[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 346

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Mathemat >>:
По поводу задачи с 1999 числами: MD, ответ правилен. Но доказательство там мутное и не такое простое.

I'm guessing. :)

 
alsu >>:

предполагается, что он гомотетичен восьмикласснику, решающему задачу. С дробями, мне кажется, и красивее, и более по-программерски что-ли:)

To assuage my conscience, the problem is for 11th grade. And with fractions - yep, very original.

For warm-up (8th):

Two different numbers x and y (not necessarily integers) are such that x2-2000x=y2-2000y. Find the sum of the numbers x and y.

P.S. I don't know what's so funny about it. It's solved in mind.

 
Mathemat >>:
Два различных числа x и y (не обязательно целых) таковы, что x2-2000x=y2-2000y. Найдите сумму чисел x и y.

x + y == 2000

 
There are 12 parties in the 100-member parliamentary elections. The parties that receive strictly more than 5% of the votes get into parliament. The seats are distributed between the parties that passed into parliament in proportion to the number of votes they received (that is, if one of the parties gets x times as many votes as the other, it will get x times as many seats in parliament). After the election each voter turned out to have voted for exactly one of the parties (there were no invalid ballots, votes "against all", etc.) and each party got an integer number of seats. In doing so, the Party of Amateur Mathematicians got 25% of the votes. What was the largest number of seats it could get in parliament? (Explain the answer.)
 
Mathemat >>:
В выборах в 100-местный парламент участвовали 12 партий. В парламент проходят партии, за которые проголосовало строго больше 5% избирателей. Между прошедшими в парламент партиями места распределяются пропорционально числу набранных ими голосов (т. е. если одна из партий набрала в x раз больше голосов, чем другая, то и мест в парламенте она получит в x раз больше). После выборов оказалось, что каждый избиратель проголосовал ровно за одну из партий (недействительных бюллетеней, голосов "против всех" и т. п. не было) и каждая партия получила целое число мест. При этом Партия любителей математики набрала 25% голосов. Какое наибольшее число мест в парламенте она могла получить? (Ответ объясните.)

Max == 50

This maximum will take place in the rare event that ten parties get strictly 5% each, and another party gets the remaining 25%.

The seats will then be divided between the two parties of 50 each.

 
Yeah, it's clear. There is a proof with formulas, but you get a credit.

The lengths of the bases of a trapezoid are m cm and n cm (m and n are natural numbers, m is not equal to n). Prove that a trapezoid can be dissected into equal triangles.

The problem is very simple, hee-hee...
 
Mathemat >>:
Да, все четко. Там, правда, доказательство с формулами, но тебе зачод.

Длины оснований трапеции равны m см и n см (m и n - натуральные числа, m не равно n). Докажите, что трапецию можно разрезать на равные треугольники.

Задачка-то простенькая совсем, хи-хи...

Mm-hmm. draw lines parallel to both sides of the trapezoid and to the bases in 1 cm. The fallece theorem says that all the lengths of the segments are integer.

 
Let f(x)=x2+12x+30. Solve the equation

f(f(f(f(f(x)))))=0.

 
Mathemat >>:
Пусть f(x)=x2+12x+30. Решите уравнение

f(f(f(f(f(x)))))=0.

f(x) = x2+12x+30 = (x + 6)^2 - 6

f(f(f(f(x))))) = (((((x + 6)^2 - 6 + 6)^2 - 6 + 6)^2 - 6 + 6)^2 - 6 = (((((x + 6)^2 )^2)^2)^2 - 6 = 0

i.e. (x + 6)^32 = 6 => x = Root(6, 32) - 6

 
Perverse, but all true, except for a small inaccuracy. There's a +- in front of the root.

A convex polygon is drawn on "cellular" paper such that all its vertices are at the vertices of the cells and none of its sides go vertically or horizontally. Prove that the sum of lengths of vertical segments of grid lines inside the polygon is equal to the sum of lengths of horizontal segments of grid lines inside the polygon.

By the way, the author of the problem is Halperin.
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