Elsa's biggest tile

Logic Level 3

Tiles = { 0 , 1 , 2 , 3 , 4 , 5 }

Dora : The sum of squares of my tiles is half the sum of squares of Elsa's tiles.

Elsa : We all have exactly two different tiles.

Farrah : The sum of squares of one of Dora's and one of Elsa's tiles is as big as the sum of my tiles.

What is Elsa's biggest tile?

The answer is 4.

2 solutions

A Former Brilliant Member
Feb 7, 2021

from itertools import permutations
# Dora tiles: d1, d2
# Elsa tiles: e1, e2
# Farrah tiles: f1, f2
for d1, d2, e1, e2, f1, f2 in permutations('012345',6):
  if int(d1)**2+int(d2)**2 == 1/2*(int(e1)**2+int(e2)**2) and int(d1)**2+int(e1)**2 == int(f1)+int(f2):
    print(int(e1), int(e2))

1
2
3

output:
2 4
2 4

Saya Suka
Jan 18, 2021

Let Dora have tiles a & b, Elsa's c & d and Farrah's e & f. Quite obviously from Dora's statement simplified to a² + b² = (c² + d²) / 2, c and d must be of the same parity to enable a division by 2. If c and d are both odds, then their sum of squares can only have a factor of two (with odd squares being in the form of 4k(k+1)+1), so after the division a² + b² must be an odd sum with a and b having different parity. On the other hand, if c and d are both evens, then their sum of squares must always have at least two factor of two (with even squares being in the form of 4k²), so after the division a² + b² must be an even sum with a and b having the same parity, which can only be 2 odds considering the flow of events (by the conditioning of c, d both being even earlier). After which, we have the first possibility of Dora's even and odd, Elsa's 2 odds and Farrah's 2 evens, or the second one with Dora's 2 odds, Elsa's 2 evens and Farrah's even and odd.
Now, let's look at Farrah's statement that may be translated into a² + c² = e + f. Considering Elsa's statement, then
0 + 0 < e + f < 5 + 5
0 < a² + c² < 10
0 ≤ a, c ≤ 3
==> Farrah's sum can only be 1, 4, 5 or 9, but 1 is made up by two mirroring squares (by 0² = 0 and 1² = 1), so that's impossible.
It's clear that if e + f = 4 (Farrah's even sum), then this must be from the first possibility where a and c must both be odds (since the two must have the same parity), which is impossible since the earlier argument of the sum of two odd squares can only have at most one factor of two, while 4 = 2². Knowing now that it must be the second possibility with Farrah's sum being 5 = 1² + 2² or 9 = 0² + 3², then (e, f) = { (0, 5) , (4, 5) }. With f = 5 is confirmed, then it's easily seen that Dora's two odds must be 1 and 3, and with Elsa's sum of squares being double that of Dora's, she would need at least one of her tiles to be bigger than Dora's biggest tile which is 3.
Answer : Elsa's biggest tile = 4.

To be fair, once you threw the first possibility out of the window, the answer of 4 is already becoming too obvious since Elsa can't have the two smallest evens as her tiles, or else neither of the two may be bigger than Dora's largest to buffer up the double sum of squares involved.

1² + 3² = (2² + 4²) / 2 = 10 ✅
0 + 5 = 5 = 1² + 2² ✅

4 + 5 = 9 = 0² + 3² ✅
1² + 3² = 10 ≠ 2 = (0² + 2²) / 2 ❎

1 + 3 = 4 = 0² + 2² ✅
0² + 5² = 25 ≠ 10 = (2² + 4²) / 2 ❎
2² + 5² = 29 ≠ 8 = (0² + 4²) / 2 ❎

0 + 1 = 1 = 0² + 1² ❎

Elsa's biggest tile

The answer is 4.

2 solutions

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