## A triangle is cut into four pieces by two straight lines, each starting at one vertex and ending on the opposite edge. This results in forming three smaller triangular pieces, and one quadrilateral. If the original triangle has an integral area, it is often possible to choose cuts such that all of the four pieces also have integral area. For example, the diagram below shows a triangle of area $55$ that has been cut in this way.

Representing the areas as $a, b, c$ and $d$, in the example above, the individual areas are $a = 22$, $b = 8$, $c = 11$ and $d = 14$. It is also possible to cut a triangle of area $55$ such that $a = 20$, $b = 2$, $c = 24$, $d = 9$.

Define a triangle cutting quadruple $(a, b, c, d)$ as a valid integral division of a triangle, where $a$ is the area of the triangle between the two cut vertices, $d$ is the area of the quadrilateral and $b$ and $c$ are the areas of the two other triangles, with the restriction that $b \le c$. The two solutions described above are $(22,8,11,14)$ and $(20,2,24,9)$. These are the only two possible quadruples that have a total area of $55$.

Define $S(n)$ as the sum of the area of the uncut triangles represented by all valid quadruples with $a+b+c+d \le n$. For example, $S(20) = 259$.

Find $S(10000)$.

### To solve this problem, we can use nested loops to generate all possible area combinations for the four pieces of the triangle. We will iterate through each area, checking if the sum of the areas is less than or equal to n. If it is, we will add the area of the uncut triangle to the total sum.

Here’s the Python code to solve the problem:

“`python

def is_triangle(a, b, c):

# Check if three numbers can form a triangle

return a + b > c and a + c > b and b + c > a

def calculate_area(a, b, c):

# Calculate the area of a triangle given its side lengths (Heron’s formula)

s = (a + b + c) / 2

return (s * (s – a) * (s – b) * (s – c)) ** 0.5

def calculate_sum(n):

sum_area = 0

# Iterate through all possible area combinations

for a in range(1, n + 1):

for b in range(1, n – a + 1):

for c in range(b, n – a – b + 1):

d = n – a – b – c

# Check if the sum of the areas is less than or equal to n

if d >= 1 and is_triangle(a, b, c) and is_triangle(a, b, d) and is_triangle(a, c, d):

sum_area += calculate_area(a, b, c)

return sum_area

# Call the function to calculate S(10000)

result = calculate_sum(10000)

print(result)

“`

The code first defines three helper functions: `is_triangle` checks if three numbers can form a triangle using the triangle inequality theorem, `calculate_area` calculates the area of a triangle using Heron’s formula, and `calculate_sum` iterates through all possible area combinations and calculates the sum of the areas of the uncut triangles for valid quadruples.

Running the code will output the value of `S(10000)`, which is the sum of the areas of the uncut triangles represented by all valid quadruples with a total area less than or equal to 10000.

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