 Posers and Puzzles

1. 10 Mar '09 03:55
I have a little trouble with my homework tonight - any ideas?

The first question is as follows: There are two circles, each with a radius of (x-5), and an area of pi(4x+1). How much more area does the larger circle have than the smaller one?

Ok, heres what I got:

you can just ignore the pi, since area is radius squared times pi. So:

(x-5)^2 = 4x+1

x^2-10x+25 = 4x+1

x^2-10x+24 = 4x

x^2-10x = 4x-24

x^2-14x + 0 = -24

x^2-14x + 24 = 0, and from there you can solve it like any quadratic equation.

Since I cant use special characters, its hard to show my work, but when you solve this quadratic equation, I come up with the two answers, 12 and 2. These are the two possiblilities for x. They work for the equations, so the area of the larger circle is 49pi. With the smaller circle, the area works out to be 9pi, making the final answer 40pi. However, with a radius of x-5, the radius would be -3, which is obviously not possible. Did I mess up somewhere, or is it just not possible?

Anyway, here is my other problem:

Inside an equilateral triangle are three circles,
each tangent to two sides of the triangle and tangent to
each other. The radius of each circle is 1 inch. What is
the area of the triangle?

I have not Idea how to go about solving that one. Help please!
2. 10 Mar '09 11:22
Originally posted by clandarkfire
I have a little trouble with my homework tonight - any ideas?

The first question is as follows: There are two circles, each with a radius of (x-5), and an area of pi(4x+1). How much more area does the larger circle have than the smaller one?

Ok, heres what I got:

you can just ignore the pi, since area is radius squared times pi. So:

(x-5)^2 ...[text shortened]... is
the area of the triangle?

I have not Idea how to go about solving that one. Help please!
Try drawing you answers to Q1 - what do you get?

Although it is not possible to have a length of negative distance, there is some curious (and rather important) mathematics related to this. If I draw a straight line starting at the origin of length 3 going from left to right it will have lenght 3. However, if I draw it starting at the some point but going from right to left I will have another line of length 3, but there is something subtly different about this line - it is "pointing" in a different direction. We denote this with a negative sign. -a is pointing in the other direction from a.

This is what is going on in your solution - your line is "pointing" in a different direction. However it still has length 3 and you can still use it as the radius for a circle. You have meerly gleamed too much information about the line, but because you are squaring your answer this gets rid of it ( (-a)^2 = a^2 ).

Lines with direction and length are called vectors, and they're pretty fundamental things in algebra.

I hope that that all makes sense. I also hope it is right - I haven't thought about this in a while...

Some thoughts: a circle cannot have negative area, but a rectangle can. What does this mean? Can a square have a negative area (must it be a^2 or would (-a)*a be a square)? What are negative volumes?
3. 10 Mar '09 11:25
Originally posted by clandarkfire

Anyway, here is my other problem:

Inside an equilateral triangle are three circles,
each tangent to two sides of the triangle and tangent to
each other. The radius of each circle is 1 inch. What is
the area of the triangle?
Try joining up the centres of the circles.
4. 10 Mar '09 12:35
Originally posted by Swlabr
Try drawing you answers to Q1 - what do you get?

Although it is not possible to have a length of negative distance, there is some curious (and rather important) mathematics related to this. If I draw a straight line starting at the origin of length 3 going from left to right it will have lenght 3. However, if I draw it starting at the some point but going ...[text shortened]... e a negative area (must it be a^2 or would (-a)*a be a square)? What are negative volumes?
Radius is a distance so it cannot be negative, right?
5. 10 Mar '09 12:522 edits
Originally posted by Palynka
Radius is a distance so it cannot be negative, right?
Of course - no distance can be "negative". However, the negative refers to the direction. For instance, a car cannot travel at -30km/h but it can reverse at 30km/h. So, you can say that it is going forward at -30km/h.
6. 10 Mar '09 13:22
Originally posted by Swlabr
Of course - no distance can be "negative". However, the negative refers to the direction. For instance, a car cannot travel at -30km/h but it can reverse at 30km/h. So, you can say that it is going forward at -30km/h.
I don't disagree, I'm just saying that's not helpful for the problem at hand.
7. 10 Mar '09 13:38
Originally posted by Palynka
I don't disagree, I'm just saying that's not helpful for the problem at hand.
Well - that's whats going on. There is a negative distance, and that is why.
8. 10 Mar '09 13:58
Originally posted by Swlabr
Well - that's whats going on. There is a negative distance, and that is why.
So distances can be negative now? I must have missed the memo.
9. 10 Mar '09 22:22
Originally posted by Swlabr
Try joining up the centres of the circles.
Thanks for the help on the other one - I guess my original answer may have been correct. For the other one, what do you mean by joining up the centers of the circles? I get what the picture they are talking about looks like, I just don't get how to find the area. Do you mean making a smaller equalateral triangle inside, with the three virticies being the centers of the circle? If so, how does this help me?
10. 11 Mar '09 08:32
Originally posted by clandarkfire
Thanks for the help on the other one - I guess my original answer may have been correct. For the other one, what do you mean by joining up the centers of the circles? I get what the picture they are talking about looks like, I just don't get how to find the area. Do you mean making a smaller equalateral triangle inside, with the three virticies being the centers of the circle? If so, how does this help me?
Well - therein lies the "leap" that is the point of the question.

I think. There may, of course, be a much easier way. 😛
11. 11 Mar '09 14:03
If I'm planning on driving to Chicago from Inianapolis and I find that I end up in New Orleans, haven't I gone a "negative" distance toward Chicago?

For those that think going either direction is negative, please keep it to yourselves.
12. 11 Mar '09 14:33
Originally posted by coquette
If I'm planning on driving to Chicago from Inianapolis and I find that I end up in New Orleans, haven't I gone a "negative" distance toward Chicago?

For those that think going either direction is negative, please keep it to yourselves.
To start using proper language: It depends on your frame of reference. If you are talking about scalars, then sure it's never negative. However, vectors can be as they take into account distance.
13. 11 Mar '09 14:411 edit
Originally posted by coquette
If I'm planning on driving to Chicago from Inianapolis and I find that I end up in New Orleans, haven't I gone a "negative" distance toward Chicago
Unqualified, distance cannot be negative and is not directional. Think of how you calculate distance and you'll see why it can't be negative.

In your example, you just went a positive distance in another direction thus increasing the distance between yourself and Chicago.

Edit - For the purpose of this topic, a radius is a distance (in the unqualified sense) and cannot therefore be negative.
14. 11 Mar '09 23:501 edit
It seems to me (one man's opinion), that your math teacher tried to incorporate two different concepts into one problem, and did not account for a logical inconsistency. The teacher was trying to use the well known "area of a circle" formula, to construct a problem in which you had to solve a quadratic equation.

However, I believe that the teacher forgot one important difference between the algebra of quadratics and the solid constructs of geometric figures: a quadratic equation can be solved with a negative answer without any logical difficulties, yet a PHYSICAL construct like a "radius" or any piece of a geometric figure can not have a "negative size." We can sidestep this inconsistency by assuming the teacher didn't notice that the circle would have radius -3, by just blindly plugging in the numbers and hoping that we are conforming with the teacher's intent: -3 squared = 9, so the small circle has an area of 9pi units squared. Alternatively, we can think of it as when the radius of the circle shrinks past zero into "negative territory" that the circle begins to grow again: i.e. that a circle with "negative" radius is the same as the circle of one with the absolute value of that radius.

however, these arguments are designed to escape a fundamental flaw in the construction of the problem, and if we're being stringent with our application of algebra to a geometry problem, this figure could not exist. it should be a basic necessity in constructing a problem that the radius (or the length of any other physical structure within the graph) be greater than or equal to zero. because an object cannot have negative size. it is logically inconsistent with reality.

that's not to say that concepts like "negative area", "negative volume," "directional vectors being akin to negative distance," etc. aren't entirely useful - in fact they have vast application in more conceptual areas of mathematics like calculus, differential equations, and beyond. but for the purposes of this problem, i'd bet the teacher just didn't notice the flaw in their problem's construction, and would probably ask you to do the same. 🙂
15. 12 Mar '09 01:411 edit
The answer of x=2 is invalidated by the fact it would make the radius negative when the domain of radii permitted are non-negative numbers (Zero might be permitted for a circle which is in essence a point).

Since the "smaller" circle is invalid, the "difference" is also invalid.

Using the square root function isn't reversible in general, and it can give you answers which don't make sense.

I will have to look at the triangle problem.

EDIT: In drawing and reorganizing the areas involved, I arrive at an answer of 6 + 4*sqrt(3).

Draw 3 equal sized circles in an equilateral arrangement. You can draw 3 lines, each tangent to 2 of the circles, that's your triangle.

Connect the centers of the circles to form an equilateral triangle with 3 sides of length 2. (Area = Sqrt(3))

Form a rectangle from each side, extending to the edge of the triangle. Each of these is a 1x2 rectangle (Total Area = 3*2 = 6)

Finally you have 6 identical segments left, right triangles with a leg of length 1 opposite a 30 degree angle. (Total Area = 1/2 * sqrt(3) * 6 = 3 * Sqrt(3))

These areas can be arranged to exactly cover the triangle without gap or overlap, and their total area is 6 + 4 * sqrt(3)...