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No, not the conspiracy theory. For example, here is an Earth-splat:

And here is an Earth-square:

And both are equal-area projections! How did I do this? Math. Math math math.

We’ll use spherical coordinates, with polar angle $\phi$ and azimuthal angle $\theta$. This means that latitude is, in radians, $\pi/2-\phi$, and longitude is $\theta$. We’ll say that the sphere has a radius of $R_s$. We can have the angle on the flat map be $\theta_f$, the distance from the center is $r_f$, and the radius of the shape at each angle is $R(\theta_f)$.

Since the radius changes, we can’t just say that $\theta_f = \theta$. Instead, use integrals to match the total area, finding a relation between the two:
\begin{aligned} A_{ratio} \int_0^{\theta_f} \int_0^{R(t)} r dr dt &= \int_0^{\theta} \int_0^{R(t)} R_s^2 \sin(\phi) d\phi dt \\ A_{ratio} \int_0^{\theta_f} \frac{1}{2} R^2(t) dt &= 2 \theta R_s^2 \\ \int_0^{\theta_f} \frac{R^2(t) A_{ratio}}{4 R_s^2 } dt &= \theta \end{aligned}

We want the area ratio to be a constant, and we know $\frac{d\theta}{d\theta_f} = \frac{R^2(\theta_f)A_{ratio}}{4 R_s^2 }$ from the integral, and that $\frac{d\theta}{dr} = 0$. This yields

\begin{aligned} A_{ratio} &= \left( \frac{d\phi}{dr} \frac{d\theta}{d\theta_f} - \frac{d\phi}{d\theta_f} \frac{d\theta}{dr} \right) \frac{\sin(\phi) R_s^2}{r} \\ &= \frac{d\phi}{dr} \frac{R^2(\theta_f)A_{ratio} }{4 }\frac{\sin(\phi) }{r} \end{aligned}

We can separate variables and integrate to find $\cos(\phi) = 1-2\frac{r^2}{R^2(\theta_f)}$.

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