John drew a triangle such that all the angles, in degrees, were perfect squares. What was the largest angle among them, in degrees?
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Great solution.
It could be cleaned up by presenting it in the following way:
1. Squares leave a remainder of 0 or 1 when divided by 4.
2. Hence, if one of the squares was odd, we cannot find 3 squares which sum to 180.
3. 144 doesn't work, because 180-144=36, and there are no solutions.
4. 4 doesn't work, because
1
8
0
−
4
=
1
7
6
=
1
1
×
1
6
, and Fermat's two square theorem tells us that since
1
1
∣
1
7
6
, hence no solutions exist.
5. Thus, we are only left with 16, 64, 100. Notice that these do add up to 180.
@Sam Reeve That's a great proof! I've converted your comment into a solution :)
Using 3 perfect squares of 100, 64, and 16, which is 1 0 2 , 8 2 , and 4 2 , they will add up to 180.
Prove that this is the only possible solution.
Why 6 4 2 ?
Starting from 169 (no triangle can have a > 180 degree angle), you can use simple proof by exhaustion in order to prove that 100 + 64 + 16 is the only possibility.
169 - No perfect squares add up to 11 (9+1 is the closest possible) 144 - No perfect squares add up to 36, as it is a perfect square itself 121 - No perfect squares add up to 59 (9 + 49 is closest possible) 100 - 64 + 16 works.
Therefore the solution is 100.
Meet In The Middle approach :v
Sum of angles of triangle is 180 degrees a^2+b^2+c^2 = 180 You know a,b,c < 14 because 14^2 = 196
Testing 13^2 ---> 180-13^2 = 11 and no two squares combine to make 11 Testing 12^2 ---> 180-12^2 = 36 and no two squares combine to make 36 Testing 11^2 ---> 180-11^2 = 59 and no two squares combine to make 59 Testing 10^2 ---> 180-10^2 = 80
a^2+b^2 = 80
80 = 4^2+8^2
Therefore, the largest angle of this triangle is 10^2 degrees = 100 degrees
My method: 180 can write into square number times constant in 2 way, they are 9*20 and 4 \times 45 .
I check out only square 1 4 9 16 25 36 .
for 9*20 I can't write 20 with sum of three square numbers.
but in case of 4*45.
I can make 45 by 4+16+25.
so, 180 = 4 45 = 4 4 +4 16 +4 25 =4^2+8^2+10^2.
therefore 10^2 = 100 is the largest angle.
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1 0 0 + 6 4 + 1 6 .
Proof that there is only this solution.
Only possible squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169.
If ( a 2 + b 2 = c 2 ) then ( c = s 2 + t 2 ) Also a square can only be of the form 4 n or 4 n + 1
Each square will leave a remainder from 180 which must be the sum of two squares.
Therefore we cannot have odd numbers
1 : odd
4 : 176 is 16*11, which has no solutions ( 4 a 2 + 4 b 2 = 4 n so 11 would have to be a sum of two sqrs)
9 : odd
16 : 164
25 : odd
36 : 144 is square. But 12 cannot be expressed as sum of two squares (since 3 can't)
49 : odd
64 : 116
81 : odd
100 : 80
121 : odd
144 : 36 is square. But 6 has no solutions
169 : odd
So it could only be 16, 64 and 100. Which do add up to 180