Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 30

New comment
 

Simple, 3 points:

How do you cut a shape from a 3x3 square of paper which is a reamer of the full surface of a single cube?

Another one, also 3 points:

You have to choose between two cylinders. Externally, the cylinders are exactly the same: they are the same size and weight and each is painted green. But one inside is hollow and made of gold, the other is solid (without hollows) and made of a non-magnetic alloy. You can't damage the cylinders or scratch the paint. Is it very easy to find out which cylinder is made of gold?

And another one with the same weight:

Let's prove that any pointed triangle is isosceles.

  1. Take an arbitrary pointed triangle ABC (see figure). Construct in it the bisector of AL and the midpoint of the side BC the point H. Raise a perpendicular from the point H to BC. Let it intersect with AL in point O. Let us draw the perpendiculars OD and OE from O to AB and AC respectively. Draw the segments BO and SO.
  2. Triangle BNO is equal to triangle CHO (by two cathetuses), so BO=CO.
  3. Triangle AOD is equal to triangle AOE (by hypotenuse and acute angle), so OD=OE and AD=AE.
  4. Triangle BDO is equal to triangle CEO (by hypotenuse and cathetus) since BO=CO(item 2) and OD=OE(item 3). Hence, BD=CE.
  5. By adding AD=AE(item 3) and BD=CE(item 4), AB=AC. Thus triangle ABC is isosceles, which is required to prove.

Find the error.

Please don't google it!

 
Mathemat:

Prove that any pointed triangle is isosceles.

  1. Take an arbitrary pointed triangle ABC (see figure). Construct in it the bisector AL and the midpoint BC point H. Raise a perpendicular to BC from point H. Let it intersect with AL in point O. Let us draw the perpendiculars OD and OE from O to AB and AC respectively. Draw the segments BO and SO.
  2. Triangle BNO is equal to triangle CHO (by two cathetuses), so BO=SO.
  3. Triangle AOD is equal to triangle AOE (by hypotenuse and acute angle), so OD=OE and AD=AE.
  4. Triangle BDO is equal to triangle CEO (by hypotenuse and cathetus) since BO=CO(item 2) and OD=OE(item 3). Hence, BD=CE.
  5. By adding AD=AE(item 3) and BD=CE(item 4), AB=AC. Thus triangle ABC is isosceles, which is required to prove.

Find the error.

The first item is not feasible because the perpendicular from the midpoint H does not intersect AL, hence the error in the proof.
 
joo: The first point is impossible because the perpendicular from the middle of the side at point H does not intersect AL, hence the error in the proof.

This is not the whole error. There will be an intersection, just in a different place - outside the triangle.

You need to find the specific place where the error is.

P.S. I also wrote about this at first, but I was told that the error has not yet been found. And they showed me the second drawing, an alternative one:


 

Prove that 1 + 1 equals two.

 
Zeleniy: Prove that 1 + 1 equals two.

Give strict definitions of the following concepts:

  • 1,
  • amounts (+),
  • 2,
  • identity equality.

And explain what you mean by a proof. Because I don't quite understand you...

P.S. You should understand that the proof of this statement can be made only within the framework of the corresponding purely private theory, in which the complete axiomatics of the set of natural numbers is stated. Therefore to operate with intuitive notions of natural numbers themselves and their addition, known at school at the level of unprovable statements, is obviously wrong.

 
Mathemat:

It's not the whole error. The intersection will be, just in another place - outside the triangle.

You need to find the specific place where the error is.

P.S. I also wrote about this at first, but I was told that the error has not yet been found. And they showed me a second picture, an alternative one:


Here. The perpendicular BH from the middle of BC does not intersect AO within the triangle, only outside the triangle. In this case triangles AOD and AOE are not right-angled, hence the condition of equality "by hypotenuse and angle" is not satisfied (item 3).

 

Andrew, the convention proves isosceles only for acutely angular ones. That's first of all. Well, yes, you have an acute-angled one...

Second, Triangles AOD and AOE cannot but be right-angled - by construction:

Опустим из О перпендикуляры OD и OE на AB и AC соответственно.

 

(5 points)

A megabrain walked into a pet shop and bought two plus half of the remaining rabbits. The second megabrain bought three plus a third of the remaining rabbits. The third megabrain bought four plus a quarter of the remaining rabbits. And so on, until it was no longer possible to divide the rabbits. How many maximum megamogs could buy rabbits?

(3 points).

In what minimum number of saws could a 3x3x3 cube be sawn into 1x1x1 composite cubes? Each cut can go through several pieces already sawn off. Justify the minima.

 
MetaDriver: Minimum number of cuts = 6, as the middle cube must be cut from six sides.
Yeah, that's right. Score.
 
MetaDriver:


All other reasoning can be disregarded

)) +100
1...232425262728293031323334353637...229
New comment