Math Contest #12 Results and Solution

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Solution:

The problem of the last contest was:
a
It seems that this problem was/seemed so hard that we only had one answer, which was founded on a guess.
So I'll show you here here how to get a result without guessing.
The equation can be seperated into 2 equations:
1: 1/x + 1/y = xy, 2:1/(x-y) = xy
1 → 3:x + y = x²y²
2 → 4:1 = xy*(x-y) → 5:xy = x²y²*(x-y)
3 in 5: xy = (x+y)*(x-y) → xy = x²-y² → x²-xy-y² = 0
→ x = y/2 ± √(y²/4 + y²) → x = y/2 ± √5 * y/2
x in 4:(y/2 ± √5 * y/2)*y*(y/2 ± √5 * y/2 - y) = 1
→ 6:y*(y/2 ± √5 * y/2)*(-y/2 ± √5 * y/2) = 1
(y/2 ± √5 * y/2)*(-y/2 ± √5 * y/2) is equivalent to the third binomial formula and can therefor be simplified to:
-y²/4 + (±√5 * y/2)² = -y²/4 + 5y²/4 = y²
6 → y³ = 1 → y = 1
↓
2 real solutions:
(½ + √5 /2, 1), (½ - √5 /2, 1)

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What is your chance of winning:

p(You Win) = 1/n, n = number of entries

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List of participants with their entries:

Namesolutions foundcomment
tonimontana@tonimontanaboth real solutionsfirst and last

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Winner draw:

Not needed.

That makes 1 SBI and 20 STEM for tonimontana@tonimontana.
I decided to remove the confirmation image. If you want to confirm the transaction you can always look at my wallet.
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The next contest starts tomorrow. Don't miss it!

Math Contest #12 Results and Solution | Ecency