \times 1\].
Solving gives \(x=3, y=2\).
Thus, \((a, b, c)=(3,4,5)\).
(iii) When \(k=3\), equation (2) becomes
\[9 x y z=4(x+y+z)\].
(7)
From equation (1), \(z^{2} \leqslant \frac{4}{3}\), which means \(z \leqslant \frac{2}{\sqrt{3}}\). Thus, \(z=1\).
Therefore, equation (7) becomes \(9 x y=4 x+4 y+4\), which means
\[(9 x-4)(9 y-4)=52=52 \times 1=26 \times 2=13 \times 4\].
This equation has no integer solutions, so \(k \neq 3\).
(iv) When \(k=5,6\), from equation (1), \(z<1\). Thus, \(k \neq 5\) and \(k \neq 6\).
(v) When \(k=4\), from equation (1),
\[m+n+p=m n p\].
From (ii), similarly, \(m=3, n=2, p=1\). In this case, \((a, b, c)=(1.5,2,2.5)\) is not an integer set, so \(k \neq 4\).
In summary, \(k=1,2\).
Thus, there are 6 triangles that satisfy the conditions, i.e., \((a, b, c, k)\)
\((6,25,29,1),(7,15,20,1),(9,10,17,1)\),
\((5,12,13,1),(6,8,10,1),(3,4,5,2)\).
(Zou Shouwen, Anhui Nanling County Experimental Middle School)