[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 305

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Mathemat >>:
Не факт.
Тут надо за что-то зацепиться. Одна зацепка есть, но что с ней делать, не знаю пока.

The lead:

S=S1+S2;
S=S3+S4;
S=S5+S6;
S=T1+T2+T3+T4+K1+K2+K3;
S1=K1+K3+T1+T4;
S2=K2+T2+T3;
S3=K1+K2+T2+T4;
S4=K3+T1+T3;
S5=K2+K3+T3+T4;
S6=K1+T1+T2; where
S - total area
S1-S6 - areas formed from section S into two parts
T1-T4 - areas of triangles
K1-K3 - areas of quadrangles,
geometric equations are missing.
 
coaster, that's not a clue, agree. An eighth grader won't solve such systems, it's an olympiad after all. He needs to find something really nifty, not this nonsense stuff :)
2 Richie:
The little 3 triangles are equal, so as a consequence they are similar. <br / translate="no">
Equalness is simply the equality of areas, not similarity. Even a square and a circle can be equal.
 
Let's first introduce notation so that we don't have to explain in words.

If we prove that V is the midpoint of CC', then we prove everything: the triangle AC'C will then be divided by the segment AV into equal parts. Since the shaded triangles inside AC'C are equal, then both quadrilaterals are equal. The other partial triangles ABA' and BCB' can be considered in the same way.
There are clues. For example, that AUVB' is a trapezium. The parallelism of its sides AU and VB' is easily proved from the homothety of the corresponding triangles - AUW and B'WV. But I don't see where to apply this fact.
And the homothety of AUW and B'WV follows from the equiaxiality of shaded triangles and application of the formula for the area of triangle through the sides and the sine of the angle between them.
P.S. The solution strikes by its brevity (probably, almost every eighth-grader can solve the problem in his mind):

But there is some hint of the golden ratio. I suspected...
 
Mathemat писал(а) >>
AUW and B'WV. But where to apply this fact - I don't see where.

I tried to apply it to calculate the lengths of VB and UA, as we know the areas of the triangles - 1 sq.cm. The side WV is easy to find. If triangle UWV is equilateral, i.e. its angles are 60g, we know all angles and it is easy to calculate the trapezoid. If we know VB and UA, which break open 4 angles into triangles, then we obtain the area of major triangle ABC and use this area to calculate the areas of 4 angles.
Yes, the answer is beautiful :))

 
Richie >>: Если треугольник UWV равносторонний

Why an equilateral?

 
Mathemat писал(а) >>

>> Why is it equilateral?


Yeah, it's not a fact. It's just easier that way. This is what I wrote above: If ABC and UWV are equilateral and the side triangles are equal (the problem condition), then these side triangles will be similar, though I may be wrong.
In general, I find it much easier to solve this problem on computer by making a system :))
Where does (Root(5)+1) come from?

 
Another hitch:

1. It suffices, for example, to prove that triangles AC'C and B'BC are equal. Well and make for similar ones.
2. How to do it? Their heights are related as AC'/AB and their bases as AC/B'C. In other words, both relations show how the points C' and B' divide the sides of the original triangle. If we prove that these relations are inverse to each other, then the former will follow.
P.S. Found a solution online, but haven't looked at it. Just made sure that no properties of the original triangle are used. It's not equilateral, not isosceles etc. But the problem is solved quite correctly. Let's put it aside for now.

Next:
Find four equilateral right triangles whose sides are natural numbers.
I hope everyone remembers the formulas for integer Pythagorean triples (2pq, pp-qq, pp+qq)?
 
Mathemat >>:
Еще одна зацепочка:

1. Достаточно, например, доказать, что треуги AC'C и B'BC равновелики. Ну и сделать для аналогичных.
2. Как это сделать? Высоты их соотносятся как AC'/AB, а основания - как AC/B'C. Другими словами, оба отношения показывают, как точками C' и B' делятся стороны исходного треуга. Если мы докажем, что эти отношения обратны друг другу, то отсюда будет вытекать первое.
P.S. Нашел в сети решение, но не смотрел. Просто убедился, что никакие свойства первоначального треуга не используются. Он не равносторонний, не равнобедренный и т.п. Но задачка решается вполне корректно. Отложим пока.

I sat for two hours, found all aspect ratios, expressed the area of required quadrilateral through the sides of small triangles (it is equal to 1 sq.cm*2*WB'/UB'), but I still haven't got a final solution. Come on, put the solution out there, or my brain will break:(

Next:
Find four equilateral right triangles whose sides are natural numbers.
I hope everyone remembers the formulas for integer Pythagorean triples (2pq, pp-qq, pp+qq)?

i.e. the problem reduces to finding four pairs of numbers p,q for which pppq-pqqq is invariant.

 
alsu >>:

Сидел два часа, нашел все отношения сторон, выразил площадь требуемого четырехугольника через стороны маленьких треугов (она равна 1кв.см*2*WB'/UB'), но окончательно нихрена так и не получилось. Давай, выкладывай решение, а то моск сломается:(

Wow, I didn't get shit at all, apart from the remarks I posted here. There may be a nice trapezoidal property involved. Here's the link: http://www.problems.ru/view_problem_details_new.php?id=55137.

There seems to be a problem with the links. OK, here we go: http://www.problems.ru/view_problem_details_new.php?id=55137

i.e. the problem is reduced to finding four pairs of numbers p,q for which pppq-pqqq is invariant.

Well, it seems so. pq(p-q)(p+q) = inv.

 
Mathemat >>:

http://www.problems.ru/view_problem_details_new.php?id=55137.

There's a bit of a mix-up with the indices in the solution.
I should have sat for another hour, I was close:) But the difficulty of the problem is clearly for eighth-graders, but not below the level of the regional Olympiad.

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