Placing Numbers in a Square 1

The sum of all numbers from 1 to 12 is $\frac{12\times13}{2} = 78$ , however, this does not take account that the corners were count twice in summing the sum of each side. Given the sum for all sides (when the corner was counted twice) is $25\times4 = 100$ , the corner shall have the sum of $100-78 = 22$

$1+2+3 ... + 12 = \frac{12\times 13}{2}=78$ , where, each number counted once. If we add the numbers along sides, e.g. consider this square as ABCD, now if we add the numbers along side AB and side DA, then the number on corner A is counted twice. Similarly if we add up the numbers along 4 sides it will include the 4 corner numbers twice and the sum becomes as $25\times 4 = 100$ . Hence the sum of numbers in 4 corners $=100-78=22$ .

I brute-forced it with:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

function addup(o)
{
    if(o[0]+o[1]+o[2]+o[3]!=25)return false;
    if(o[8]+o[9]+o[10]+o[11]!=25)return false;
    if(o[0]+o[4]+o[6]+o[8]!=25)return false;
    if(o[3]+o[5]+o[7]+o[11]!=25)return false;
    return true;
}


var o=[1,2,3,4,5,6,7,8,9,10,11,12];
var foundIt=false;
for(var i=0;i<100000;i++)
{
    o.sort(function(){return [-1,1][Math.floor(Math.random()*2)];})
    if(addup(o)){foundIt=true;break;}
}
if(foundIt)
{
    alert('found it!\n'+o.join(','));
}else{
    alert('no luck... try refreshing page');
}

easy. 1+4+7+10=22. hahaha

I came up with arrangement 12 8 4 1 on top, 12 2 5 6 going down left side, 1 10 11 3 going down right side and 6 9 7 3 on bottom.