[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 304
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F(1/y, 1/x) = min( 1/y, 1/x + y, x ) = F(x,y).
Hence, the minimum if you replace y with 1/x and x with 1/y will not change. Y = 1/x.
So F(x,1/x) = min( x, 2/x, x ) = min( x, 2/x ). It equals x if x < sqrt(2), and 2/x otherwise.
Draw both curves y=x and y=2/x. Obviously, the maximum is exactly at the point of their intersection and equals sqrt(2).
The solution in the problem book is rather vague, I don't like it:
Next (8th):
This part is trivially constructed. Let's leave the intrigue behind.
The second part of the problem (also 8th):
Simplified the figure:
Richie, how come there are equal angles in a shaded triangle?
By the way, the problem statement doesn't say anything about the original triangle being equilateral. Although it is drawn like an equilateral one.
Come on. Prove it.
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Well, anyway, the idea was this:
The area of a 4x triangle is equal to one third of the difference of the area of the large triangle and the 4 small triangles, i.e. 4 sq.cm.
To find the area of the large triangle you need to find its side (in the figure - A).
Find the side of the central triangle by area, knowing that it is equilateral is not a problem, it is equal to sqr(4*S/sqr(3)).
В центральном? Это очевидно.
Only if the three triangles (apart from the central one) are the sameBut that's not a fact according to the conditions
There has to be something to hold on to. There is one lead, but I don't know what to do with it yet.
Only if the three triangles (apart from the central one) are identical.
But that's not a fact.
Well you've got me completely confused.
I thought the big triangle was equilateral. The small 3 triangles are equilateral, so as a consequence they are similar.