MHB Can a Triangle with Area 1 Have a Side Less Than √2?

[anemone]

Let $p,\,q,\,r$ be the sides of triangle $PQR$ with $p\ge q\ge r$. Prove that if the area of triangle $PQR$ is 1, then $q\ge \sqrt{2}$.

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[anemone]

Congratulations to greg1313 for his correct solution. :)

Here is the step-by-step solution:

Spoiler

We're given that the area of the triangle $PQR$ is 1, so we have

$\dfrac{pq\sin R}{2}=1\,\,\,\implies\,\,\sin R=\dfrac{2}{pq}$

By using the trigonometric identity that says $\sin^2 R+\cos^2 R=1$, we then have

$\cos R=\dfrac{\sqrt{p^2q^2-4}}{pq}$---(1)

By the Law of Cosines we have

$\begin{align*}r^2&=p^2+q^2-2pq\cos R\\&=p^2+q^2-2pq\left(\dfrac{\sqrt{p^2q^2-4}}{pq}\right)---(\text{from (1)})\\&=p^2+q^2-2\sqrt{p^2q^2-4}\end{align*}$

Since $q\ge r$, $q^2\ge r^2$ so we have

$q^2\ge p^2+q^2-2\sqrt{p^2q^2-4}$

$2\sqrt{p^2q^2-4}\ge p^2$

$4p^2q^2-16\ge p^4$

$0\ge p^4-4p^2q^2+16=(p^2-2q^2)^2+16-4q^4$

In order for the above inequality to be true, $16-4q^4$ cannot be positive, therefore

$16-4q^4\le 0$

$q\ge \sqrt{2}$ (Q.E.D.)