You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 17,466

1. Connect the nine points given below with four straight lines. All the points should be connected. When you start connecting the points with a pencil, don't take the pencil off the paper until you have finished.

* * *

* * *

* * *

2. The following area (ABCDEFGHIJ) is made up of five squares, each square of side one unit and area one square unit, and the total area is five square units. How would you divide this into exactly two parts with one straight line?

A B C

........................................

. . .

. . .

. . .

. . .

. . .D

J...........................................................E

. . . .

. . . .

. . . .

. . . .

. . . .

...........................................................F

I H G

Character is who you are when no one is looking.

Offline

**justlookingforthemoment****Moderator**- Registered: 2005-05-26
- Posts: 2,161

Ooh, ooh!

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 17,466

Character is who you are when no one is looking.

Offline

**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

Errr ...

question 2 is really going to bother me for a while I think

Offline

**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

My solution to 1 was slightly different from JLFM, but the same principle. It is, of course, a perfect example of thinking "outside the box"!!

Questions 2...... hmmm, need some thought....

Offline

**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

2. Grammar check, please! I presume that the question should be "exactly 2 EQUAL parts" (ie, 2.5 square units each).

If so, I have 2 answers, both a little lateral and probably not what you are looking for:

a) a horizontal line 5/6 units up from the bottom giving 5/6 of 3 squares below the line = 2.5 units

b) I would employ physics.... cut the shape from a homogenous material, suspend it from any point along the shape's edge, and run a plumbline to find the centre of gravity line! As the material is homogeneous, the area each side of that line would be 2.5 square units. I would guess that there would be many solutions where this COG line would pass entirely within the shape to give a valid solution.

Offline

**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

I worked out that this can be done using a straightedge only. Ashwil's solution a) can be accomplished as well, but I think you'd need some other tool like a protractor as well. We'll see if I'm right

My solution using only a straightedge was

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

I've seen the problem 2 with circles.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

fgarb, my reasoning for choosing the 5/6 solution was based on the way ganesh had drawn the diagram. Note, the vertical lines are conveniently punctuated with dots at 1/6 intervals. Therefore, a straight line joining the dot below J with the dot below E would do the job just fine!

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Anyone know how to do piecewise function with latex?

And can some explain why finding the center of mass of this piecewise function:

f(x) = {2 when 0≤x≤2; 1 when 2<x≤3}

And then using that point to divide it doesn't work?

Offline

**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

No idea how to do piecewise functions in LaTeX, however the center of mass represents the point at which the object could balance, which does not gaurantee that drawing a line through it would split the area equally.

Say you were drawing a vertical line, for example. The center of mass on the x axis is 1.3 unless I messed up my calculation, whereas what we might call the "center of area is 1.25". The explanation for this has to do with torque, which is getting more into physics than math. Basically, the farther a mass is from the pivot point, the more effect it has on the making the object want to pivot in that direction. That's why longer levers are more effective. As a result, since there is less material farther out on the right side than on the left, the center of mass needs to be slightly to the right of the "center of area". That's how I understand it anyway, I'd be happy to be corrected.

krassi, what's this circle problem you're mentioning? I can't really picture a circular version of this puzzle.

Offline

**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

OK, scrub the plumbline - the moments don't work like that for an irregular shape! My mistake. Soooorryy!

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 17,466

Well, a simple solution, but which requires 'thinking differently'!

Join A and midpoint of G and F.

Two parts each of 2.5 square units each!

Character is who you are when no one is looking.

Offline

**Ikcelaks****Member**- Registered: 2006-03-13
- Posts: 8

My solution was to connect D to the midpoint of IJ, for the exact same reason.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 17,466

**Ikcelaks, that is better than my solution!**

Character is who you are when no one is looking.

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

ganesh wrote:

How would you divide this into exactly two parts with one straight line?

Is it only the areas that have to be equal, or do the shapes have have to be identical as well? If only the areas need to be equal (not the shapes), then this is too easy. You just have to slice off an area of exactly 2.5 from the figure.

One way to do it is to join the points K and L on AB and IF respectively, where BK = HL = 0.25; Then the area of rectangle AKLI is 2×1.25 = 2.5 exactly half the area of the whole thing.

Or you can join points M and N on EF and JI respectively such that MF = NI = 5⁄6.

On the other hand, if you want the shapes to be identical as well

*Last edited by JaneFairfax (2007-03-12 18:11:56)*

Offline

Pages: **1**