Canadian Mathematical Olympiad

Solution:
Let the end-points of the segment be $A(a, a)$ on $y = x$ and $B(b, 2b)$ on $y = 2x$.

The length of $AB$ is $4$:
$$AB = \sqrt{(b - a)^2 + (2b - a)^2} = 4$$
So,
$$(b - a)^2 + (2b - a)^2 = 16$$
Expanding:
$$(b - a)^2 + (2b - a)^2 = (b - a)^2 + (4b^2 - 4ab + a^2) = (b - a)^2 + 4b^2 - 4ab + a^2$$
But $(2b - a)^2 = 4b^2 - 4ab + a^2$, so:
$$(b - a)^2 + (2b - a)^2 = (b^2 - 2ab + a^2) + (4b^2 - 4ab + a^2) = 5b^2 - 6ab + 2a^2 = 16$$
Let the midpoint $M$ of $AB$ be $\left(\frac{a + b}{2}, \frac{a + 2b}{2}\right)$.
Let $x = \frac{a + b}{2}$ and $y = \frac{a + 2b}{2}$.

Then:
$$a = 2x - b$$
$$y = \frac{a + 2b}{2} = \frac{(2x - b) + 2b}{2} = \frac{2x + b}{2}$$
So,
$$2y = 2x + b \implies b = 2y - 2x$$
Substitute $b$ into $a$:
$$a = 2x - (2y - 2x) = 2x - 2y + 2x = 4x - 2y$$
Now, substitute $a$ and $b$ in the length equation:
$$5b^2 - 6ab + 2a^2 = 16$$
Plug in $a = 4x - 2y$, $b = 2y - 2x$:

First, compute $b^2$:
$$b^2 = (2y - 2x)^2 = 4(y - x)^2$$
$a^2$:
$$a^2 = (4x - 2y)^2 = 16x^2 - 16x y + 4y^2$$
$ab$:
$$ab = (4x - 2y)(2y - 2x) = 4x(2y - 2x) - 2y(2y - 2x) = 8x y - 8x^2 - 4y^2 + 4x y = (8x y + 4x y) - 8x^2 - 4y^2 = 12x y - 8x^2 - 4y^2$$
But let's expand directly:
$$ab = (4x - 2y)(2y - 2x) = 4x \cdot 2y - 4x \cdot 2x - 2y \cdot 2y + 2y \cdot 2x = 8x y - 8x^2 - 4y^2 + 4x y = (8x y + 4x y) - 8x^2 - 4y^2 = 12x y - 8x^2 - 4y^2$$
Now, plug into the equation:
$$5b^2 - 6ab + 2a^2 = 5 \cdot 4(y - x)^2 - 6(12x y - 8x^2 - 4y^2) + 2(16x^2 - 16x y + 4y^2)$$
Compute each term:
$5b^2 = 20(y - x)^2$
$-6ab = -6(12x y - 8x^2 - 4y^2) = -72x y + 48x^2 + 24y^2$
$2a^2 = 32x^2 - 32x y + 8y^2$

Sum:
$$20(y - x)^2 + 48x^2 + 24y^2 - 72x y + 32x^2 - 32x y + 8y^2 = 16$$
But $20(y - x)^2 = 20(y^2 - 2x y + x^2) = 20y^2 - 40x y + 20x^2$
So sum all $x^2$ terms:
$20x^2 + 48x^2 + 32x^2 = 100x^2$
$y^2$ terms: $20y^2 + 24y^2 + 8y^2 = 52y^2$
$-x y$ terms: $-40x y - 72x y - 32x y = -144x y$

So the equation is:
$$100x^2 + 52y^2 - 144x y = 16$$
Or, dividing both sides by $4$:
$$25x^2 + 13y^2 - 36x y = 4$$
Thus, the equation of the locus is:
$$25x^2 + 13y^2 - 36x y = 4$$