Brainteaser 1798: Trisection
From The Sunday Times, 2nd March 1997 [link]
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 is included in the book Brainteasers (2002). The puzzle text above is taken from the book.
[teaser1798]
S2T2 
Jim Randell 8:17 am on 17 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 3x and 8x, 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