The Louvre

Suppose that the length $GH = x$ , and that $\angle ACF = \theta$ . Applying the Sine Rule to the triangle $BCG$ , we see that $(1 + x)\sin\theta \; = \; \sin(120^\circ - \theta)$ If $X$ is the length of the large equilateral triangle, then $X \; = \; 1 + AC \; = \; 1 + \frac{(2+x)\sin(120^\circ - \theta)}{\sin60^\circ}$ using the Sine Rule on $ACF$ . On the other hand $X \; = \; 2 + AF \; = \; 2 + \frac{(2+x)\sin\theta}{\sin60^\circ}$ Putting these last equations together, $\sin60^\circ \; = \; (2+x)\big[\sin(120^\circ - \theta) - \sin\theta\big] \; = \; (2+x)\sin(60^\circ-\theta)$ and hence we obtain the equation $\begin{aligned} \frac{\sin60^\circ}{\sin(60^\circ - \theta)} & = \; 2+x \; = \; \frac{\sin(120^\circ - \theta)}{\sin\theta} + 1 \; = \; \frac{\sin(60^\circ+\theta)}{\sin\theta} + 1 \\ \sin60^\circ\sin\theta & = \; \sin(60^\circ-\theta)\sin(60^\circ+\theta) + \sin\theta\sin(60^\circ-\theta) \; = \; \tfrac34\cos2\theta + \tfrac14\sqrt{3}\sin2\theta \\ \sin\theta & = \; \sin(2\theta + 60^\circ) \end{aligned}$ and hence we have $2\theta + 60^\circ = 180^\circ - \theta$ , and hence $\theta = 40^\circ$ . From this we can now calculate that $x = \tfrac12\sec20^\circ$ and that $X = 2 + 2\cos20^\circ$ , making the answer $\boxed{20^\circ}$ .