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"
(→‎Trivial Math Proofs)
Line 10: Line 10:
 
(My gallery is now the Animation Studio)
 
(My gallery is now the Animation Studio)
  
 +
----
 
==Trivial Math Proofs==
 
==Trivial Math Proofs==
 +
 +
Proof that the area of an equilateral triangle with side length <math>s</math> is <math>\dfrac{s^2\sqrt {3}}{4}</math>:
 +
 +
Let's say that there is an equilateral triangle that has a side length of <math>s</math>. We can then draw the following figure:
 +
 +
<center>
 +
<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>
 +
</center>
 +
 +
Now let's figure out the altitude so we can complete the triangle area forumla of <math>\dfrac{bh}{2}</math>:
 +
 +
<center>
 +
<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>
 +
</center>
 +
 +
We can now use the pythagorean theorem to find the length of the altitude:
 +
 +
<center>
 +
<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>
 +
</center>
 +
 +
Since we know that this is a <math>30 - 60 - 90</math> triangle, we can use proportions to find the altitude <math>a</math> in terms of side lenth <math>s</math>:
 +
 +
<math>\begin{align*} \dfrac{2}{\sqrt {3}} & = \dfrac{s}{a} \\
 +
\sqrt {3}s & = 2a \\
 +
\dfrac{\sqrt {3}}{2}s & = a \end{align*}</math>
 +
 +
Now plugging this altitude into the triangle area forumla gives us:
 +
 +
<math>\dfrac{\frac {\sqrt {3}}{2}s\times s}{2} = \dfrac{\frac {s^2\sqrt {3}}{2}}{2} = \boxed{\dfrac{s^2\sqrt {3}}{4}}</math>
 +
 +
Proof can be found on [http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=273429 this] post of my blog.
 +
 +
----
  
 
==Interests==
 
==Interests==

Revision as of 22:10, 25 April 2009

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

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.

Interests

Here are some of the things I do in my spare time:

Math

Classes taken, in order:

1) Introduction to Geometry

2) MATHCOUNTS problem series

3) Intermediate Algebra.

4) AMC 10

Classes to be taken:

1) Introduction to Counting and Probability

2) Introduction to Number Theory

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 - 550

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

Compositions/Arrangements:

Invention 13

The Journey

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=31473"