## Brainteaser 1798: Trisection

**From The Sunday Times, 2nd March 1997**

We have taken a piece of paper in the shape of an equilateral triangle and folded it so that one of the corners lies at some point on the opposite side:

In the figure triangle A has an area of 36 cm² and triangle B has an area of 16 cm².

What is the area of the unfolded piece of paper?

This puzzle was included in the book *Brainteasers* (2002, edited by Victor Bryant). The puzzle text above is taken from the book.

[teaser1798]

## Jim Randell 8:17 am

on17 March 2020 Permalink |(See also:

Enigma 1402).We note that triangles A and B are similar (they both have angles of (60°, θ, 120° – θ)).

And their areas are in the ratio 36:16 = 6²:4², so their sides are in the ratio 6:4 = 3:2.

We can label the sides of A as

(3x, 3y, 3z)and of B as(2x, 2y, 2z).Looking at the base of the original triangle we see it has a length of:

(2x + 3y)And so the the missing lengths on the other sides are:

(3y – x)and(2x + y)But these are also the lengths of the remaining sides of triangles A and B, hence:

So triangle A has sides of

3xand8x, separated by an angle of 60°, and it has an area of 36, so:And the base of the original triangle is

8x + 2x = 10x, so the total area is:Hence:

Solution:The area of the original triangle is 150 cm².LikeLike