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| − | My name is '''Dojo''' and I currently am 13, and live in Washington.
| + | <div id="content" style="width: 600px; margin-left: auto; margin-right: auto; background-color: #cfffbe; border: 5px #6eac68 solid; "> |
| − | | + | <h1 style="font-family: 'ITC Avant Garde Gothic Std','URW Gothic L','Century Gothic','Avant Garde','Trebuchet MS',sans-serif; margin-left: 10px;"> Dojo's Wiki</h1> |
| − | My interests are math, technology, solving rubiks cubes, cello, piano, composing, track, cross country and tennis, just to name a few.
| + | <div id="inner" style="width: 500px; margin-left: auto; margin-right: auto;"> |
| − | | + | Please visit my website: [http://www.dgkim.com dgkim.com] |
| − | ==The Spinning Sphere== | + | </div> |
| − | | + | </div> |
| − | Yes yes,[big voice] I am the creator of the almighty spinning sphere!!! [/end big voice]
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| − | Yeah well anyway, for anyone interested, I have created a gallery of these spheres:
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| − | [http://www.artofproblemsolving.com/Forum/album.php?t=234501 My Gallery]
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| − | (My gallery is now the Animation Studio)
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| − | | |
| − | ----
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| − | ==Trivial Math Proofs==
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| − | | |
| − | Suggest your trival proofs you want [http://www.artofproblemsolving.com/Wiki/index.php/User_talk:Dojo here].
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| − | | |
| − | ===Equilateral Triangle Area===
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| − | Proof that the area of an equilateral triangle with side length <math>s</math> is <math>\dfrac{s^2\sqrt {3}}{4}</math>:
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| − | 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:
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| − | <center>
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| − | <asy>
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| − | draw((0,0)--(1,sqrt(3)),linewidth(1));
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| − | add(pathticks((0,0)--(1,sqrt(3)),1,green+linewidth(1)));
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| − | draw((2,0)--(1,sqrt(3)),linewidth(1));
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| − | add(pathticks((2,0)--(1,sqrt(3)),1,green+linewidth(1)));
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| − | draw((2,0)--(0,0),linewidth(1));
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| − | add(pathticks((2,0)--(0,0),1,green+linewidth(1)));
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| − | label("$s$",(1,0),S);
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| − | </asy>
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| − | </center> | |
| − | | |
| − | Now let's figure out the altitude so we can complete the triangle area forumla of <math>\dfrac{bh}{2}</math>:
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| − | <center>
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| − | <asy>
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| − | draw((0,0)--(1,sqrt(3)),linewidth(1));
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| − | add(pathticks((0,0)--(1,sqrt(3)),1,green+linewidth(1)));
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| − | draw((2,0)--(1,sqrt(3)),linewidth(1));
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| − | add(pathticks((2,0)--(1,sqrt(3)),1,green+linewidth(1)));
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| − | draw((2,0)--(0,0),linewidth(1));
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| − | add(pathticks((2,0)--(0,0),1,green+linewidth(1)));
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| − | label("$s$",(1,0),S);
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| − | draw((1,0)--(1,sqrt(3)),dashed+linewidth(1));
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| − | </asy>
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| − | </center>
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| − | | |
| − | We can now use the pythagorean theorem to find the length of the altitude:
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| − | <center>
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| − | <asy>
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| − | draw((0,0)--(0,sqrt(3))--(1,0)--cycle,linewidth(1));
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| − | draw(rightanglemark((0,sqrt(3)),(0,0),(1,0)),red+linewidth(1));
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| − | </asy>
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| − | </center>
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| − | | |
| − | 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>:
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| − | <math>\begin{align*} \dfrac{2}{\sqrt {3}} & = \dfrac{s}{a} \\ | |
| − | \sqrt {3}s & = 2a \\
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| − | \dfrac{\sqrt {3}}{2}s & = a \end{align*}</math>
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| − | | |
| − | Now plugging this altitude into the triangle area forumla gives us:
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| − | <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>
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| − | Proof can be found on [http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=273429 this] post of my blog.
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| − | | |
| − | ===Diagonal Forumla===
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| − | Proof that the number of diagonals in a polygon with <math>n</math> sides is <math>\dfrac{n(n-3)}{2}</math>:
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| − | | |
| − | First, lets see the hexagon:
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| − | <asy>
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| − | size(200);
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| − | for(int i=0; i<6; ++i)
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| − | for(int j=i+1; j<6; ++j)
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| − | draw(dir(60*i)--dir(60*j));
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| − | </asy>
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| − | | |
| − | If you count carefully, you'll see that there are 9 diagonals.
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| − | | |
| − | Now we need to see how we can derive a forumla for the number of diagonals.
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| − | For any polygon with <math>n</math> sides, we see that there are <math>n</math> vertecies. To create a diagonal, we need one other point, which can be selected from a pool of <math>n - 3</math> 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 <math>n(n - 3)</math> diagonals, right? Wrong. Let's say that two of the points are <math>A</math> and <math>B</math>. Using the above method, both the diagonals <math>AB</math> and <math>BA</math> would be counted. Therefore, we must divide the forumla by 2, giving us the diagonal forumla:
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| − | <math>\boxed{\dfrac{n(n - 3)}{2}}</math>
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| − | Proof can be found on [http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=273655 this] post of my blog.
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| − | | |
| − | ===Handshake formula===
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| − | Proof that the number of handshakes occuring in a group of <math>n</math> where each person shakes each other's hand is <math>\dfrac{n(n-1)}{2}</math>.
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| − | 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 <math>n-1</math> people (Everyone except themselves) and there are <math>n</math> people, we get a doubled <math>n(n-1)</math> handshakes. Once again, we must discount the doubled handshakes occuring between say person A and B. Therefore, we have <math>\boxed{\dfrac{n(n-1)}{2}}</math>.
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| − | Proof can be found on [http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=275015 this] post of my blog.
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| − | | |
| − | ==Interests==
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| − | Here are some of the things I do in my spare time:
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| − | | |
| − | ===Math===
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| − | Classes taken, in order:
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| − | 1) Introduction to Geometry
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| − | 2) MATHCOUNTS problem series
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| − | 3) Intermediate Algebra.
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| − | 4) AMC 10
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| − | Classes to be taken:
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| − | 1) Introduction to Counting and Probability
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| − | 2) Introduction to Number Theory
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| − | | |
| − | =My current, sad accomplishments:=
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| − | Best:
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| − | AMC8: 23
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| − | AMC10 A: 114.0
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| − | AMC10 B: 106.5
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| − | AMC12 A: n/a (untaken.)
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| − | AMC12 B: n/a (untaken.)
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| − | AIME: n/a (untaken.)
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| − | USAMO: n/a (untaken.)
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| − | IMO: n/a (untaken.)
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| − | SAT:
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| − | Mathematics - 690
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| − | Critical Reading - 570
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| − | Writing - 610
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| − | Essay - 8
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| − | All:
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| − | KSEA:
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| − | 6th grade: 2nd place locally.
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| − | 7th grade: 2nd place locally.
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| − | Local Math is Cool competition:
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| − | 4th grade-
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| − | 4th grade competition – 2nd place
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| − | 5th grade-
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| − | 5th grade competition – 2nd place
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| − | 7th grade competition – 5th place
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| − | 6th grade –
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| − | 6th grade competition – 9th place
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| − | 7th grade competition – 5th place
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| − | 7th grade –
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| − | 7th grade competition – 1st place
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| − | 9th grade competition – 7th place
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| − | ===Music===
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| − | My musical side?
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| − | 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.)
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| − | For all you less musically knowing, a cello is well described [http://en.wikipedia.org/wiki/Cello here].
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| − | Lets hope you know what a piano is. :)
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| − | Masterclasses taken with:
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| − | Amy Sue Barston
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| − | Alisha Weiserstien
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| − | Compositions/Arrangements:
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| − | [http://www.youtube.com/watch?v=F5M_TpzTppQ&feature=channel_page Invention 13]
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| − | [http://www.youtube.com/watch?v=wEEkxsP1CpU&feature=channel_page The Journey]
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| − | ===Athletics===
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| − | 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.
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| − | 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.
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| − | Tennis. Well I didn't have the best hand-eye around, but I manage to play tennis, relatively well and have lessons every sunday...
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| − | | |
| − | ==Contact==
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| − | Some ways you can reach me:
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| − | *Email. dojothegreat@gmail.com
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| − | *PM [http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=37836 Profile]
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| − | *My [http://www.artofproblemsolving.com/Forum/weblog.php?w=1355 Blog]
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| − | * [http://www.cyneer.com/phpBB3/index.php Forum]
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