Problem 16

ThePerfectHacker · #1 $Old$ January 22nd, 2007, 01:02 PM

1)Given a triangle. From a vertex draw a line to opposite side such that the two smaller triangles have same perimeter. Do the same with the second and third vertex. Show that the three lines intersect in a single point called "Perimecenter"

Here is Diagram from Quick:

Sorry, your Web browser is not Java compatible. This prototype will not be of much interest to you.

2)Given a triangle. From the vertex draw a line to opposite side such that the areas of the two smaller triangles are equal. Do the same with second and third. Show that the three lines intersect in a single point.

3)A physicist trys to show off to a mathemation. He says I found an error in mathematics. Thus, the mathemation says, "What?".
And he does the following:
Consider the set of rationals $\mathbb{Q}$ .
Assume that they are countable then we can write all of them in a row (for example),
.123123123123123...
.454545454545454...
.121212121212121...
.666666666666666...
................................
Define a new number whose in the following way.
If the bold number is "1" then it is "5" if it is not "1" then it is "4".
Then we obtain,
.5454..................................
This new number is not the first one (because it has a different 1st digit).
This new number is not the second one (because it has a different 2nd digits).
And thus on.
Thus, this number is not contained in the list.
Thus, $\mathbb{Q}$ is uncountable.

The physicists is all smiling thinking he is right. But the mathemation is looking at him like an idiot, why?

ThePerfectHacker · #2 $Old$ January 29th, 2007, 09:14 AM

All these three problems are my own and I hope you liked them. For reasons I shall not say I cannot fully respond to all three answers right now, but soon will.

1)For the "perimecenter", I solved it by cheating, I used Ceva's Theorem for coincidence. Use it as a hint, before I respond.

2)This one is simple, when you draw a line as in #1 to create equal area, the heights of the two triangle are the same. That means the bases must be equal for there to be equal area. Thus, that line is acutally a median. And it is a known fact that all medians pass through a common point. Again you can use Ceva's Theorem.

3)The physicist is assuming that the new number that he formed from using the diagnol argument is contained within the set of rationals. He never confirmed this. In the original diagnol argument the number formed has a decimal expansion and is therefore real and is thus contained in the set. The same cannot be said here for it is not contained within the set.

topsquark · #3 $Old$ January 29th, 2007, 10:22 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

3)The physicist is assuming that the new number that he formed from using the diagnol argument is contained within the set of rationals. He never confirmed this. In the original diagnol argument the number formed has a decimal expansion and is therefore real and is thus contained in the set. The same cannot be said here for it is not contained within the set.

That was what I had come up with, but it seemed too simple.

-Dan

ThePerfectHacker · #4 $Old$ January 29th, 2007, 01:00 PM

Here is the solution to the "perimecenter".
From the diagram we have,

1)AB+BD=AC+DC
2)AB+AE=BC+EC
3)From #1: AB=AC+DC-BD
4)From #2: AB=BC+EC-AE
5)From #3 and #4: AC+DC-BD=BC+EC-AE
6)From #5: AE+EC+DC-BD=BD+DC+EC-AE
7)From #6: 2AE=2BD
8)From #7: AE=BD
9)Without lose of generality: AF=DC and FB=EC
10)Marks appear on the triangle.

Now, a direct reference to Ceva's Theorem. (one of my favorites from Geometry)
We have,

$\frac{AE}{AF}\cdot \frac{FB}{BD}\cdot \frac{DC}{EC}=1$ .
Thus, the 3 lines are concurrent.

Quote:

Originally Posted by topsquark

That was what I had come up with, but it seemed too simple.

Good for you.