1. ## 3 questions

I got 3 questions to solve here.

1) if: x^2 + x + 1 = 0 then calculate: x^3333 + x^333+ x^33 + x^3 + 3 = ?

2) a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da

Prove: a=b=c=d

3) You are given a circles diameter and a point that lies above the diameter. without using a protractor how would one find the EXACT perpendicular line (EXACTLY 90 degrees) from the point to the diameter.

Thanksies

2. Originally Posted by Freaky-Person
I got 3 questions to solve here.

1) if: x^2 + x + 1 = 0 then calculate: x^3333 + x^333+ x^33 + x^3 + 3 = ?
Solve,
$x^2+x+1=0$
$x=\frac{-1\pm \sqrt{1-4} }{2}$
$x=-1/2+i/2\cdot \sqrt{3}, -1/2-i/2\cdot \sqrt{3}$
---
There are 2 ways. But I found a more elegant approach.
The expected way is through de Moiver's theorem.
---
We know that,
$x^2=-x-1$ (1)
Multiply by $x$,
$x^3=-x^2-x$
Substitute equation (1),
$x^3=-(-x-1)-x=1$.
Thus,
$x^3=1$
Raise by power of: 11,111,1111 respectively to get,
$x^{33}=1$
$x^{333}=1$
$x^{3333}=1$
Thus,
$1+1+1+1+3=7$

3. Originally Posted by Freaky-Person

2) a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da

Prove: a=b=c=d
This is the Cauchy-Swartzh inequality.
Consider,
$\bold{x}=$
$\bold{y}=$
Then,
$|\bold{x}\cdot \bold{y}|\geq ||\bold{x} ||\cdot ||\bold{y}||$
Furthermore, we only have equality when,
$\bold{x}=\bold{y}$
Thus,
$a=b=c=d$

4. Originally Posted by Freaky-Person
3) You are given a circles diameter and a point that lies above the diameter. without using a protractor how would one find the EXACT perpendicular line (EXACTLY 90 degrees) from the point to the diameter.
I got a problem with this question. I think in projective geometry it is shown that any construction with a straightedge and compass can be done with the compass alone. But they are certainly talking about something else there. For example, you cannot draw a line with a compass! That is what you are asking, no?

5. Originally Posted by ThePerfectHacker
This is the Cauchy-Swartzh inequality.
Consider,
$\bold{x}=$
$\bold{y}=$
Then,
$\bold{x}\cdot \bold{y}\geq ||\bold{x} ||\cdot ||\bold{y}||$
Furthermore, we only have equality when,
$\bold{x}=\bold{y}$
Thus,
$a=b=c=d$
....

what?

6. Originally Posted by ThePerfectHacker
I got a problem with this question. I think in projective geometry it is shown that any construction with a straightedge and compass can be done with the compass alone. But they are certainly talking about something else there. For example, you cannot draw a line with a compass! That is what you are asking, no?
no no, what I mean, is I need to find how to make a perfect 90 degrees angle from the diameter. The line doesn't matter really, I just need exactly 90 degrees without a compass. Just how it could be done.

7. Originally Posted by Freaky-Person
....

what?
The Cauchy-Swartzh Inequality says,
$(a\cdot b+b\cdot c+c\cdot d+d\cdot a)\geq \sqrt{a^2+b^2+c^2+d^2}\cdot \sqrt{b^2+c^2+d^2+a^2}$
Thus,
$ab+bc+cd+ad\geq a^2+b^2+c^2+d^2$
And the only way we have equality (like here).
Is when, (all are equal)
$a=b=c=d$

8. Originally Posted by Freaky-Person
no no, what I mean, is I need to find how to make a perfect 90 degrees angle from the diameter. The line doesn't matter really, I just need exactly 90 degrees without a compass. Just how it could be done.
Let me butt in.

Can you use a ruler or any straigth edge and a measuring device for distances if the straight edge has none?
If yes, then draw two equal line segments from the point to the diameter. Each of these equally long line segments (rays) must be longer than the shortest distance from the point to the diameter. [You can approximate that, I'm sure.]. Then get the midpoint of the line segment on the diameter that is cut by these two rays using the measuring device. Then connect this midpoint to the given point. There, you have a 90-degree angle.

[The perpendicular bisector of the base in an isosceles triangle passes through the apex of the triangle.]

9. Hello, Freaky-Person!

I solved #1 in a manner similar to ThePerfectHacker.

1) If $x^2 + x + 1 \:=\:0$, then calculate: $x^{3333} + x^{333} + x^{33} + x^3 + 3$

We are given: . $x^2 + x + 1 \:=\:0$

Multiply both sides by $(x - 1)\!:\;\;(x - 1)(x^2 + x + 1) \:=\:0(x-1)$

. . and we have: . $x^3 - 1 \:=\:0\;\;\Rightarrow\;\;x\:=\:\sqrt[3]{1}$

Hence, $x$ = the three cube roots of $1.$
. . They happen to be: . $x_1\,=\,1,\;x_2\,=\,\frac{-1 + i\sqrt{3}}{2},\;x_3\,=\,\frac{-1-i\sqrt{3}}{2}$
But we really don't care about their exact values,
. . we only care that: . $x_i^3\,=\,1$

So we have: . $x^{3333} + x^{333} + x^{33} + x^3 + 3 \;= \;\left(x^3\right)^{1111} +\left(x^3\right)^{111} + \left(x^3\right)^{11} + \left(x^3\right) + 3$

. . . . . . $= \;1^{1111} + 1^{111} + 1^{11} + 1 + 3 \;=\;1 + 1 + 1 + 1 + 3 \;=\;\boxed{7}$

10. Originally Posted by Freaky-Person
3) You are given a circles diameter and a point that lies above the diameter. without using a protractor how would one find the EXACT perpendicular line (EXACTLY 90 degrees) from the point to the diameter.
You can use neither a protractor nor a compass? (You didn't mention you couldn't use a compass in your problem statement.) I presume you can't use a ruler, either, but I'm hoping you are at least allowed a straight-edge.

I was reviewing The Elements. I don't think this can be done without a compass and a straight-edge.

-Dan

11. Here is another way to solve #1.

We can write,
$x=-\cos \frac{\pi}{3}\pm i\sin \frac{\pi}{3}$
When we cube it by de Moivers theorem,
$x^3=-\cos \pi \pm i\sin \pi=1$
The rest is trivial.

12. Originally Posted by topsquark
You can use neither a protractor nor a compass? (You didn't mention you couldn't use a compass in your problem statement.) I presume you can't use a ruler, either, but I'm hoping you are at least allowed a straight-edge.
As I said any construction that is doable with a straightedge and compass can be done with a compass along. Thus, if you can do it with both then you can do it with a single one.
I was reviewing The Elements. I don't think this can be done without a compass and a straight-edge.
You know there is a way to prove impossibility.

13. you can use a straight edge, but the thing is, my teacher said that not all edges we are given are exactly 90 degrees and he said we're using bad Canadian rulers with all the edges rounded XD

14. Originally Posted by Freaky-Person
you can use a straight edge, but the thing is, my teacher said that not all edges we are given are exactly 90 degrees and he said we're using bad Canadian rulers with all the edges rounded XD
I see this was last edited by Hacker...I wonder what he changed

Also!

You could use LOCI construction of the right angles if you had a compass. Maybe you could fold paper

15. Originally Posted by anthmoo
I see this was last edited by Hacker...I wonder what he changed
We have the most strict proper language policy than any other forum on the internet.

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