Difference between revisions of "AoPS Wiki:Sandbox"

Revision as of 19:28, 2 April 2014

Welcome to the sandbox, a location to test your newfound wiki-editing abilities.

Please note that all contributions here may be deleted periodically and without warning.

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

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.

$[asy] unitsize(0.6inch); path tt, tri; real x, y; y = 1+sqrt(2); x = y+(6/1.7); tt=(0,0)..(y,1)--(y-1, 0)--(y,-1)..cycle; tri=(y, 1)--(y,-1); draw(tt); fill(tt, gray(0.4)); label("1.7", (y, 0), E); label("C", (y-1, -0.1), S); [/asy]$

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 $\frac{1.7}{\sqrt{2}} = \frac{1.7 \sqrt{2}}{2}$ .

Hence, the area of the whole circle is $\pi r^2$ , and the area of the 3/4 circle is $\frac{3}{4} * \pi * (\frac{1.7 \sqrt{2}}{2})^2$ .

Adding it all up, we find the answer to be $\frac{3}{4} \pi (\frac{1.7 \sqrt{2}}{2})^2 + 0.7225 + 5.1$ .

Just plug it into your calculator. And please understand where all the numbers came from before writing the answer down.

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Difference between revisions of "AoPS Wiki:Sandbox"

Revision as of 19:28, 2 April 2014

$\LaTeX$ test

@@ Line 9: / Line 9: @@
 ==<math>\LaTeX</math> 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:
@@ Line 74: / Line 71: @@
 <asy>
-unitsize(0.8inch);
+unitsize(0.6inch);
 path tt, tri;
 real x, y;
 y = 1+sqrt(2);
 x = y+(6/1.7);
-tt=(0,0)..(y,1)--(y,-1)..cycle;
+tt=(0,0)..(y,1)--(y-1, 0)--(y,-1)..cycle;
-tri=(y,-1)--(y-1,0)--(y,1);
+tri=(y, 1)--(y,-1);
 draw(tt);
-draw(tri);
+fill(tt, gray(0.4));
 label("1.7", (y, 0), E);
 label("C", (y-1, -0.1), S);
@@ Line 88: / Line 85: @@
 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> .
+Hence, the area of the whole circle is <math>\pi r^2</math>, and the area of the 3/4 circle is <math>\frac{3}{4} * \pi * (\frac{1.7 \sqrt{2}}{2})^2 </math>.
+Adding it all up, we find the answer to be <math>\frac{3}{4} \pi (\frac{1.7 \sqrt{2}}{2})^2 + 0.7225 + 5.1</math> .
+Just plug it into your calculator. And please understand where all the numbers came from before writing the answer down.