Do you mean these questions?

I have a pizza. The radius is 10 inches long. The pizza was cut into 16 equal slices. When 1 slice was left, my sister and I both wanted it, so we agreed to cut it in half, but I like the crust more than she does, so we decided to cut it the "other way." In other words, the two pieces would not be symmetrical. The inside piece would contain all topping, and the outer piece would contain some topping and some crust.

1. Find the area of the whole pizza.

2. What is the area of one piece of pizza?

3. What is the area of a half-piece?

4. What would the area of the whole pizza be if it were made of half pieces?

The lengths are inches so the areas will be square inches or in.²

Bob

]]>I wouldn't say it's a paradox; just a confusing statement.

Here is a paradoxical statement:

This statement is FALSE.

Bob

]]>http://www.mathisfunforum.com/viewtopic.php?id=18391

B

]]>BC = BD = AD = x

So AC = AB = x + y

Using the similar triangles

Divide all by x^2 and replace y/x by r, leading to a quadratic in r, which you will need to solve using the formula (or completing the square).

Two solutions, but y/x is obviously positive so you can reject the negative solution.

You can use the cosine rule or split a 36, 72, 72 triangle along its line of symmetry to get cos36 and cos72, firstly in terms of x and y, and then in terms of r, to finish the question.

Bob

]]>I think this works:

As AGB = 90, then a circle centred on the midpoint, D, of AB will go through G.

(EDIT: As I read this through, I see I haven't used the circle at all. Oh well, not to worry.)

Extend AY to E, where AG = GE, and BX to F, where BG = GF.

Join CE and CF.

As G is the median BG = 2GX => GX = XF.

Triangles AGX and CFX are congruent, (SAS) so FC = AG.

Similarly CE = BG.

So triangles AGB, CFG and CEG are congruent (SSS)

So CG = AB.

Bob

]]>In triangle ABD by cosine rule:

and in triangle ACD:

But those angles add up to 180 so

So adding

Bob

]]>Welcome to the forum.

That circle is called the nine-point-circle of ABC. It goes through D, E, F and Z.

So EFD = EZD.

My software Sketchpad labels the points in alphabetical order so the point Z shows up as G.

I changed it on the diagram but forgot to alter the measurement statement m<MGD to m<MZD.

It's 70 whatever you call it.

I've got a proof somewhere which I'll research if you post back.

Bob

]]>a=(pi)(10^2), =314

2. What is the area of one piece of pizza?

(pi)(r^2)/16, 100(pi)/16=19.6

3. What is the area of a half-piece?

100(pi)/36=10

4. What would the area of the whole pizza be if it were made of half pieces?

a=10(pi), a=100/10, a=10

5. What is the radius of a half-piece? (ie, where do I need to cut to make two equal halves out of a piece?)

x=7.158

PLEASE HELP ME CORRECT THIS

]]>http://www.mathisfunforum.com/viewtopic.php?id=19905

If there is something about that explanation that you don't understand, please use that thread to say what the problem is and I'll try to help you.

Bob

]]>13) C

14) D

]]>I found an 81-move solution which betters the 83-move record, but don't know how to have the website recognise it:

Bob

]]>Let the point where the max tangent hits the circle be P, and the centre of the circle be C, origin O.

Then OC^2 = CP^2 + PO^2 => OP = root 12.

So P is on a circle x^2 + y^2 = 12.

Put this with the other circle to get x + y = 4. Substitute y = 4 - x into the first circle and solve for x.

You'll get two answers. One is the max and one the min.

Find y and compute y/x

Bob

]]>