# Easy but very tricky geometry problem!

• Aug 14th 2013, 03:05 PM
ChessTal
Easy but very tricky geometry problem!
The image is self explanatory. The problem is really of geometry and not one in the probability area(since the probability matter of the problem is trivial).

http://i.minus.com/i0k95R4p1Zj1y.png
• Aug 14th 2013, 03:44 PM
Shakarri
Re: Easy but very tricky geometry problem!
Surely there has to be some restriction on the way in which the insect moves
Edit: nevermind, I missread the question.
• Aug 15th 2013, 01:20 AM
ChessTal
Re: Easy but very tricky geometry problem!
Quote:

Originally Posted by Shakarri
Surely there has to be some restriction on the way in which the insect moves
Edit: nevermind, I missread the question.

The probability is(not the solution but a hint)
Spoiler:
just the "area of ABCDEFG polygon" / "area of AHJK rectangle" so the real problem is to find these areas....
• Aug 15th 2013, 01:23 PM
Shakarri
Re: Easy but very tricky geometry problem!
Quote:

Originally Posted by ChessTal
The probability is(not the solution but a hint)
Spoiler:
just the "area of ABCDEFG polygon" / "area of AHJK rectangle" so the real problem is to find these areas....

Spoiler:
I thought that was the trick to it but you'd have to assume that the time is long enough for initial conditions to be unimportant (I'm thinking of an insect which moves at a finite speed between points, I suppose an insect that can jump between spots would be modeled as instantly moving between points).
I still have no idea how to get the area with only 3 measurements or how the inequalities help :(
• Aug 16th 2013, 07:44 AM
ChessTal
Re: Easy but very tricky geometry problem!
Quote:

Originally Posted by Shakarri
Spoiler:
I thought that was the trick to it but you'd have to assume that the time is long enough for initial conditions to be unimportant (I'm thinking of an insect which moves at a finite speed between points, I suppose an insect that can jump between spots would be modeled as instantly moving between points).
I still have no idea how to get the area with only 3 measurements or how the inequalities help :(

Well:
Spoiler:
Well it is a very beautiful and relatively unknown feature of compass and straightedge that when you have a field that consists of straight lines and with one right angle(90°) and by knowing the one side that forms the right angle then you can find its area no matter how complicated the shape is with just 1(yes one) measurement!!
E.g for this:
http://img713.imageshack.us/img713/9825/fmuj.png

We can find its area by using compass and straightedge as many times as we want and with just one measurement with a ruler.

Hint: For the construction/proof of that you will have to think that a trapezoid's diagonals create 2 equal area triangles.
• Aug 17th 2013, 05:36 AM
ebaines
Re: Easy but very tricky geometry problem!
Similar to the previous post, I have a way to solve this with only two measurements, not three, and without using the 7cm length information. It involves breaking the polygon into five adjacent trapezoids, each with vertical sides, and using the base of the rectangle as its base, and with slanted top edges, and then combining two adjacent trapezoids into a single trapezoid of equal area. This combining of trapezoids requires use of the compass and straightedge - the process is to find the midpoint of the upper edge of both adjacent trapezoids and then draw a straight line through those two points, creating a single trapezoid with combined width of the original two trapezoids, which now has the same area as the original two. Repeat this process four times and you end up with a single trapezoid with base length equal to the original large rectangle. The only trick is that for the right-most trapezoid, which has the triangle area cut out of the bottom, you can move that triangle to an equivalent area at the top - simply copy its height using the compass to extend the upper portion down an equal amount. After all five sections have been converted into one large trapezoid the last step is to convert that trapezoid into a rectangle - simply find the midpoint of its top slanted edge and draw a horizontal line through it. So now the original polygon has been converted into a rectangle with width equal to the large rectangle. Use the ruler to measure the height of the inner rectangle that has been converted from the original polygon (call this height x) and the height of the large rectangle (call this y), and the probability of the bug being in the original polygon is the same as the probability of it being in the equivalent rectangle, which is x/y. So again - only two measurements are needed.
• Aug 17th 2013, 06:48 AM
ChessTal
Re: Easy but very tricky geometry problem!
Quote:

Originally Posted by ebaines
Similar to the previous post, I have a way to solve this with only two measurements, not three, and without using the 7cm length information. It involves breaking the polygon into five adjacent trapezoids, each with vertical sides, and using the base of the rectangle as its base, and with slanted top edges, and then combining two adjacent trapezoids into a single trapezoid of equal area. This combining of trapezoids requires use of the compass and straightedge - the process is to find the midpoint of the upper edge of both adjacent trapezoids and then draw a straight line through those two points, creating a single trapezoid with combined width of the original two trapezoids, which now has the same area as the original two. Repeat this process four times and you end up with a single trapezoid with base length equal to the original large rectangle. The only trick is that for the right-most trapezoid, which has the triangle area cut out of the bottom, you can move that triangle to an equivalent area at the top - simply copy its height using the compass to extend the upper portion down an equal amount. After all five sections have been converted into one large trapezoid the last step is to convert that trapezoid into a rectangle - simply find the midpoint of its top slanted edge and draw a horizontal line through it. So now the original polygon has been converted into a rectangle with width equal to the large rectangle. Use the ruler to measure the height of the inner rectangle that has been converted from the original polygon (call this height x) and the height of the large rectangle (call this y), and the probability of the bug being in the original polygon is the same as the probability of it being in the equivalent rectangle, which is x/y. So again - only two measurements are needed.

Interesting but can you elaborate more into how exactly to do what you are saying? I mean can you draw an image or be more specific on your description?
• Aug 17th 2013, 03:35 PM
ebaines
Re: Easy but very tricky geometry problem!
I hope the attached helps explain it. I break the original polygon into 4 trapezoids and a triangle, labelled A-E. In each step I combine a trapezoid with its neighbor, keeping the same total area. After I get a trapezod for ABCD I convert it to a rectangle of same area. Then for triange E I first convert it to a right angle trianangle of equal area as E, and then combine with rectangle ABCD to create trapezoid ABCDE. Finally I turn that into a rectangle. The percentage of the area of the original polygon to the overall reactangle is then x/y.

Attachment 29011