Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "User:Dojo"

(→‎Wiki pages I created)
Line 1: Line 1:
 +
{| class="wikitable" style="border:1px solid black; background:#CEE0F2;text-align:center;"
 +
|-
 +
|style="background:#B0C4DE;"|'''[[User:Dojo|Dojo]]'''
 +
|-
 +
|[[User:Dojo#The_Spinning_Sphere|The Spinning Sphere]] - [[User:Dojo#Trivial_Math_Proofs|Trivial Math Proofs]] - [[User:Dojo#AoPS_activity|AoPS activity]] - [[User:Dojo#My_current.2C_sad_accomplishments:|My sad accomplishments]] - [[User:Dojo#Contact|Contact]]
 +
|}
 +
 
My name is '''Dojo''' and I currently am 13, and live in Washington.
 
My name is '''Dojo''' and I currently am 13, and live in Washington.
  

Revision as of 02:36, 8 January 2010

Dojo
The Spinning Sphere - Trivial Math Proofs - AoPS activity - My sad accomplishments - Contact

My name is Dojo and I currently am 13, and live in Washington.

My interests are math, technology, solving rubiks cubes, cello, piano, composing, track, cross country and tennis, just to name a few.

The Spinning Sphere

Yes yes,[big voice] I am the creator of the almighty spinning sphere!!! [/end big voice] Yeah well anyway, for anyone interested, I have created a gallery of these spheres: My Gallery (My gallery is now the Animation Studio)

Trivial Math Proofs

Suggest your trival proofs you want here.

Equilateral Triangle Area

Proof that the area of an equilateral triangle with side length $s$ is $\dfrac{s^2\sqrt {3}}{4}$:

Let's say that there is an equilateral triangle that has a side length of $s$. We can then draw the following figure:

[asy] draw((0,0)--(1,sqrt(3)),linewidth(1)); add(pathticks((0,0)--(1,sqrt(3)),1,green+linewidth(1))); draw((2,0)--(1,sqrt(3)),linewidth(1)); add(pathticks((2,0)--(1,sqrt(3)),1,green+linewidth(1))); draw((2,0)--(0,0),linewidth(1)); add(pathticks((2,0)--(0,0),1,green+linewidth(1))); label("$s$",(1,0),S); [/asy]

Now let's figure out the altitude so we can complete the triangle area forumla of $\dfrac{bh}{2}$:

[asy] draw((0,0)--(1,sqrt(3)),linewidth(1)); add(pathticks((0,0)--(1,sqrt(3)),1,green+linewidth(1))); draw((2,0)--(1,sqrt(3)),linewidth(1)); add(pathticks((2,0)--(1,sqrt(3)),1,green+linewidth(1))); draw((2,0)--(0,0),linewidth(1)); add(pathticks((2,0)--(0,0),1,green+linewidth(1))); label("$s$",(1,0),S); draw((1,0)--(1,sqrt(3)),dashed+linewidth(1)); [/asy]

We can now use the pythagorean theorem to find the length of the altitude:

[asy] draw((0,0)--(0,sqrt(3))--(1,0)--cycle,linewidth(1)); draw(rightanglemark((0,sqrt(3)),(0,0),(1,0)),red+linewidth(1)); [/asy]

Since we know that this is a $30 - 60 - 90$ triangle, we can use proportions to find the altitude $a$ in terms of side lenth $s$:

$\begin{align*} \dfrac{2}{\sqrt {3}} & = \dfrac{s}{a} \\ \sqrt {3}s & = 2a \\ \dfrac{\sqrt {3}}{2}s & = a \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Now plugging this altitude into the triangle area forumla gives us:

$\dfrac{\frac {\sqrt {3}}{2}s\times s}{2} = \dfrac{\frac {s^2\sqrt {3}}{2}}{2} = \boxed{\dfrac{s^2\sqrt {3}}{4}}$

Proof can be found on this post of my blog.

Diagonal Forumla

Proof that the number of diagonals in a polygon with $n$ sides is $\dfrac{n(n-3)}{2}$:

First, lets see the hexagon:

[asy] size(200); for(int i=0; i<6; ++i) for(int j=i+1; j<6; ++j) draw(dir(60*i)--dir(60*j)); [/asy]

If you count carefully, you'll see that there are 9 diagonals.

Now we need to see how we can derive a forumla for the number of diagonals.

For any polygon with $n$ sides, we see that there are $n$ vertecies. To create a diagonal, we need one other point, which can be selected from a pool of $n - 3$ points. We must exclude 3 points because the point connecting to the point itself doesn't count as a diagonal, and connecting to the 2 adjecent points don't count because they have already been "drawn in" as the sides of the polygon. We would then assume that there are $n(n - 3)$ diagonals, right? Wrong. Let's say that two of the points are $A$ and $B$. Using the above method, both the diagonals $AB$ and $BA$ would be counted. Therefore, we must divide the forumla by 2, giving us the diagonal forumla:

$\boxed{\dfrac{n(n - 3)}{2}}$

Proof can be found on this post of my blog.

Handshake formula

Proof that the number of handshakes occuring in a group of $n$ where each person shakes each other's hand is $\dfrac{n(n-1)}{2}$.

This proof is similar to the diagonal proof, with one exception. Each "vertex" is replaced by a person and people can shake hands to adjacent people. Since each person will shake hands with $n-1$ people (Everyone except themselves) and there are $n$ people, we get a doubled $n(n-1)$ handshakes. Once again, we must discount the doubled handshakes occuring between say person A and B. Therefore, we have $\boxed{\dfrac{n(n-1)}{2}}$.

Proof can be found on this post of my blog.

AoPS activity

AoPS activity.

Blog

http://www.artofproblemsolving.com/Forum/weblog.php?w=1355

Wiki pages I created

Drawing

Drawing part 2

Labeling

Marking Angles

Graphing

Olympiad Package Part 1: Value Setting

Example 1

My current, sad accomplishments:

Best:

AMC8: 23

AMC10 A: 114.0

AMC10 B: 106.5

AMC12 A: n/a (untaken.)

AMC12 B: n/a (untaken.)

AIME: n/a (untaken.)

USAMO: n/a (untaken.)

IMO: n/a (untaken.)

SAT:

Mathematics - 690

Critical Reading - 570

Writing - 610

Essay - 8

All:

KSEA:

6th grade: 2nd place locally.

7th grade: 2nd place locally.

Local Math is Cool competition:

4th grade-

4th grade competition – 2nd place

5th grade-

5th grade competition – 2nd place

7th grade competition – 5th place

6th grade –

6th grade competition – 9th place

7th grade competition – 5th place

7th grade –

7th grade competition – 1st place

9th grade competition – 7th place


Music

My musical side?

At a young age, I was not the most talented musician. I couldn't sing, I couldn't move my fingers seperately but here I am now. Playing the cello and piano with (in my opinion) very fluid actions. I have perfect pitch and when I sing, I sing in tune. Its just the quality that is... less than perfect. (Sounds like a duck that swallowed a harmonica.)

For all you less musically knowing, a cello is well described here.

Lets hope you know what a piano is. :)

Masterclasses taken with:

Amy Sue Barston

Alisha Weiserstien

Athletics

It is generally assumed that atheletics is not a great part of an AoPSer's life. I mean what kind of athelete would be sitting here writing this wiki page? Well I follow with that, in moderation.

I love to run. It is something that I discovered this year. Cross country, track. Recently the season has ended and I find myself itching to run.

Tennis. Well I didn't have the best hand-eye around, but I manage to play tennis, relatively well and have lessons every sunday...

Contact

Some ways you can reach me:

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=User:Dojo&oldid=33180"