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We have a triangle $ABC$. It is given that $\sin^2A + \sin^2B = 5\sin^2C$. Prove that $\sin C \leq \frac{3}{5}$.

Let's say that $BC = a$, $AC=b$, $AB=c$.
According to the sine law,
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$$ then
$\sin A = \frac{a}{2R}\implies \sin^2A = \frac{a^2}{4R^2}$
$\sin B = \frac{b}{2R}\implies \sin^2B = \frac{b^2}{4R^2}$
$\sin C = \frac{c}{2R}\implies \sin^2C = \frac{c^2}{4R^2}$
Then we get:
$$\frac{a^2}{4R^2} + \frac{b^2}{4R^2} = 5\frac{c^2}{4R^2}.$$
Since $4R^2 > 0$, we get that $a^2 + b^2 = 5c^2$
Guys, is that correct? Even if it is, do you have any ideas what shall I do next?

R_D
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thomas21
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    I believe your question was closed because it was not apparent that you had provided your attempt. So I have included a link to it from your question, and voted to reopen. Next time at least state that your attempt is below. But if you do not have a full solution (like here), it is better to put your attempts in your question instead. – user21820 Mar 21 '18 at 05:27

2 Answers2

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Since $A + B + C = \pi$, we have $\sin (A + B) = \sin C$.

Also notice that

$$\cos^2 A + \cos^2 B = 1 - \sin^2A + 1 - \sin^2B = 2 - 5\sin^2(A + B)$$

Proceeding by CSB, we get:

\begin{align} \sin(A + B) &= \sin A \cos B + \sin B \cos A \\ &\le \sqrt{\sin^2 A + \sin ^2 B} \sqrt{\cos^2A + \cos^2B} \\ &= \sqrt{5} \sin(A + B) \sqrt{2 - 5\sin^2(A + B)} \end{align}

Therefore $$1 \le 5(2 - 5\sin^2(A + B)) = 10 - 25\sin^2(A + B)$$ which yields $$\sin C = \sin(A + B) \le \frac35$$

mechanodroid
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By the law of sines and AM-GM we obtain $$5c^2=a^2+b^2\geq2ab,$$ which gives $$\frac{c^2}{ab}\geq\frac{2}{5}.$$ In another hand, by the law of cosines we obtain: $$\cos{C}=\frac{a^2+b^2-c^2}{2ab}=\frac{2c^2}{ab}\geq\frac{4}{5}.$$ Id est, $$\sin{C}=\sqrt{1-\cos^2C}\leq\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}.$$

Michael Rozenberg
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