Difference between revisions of "AoPS Wiki:Sandbox"

Revision as of 18:16, 2 April 2014

Template:AoPSWiki:Sandbox/header In the computer world, a sandbox is a place to test and experiment -- essentially, it's a place to play.

This is the AoPSWiki Sandbox. Feel free to experiment here.

Warning: anything you place here is subject to deletion without notice.

[This was deleted due to its inappropriateness.]

$\LaTeX$ test

---experimenthere-----

Please do not delete from this point on until 5:00 PST on 4/2!

We want to find the area of this figure:

$[asy] path rt,tt, tri; real x, y; y = 1+sqrt(2); x = y+(6/1.7); tt=(0,0)..(y,1)--(y,-1)..cycle; rt=(y,-1)--(x,-1)--(x,1)--(y,1); tri=(y,-1)--(y-1,0)--(y,1); draw(rt); draw(tt); draw(tri); label("1.7", (x, 0), E); label("3", (y+(3/1.7), -1), S); label("C", (y-1, -0.1), S); [/asy]$

We label the circle as circle C. We can break the figure into three parts, shown as the 3/4 circle, the triangle, and the rectangle.

Lets first take a look at the rectangle.

$[asy] path rt; real x, y; y = 1+sqrt(2); x = y+(6/1.7); rt=(y,-1)--(x,-1)--(x,1)--(y,1)--cycle; draw(rt); label("1.7", (x, 0), E); label("3", (y+(3/1.7), -1), S); [/asy]$

It has an area of $3 * 1.7 = 5.1$ .

Lets now take a look at the triangle, after drawing the height.

$[asy] unitsize(0.8inch); path tri, lin; real x, y; y = 1+sqrt(2); x = y+(6/1.7); tri=(y,-1)--(y-1,0)--(y,1)--cycle; lin=(y-1, 0)--(y,0); draw(tri); draw(lin); label("1.7", (y, 0), E); label("C", (y-1, -0.1), S); [/asy]$

We see that both the radii are the two shorter sides of the triangle, making this a isosceles 45-45-90 triangle.

We also see that the height that we drew is half the hypotenuse(note the two smaller 45-45-90 isosceles triangles).

Hence, the area of the triangle is $\frac{1.7 * \frac{1.7}{2}}{2} = 0.7225$ .

Now, let's take a look at the 3/4 circle. We know it is 3/4 because there is a 90 degree triangle cut out of it.

$[asy] unitsize(0.8inch); path tt, tri; real x, y; y = 1+sqrt(2); x = y+(6/1.7); tt=(0,0)..(y,1)--(y,-1)..cycle; tri=(y,-1)--(y-1,0)--(y,1); draw(tt); draw(tri); label("1.7", (y, 0), E); label("C", (y-1, -0.1), S); [/asy]$

We find the radius using the 45-45-90 triangle. Since the ratios of the sides are 1:1: $\sqrt{2}$ , we can find the radius to be $\frac{1.7}{\sqrt{2}} = \frac{1.7 \sqrt{2}}{2}$ .

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Difference between revisions of "AoPS Wiki:Sandbox"

Revision as of 18:16, 2 April 2014

$\LaTeX$ test

@@ Line 14: / Line 14: @@
 We want to find the area of this figure:
 <asy>
-path rt,tt;
+path rt,tt, tri;
-int x;
+real x, y;
-x = 1+sqrt(2)+(6/1.7);
+y = 1+sqrt(2);
-tt=(0,0)..(1+sqrt(2),1)--(1+sqrt(2),-1)..cycle;
+x = y+(6/1.7);
-rt=(1+sqrt(2),-1)--(1+sqrt(2)+(6/1.7),-1)--(1+sqrt(2)+(6/1.7),1)--(1+sqrt(2),1);
+tt=(0,0)..(y,1)--(y,-1)..cycle;
+rt=(y,-1)--(x,-1)--(x,1)--(y,1);
+tri=(y,-1)--(y-1,0)--(y,1);
 draw(rt);
 draw(tt);
+draw(tri);
 label("1.7", (x, 0), E);
+label("3", (y+(3/1.7), -1), S);
+label("C", (y-1, -0.1), S);
 </asy>
+We label the circle as circle C. We can break the figure into three parts, shown as the 3/4 circle, the triangle, and the rectangle.
+Lets first take a look at the rectangle.
+<asy>
+path rt;
+real x, y;
+y = 1+sqrt(2);
+x = y+(6/1.7);
+rt=(y,-1)--(x,-1)--(x,1)--(y,1)--cycle;
+draw(rt);
+label("1.7", (x, 0), E);
+label("3", (y+(3/1.7), -1), S);
+</asy>
+It has an area of <math> 3 * 1.7 = 5.1</math> .
+Lets now take a look at the triangle, after drawing the height.
+<asy>
+unitsize(0.8inch);
+path tri, lin;
+real x, y;
+y = 1+sqrt(2);
+x = y+(6/1.7);
+tri=(y,-1)--(y-1,0)--(y,1)--cycle;
+lin=(y-1, 0)--(y,0);
+draw(tri);
+draw(lin);
+label("1.7", (y, 0), E);
+label("C", (y-1, -0.1), S);
+</asy>
+We see that both the radii are the two shorter sides of the triangle, making this a isosceles 45-45-90 triangle.
+We also see that the height that we drew is half the hypotenuse(note the two smaller 45-45-90 isosceles triangles).
+Hence, the area of the triangle is <math>\frac{1.7 * \frac{1.7}{2}}{2} = 0.7225</math> .
+Now, let's take a look at the 3/4 circle. We know it is 3/4 because there is a 90 degree triangle cut out of it.
+<asy>
+unitsize(0.8inch);
+path tt, tri;
+real x, y;
+y = 1+sqrt(2);
+x = y+(6/1.7);
+tt=(0,0)..(y,1)--(y,-1)..cycle;
+tri=(y,-1)--(y-1,0)--(y,1);
+draw(tt);
+draw(tri);
+label("1.7", (y, 0), E);
+label("C", (y-1, -0.1), S);
+</asy>
+We find the radius using the 45-45-90 triangle. Since the ratios of the sides are 1:1:<math>\sqrt{2}</math>, we can find the radius to be <math>\frac{1.7}{\sqrt{2}} = \frac{1.7 \sqrt{2}}{2}</math> .