Maths problem of Today

[Markus]

New problem:
In every field of a 4x4 grid, you can order the numbers 12,3,...,16 in such a way, that every of the 2x2 sub-grids will have the same sum, as in:

4| 6| 3| 5
------------
11|13|12|14
-----------
2| 8| 1| 7
------------
9|15|10|16


(like, 4+6+11+13 = 1+7+10+16 = 13+12+8+1 etc)

There are many other ways like this. One way starts like this:

| 7| | 3
------------
9| 2| |
-----------
| | |
------------
1| | | ?


What's the value of the number on the bottom-right square?

No SCAR/Simba/Calculator allowed! Just pencil and paper, and I'd like to see your exact logics in solving it (it can be done 100% logically, I did it myself).

The first correct answer was posted by Boreas (original problem there too, it's a geometry one):
http://villavu.com/forum/showthread.php?p=737298#post737298

[Cstrike]

I love these kinds of questions, I just forget how to get the area of U'AU. If I learned how to get that, I could do it ;)

[Markus]

Trust me, you won't need to do that ;) It's a lot easier than it looks.

Oh and if you really like these, I got tons of others like this one and I'll post them all if you want :p

[Shuttleu]

can i just take a guess and say somewhere around 0.45?
:p

~shut

[Markus]

Close but not correct.

[Dynamite]

0.64

[Markus]

Any reason behind that? (it's not 0.64)

[Nava2]

I see how to do it, but haven't the patience to do it. Heh.

Area of CC'V == AC'U, the angles combine to 90, if you have an area of 1, you can find the radius which is C'V.

Is the solution http://mathbin.net/equation_previews/6260_0.png

nvm, answer is incorrect found flaws in my solution.

[Markus]

The solution can be written as a regular decimal number. But you're on the right track Nava, but your current solution is incorrect.
Besides 72/pi = ±23, and the surface of the half-circle is 1 ;)

[Dynamite]

Any reason behind that? (it's not 0.64)

None at all, I just wanted to join in the fun.
(I am awful at maths)

[Boreas]

2/5 but pretty sure I used way too many steps.

[Nava2]

Best I've gotten is 73/pi. :<

Which I *know* is not correct.

[Markus]

2/5 but pretty sure I used way too many steps.

We got a winner! :p

My solution:
Trig C'U'U = C'V'V
Using this, we can combine two parts to this:
http://i54.tinypic.com/aayuxi.png

As you can see, the angle is 36 degrees (= 10% of 360 degrees = 10% of the area of a circle). We got 20% of an entire circle covered, = 40% of a half-circle. 40% of 1 = 0,40.


Original problem:
http://i53.tinypic.com/2qkqo3r.png

The area of the half-circle is 1. What's the area of the two grey parts?

Note: No calculators allowed. I had to do this with only a piece of paper and limited time (few minutes), and I'd like to see if you can do it :)

Semi-small hint (highlight beneath): (I didn't get this one, but some of you might like it).
The angles were carefully picked. Now to why they are like this..

Cookies for the first one to post the right solution + method.

[Markus]

New problem:
In every field of a 4x4 grid, you can order the numbers 12,3,...,16 in such a way, that every of the 2x2 sub-grids will have the same sum, as in:

4| 6| 3| 5
------------
11|13|12|14
-----------
2| 8| 1| 7
------------
9|15|10|16


(like, 4+6+11+13 = 1+7+10+16 = 13+12+8+1 etc)

There are many other ways like this. One way starts like this:

| 7| | 3
------------
9| 2| |
-----------
| | |
------------
1| | | ?


What's the value of the number on the bottom-right square?

No SCAR/Simba/Calculator allowed! Just pencil and paper, and I'd like to see your exact logics in solving it (it can be done 100% logically, I did it myself).

The first correct answer was posted by Boreas (original problem there too, it's a geometry one):
http://villavu.com/forum/showthread.php?p=737298#post737298

[bolshak25]

the answer is 14

[Markus]

Correct! Did you try or did you actually solve it? :p

[bolshak25]

actually solve. i used your other one as an example and realized they always have to add up to 34 so i went from there

[Cstrike]

how would you go about logically solving these without trial and error?

I tried using equations and setting each box as its own variable. Solving it just made things worse.

Is there ways to say "well we know the top left box HAS to be a huge ass number"?

I think I'm missing something, does it only depend on the 2x2 sub-boxes adding up to the same number? What about rows? Cause technically if you can put any number from a 2x2 point in another place, would that not give 4 answers to "?"... ?

[bolshak25]

well if you add all 16 numbers up you get 136. so each of the four
corners groups of 2x2 have to be 34 (136/4) so you logically get the top left
number as 16.


then you use the 9 and the 2 with the empty spaces below. you add 9 and 2
getting 11, therefore the numbers below them have to add up to 23 to get
the total of 34. so the possibilities are 7,16 8,15 9,14 10,13 11,12 rule out
7,16 and 9,14. leaving you with now 8,15 10,13 11,12. but when you add
them with the 1 in the bottom left you get 24, forcing that other number to
be the 10. 8,15 and 11,12 only work now.


now look at the 7 and 2. they add up to 9 meaning the other 2 have to add
to 25. combos that work are 11,14 and 12,13.


neither of those combos work with the 11,12 for the first set we looked at,
because they both have one of the numbers, making the first set the 8 and
15.


16| 7 | 12 | 3
------------
9 | 2 | 13 | 6
----------- you know the numbers to the left of the 3 add to
8 | 15| 4 | 11 25, so those plus 3 is 28, making the other 6.
------------
1 | 10| 5 | ? 14


or


16| 7 | 13 | 3
------------
9 | 2 | 12 | 6
-----------
8 | 15| 5 | 11
------------
1 | 10| 4 | ? 14


or


16| 7 | 12 | 3
------------
9 | 2 | 13 | 6
-----------
15| 8 | 11 | 4
------------
1 | 10| 5 | ? 14


or


16| 7 | 12 | 3
------------
9 | 2 | 13 | 6
-----------
8| 15 | 4 | 11
------------
1 | 10| 5 | ? 14


as you see theres alot of solutions, and im probably missing some. but if you
notice its usually flipping sets of numbers in a row. you cant flip unless the
whole row is empty at the beginning. thats why the 14 cant move. it is in 2
rows with numbers forcing it to stay

that took forever so theres prolly some mistakes or missing parts

[Markus]

Your method is correct, and yes, there are four ways of solving it :)
New problem later today, gtg to school now