[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 129

 
Proof of the impossibility of constructing a triangle by three bisectors
 

OK, the third one has been dealt with. And on the two sides and the bisector between, I hope you can?

 
Mathemat >>:

ОК, с третьей разделались. А по двум сторонам и биссектрисе между, надеюсь, можно?

my head is already broken:))))

 
Mathemat >>:

ОК, с третьей разделались. А по двум сторонам и биссектрисе между, надеюсь, можно?

Yes, a bit more complicated than the first two.

 
The fifth point tells me that without knowing any of the angles dealing with bisectors is difficult. I would purely intuitively suggest that the problem also has no solution, perhaps it can even be reduced to a third.
 

There is a similar problemhere:

1.4.05. В треугольнике известны длины двух его сторон и биссектриса угла между ними. Найти длину третьей стороны.

The idea is that ours should be solvable as well.

 
Mathemat >>:

Тут есть похожая задача:

По идее должна быть решаема и наша.

This problem is not a construction problem. The missing side c is determined from the ratio


l=sqrt(ab(a+b+c)(a+b-c))/(a+b)


It does not follow from the unambiguity of the answer that it is possible to construct:)

 

And here I found what I was looking for, though without a solution. Looks like my intuition failed me:)))


169. Construct a triangle knowing its two sides and the bisector of the angle enclosed between them.

 
Mathemat >>:

Тут есть похожая задача:

По идее должна быть решаема и наша.

This problem is solved quite easily by the already mentioned property of dividing the third side into segments proportional to the original sides.

But I would solve it algebraically, geometrically it reduces to ours.

And ours is solvable, I think. But I haven't solved it yet. :)

By the way, I made an observation: for any two non-equal segments there is always a triangle with two sides equal to the original segments and the bisector of the angle between them equal to the smaller of the two original segments. Nice.

// Only how to at least build it... ?-) Seems to be a special case, and I can't even get it right yet.

 

(a+b)^2 * (1 - l^2/(ab) ) = c^2

The side c is constructible, bastard. But I don't dare to use such a formula, and it's not nice either.

It is enough to construct a right-angled triangle with hypotenuse (a+b) and cathetus l*(a+b)/sqrt(ab). The hypotenuse is easy to build, but the cathetus is a bit more complicated.