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Difference between revisions of "Geometry Solutions"

m (→‎Probability 1)
(→‎Probability 1)
Line 520: Line 520:
  
 
Thus, our final probability is <math>\frac{4y^2}{20y^2}=\frac{1}{5}</math>
 
Thus, our final probability is <math>\frac{4y^2}{20y^2}=\frac{1}{5}</math>
 +
 +
-----------------
 +
------------
 +
 +
Another method to do the same problem is rearrangement, when you move pieces of the figure around to make a nicer figure.
 +
 +
 +
<asy>
 +
path squ;
 +
squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle;
 +
fill(squ, gray(0.6));
 +
 +
path bigs;
 +
bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle;
 +
draw(bigs);
 +
 +
draw( (0,0)--(1,2) );
 +
draw( (2,0)--(0,1) );
 +
draw( (2,2)--(1,0) );
 +
draw( (0,2)--(2,1) );
 +
 +
label ("x",(2, 0.5), E);
 +
label ("x",(2, 1.5), E);
 +
</asy>
 +
 +
<asy>
 +
//path squ;//left,    bottom        right        top
 +
//squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle;
 +
//fill(squ, gray(0.6));               
 +
 +
path bigtri;
 +
bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle;
 +
//fill(bigtri, gray(0.7));
 +
 +
path trtri;
 +
trtri = (2,2)--(1.6,1.2)--(2,1)--cycle;
 +
fill(trtri, gray(0.6));
 +
 +
path bigs;
 +
bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle;
 +
draw(bigs);
 +
 +
draw( (0,0)--(1,2) );
 +
draw( (2,0)--(0,1) );
 +
draw( (2,2)--(1,0) );
 +
draw( (0,2)--(2,1) );
 +
draw((2,2)--(1.6,1.2));
 +
draw((2,0)--(1.2,0.4));
 +
 +
label ("x",(2, 0.5), E);
 +
label ("x",(2, 1.5), E);
 +
//label ("y", (1.8, 1.1), SW);
 +
//label ("2y", (1.8, 1.6), NW);
 +
//label ("2y", (1.6, 0.2), SW);
 +
//label ("2y", (1.4, 0.8), NW);
 +
</asy>
 +
 +
We are aiming to make figures that look most like our desired figure(the small square), so we think about moving the small triangles around.
 +
 +
 +
<asy>
 +
//path squ;//left,    bottom        right        top
 +
//squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle;
 +
//fill(squ, gray(0.6));               
 +
 +
path bigtri;
 +
bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle;
 +
//fill(bigtri, gray(0.7));
 +
 +
path trtri;
 +
trtri = (2,2)--(1.6,1.2)--(2,1)--cycle;
 +
fill(trtri, gray(0.6));
 +
 +
path bigs;
 +
bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle;
 +
draw(bigs);
 +
 +
path newtri;
 +
newtri = (2.4, 0.8)--(2,0)--(2,1)--cycle;
 +
fill(newtri, gray(0.6));
 +
draw(newtri);
 +
 +
draw( (0,0)--(1,2) );
 +
draw( (2,0)--(0,1) );
 +
draw( (2,2)--(1,0) );
 +
draw( (0,2)--(2,1) );
 +
draw((2,2)--(1.6,1.2));
 +
draw((2,0)--(1.2,0.4));
 +
 +
label ("x",(2, 0.5), W);
 +
label ("x",(2, 1.5), E);
 +
//label ("y", (1.8, 1.1), SW);
 +
//label ("2y", (1.8, 1.6), NW);
 +
//label ("2y", (1.6, 0.2), SW);
 +
//label ("2y", (1.4, 0.8), NW);
 +
</asy>
 +
 +
You can skip the next few lines if your diagram is accurate enough to show you that the two triangles are congruent.
 +
 +
However, we will still prove it.
 +
 +
We can see, by vertical angles, that there is one congruent angle.
 +
 +
We can also see, because all right angles are congruent, that there is another pair of congruent angles.
 +
 +
Last, the hypotenuse of both triangles has length x.
 +
 +
By AAS congruence, we have proved that we can move the top shaded triangle to the bottom shaded triangle.
 +
 +
We then perform the same step for every small triangle in the figure.
 +
 +
<asy>
 +
//path squ;//left,    bottom        right        top
 +
//squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle;
 +
//fill(squ, gray(0.6));               
 +
 +
path bigtri;
 +
bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle;
 +
//fill(bigtri, gray(0.7));
 +
 +
path trtri;
 +
trtri = (2,2)--(1.6,1.2)--(2,1)--cycle;
 +
//fill(trtri, gray(0.6));
 +
 +
path bigs;
 +
bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle;
 +
draw(bigs);
 +
 +
path newtri;
 +
newtri = (2.4, 0.8)--(2,0)--(2,1)--cycle;
 +
fill(newtri, gray(0.6));
 +
draw(newtri);
 +
 +
path newtri2;
 +
newtri2 = (0,0)--(1,0)--(0.8,-0.4)--cycle;
 +
fill(newtri2, gray(0.6));
 +
draw(newtri2);
 +
 +
path newtri3;
 +
newtri3 = (-0.4, 1.2)--(0,1)--(0,2)--cycle;
 +
fill(newtri3, gray(0.6));
 +
draw(newtri3);
 +
 +
path newtri4;
 +
newtri4 = (1.2, 2.4)--(1,2)--(2,2)--cycle;
 +
fill(newtri4, gray(0.6));
 +
draw(newtri4);
 +
 +
draw( (0,0)--(1,2) );
 +
draw( (2,0)--(0,1) );
 +
draw( (2,2)--(1,0) );
 +
draw( (0,2)--(2,1) );
 +
draw((2,2)--(1.6,1.2));
 +
draw((2,0)--(1.2,0.4));
 +
 +
label ("x",(2, 0.5), W);
 +
label ("x",(2, 1.5), E);
 +
//label ("y", (1.8, 1.1), SW);
 +
//label ("2y", (1.8, 1.6), NW);
 +
//label ("2y", (1.6, 0.2), SW);
 +
//label ("2y", (1.4, 0.8), NW);
 +
</asy>
 +
 +
After removing unnecessary lines...
 +
 +
<asy>
 +
//path squ;//left,    bottom        right        top
 +
//squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle;
 +
//fill(squ, gray(0.6));               
 +
 +
path bigtri;
 +
bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle;
 +
//fill(bigtri, gray(0.7));
 +
 +
path trtri;
 +
trtri = (2,2)--(1.6,1.2)--(2,1)--cycle;
 +
//fill(trtri, gray(0.6));
 +
 +
path bigs;
 +
bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle;
 +
//draw(bigs);
 +
 +
draw((0,0)--(1,0));
 +
draw((2,0)--(2,1));
 +
draw((1,2)--(2,2));
 +
draw((0,1)--(0,2));
 +
 +
path newtri;
 +
newtri = (2.4, 0.8)--(2,0)--(2,1)--cycle;
 +
fill(newtri, gray(0.6));
 +
draw(newtri);
 +
 +
path newtri2;
 +
newtri2 = (0,0)--(1,0)--(0.8,-0.4)--cycle;
 +
fill(newtri2, gray(0.6));
 +
draw(newtri2);
 +
 +
path newtri3;
 +
newtri3 = (-0.4, 1.2)--(0,1)--(0,2)--cycle;
 +
fill(newtri3, gray(0.6));
 +
draw(newtri3);
 +
 +
path newtri4;
 +
newtri4 = (1.2, 2.4)--(1,2)--(2,2)--cycle;
 +
fill(newtri4, gray(0.6));
 +
draw(newtri4);
 +
 +
draw( (0,0)--(1,2) );
 +
draw( (2,0)--(0,1) );
 +
draw( (2,2)--(1,0) );
 +
draw( (0,2)--(2,1) );
 +
draw((2,2)--(1.6,1.2));
 +
draw((2,0)--(1.2,0.4));
 +
 +
label ("x",(2, 0.5), W);
 +
label ("x",(2, 1.5), E);
 +
//label ("y", (1.8, 1.1), SW);
 +
//label ("2y", (1.8, 1.6), NW);
 +
//label ("2y", (1.6, 0.2), SW);
 +
//label ("2y", (1.4, 0.8), NW);
 +
</asy>
 +
 +
It's five congruent squares!
 +
 +
Since we want the area of the square in the middle compared to the entire cross, our final answer is <math>\frac{1}{5}</math>.
  
 
==Probability 2==
 
==Probability 2==

Revision as of 11:23, 5 April 2014

Geometry Explained

This page is meant for explaining the most difficult problems in the geometry packet.

This page can be accessed by tinyurl.com/teachgeo.

Composite Figure

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); draw(tri); fill(tt, gray(0.6)); 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 $\frac{270}{360} * \pi * (\frac{1.7 \sqrt{2}}{2})^2 = \frac{3}{4} * \pi * \frac{2.89}{2}$.

Adding it all up, we find the answer to be $\frac{3}{4} * \pi * \frac{2.89}{2} + 0.7225 + 5.1$ .

Just plug it into your calculator.

Spotlight!

We want to find the area of the intersection of the circles in this figure:

[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path inter; inter=(1,-sqrt(3))..(2,0)..(1, sqrt(3))--cycle; draw(inter); fill(inter, gray(0.6)); path inter2; inter2=(1,-sqrt(3))..(0,0)..(1, sqrt(3))--cycle; draw(inter2); fill(inter2, gray(0.6)); [/asy]

Lets call the radius of each of the circles 1, because we are calculating probability. To do this, we need to divide the intersection area into smaller parts that we can find the area of.

[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0); c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); fill(a, gray(0.6)); draw(b); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1.5,sqrt(3)/2), NE); label("1", (1,0), S); [/asy]

Lets examine the shaded $\triangle ABC$. Since each of its sides are radii, it is equilateral.

Hence, each of its angles measure 60 degrees. Namely, $\angle ABC$ measures 60 degrees.

Important note: The area of an equilateral triangle is calculated by $\frac{s^2\sqrt{3}}{4}$, where $s$ is the side length of a triangle. This is an EXTREMELY important formula to know, and it it also extremely simple to prove and memorize.

After this, we can deduce that the area of triangle ABC with side length 1 is $\frac{1^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}$

Let us now look at the encompassing sector.

[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0); c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); fill(a, gray(0.6)); draw(c); fill(c, gray(0.6)); draw(b); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1.5,sqrt(3)/2), NE); label("1", (1,0), S); [/asy] We then find that the area of the entire sector $ABC$ is $\frac{60}{360} * \pi * 1^2 = \frac{\pi}{6}$ .

Subtracting the area of $\triangle ABC$ from sector $ABC$ gives us the shaded part below:

[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0); c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); draw(c); fill(c, gray(0.6)); draw(b); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1.5,sqrt(3)/2), NE); label("1", (1,0), S); [/asy]

Plugging in the values we know, we find that this has area $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$ .

We see that there are 4 such small curved areas.

[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c, d, e, f; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0); c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; d = (0,0)..(2-sqrt(3), -1)..(1, -sqrt(3))--cycle; e = (2,0)..(sqrt(3), -1)..(1, -sqrt(3))--cycle; f = (2,0)..(sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); //draw(c); fill(c, gray(0.6)); //draw(d); fill(d, gray(0.6)); //draw(e); fill(e, gray(0.6)); //draw(f); fill(f, gray(0.6)); draw(b); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1,0), S); [/asy]

Multiplying the value we got for an individual small curved part by 4, we find the area of the above shaded region to be $\frac{4\pi}{6} - \frac{4\sqrt{3}}{4}$ .

We then need to add back the areas of the equilateral triangles.

[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c, d, e, f; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0)--cycle; c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; d = (0,0)..(2-sqrt(3), -1)..(1, -sqrt(3))--cycle; e = (2,0)..(sqrt(3), -1)..(1, -sqrt(3))--cycle; f = (2,0)..(sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); fill(a, gray(0.6)); //draw(c); //fill(c, gray(0.6)); //draw(d); //fill(d, gray(0.6)); //draw(e); //fill(e, gray(0.6)); //draw(f); //fill(f, gray(0.6)); draw(b); fill(b, gray(0.6)); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1,0), S); [/asy]

Recall that the area of each equilateral triangle was $\frac{\sqrt{3}}{4}$.

Hence the area of the above shaded region is $\frac{2\sqrt{3}}{4}$ .

We then look at the total.

[asy] //import graph; draw(circle((0,0), 2)); draw(circle((2,0), 2)); path a, b, c, d, e, f; a = (0,0)--(1, sqrt(3))--(2,0)--cycle; b = (0,0)--(1, -sqrt(3))--(2,0)--cycle; c = (0,0)..(2-sqrt(3), 1)..(1, sqrt(3))--cycle; d = (0,0)..(2-sqrt(3), -1)..(1, -sqrt(3))--cycle; e = (2,0)..(sqrt(3), -1)..(1, -sqrt(3))--cycle; f = (2,0)..(sqrt(3), 1)..(1, sqrt(3))--cycle; draw(a); fill(a, gray(0.6)); draw(c); fill(c, gray(0.6)); draw(d); fill(d, gray(0.6)); draw(e); fill(e, gray(0.6)); draw(f); fill(f, gray(0.6)); draw(b); fill(b, gray(0.6)); label("A", (0,0), W); label("B", (2,0), E); label("C", (1,sqrt(3)), N); label("D", (1,-sqrt(3)), S); label("1", (1,0), S); [/asy]

Hence the area of the total shaded region is $\frac{2\sqrt{3}}{4} + \frac{4\pi}{6} - \frac{4\sqrt{3}}{4} = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$ .

We plug this into the calculator to find the final answer.

In order to find the union(total) of the two circles, we examine the figure.

We want to find the area of the intersection of the circles in this figure:

[asy] //import graph; fill((circle((2,0), 1.98)), gray(0.6)); fill((circle((0,0), 1.98)), gray(0.6)); draw(circle((0,0), 2)); draw(circle((2,0), 2)); draw((0,0)--(2,0)); label("1", (1,0), S); [/asy]

Using $\pi r^2$, the area of the circle on the left is $\pi$. We also know that the area of the circle on the right is $\pi$. When we add these together, we are counting the entire area, but we are double-counting the overlap area. We thus need to subtract the overlap from the $2\pi$ in order to find the union(total) of the figure.

Hence, the total is $2\pi - (\frac{2\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$.

Probability 1

We want to find the probability that a point lies inside the shaded square below, given that it lies in the larger rectangle.

[asy] path squ; squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; fill(squ, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; draw(bigs); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); label ("x",(2, 0.5), E); label ("x",(2, 1.5), E); [/asy]

There are several ways to do this question.

We will examine 3, including similar triangles, rearranging pieces, and analytic(coordinate) geometry.

First, we can look at the solution with similar triangles.

[asy] //path squ;//left, bottom right top //squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; //fill(squ, gray(0.6)); path trtri; trtri = (2,2)--(1.6,1.2)--(2,1)--cycle; fill(trtri, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; draw(bigs); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); label ("x",(2, 0.5), E); label ("x",(2, 1.5), E); label ("y", (1.8, 1.1), SW); label ("z", (1.8, 1.6), NW); [/asy]

We label these 3 sides in the right triangle in the top right of the square.

We know it is right because one angle is vertical to the square's 90 degree angle.

[asy] //path squ;//left, bottom right top //squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; //fill(squ, gray(0.6)); path bigtri; bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle; fill(bigtri, gray(0.7)); path trtri; trtri = (2,2)--(1.6,1.2)--(2,1)--cycle; fill(trtri, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; draw(bigs); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); label ("x",(2, 0.5), E); label ("x",(2, 1.5), E); label ("y", (1.8, 1.1), SW); label ("z", (1.8, 1.6), NW); [/asy]

We examine the smaller shaded triangle and the larger shaded triangle.

These are similar by AA similarity(angle at the top, and right angle).

We can see, by the $x$'s on the left, that the sides are in ratio 1:2.

Thus, the bottom of the larger shaded triangle has length $2y$, and the side length of the smaller square has side length $z$.

However, because of rotational symmetry, the $z$ on the top is equal to that $2y$ on the bottom!

For clarity, here is a diagram.

[asy] //path squ;//left, bottom right top //squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; //fill(squ, gray(0.6)); path bigtri; bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle; //fill(bigtri, gray(0.7)); path trtri; trtri = (2,2)--(1.6,1.2)--(2,1)--cycle; //fill(trtri, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; draw(bigs); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); draw((2,2)--(1.6,1.2), linewidth(2.5)); draw((2,0)--(1.2,0.4), linewidth(2.5)); label ("x",(2, 0.5), E); label ("x",(2, 1.5), E); label ("y", (1.8, 1.1), SW); label ("z", (1.8, 1.6), NW); label ("2y", (1.6, 0.2), SW); label ("z", (1.4, 0.8), NW); [/asy]

Since we only want one variable, let's express x in terms of y using Pythagorean theorem in the top right triangle.

[asy] //path squ;//left, bottom right top //squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; //fill(squ, gray(0.6)); path bigtri; bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle; //fill(bigtri, gray(0.7)); path trtri; trtri = (2,2)--(1.6,1.2)--(2,1)--cycle; fill(trtri, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; draw(bigs); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); draw((2,2)--(1.6,1.2), linewidth(2.5)); draw((2,0)--(1.2,0.4), linewidth(2.5)); label ("x",(2, 0.5), E); label ("x",(2, 1.5), E); label ("y", (1.8, 1.1), SW); label ("2y", (1.8, 1.6), NW); label ("2y", (1.6, 0.2), SW); label ("2y", (1.4, 0.8), NW); [/asy]

We get that

$y^2$ + $(2y)^2$ = $x^2$

$y^2$ + $4y^2$ = $x^2$

$5y^2$ = $x^2$

Square rooting both sides,

$y\sqrt{5}$ = $x$

[asy] path squ;//left, bottom right top squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; fill(squ, gray(0.6)); path bigtri; bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle; //fill(bigtri, gray(0.7)); path trtri; trtri = (2,2)--(1.6,1.2)--(2,1)--cycle; //fill(trtri, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; draw(bigs); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); //draw((2,2)--(1.6,1.2), linewidth(2.5)); //draw((2,0)--(1.2,0.4), linewidth(2.5)); label ("$y\sqrt{5}$",(2, 0.5), E); label ("$y\sqrt{5}$",(2, 1.5), E); label ("2y", (1.8, 1.6), NW); label ("2y", (1.4, 0.8), NW); [/asy]

The area of the small square is $(2y)^2 = 4y^2$.

The area of the large square is $(y\sqrt{5}+y\sqrt{5})^2 = (2y\sqrt{5})^2 = 20y^2$

Thus, our final probability is $\frac{4y^2}{20y^2}=\frac{1}{5}$

Another method to do the same problem is rearrangement, when you move pieces of the figure around to make a nicer figure.


[asy] path squ; squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; fill(squ, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; draw(bigs); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); label ("x",(2, 0.5), E); label ("x",(2, 1.5), E); [/asy]

[asy] //path squ;//left, bottom right top //squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; //fill(squ, gray(0.6)); path bigtri; bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle; //fill(bigtri, gray(0.7)); path trtri; trtri = (2,2)--(1.6,1.2)--(2,1)--cycle; fill(trtri, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; draw(bigs); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); draw((2,2)--(1.6,1.2)); draw((2,0)--(1.2,0.4)); label ("x",(2, 0.5), E); label ("x",(2, 1.5), E); //label ("y", (1.8, 1.1), SW); //label ("2y", (1.8, 1.6), NW); //label ("2y", (1.6, 0.2), SW); //label ("2y", (1.4, 0.8), NW); [/asy]

We are aiming to make figures that look most like our desired figure(the small square), so we think about moving the small triangles around.


[asy] //path squ;//left, bottom right top //squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; //fill(squ, gray(0.6)); path bigtri; bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle; //fill(bigtri, gray(0.7)); path trtri; trtri = (2,2)--(1.6,1.2)--(2,1)--cycle; fill(trtri, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; draw(bigs); path newtri; newtri = (2.4, 0.8)--(2,0)--(2,1)--cycle; fill(newtri, gray(0.6)); draw(newtri); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); draw((2,2)--(1.6,1.2)); draw((2,0)--(1.2,0.4)); label ("x",(2, 0.5), W); label ("x",(2, 1.5), E); //label ("y", (1.8, 1.1), SW); //label ("2y", (1.8, 1.6), NW); //label ("2y", (1.6, 0.2), SW); //label ("2y", (1.4, 0.8), NW); [/asy]

You can skip the next few lines if your diagram is accurate enough to show you that the two triangles are congruent.

However, we will still prove it.

We can see, by vertical angles, that there is one congruent angle.

We can also see, because all right angles are congruent, that there is another pair of congruent angles.

Last, the hypotenuse of both triangles has length x.

By AAS congruence, we have proved that we can move the top shaded triangle to the bottom shaded triangle.

We then perform the same step for every small triangle in the figure.

[asy] //path squ;//left, bottom right top //squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; //fill(squ, gray(0.6)); path bigtri; bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle; //fill(bigtri, gray(0.7)); path trtri; trtri = (2,2)--(1.6,1.2)--(2,1)--cycle; //fill(trtri, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; draw(bigs); path newtri; newtri = (2.4, 0.8)--(2,0)--(2,1)--cycle; fill(newtri, gray(0.6)); draw(newtri); path newtri2; newtri2 = (0,0)--(1,0)--(0.8,-0.4)--cycle; fill(newtri2, gray(0.6)); draw(newtri2); path newtri3; newtri3 = (-0.4, 1.2)--(0,1)--(0,2)--cycle; fill(newtri3, gray(0.6)); draw(newtri3); path newtri4; newtri4 = (1.2, 2.4)--(1,2)--(2,2)--cycle; fill(newtri4, gray(0.6)); draw(newtri4); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); draw((2,2)--(1.6,1.2)); draw((2,0)--(1.2,0.4)); label ("x",(2, 0.5), W); label ("x",(2, 1.5), E); //label ("y", (1.8, 1.1), SW); //label ("2y", (1.8, 1.6), NW); //label ("2y", (1.6, 0.2), SW); //label ("2y", (1.4, 0.8), NW); [/asy]

After removing unnecessary lines...

[asy] //path squ;//left, bottom right top //squ = (0.4, 0.8)--(1.2, 0.4)--(1.6,1.2)--(0.8,1.6)--cycle; //fill(squ, gray(0.6)); path bigtri; bigtri = (2,2)--(1.2,0.4)--(2,0)--cycle; //fill(bigtri, gray(0.7)); path trtri; trtri = (2,2)--(1.6,1.2)--(2,1)--cycle; //fill(trtri, gray(0.6)); path bigs; bigs = (0,0)--(2,0)--(2,2)--(0,2)--cycle; //draw(bigs); draw((0,0)--(1,0)); draw((2,0)--(2,1)); draw((1,2)--(2,2)); draw((0,1)--(0,2)); path newtri; newtri = (2.4, 0.8)--(2,0)--(2,1)--cycle; fill(newtri, gray(0.6)); draw(newtri); path newtri2; newtri2 = (0,0)--(1,0)--(0.8,-0.4)--cycle; fill(newtri2, gray(0.6)); draw(newtri2); path newtri3; newtri3 = (-0.4, 1.2)--(0,1)--(0,2)--cycle; fill(newtri3, gray(0.6)); draw(newtri3); path newtri4; newtri4 = (1.2, 2.4)--(1,2)--(2,2)--cycle; fill(newtri4, gray(0.6)); draw(newtri4); draw( (0,0)--(1,2) ); draw( (2,0)--(0,1) ); draw( (2,2)--(1,0) ); draw( (0,2)--(2,1) ); draw((2,2)--(1.6,1.2)); draw((2,0)--(1.2,0.4)); label ("x",(2, 0.5), W); label ("x",(2, 1.5), E); //label ("y", (1.8, 1.1), SW); //label ("2y", (1.8, 1.6), NW); //label ("2y", (1.6, 0.2), SW); //label ("2y", (1.4, 0.8), NW); [/asy]

It's five congruent squares!

Since we want the area of the square in the middle compared to the entire cross, our final answer is $\frac{1}{5}$.

Probability 2

We want to find the probability that a point lies inside the shaded square below, given that it lies inside the larger rectangle.

[asy] path r; r = (0,0)--(4*sqrt(2), 0)--(4*sqrt(2), 3*sqrt(2))--(0, 3*sqrt(2))--cycle; draw(r); path tot;//bottom right top left tot = (2*sqrt(2),1*sqrt(2))--(2.5*sqrt(2), 1.5*sqrt(2))--(2*sqrt(2), 2*sqrt(2))--(1.5*sqrt(2),1.5*sqrt(2))--cycle; fill(tot, gray(0.6)); path sq; sq = (0,0)--(2*sqrt(2), 2*sqrt(2))--(4*sqrt(2), 0); draw(sq); path sq2; sq2 = (0,3*sqrt(2))--(2*sqrt(2), sqrt(2))--(4*sqrt(2), 3*sqrt(2)); draw(sq2); label("x", (2.25*sqrt(2),1.25*sqrt(2)), SE); label("x", (1.75*sqrt(2),1.25*sqrt(2)), SW); label("3x", (3*sqrt(2),2*sqrt(2)), SE); label("3x", (1*sqrt(2),2*sqrt(2)), SW); label("3x", (3*sqrt(2),1*sqrt(2)), NE); label("3x", (1*sqrt(2),1*sqrt(2)), NW); [/asy]


We look at the the 45-45-90 triangle on the right.

[asy] path r; r = (0,0)--(4*sqrt(2), 0)--(4*sqrt(2), 3*sqrt(2))--(0, 3*sqrt(2))--cycle; draw(r); path trir; trir = (2.5*sqrt(2), 1.5*sqrt(2))--(4*sqrt(2), 3*sqrt(2))--(4*sqrt(2),0)--cycle; fill(trir, gray(0.6)); path sq; sq = (0,0)--(2*sqrt(2), 2*sqrt(2))--(4*sqrt(2), 0); draw(sq); path sq2; sq2 = (0,3*sqrt(2))--(2*sqrt(2), sqrt(2))--(4*sqrt(2), 3*sqrt(2)); draw(sq2); path tot;//bottom right top left tot = (2*sqrt(2),1*sqrt(2))--(2.5*sqrt(2), 1.5*sqrt(2))--(2*sqrt(2), 2*sqrt(2))--(1.5*sqrt(2),1.5*sqrt(2))--cycle; draw(tot); label("3x", (3*sqrt(2),1*sqrt(2)), NE); label("3x", (3*sqrt(2),2*sqrt(2)), SE); [/asy]

We know its 45-45-90 because it is isosceles (two sides of length 3x) and one of its angles is 90(vertical to the 90 degrees in the square). Since we know that the sides are in the ratio 1:1:$\sqrt(2)$, the right side must have a length of $3x \sqrt{2}$.

We then look at the larger 45-45-90 triangle.

[asy] path trir; trir = (2*sqrt(2), 2*sqrt(2))--(0,0)--(4*sqrt(2),0)--cycle; fill(trir, gray(0.6)); path r; r = (0,0)--(4*sqrt(2), 0)--(4*sqrt(2), 3*sqrt(2))--(0, 3*sqrt(2))--cycle; draw(r); path sq; sq = (0,0)--(2*sqrt(2), 2*sqrt(2))--(4*sqrt(2), 0); draw(sq); path sq2; sq2 = (0,3*sqrt(2))--(2*sqrt(2), sqrt(2))--(4*sqrt(2), 3*sqrt(2)); draw(sq2); path tot;//bottom right top left tot = (2*sqrt(2),1*sqrt(2))--(2.5*sqrt(2), 1.5*sqrt(2))--(2*sqrt(2), 2*sqrt(2))--(1.5*sqrt(2),1.5*sqrt(2))--cycle; draw(tot); label("3x", (3*sqrt(2),1*sqrt(2)), NE); label("3x", (1*sqrt(2),1*sqrt(2)), NW); label("x", (2.25*sqrt(2),1.75*sqrt(2)), NE); label("x", (1.75*sqrt(2),1.75*sqrt(2)), NW); [/asy]

We know its 45-45-90 because it is isosceles (two sides of length 4x) and one of its angles is 90(vertical to the 90 degrees in the square). Since we know that the sides are in the ratio 1:1:$\sqrt{2}$, the bottom side must have a length of $4x \sqrt{2}$.

We want to find the probability that a point lies inside the shaded square below, given that it lies inside the larger rectangle.

[asy] path r; r = (0,0)--(4*sqrt(2), 0)--(4*sqrt(2), 3*sqrt(2))--(0, 3*sqrt(2))--cycle; draw(r); path tot;//bottom right top left tot = (2*sqrt(2),1*sqrt(2))--(2.5*sqrt(2), 1.5*sqrt(2))--(2*sqrt(2), 2*sqrt(2))--(1.5*sqrt(2),1.5*sqrt(2))--cycle; fill(tot, gray(0.6)); path sq; sq = (0,0)--(2*sqrt(2), 2*sqrt(2))--(4*sqrt(2), 0); draw(sq); path sq2; sq2 = (0,3*sqrt(2))--(2*sqrt(2), sqrt(2))--(4*sqrt(2), 3*sqrt(2)); draw(sq2); label("x", (2.25*sqrt(2),1.25*sqrt(2)), SE); label("x", (1.75*sqrt(2),1.25*sqrt(2)), SW); label("3x", (3*sqrt(2),2*sqrt(2)), SE); label("3x", (1*sqrt(2),2*sqrt(2)), SW); label("3x", (3*sqrt(2),1*sqrt(2)), NE); label("3x", (1*sqrt(2),1*sqrt(2)), NW); [/asy]

Remember that we want the shaded over the total. Shaded is a square with area $x^2$. Total is a rectangle with area $4x\sqrt{2}*3x\sqrt{2}=24x^2$

Thus, our answer is $\frac{x^2}{24x^2} = \frac{1}{24}$.

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