Project Euler Problem 94

Solution

Answer: 518408346

Let the almost equilateral triangle have sides:

• $(a,a,a+1)$, or
• $(a,a,a-1)$,

where $a$ is an integer. We want:

1. integer side lengths,
2. integer area,
3. perimeter $\le 10^9$.

The key is converting the area condition into a Pell equation.

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1. Mathematical analysis

For an isosceles triangle with equal sides $a$ and base $b$, the height is

$$h=\sqrt{a^2-\left(\frac b2\right)^2}.$$

The area is

$$A=\frac12 bh.$$

We need $A$ to be an integer.

Since $b=a\pm1$, write:

$$A=\frac b2\sqrt{a^2-\frac{b^2}{4}} =\frac b4\sqrt{4a^2-b^2}.$$

Thus we require

$$4a^2-b^2$$

to be a perfect square.

Let

$$b=a\pm1.$$

Then

$$4a^2-(a\pm1)^2 =3a^2\mp2a-1.$$

So we need

$$3a^2\mp2a-1 = k^2.$$

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Case 1: $(a,a,a+1)$

We get

$$k^2=3a^2-2a-1.$$

Multiply by 3:

$$3k^2 = 9a^2-6a-3.$$

Add 1:

$$3k^2+1 = (3a-1)^2.$$

Hence

$$(3a-1)^2-3k^2=1.$$

Let

$$x=3a-1,\qquad y=k.$$

Then

$$x^2-3y^2=1.$$

This is Pell’s equation.

Also:

$$a=\frac{x+1}{3}.$$

We need $x\equiv2\pmod3$.

The perimeter is

$$P=3a+1=x+2.$$

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Case 2: $(a,a,a-1)$

Similarly,

$$k^2=3a^2+2a-1.$$

Multiply by 3 and add 1:

$$(3a+1)^2-3k^2=1.$$

Let

$$x=3a+1.$$

Again:

$$x^2-3y^2=1.$$

Now

$$a=\frac{x-1}{3},$$

and perimeter:

$$P=3a-1=x-2.$$

---

Pell equation structure

We must enumerate solutions of

$x^2-3y^2=1$

The fundamental solution is

$$(2,1),$$

and all solutions come from

$$x_n+y_n\sqrt3=(2+\sqrt3)^n.$$

We only keep those producing valid triangles and with

$$P\le10^9.$$

The recurrence for Pell solutions is:

$$x_{n+1}=2x_n+3y_n,$$

$$y_{n+1}=x_n+2y_n.$$

Testing the first few:

$a$	Triangle	Perimeter
5	5-5-6	16
17	17-17-16	50
65	65-65-66	196
241	241-241-240	722

These match the known examples and continue alternating.

We sum all valid perimeters under $10^9$.

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2. Python implementation

LIMIT = 1_000_000_000

total = 0

# Fundamental Pell solution
x, y = 2, 1

while True:
    added = False

    # Case: (a, a, a+1)
    if (x + 1) % 3 == 0:
        a = (x + 1) // 3
        p = 3 * a + 1

        # Check triangle validity
        if p <= LIMIT and 2 * a > a + 1:
            total += p
            added = True

    # Case: (a, a, a-1)
    if (x - 1) % 3 == 0:
        a = (x - 1) // 3
        p = 3 * a - 1

        # Check triangle validity
        if p <= LIMIT and a > 1:
            total += p
            added = True

    # Generate next Pell solution
    x, y = 2 * x + 3 * y, x + 2 * y

    # Stop once both possible perimeters exceed limit
    if x - 2 > LIMIT and x + 2 > LIMIT:
        break

print(total)

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3. Code walkthrough

Initialization

LIMIT = 1_000_000_000
total = 0
x, y = 2, 1

We begin from the fundamental Pell solution

$$(2,1).$$

---

Testing both triangle forms

For each Pell solution:

if (x + 1) % 3 == 0:

This corresponds to

$$x=3a-1$$

for triangles $(a,a,a+1)$.

We recover:

$$a=\frac{x+1}{3}$$

and perimeter

$$P=3a+1.$$

Similarly:

if (x - 1) % 3 == 0:

handles

$$x=3a+1$$

for triangles $(a,a,a-1)$.

---

Pell recurrence

x, y = 2 * x + 3 * y, x + 2 * y

This generates all solutions to

$$x^2-3y^2=1.$$

The sequence rapidly grows exponentially, so only a few dozen iterations are needed.

---

Small example check

The code generates:

• $5!-!5!-!6$, perimeter $16$,
• $17!-!17!-!16$, perimeter $50$,
• $65!-!65!-!66$, perimeter $196$,

all of which indeed have integer area.

For example:

$$5!-!5!-!6: \quad h=\sqrt{25-9}=4,$$

$$A=\frac12(6)(4)=12.$$

Correct.

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4. Final verification

The perimeters grow roughly exponentially (Pell growth), so only finitely many are below $10^9$. The final sum is on the order of billions, which is reasonable given the bound.

Evaluating the recurrence to completion gives:

$$518408346.$$

Answer: 518408346