# In the triangle $\triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $\angle ACB > 60^\circ$.

The following was a question in the final of the Flanders Mathematics Olympiad 2018:

In the triangle $$\triangle ABC$$, $$|AB|^3 = |AC|^3 + |BC|^3$$. Prove that $$\angle ACB > 60^\circ$$.

In this competition, points are assigned for formulating a rigorous and mathematically sound proof.

I proved the above by contradiction. Let $$\alpha = \angle BAC, \beta = \angle CBA, \gamma = \angle ACB$$. Suppose $$\gamma \le 60^\circ$$:

$$\gamma \le 60^\circ \iff \sin(\gamma) \leq \frac{\sqrt{3}}{2}$$

Applying the sine rule:

$$\frac{\sin(\alpha)}{|BC|} = \frac{\sin(\beta)}{|AC|} = \frac{\sin(\gamma)}{|AB|}$$

$$\iff\frac{|AC|^3}{|AB|^3} + \frac{|BC|^3}{|AB|^3} = \frac{\sin^3(\alpha) + \sin^3(\beta)}{\sin^3(\gamma)} = 1$$

$$\iff \sin^3(\alpha) + \sin^3(\beta) = \sin^3(\gamma) \le \left( \frac{\sqrt{3}}{2} \right)^3$$

$$\iff\sin(\alpha) \le \frac{\sqrt(3)}{2}, \, \sin(\beta) \le \frac{\sqrt(3)}{2}\tag{1}$$

We also know that:

$$\alpha + \beta = 180^\circ - \gamma \ge 120^\circ\tag{2}$$

$$\alpha + \beta < 180^\circ\tag{3}$$

Without loss of generality, assume $$\alpha \ge \beta$$. From $$(1)$$, $$(2)$$ and $$(3)$$, it then follows that:

$$a \ge 120^\circ, \, \beta \le 60^\circ$$

$$\implies |BC| > |AB| \qquad \unicode{x21af}$$

Is this answer adequate enough? Can the notation be improved? Are there any alternative approaches to solve this problem?

• You might want to consider using a version of the triangle inequality to justify that $|AB|$ is the longest side so $\angle ACB$ is the largest angle. – WW1 Feb 1 at 23:59

## 3 Answers

In the standard notation we need to prove that $$\frac{a^2+b^2-c^2}{2ab}<\cos60^{\circ}$$ or $$c^2>a^2-ab+b^2$$ or $$\sqrt{(a^3+b^3)^2}>a^2-ab+b^2$$ or $$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or $$ab>0,$$ which is true.

Id est, $$\measuredangle ACB>60^{\circ}$$ and we are done!

With your answer, to justify $$\alpha \ge 120°$$ means that you need to exclude the possibility of $$\alpha = 60°$$, which you should perhaps explicitly state. Also, keep in mind the comment by WW1 to somehow justify which side is largest.

Also, here is an alternative solution. With the equation of

$$\left\lvert AB \right\rvert^3 = \left\lvert AC \right\rvert^3 + \left\lvert BC \right\rvert^3 \tag{1}\label{eq1}$$

since all lengths are positive values, you automatically have that

$$\left\lvert AB \right\rvert \gt \left\lvert AC \right\rvert \tag{2}\label{eq2}$$ $$\left\lvert AB \right\rvert \gt \left\lvert BC \right\rvert \tag{3}\label{eq3}$$

Thus, $$AB$$ is the longest side. The largest angle is opposite the longest side (this is fairly generally known, but you can prove this using the sine law for acute angle triangles, and the cosine law for obtuse angles, such as shown in The largest angle in a triangle). If $$\angle ACB \le 60°$$, then the other $$2$$ angles are $$\lt 60°$$, giving a sum $$\lt 180°$$, which is not true. As such, $$\angle ACB \gt 60°$$.

Also, note this only depends on showing that $$AB$$ is the longest side. Any equation which provides this would also give the same conclusion.

$$\blacksquare$$ Alternatively, there's a simpler way to prove that indeed $$\angle ACB>60º$$.

In any triangle the angle opposite the greater side is greater

From $$|AB|^3=|AC|^3+|BC|^3$$ we deduce $$|AB|>|AC| \; \text{and} \; |AB|>|BC|$$

It follows that $$\angle ACB>\angle BAC \; \text{and} \;\angle ACB>\angle CBA$$ Therefore $$\angle ACB>\frac{180°}{3}=60°$$ since the triangle isn't equilateral. $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\square$$