$3+$ [ Trigonometric ratios in a right triangle ]

Two vertices of a square are located on the base of an isosceles triangle, and the other two are on its lateral sides. Find the side of the square if the base of the triangle is $a$, and the angle at the base is $30^{\circ}$.
Let the vertices $M$ and $N$ of the square $M N K L$ be located on the lateral sides $A B$ and $B C$ of the isosceles triangle $A B C$, and the vertices $K$ and $L$ on the base $A C=a$. Denote the side of the square by $x$.

From the right triangles $A M L$ and $C N K$, we find that

$$
A L=K C=K N \cot 30^{\circ}=x \sqrt{3} .
$$

Since $A C=A L+L K+K C$, we have the equation

$$
a=x \sqrt{3}+x+x \sqrt{3},
$$

from which we find that

$$
x=\frac{a}{2 \sqrt{3}+1}=\frac{a(2 \sqrt{3}-1)}{11}
$$

Let the vertices $M$ and $N$ of the square $M N K L$ be located on the lateral sides $A B$ and $B C$ of the isosceles triangle $A B C$, and the vertices $K$ and $L$ on the base $A C=a$. Denote the side of the square by $x$.

From the right triangles $A M L$ and $C N K$, we find that

$$
A L=K C=K N \cot 30^{\circ}=x \sqrt{3} .
$$

Since $A C=A L+L K+K C$, we have the equation

$$
a=x \sqrt{3}+x+x \sqrt{3},
$$

from which we find that

$$
x=\frac{a}{2 \sqrt{3}+1}=\frac{a(2 \sqrt{3}-1)}{11}
$$

Let the vertices $M$ and $N$ of the square $M N K L$ be located on the lateral sides $A B$ and $B C$ of the isosceles triangle $A B C$, and the vertices $K$ and $L$ on the base $A C=a$. Denote the side of the square by $x$.

From the right triangles $A M L$ and $C N K$, we find that

$$
A L=K C=K N \cot 30^{\circ}=x \sqrt{3} .
$$

Since $A C=A L+L K+K C$, we have the equation

$$
a=x \sqrt{3}+x+x \sqrt{3},
$$

from which we find that

$$
x=\frac{a}{2 \sqrt{3}+1}=\frac{a(2 \sqrt{3}-1)}{11}
$$

## Answer

$\frac{a}{2 \sqrt{3}+1}=\frac{a(2 \sqrt{3}-1)}{11}$.