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| − | {| class="wikitable" style="border:1px solid black; background:#CEE0F2;text-align:center;"
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| − | |style="background:#B0C4DE;"|'''[[User:Dojo|Dojo]]'''
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| − | |[[User:Dojo#The_Spinning_Sphere|The Spinning Sphere]] - [[User:Dojo#Trivial_Math_Proofs|Trivial Math Proofs]] - [[User:Dojo#AoPS_activity|AoPS activity]] - [[User:Dojo#Contact|Contact]]
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| − | |}
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| − | My name is '''Dojo''' and I currently am 14, and live in Washington.
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| − | My interests are math, technology, solving rubiks cubes, cello, piano, composing, track, cross country and tennis, just to name a few.
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| − | ==The Spinning Sphere==
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| − | 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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| − | ----
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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>
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| − | 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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| − |
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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} \\
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| − | \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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| − |
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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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| − |
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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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| − |
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| − | ==AoPS activity==
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| − | AoPS activity.
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| − | ===Forums I moderate===
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| − | http://www.artofproblemsolving.com/Forum/index.php?f=624
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| − |
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| − | ===Blog===
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| − | http://www.artofproblemsolving.com/Forum/weblog.php?w=1355
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| − |
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| − | ===Wiki pages I created===
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| − | [[Asymptote: Drawing|Drawing]]
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| − | [[Asymptote: Drawing part 2|Drawing part 2]]
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| − | [[Asymptote: Labeling|Labeling]]
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| − | [[Asymptote: Marking Angles|Marking Angles]]
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| − | [[Asymptote: Graphing|Graphing]]
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| − | [[Asymptote: Olympiad Package Part 1|Olympiad Package Part 1: Value Setting]]
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| − | [[Asymptote: Example 1|Example 1]]
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| − |
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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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| − | *[http://www.artofproblemsolving.com/Forum/privmsg.php?mode=post&u=37836 PM]
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| − | *My [http://www.artofproblemsolving.com/Forum/weblog.php?w=1355 Blog]
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