Very Hard MATH Question (insanely hard)?

Question

A circle with radius of 1 is randomly placed so that it lies entirely inside a 15 x 36 rectangle ABCD. Find the probability that it does not meet the diagonal AC.

The answer is 375/442

I need the SOLUTION

OK YOU PEOPLE. YOU THINK ITS SO SIMPLE THAT YOU JUST SAY, THE ANSWER IS THE ALLOWABLE AREA OVER THE TOTAL AREA. I FRIGGING KNOW THAT.. THE F*ING HARD PART IS SOLVING IT CORRECTLY, WHICH I BET NONE OF YOU IGNORANT PEOPLE CAN DO. SHEESH, SUCH MEDIOCRE ANSWERS.

Answer 1

This is obtained by finding the area of all the space occupied by the center of the selected circles divided by the area of the space occupied by all circles that fit within the rectangle.

Let's do the simpler one, the denominator:
The center has to be at least 1 from each edge. This describes a rectangle 13 * 34 = 442, which is denominator.

The numerator is the sum of two identical rt triangles (one on each side of the diagonal) with the following specification:

The two legs are parallel to the two adjacent rectangle sides and 1 unit inside the rectangle.
The hypotenuse is 1 unit away and parallel to the diagonal AC.
The two ends of the hypotenuse have the characteristic of being 1 unit away from the edge and the hypotenuse. Call them points P and Q, and R is the right angle point which is (1,1) from the corner B.

Let's do some calculations.
The half rectangle is similar to the 5-12-13 rt angle triangle, ie., 15-36-39. Diagonal is 39.
The angles are C: arcsin 15/39 and A: arcsine 36/39

Now this gets a bit tricky.
Draw a line from one end of the diagonal to points P, and from the other end to point Q. Drop perpendiculars from points P and Q to both the rectangle side and the diagonal.
These give you triangles with the angle being half the angles of the earlier big triangles because the line from the end of the diagonal to P and Q bisect the angle.

let these half angles be p and q

tan p = (1-cos2p)/sinp (half angle formula)

plugging in sin p = 36/39, cosp = 15/39
and sin q = 15/39 and cos q = 36/39

we get tan p = 2/3 and tan q = 1/5
Since the opposite side is a length of 1, the intercept on the sides of the tiny right triangles are:
1/tan p = 3/2 and 1/tanq = 5

Now we can calculate the size of the right angle triangle PQR
Leg PR: R is 1 unit from corner B, and P is 3/2 units from A
therefore PR = 15 - 1 - 3/2 = 12.5
Leg QR: R is 1 unit from corner B and Q is 5 units from C.
QR = 36 - 1 - 5 = 30
Area of PQR = 1/2 * 12.5 * 30
Now there's another identical triangle on the other side of the diagonal.
Therefore total area of complying circles is two times area of triangle PQR = 12.5 * 30 = 375

Therefore probability is 375/442

Answer 2

Yeah, the answer is 375/442, that's correct.

The way to solve that is to draw an interior rectange 13 x 34, which represents the area where the center of the circle can be found. That's the 442 part. Then we draw the diagonal, and two more parallel lines either side of the diagonal. Those lines form triangles with the interior rectangle, which represents the area where the center of the circle can be found and NOT meet the main diagonal. This has the area of 375.

To figure this out, carefully draw 2 triangles at each corner (right triangles only, with one side = 1), which will help you determine the dimensions of these triangles. It helps that the hypotenuse of the 15 x 36 triangle is 39 exactly, so using these proportions, we find that the sides of the small triangles are:

A = 15 - 1 - 39/36 - 15/36 = 25/2
B = 36 - 1 - 39/15 - 36/15 = 30

The inside area is (25/2) (30) = 375, so there you go.

Answer 3

Don't look at the circle. Look at where in the rectangle the center of the circle can be.

You have a 15 by 36 rectangle and a circle with radius 1. The center of the circle cannot be closer than a distance of 1 from the edge of the rectangle. That leaves a 13 by 34 rectangle that could contain the center of the circle. This is 13*34 = 442 units squared, which is the denominator.

The circle will touch the diagonal if the center of the circle is within a distance of 1 from the diagonal. So take the area of a diagonal swath of width 2 (distance of 1 on either side of the diagonal) and divide it by the area of the 13 by 34 rectangle above. Subtract that fraction from 1 and you have your answer.

Scythian, below, gives a very good explanation with the two triangles in the corner.

Answer 4

probability is a ratio: the number of successful outcomes to the total possible number of outcomes

calculate the area of the rectangle that could be a possible outcome (i.e. the area where such a circle could be complete within the rectangle). this is obviously a smaller rectangle (13 x 34 = 442). next, ...well i'm sure you get the idea - find the area of the four "permissible" triangles.


alternatively, you could calculate 1 minus the ratio of the number of failed outcomes to the total possible number of outcomes.

Answer 5

14

Answer 6

Have you actually tried to draw the regions where:

A. The center of the circle can lie and still have the circle inside the rectangle?

B. All the conditions of the problem are met?

If you do, I think it will look a lot easier. It will be all about rectangles and triangles, and circles will be gone from the problem.

Answer 7

I'm not sure why you are so angry with people that are trying their best to answer YOUR homework question for you. At least three of these answers explain step by step how to reach the same answer that you have already provided from the back of the book but because they didn't write it out with the numbers they are ignorant? I think you might want to look up the meaning of the word. Instead of just trying to get the answer typed out for you, follow their advice and draw the problem out step by step. When you get to the end and have the same answer, guess what? You just wrote out the solution and hopefully learned something in the process!

Answer 8

the probability that it does not touch the diagnal is the probability that its center does not lie within a stripe one radius wide from the diagnol. In addition, since the circle must lie entirely within the rectacgle a stripe around the border is excluded. First "contract" the rectangle to the non excluded portion. The center of the circle is just as likely to have its center fall anywhere in there . Thus the probability of the circle not touching the diagnol is the contracted area minus the diagnol stripe within the contracted area divided by the contracted area.

Answer 9

The solution is 375/442.

If you want to know the steps to getting this solution, I have no idea.

Answer 10

go 2 tutor.com. they have online help