Shen A.H.

A regular triangle $ABC$ cut out of plywood was placed on the floor. Three nails were hammered into the floor (one next to each side of the triangle) in such a way that the triangle cannot be rotated without lifting it off the floor. The first nail divides side $AB$ in the ratio $1:3$, counting from vertex $A$, the second nail divides side $BC$ in the ratio $2:1$, counting from vertex $B$. In what ratio does the third nail divide side $AC$?
$1^{0}$. Let's first solve the "one nail problem". Suppose one nail is hammered in, touching the triangle at point $M$ on side $A C$. We fix the center of the supposed rotation (point $O$). Can the triangle be rotated around point $O$ by a small angle, and if so, in which direction?

Suppose the triangle is positioned as shown in the figure. Draw a perpendicular to line $A C$ at point $M$. This perpendicular will divide the plane into two half-planes. If point $O$ 

![](https://cdn.mathpix.com/cropped/2024_05_06_7099908082d8b9f44e68g-19.jpg?height=406&width=414&top_left_y=1652&top_left_x=-1)

lies in the same half-plane as $C$, then the triangle can be rotated around point $O$ counterclockwise. If point $O$ lies in the other half-plane, then the triangle can be rotated clockwise.

If point $O$ lies on the perpendicular, then if points $O$ and $B$ lie on the same side of line $A C$ (which is always true if point $O$ lies inside the triangle), the triangle cannot be rotated in any direction. If points $O$ and $B$ lie on opposite sides of line $A C$, the triangle can be rotated in either direction.

$2^{\circ}$. Let's return to the problem of three nails. Draw three corresponding perpendiculars. We will prove that if they do not intersect at one point, then the triangle can be rotated. Indeed, the perpendiculars divide the plane into seven regions. We can associate each region with a string of three symbols, either + or -. The first plus indicates that the nail on side $A B$ does not prevent the triangle from rotating around points in this region counterclockwise, and so on (see figure). Different regions cannot correspond to the same string. There are eight possible strings.

Therefore, there will definitely be either the string + + + or the string - - -. In the first case, the triangle can be rotated counterclockwise, and in the second case, clockwise.

$3^{\circ}$. Suppose the perpendiculars intersect at one point. Then, if the point does not lie on any of the perpendiculars, the corresponding string will have at least one + (prohibiting rotation clockwise) and at least one - (prohibiting rotation counterclockwise).

If the point lies on the perpendicular to one of the nails, and inside the triangle, then, as shown above,

![](https://cdn.mathpix.com/cropped/2024_05_06_7099908082d8b9f44e68g-20.jpg?height=412&width=408&top_left_y=797&top_left_x=-1)

this nail prevents the triangle from rotating in any direction. Finally, if the point lies on the perpendicular outside the triangle, then, as can be easily seen, the two remaining nails prevent the triangle from rotating.

$4^{0}$. Thus, it remains to determine where to hammer the third nail so that the perpendiculars to the nails intersect at one point. The points where the nails are hammered on sides $A B, B C$, and $A C$ are denoted by $D, E$, and $F$ (see figure). Suppose the perpendicular to $A B$ at point $D$ intersects $A C$ at point $K$, and the perpendicular to $B C$ at point $E$ intersects $A C$ at point $H$, and these perpendiculars intersect at point $R$. Then $A K = \frac{A D}{\cos 60^{\circ}} = \frac{A C}{2}$ and $C H = \frac{C E}{\cos 80^{\circ}} = \frac{2}{3} A C$.

Triangle $H R K$ is isosceles because angles $H K R$ and $K H R$ are both $30^{\circ}$. From this, we get that $H F = F K$ and $F$ divides side $A C$ in the ratio $5: 7$, counting from vertex $A$.

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