This is a problem from “Math World”:
A triangle has three sides of integer length and a perimeter of 1000. One of its interior angles is twice another interior angle. Find the lengths of the three sides.
At first glance, this appears to be a geometry problem, but it is actually an elementary number theory problem, specifically involving Diophantine equations. Similar problems often appear in mathematics competitions, and I would like to explore it with you here. Looking back, I actually liked number theory very much when I was a child; in elementary and middle school, I often studied terms like “prime numbers,” “perfect numbers,” “amicable numbers,” and “large number factorization.” Now that I have studied calculus and other subjects, my interest has gradually shifted toward more practical mathematics, so my level in number theory is not high—please excuse my limited knowledge.
Let the three sides be a, b, c, with corresponding angles A, B, C, and let B = 2A. According to the given conditions, we have: a + b + c = 1000
According to the Law of Cosines, we have: \cos A = \frac{b^2 + c^2 - a^2}{2bc}
According to the Law of Sines, we have: \frac{a}{\sin A} = \frac{b}{\sin 2A} = \frac{b}{2 \sin A \cos A}
Substituting the expression for \cos A into the above equation, we get: a = \frac{b}{2 \left( \frac{b^2 + c^2 - a^2}{2bc} \right)} which simplifies to: \frac{c}{a} = 1 + \left( \frac{c}{b} \right)^2 - \left( \frac{a}{b} \right)^2
Some readers might be confused as to why it is written in this form. In fact, BoJone’s idea was that when studying a problem, one should write the expression as concisely as possible and minimize the number of unknowns. In the expression above, \frac{c}{a} can be written as \frac{c/b}{a/b}. By letting \frac{a}{b} = m and \frac{c}{b} = n, we can transform the equation into one with only two unknowns: \frac{n}{m} = 1 + n^2 - m^2
From a physical perspective, the reason this convenience arises is that the rewritten \frac{a}{b} and \frac{c}{b} are “dimensionless” quantities. As for how to quickly think of this step, BoJone has no special method other than saying it requires a certain amount of practice, intuition, and constant trial and error.
The above equation can be further transformed into: \frac{n-m}{m} = (n-m)(n+m) It is easy to verify that n \neq m, so we have 1 = (n+m)m. Substituting the original variables back, we get: a^2 + ac = b^2 At this point, the most direct method would be to use computer programming to enumerate the possibilities. Of course, this cannot be done in a math competition, and some readers may prefer an analytical process, so it is worth exploring the calculation process further.
After several attempts, I thought of the following approach, which is relatively easy to calculate. By adding (\frac{1}{2}c)^2 to both sides of a^2 + ac = b^2, we get: (a + \frac{1}{2}c)^2 = b^2 + (\frac{1}{2}c)^2 This becomes a Pythagorean triple problem (though we do not yet know if c is odd or even). We recall that all Pythagorean triples can be expressed in the form (x^2+y^2)^2 = (x^2-y^2)^2 + (2xy)^2. Let us set: x^2+y^2 = a + \frac{1}{2}c, \quad x^2-y^2 = b, \quad 2xy = \frac{1}{2}c (There is a small question to consider here: why not x^2-y^2 = \frac{1}{2}c and 2xy = b?)
Given a + b + c = 1000, we have: x^2 + xy = 500 \Rightarrow y = \frac{500 - x^2}{x} \begin{aligned} x > y &\Rightarrow x > \sqrt{250} \approx 15.8 \\ y > 0 &\Rightarrow x < \sqrt{500} \approx 22.36 \end{aligned}
Since x^2 + xy = 500, x and y cannot both be of the form “.5”, so c must be an even number. By testing integer values for x from 16 to 22, we find that x=20 and y=5. This leads to the final result: a = 225, \quad b = 375, \quad c = 400
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