Orthogonal projections of triangle $A B C$ onto two mutually perpendicular planes are equilateral triangles with sides of 1. Find the perimeter of triangle $A B C$, given that $A B=$ $\frac{\sqrt{5}}{2}$

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Let two mutually perpendicular planes $\alpha$ and $\beta$ intersect along a line $l$ (we can assume that point $A$ lies on this line); $B_1$ and $C_1$ are the orthogonal projections of points $B$ and $C$ onto plane $\alpha$, and $B_2$ and $C_2$ are the orthogonal projections of points $B$ and $C$ onto plane $\beta$; $M$ and $N$ are the orthogonal projections of points $B_1$ and $C_1$ respectively onto line $l$. From the right triangle $ABB_2$, we find that

$$
B B_2 = \sqrt{A B^2 - A B_2^2} = \sqrt{\frac{5}{4} - 1} = \frac{1}{2}
$$

Then $B_1 M = B B_2 = \frac{1}{2}$, so in the right triangle $AB_1M$, the leg $B_1 M$ is half the hypotenuse $A B_1$. Therefore,

$$
\angle B_1 A M = 30^\circ, \angle C_1 A N = 180^\circ - \angle B_1 A C_1 - \angle B_1 A M = 180^\circ - 60^\circ - 30^\circ = 90^\circ
$$

Thus, point $N$ coincides with point $A$ and $C C_2 = C_1 A = 1$. From the right triangle $ACC_2$, we find that

$$
A C = \sqrt{A C_2^2 + C C_2^2} = \sqrt{1 + 1} = \sqrt{2}
$$

Consider the right trapezoid $B B_2 C_2 C$. Drop a perpendicular $B K$ from vertex $B$ to the base $C C_2$. Then

$$
B K = B_2 C_2 = 1, C K = C C_2 - K C_2 = C C_2 - B B_2 = \frac{1}{2}
$$

From the right triangle $B C K$, we find that

$$
B C = \sqrt{B K^2 + C K^2} = \sqrt{1 + \frac{1}{4}} = \frac{\sqrt{5}}{2}
$$

Therefore,

$$
A C + A B + B C = \sqrt{2} + \frac{\sqrt{5}}{2} + \frac{\sqrt{5}}{2} = \sqrt{2} + \sqrt{5}
$$

## Answer

$\sqrt{2} + \sqrt{5}$