The Pythagoras Patisserie sells triangular cakes at 39p and square buns at 23p each

Let's assume Helen bought x triangular cakes and y square buns.
The cost of x triangular cakes is 39x pence.
The cost of y square buns is 23y pence.
According to the problem, the total cost of the cakes and buns is 512 pence.
So, 39x + 23y = 512. ...(1)
We also know that £1 = 100p.
Therefore, £5.12 = 5 * £1 + 12p = 5 * 100p + 12p = 512 pence.
So, the equation (1) can be written as 39x + 23y = £5.12 * 100 = 512 pence.
Now, we need to find the values of x and y that satisfy this equation.
To solve this equation, we can use the method of trial and error or substitution.
Let's use the method of trial and error.
Let's assume x = 0 and find the value of y.
When x = 0, the equation becomes 23y = 512.
Dividing both sides by 23, we get y = 512/23 = 22.26 (approximately).
Since y cannot be a fraction, x cannot be 0.
Let's assume x = 1 and find the value of y.
When x = 1, the equation becomes 39 + 23y = 512.
Subtracting 39 from both sides, we get 23y = 512 - 39 = 473.
Dividing both sides by 23, we get y = 473/23 = 20.57 (approximately).
Since y cannot be a fraction, x cannot be 1.
Let's assume x = 2 and find the value of y.
When x = 2, the equation becomes 2 * 39 + 23y = 512.
Simplifying the equation, we get 78 + 23y = 512.
Subtracting 78 from both sides, we get 23y = 512 - 78 = 434.
Dividing both sides by 23, we get y = 434/23 = 18.87 (approximately).
Since y cannot be a fraction, x cannot be 2.
Let's assume x = 3 and find the value of y.
When x = 3, the equation becomes 3 * 39 + 23y = 512.
Simplifying the equation, we get 117 + 23y = 512.
Subtracting 117 from both sides, we get 23y = 512 - 117 = 395.
Dividing both sides by 23, we get y = 395/23 = 17.17 (approximately).
Since y cannot be a fraction, x cannot be 3.
Let's assume x = 4 and find the value of y.
When x = 4, the equation becomes 4 * 39 + 23y = 512.
Simplifying the equation, we get 156 + 23y = 512.
Subtracting 156 from both sides, we get 23y = 512 - 156 = 356.
Dividing both sides by 23, we get y = 356/23 = 15.48 (approximately).
Since y cannot be a fraction, x cannot be 4.
Let's assume x = 5 and find the value of y.
When x = 5, the equation becomes 5 * 39 + 23y = 512.
Simplifying the equation, we get 195 + 23y = 512.
Subtracting 195 from both sides, we get 23y = 512 - 195 = 317.
Dividing both sides by 23, we get y = 317/23 = 13.78 (approximately).
Since y cannot be a fraction, x cannot be 5.
Let's assume x = 6 and find the value of y.
When x = 6, the equation becomes 6 * 39 + 23y = 512.
Simplifying the equation, we get 234 + 23y = 512.
Subtracting 234 from both sides, we get 23y = 512 - 234 = 278.
Dividing both sides by 23, we get y = 278/23 = 12.09 (approximately).
Since y cannot be a fraction, x cannot be 6.
Let's assume x = 7 and find the value of y.
When x = 7, the equation becomes 7 * 39 + 23y = 512.
Simplifying the equation, we get 273 + 23y = 512.
Subtracting 273 from both sides, we get 23y = 512 - 273 = 239.
Dividing both sides by 23, we get y = 239/23 = 10.39 (approximately).
Since y cannot be a fraction, x cannot be 7.
Let's assume x = 8 and find the value of y.
When x = 8, the equation becomes 8 * 39 + 23y = 512.
Simplifying the equation, we get 312 + 23y = 512.
Subtracting 312 from both sides, we get 23y = 512 - 312 = 200.
Dividing both sides by 23, we get y = 200/23 = 8.70 (approximately).
Since y cannot be a fraction, x cannot be 8.
Let's assume x = 9 and find the value of y.
When x = 9, the equation becomes 9 * 39 + 23y = 512.
Simplifying the equation, we get 351 + 23y = 512.
Subtracting 351 from both sides, we get 23y = 512 - 351 = 161.
Dividing both sides by 23, we get y = 161/23 = 7 (approximately).
Since y is an integer, x = 9 is a valid solution.
Therefore, Helen bought 9 triangular cakes and 7 square buns. Answer: \boxed{9, 7}.