# Thread: Hard construction dealing with area of a triangle

1. ## Hard construction dealing with area of a triangle

Construct an equilateral triangle whose area is equal to the area of any scalene triangle ABC.

2. Originally Posted by brandonb122
Construct an equilateral triangle whose area is equal to the area of any scalene triangle ABC.
I would have thought this would be impossible without knowing the scalene triangle. To find the area of an equilateral triangle would be either using sin or Pythagoras. Let the triangle have sides a=b=c.

sine method:

$A = \frac{1}{2}absin(C) = \frac{1}{2}a^2sin(60) = \frac{\sqrt3 a^2}{4}$

Pythagoras:

Bisect the apex of the triangle so that it meets the base perpendicularly. The new base will then be $\frac{a}{2}$

To find the height of the line we just drew (h) use Pythagoras' theorem:

$h = \sqrt{a^2 - \frac{a^2}{4}} = \sqrt{\frac{3a^2}{4}} = \frac{a \sqrt3}{2}$.

We can then use this height in $A = \frac{1}{2}bh = \frac{1}{2}\frac{a \sqrt3}{2}a = \frac{a^2\sqrt3}{4}$

Once you know a side a then you can construct a scalene with the same area.

3. Originally Posted by brandonb122
Construct an equilateral triangle whose area is equal to the area of any scalene triangle ABC.
I'm assuming that "Construct" means "Construct in the sense of Euclidean geometry". So given the scalene triangle ABC, we want a ruler-and-compass construction of an equilateral triangle with the same area as ABC.

We know the base b of ABC, and we can construct its vertical height h. If the equilateral triangle has side a then (see e^(i*pi)'s comment above) $\tfrac12bh = \tfrac14\sqrt3a^2$. Write this as $a^2 = \frac{2bh}{\sqrt3} = \tfrac23\sqrt3bh$.

The first step is to construct a line of length $\tfrac23\sqrt3b$. Draw a circle of radius b together with a diameter. Mark a point on the circumference at a distance b from one end of the diameter. Joint this point to the other end of the diameter. You then have a right-angled triangle, two of whose sides are b and 2b. So the length of the third side is $\sqrt3b$ (by Pythagoras). There is a standard construction for a line whose length is a given rational multiple of a known length. So you can now construct a line PQ, say, of length $\tfrac23\sqrt3b$.

Extend this line to a line PQR, where QR has length h, and draw a circle with diameter PR. Construct a perpendicular line through Q, meeting the circle at X and Y. Then there's a theorem telling you that $QX^2 = PQ\times QR$. So we can take a to be the length of QX.