1993_Question_1

1993_Question_1

Solution:
First let’s write down some equations that we get by using the fact that the angles in a triangle add up to a constant amount (at that the angles that make up a line also adds up to that amount).

a+g+h=f+e+d=b+i+j=k+l+c
=g+f+i=h+e+l=j+d+k=a+b+c.

Now assume that all the angles (12) in the diagram add up to some number, x.
Then the lowest value of x= 1+2+3+...+12=78

Does this value work? Turns out it doesn’t.
That’s because in the long strand of equations we wrote, each angle appears twice. And setting a+b+c=y we get that 8y=2x \longrightarrow 4y=x
So thus x has to be a multiple of 4 and the lowest possible value is 80.

From here we have two possible sets of integers:
{{1, 2, 3,... , 10, 11, 14}} or {1, 2, 3, ... , 9, 10, 12, 13}

I wish there was an easy way from here but I just guess and checked to get:
a = 1, b = 6, c = 13, d = 8, e = 3, f = 9
g = 7, h = 12, i = 4, j = 10, k = 2, l = 5.

Thus the lowest possible value of a+b+c=20.

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