2018_Question1_h

2018_Question1_h

Here’s my attempt on this problem

\frac{4s^2+t^2}{5st} simplifies as \frac{4s}{5t}+\frac{t}{5s}

Since the areas of triangles are positive we can use AM-GM.
\frac{\frac{4s}{5t}+\frac{t}{5s}}{2} \geq \frac{2}{5}

Multiplying both sides by 2 we get \frac{4s}{5t}+\frac{t}{5s} \geq \frac{4}{5}

The lowest value attainable is \frac{4}{5} so the answer is (d).

Referred by @AwesomeLife_Math

Hope this helps!

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Wonderful work @gsny !

However, I think you have forgotten to include the equality case (it can easily be fixed though).

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Ah yes, equality is obtained when \frac{4s}{5t}=\frac{t}{5s} \to 20s^2=5t^2 \to 4s^2=t^2 \to 2s = t.

This works as s < t and it is possible to draw such two triangles.

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