[Guest]

What shape will make the scale balanced?

Remember: the bars have a weight and the distance away from the center matters

The reason the thing going down immediately to the right of the first vertical bar is supposed to be one eighth of the long bar and it is only longer so it wont overlap with the other bar right next to it

a circle

a bar=1

triangle=2

trapezoid=4

star=7

circle=15

square=40

[Guest]

unless we get the key to the weight of the shapes, theres really no way to figure it out without guessing

[Guest]

unless we get the key to the weight of the shapes, theres really no way to figure it out without guessing

I think that's the whole point of the riddle.

You're supposed to find out what shapes weigh what based on the information given.

Very nice riddle by the way eighthth.

Edited August 31, 2008 by ThunderChicken

[Guest]

My calculations were:

Bar=1

Triangle=2

Star= 4

Square= 8

Circle= 10

Lol and I couldn't figure out how much the trapezoid was. No wonder why

[Guest]

I would say the riddle was awesome! Except you failed to mention that the vertical bars don't count. WTF!

[Guest]

How did you get your circle result for this? I've worked the math a couple times and keep getting the answer to be four squares...but I get all the same results for all the rest of the weights...

[Guest]

i think that i messed up and the triangle all the way to the left should actually be a star

[Guest]

this is the correct picture (i think)

Edited September 1, 2008 by eighthth

[Guest]

first start at the very bottom where the triangle is balanced with two bars. bar=1 triangle=2.

Then work your way up to see a star is two triangles and three bars. star=7. then the circle is equal to a bar+two stars (if it is balanced both weights are equal)so circle=31 (this is kind of a 2x+1 effect)

next you know that two circles and a bar equal a square a circle and a star times two (its only half the distance from the side and the middle so 31*2=square+15+7; square=40

you know that the left side of the big bar =63(2(31)+1) so to be balanced the right side does too, the square at the end is 40 so the other two must add up to 23

next work on the trapezoids. you know that a star+two bars+two trapezoids=1 circles and .5 trapezoids

so 7+2+2x=15+.5x or 1.5x=6 so trapezoid=4

then take the total of that whole thing: 2 bars+2 circles +1 trapezoid or 2+30+4=36 divide it by 2 (it is midway between middle and end) and it is a total weight of 18 subtract eighteen from 23, and that is the total weight of the last unsolved part

first recognize that the total should be multiplied by 4 because it is 1/4 of the distance from middle to end. therefore the total of that part is 5*4=20.

you are given a bar, 1 and a trapezoid, 4. 20-4-1=15=circle

circle is the answer

[Prof. Templeton]

first start at the very bottom where the triangle is balanced with two bars. bar=1 triangle=2.

Then work your way up to see a star is two triangles and three bars. star=7. then the circle is equal to a bar+two stars (if it is balanced both weights are equal)so circle=31 (this is kind of a 2x+1 effect)

next you know that two circles and a bar equal a square a circle and a star times two (its only half the distance from the side and the middle so 31*2=square+15+7; square=40

you know that the left side of the big bar =63(2(31)+1) so to be balanced the right side does too, the square at the end is 40 so the other two must add up to 23

next work on the trapezoids. you know that a star+two bars+two trapezoids=1 circles and .5 trapezoids

so 7+2+2x=15+.5x or 1.5x=6 so trapezoid=4

then take the total of that whole thing: 2 bars+2 circles +1 trapezoid or 2+30+4=36 divide it by 2 (it is midway between middle and end) and it is a total weight of 18 subtract eighteen from 23, and that is the total weight of the last unsolved part

first recognize that the total should be multiplied by 4 because it is 1/4 of the distance from middle to end. therefore the total of that part is 5*4=20.

you are given a bar, 1 and a trapezoid, 4. 20-4-1=15=circle

circle is the answer

It looks to me as if everything on the left side of the big scale equals 94. I don't see how you get 63.

1-square = 40

2-circles = 30

2-stars = 14

2-triangles = 4

6-bars = 6

Total = 94

You're going to need a lot more than 1 circle on the right.

[Guest]

I think you're right Templeton...

Here is my derivation of this sweet solution:

Let's assume that the horizontal bars contain mass as do all of the fun shapes but the vertical bars are like strings so weightless (this isn't really necessary, but easier to work with).

Say each bar is 1 unit long making the long bar at the top a total of 4.5 units long with 2.25 units on each side of the center axis. Thus, all the mass to the left of center is hanging 2.25 units from the center, while the unknown is hanging 0.375 to the right, the next 1.125 (half way) from the center, and the final square 2.25 units from the center. These distances will be important when finding the ratio of weights to balance the system as weights farther out create more torque than weights close to the center.

The torque on each side of a pivot position needs to be the same to cause a balance.

torque = weight*(distance from pivot)

weight1*distance1 = weight2*distance2 for a balanced system

Now solving for the weights starting with the triangle and working upward (eighthth did a good job of this but I'm going to repeat a bit):

Define a bar to equal 1 in weight.

bar = 1

A triangle equals 2 bars so

triangle =2

The star equals 2 bars + 2 triangles + 1 more bar = 3 triangles +1 bar = 3*2 + 1 = 7

star = 7

The circle equals 2 stars + 1 bar = 2*7 + 1 = 15

circle = 15

Solving for the square using the above torque relationships:

(square + circle + star)*(0.25) = (2 circles + 1 bar)*(0.5)

square + 15 + 7 = (30 +1)*2

square = 31*2 - 22 = 62 - 22 = 40

square = 40

Now we have all the weight on the left side of the system:

weight leftside = square + circle + star + 1 bar + 2 circles + 1 bar = 40 + 15 + 7 + 30 + 2 = 94

weight leftside = 94

Solving for the trapezoid from the middle section of the right side of the system using the above torque equations once again:

(2 circles + 1 trap)*(0.25) = (1 star + 2 bar + 2 traps)*(0.5)

2*15 + 1 trap = (7 + 2 + 2 traps)*2

30 + 1 trap = 18 + 4 traps

3 traps = 12

trapezoid = 4

Now for the final reckoning:

Summing up the torques to balance the system:

(weight leftside)*(2.25 from center) = (square)*(2.25) + (3 traps + 4 bars + 1 star + 2 circles)*(1.125) + (1 bar + 1 trap + unknown)*(0.375)

94*2.25 = 40*2.25 + (12 + 4 + 7 + 30)*1.125 + (1 + 4 + unknown)*0.375

94*2.25 = 40*2.25 + 53*1.125 + (5 + unknown)*0.375

Divide thru by 0.375:

94*6 = 40*6 + 53*3 + 5 + unknown

564 = 240 + 159 + 5 + unknown

unknown = 564 - 240 - 159 - 5 = 160

unknown = 160

So this would equal 4 squares in its easiest form.

The answer!

unknown = 4 squares

Edited September 1, 2008 by slomic

[Guest]

It looks to me as if everything on the left side of the big scale equals 94. I don't see how you get 63.

1-square = 40

2-circles = 30

2-stars = 14

2-triangles = 4

6-bars = 6

Total = 94

You're going to need a lot more than 1 circle on the right.

Remember: the bars have a weight and the distance away from the center matters

The square circle and triangle are half the weight they would normally be, because they are closer to the center

for example if you wanted to balance a 1 pound object and a 4 pound object, you would put the 1 pound object on the end, and the 4 pound object 1/8 (of the total length of the bar) away from the center. its all proportions

[Guest]

The square circle and triangle are half the weight they would normally be, because they are closer to the center

for example if you wanted to balance a 1 pound object and a 4 pound object, you would put the 1 pound object on the end, and the 4 pound object 1/8 (of the total length of the bar) away from the center. its all proportions

I'm going to have to disagree with you on this point. While it is true that a one pound object at an end and a four pound object 1/8 of the total bar length from the center of the bar will cause the objects to be in balance, the total weight felt by the pivot point will be the summed weight of the objects, in this case 5 pounds. The proportions part of it is strictly for the balancing, not for the total weight. From what I can tell, this the only methodological error in your solution.

I did enjoy the puzzle very much and had fun working out all the weights and balances...

[Guest]

I'm going to have to disagree with you on this point. While it is true that a one pound object at an end and a four pound object 1/8 of the total bar length from the center of the bar will cause the objects to be in balance, the total weight felt by the pivot point will be the summed weight of the objects, in this case 5 pounds. The proportions part of it is strictly for the balancing, not for the total weight. From what I can tell, this the only methodological error in your solution.

i guess that makes a lot more sense than my logic

if a thing was hanging from the middle it wouldnt be zero pounds would it