From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is $s$. Then the area of triangle is

We have an equilateral triangle with an interior point. From this point, we draw perpendiculars to all three sides. The sum of these three perpendicular lengths equals $s$, and we need to find the triangle's area.

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The brilliant approach here is to calculate the triangle's area in two different ways and set them equal.

For any equilateral triangle with side length $a$:

$\text{Area} = \dfrac{\sqrt{3}}{4}a^2$

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When we have an interior point $O$ with perpendiculars to the three sides, we can divide our big triangle into three smaller triangles.

Let's call the perpendicular lengths $h_1$, $h_2$, and $h_3$, where:

$h_1 + h_2 + h_3 = s$ (given)

Each smaller triangle has base as one side of the original triangle (length $a$) and height as one of our perpendiculars.

So the areas are:

Triangle 1: $\dfrac{1}{2} \times a \times h_1$

Triangle 2: $\dfrac{1}{2} \times a \times h_2$

Triangle 3: $\dfrac{1}{2} \times a \times h_3$

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The total area equals the sum of these three parts:

$\text{Total Area} = \dfrac{1}{2}a \times h_1 + \dfrac{1}{2}a \times h_2 + \dfrac{1}{2}a \times h_3$

$\text{Total Area} = \dfrac{1}{2}a(h_1 + h_2 + h_3)$

Since $h_1 + h_2 + h_3 = s$:

$\text{Total Area} = \dfrac{1}{2}as$

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Now we have two expressions for the same area:

$\dfrac{\sqrt{3}}{4}a^2 = \dfrac{1}{2}as$

$\sqrt{3}a^2 = 2as$

$\sqrt{3}a = 2s$

$a = \dfrac{2s}{\sqrt{3}}$

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We substitute this value of $a$ back into our area formula:

$\text{Area} = \dfrac{\sqrt{3}}{4}a^2 = \dfrac{\sqrt{3}}{4} \times \left(\frac{2s}{\sqrt{3}}\right)^2$

$= \dfrac{\sqrt{3}}{4} \times \dfrac{4s^2}{3}

= \dfrac{\sqrt{3} \times 4s^2}{4 \times 3}

= \dfrac{s^2\sqrt{3}}{3}$

We can also write this as:

$\text{Area} = \dfrac{s^2}{\sqrt{3}}$

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This problem showcases a fundamental principle in geometry: the sum of perpendiculars from any interior point to the sides of an equilateral triangle is constant. This constant equals the altitude of the triangle!

This technique of equating two area expressions is incredibly powerful and appears in many competitive math problems.