$\angle B A C$ is said to be "laceable" if distinct points $X_{1}, X_{2}, \ldots, X_{2 n}$ can be found so that

- $X_{2 k-1}$ is on $A C$ for each value of $k$,
- $X_{2 k}$ is on $A B$ for each value of $k$, and
- $A X_{1}=X_{1} X_{2}=X_{2} X_{3}=\cdots=X_{2 n-1} X_{2 n}=X_{2 n} A$.

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-125.jpg?height=204&width=388&top_left_y=337&top_left_x=1324)

For example, the angle $20^{\circ}$ is laceable, as shown. The number of laceable acute angles, whose sizes in degrees are integers, is
(A) 3
(B) 4
(D) 6
(E) 7
(C) 5
Suppose that the angle $\theta$ is an acute laceable angle with $2 k$ points in the lacing. We need to determine what values of $\theta$ are possible.

First, we can note that the diagram must be symmetrical, since

$$
A X_{1}=X_{1} X_{2}=X_{2 k-1} X_{2 k}=X_{2 k} A
$$

and so the two triangles $\triangle A X_{1} X_{2}$ and $\triangle A X_{2 k} X_{2 k-1}$ are isosceles with equal base angles and equal legs, and thus congruent, so $A X_{2}=A X_{2 k-1}$.

Continuing this, we can show that each pair of corresponding points on the rays $A B$ and $A C$ are the same distance from $A$.

In particular, $A X_{k}=A X_{k+1}$. Thus $\triangle A X_{k} X_{k+1}$ is isosceles and so $\angle A X_{k} X_{k+1}=\angle A X_{k+1} X_{k}=\frac{1}{2}\left(180^{\circ}-\theta\right)$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-228.jpg?height=273&width=513&top_left_y=243&top_left_x=1359)

Next, we will develop a second expression for one of these two angles involving $\theta$. To do this, we need the fact that if we know two angles of a triangle, then we can calculate the "external angle" of the triangle, ie. in the diagram, $\angle P R S=x+y$, since $\angle P R S=180^{\circ}-\angle P R Q=180^{\circ}-\left(180^{\circ}-x-y\right)=x+y$.

Since $\triangle A X_{1} X_{2}$ is isosceles, then $\angle X_{1} A X_{2}=\angle A X_{2} X_{1}=\theta$ and so by the external angle, $\angle X_{2} X_{1} C=2 \theta$.

Since $\Delta X_{1} X_{2} X_{3}$ is isosceles, then $\angle X_{2} X_{1} X_{3}=\angle X_{2} X_{3} X_{1}=2 \theta$, and so by the external angle in $\triangle A X_{2} X_{3}, \angle X_{3} X_{2} C=3 \theta$.

Continuing in this way, $\angle X_{3} X_{2} X_{4}=\angle X_{3} X_{4} X_{2}=3 \theta$, and so on, eventually reaching

$$
\angle X_{k} X_{k-1} X_{k+1}=\angle X_{k} X_{k+1} X_{k-1}=k \theta
$$

(Try this out in the diagram given in the problem.) Therefore, comparing our two equations for angle $\angle A X_{k+1} X_{k}=\angle X_{k-1} X_{k+1} X_{k}$, we obtain

$$
\begin{aligned}
\frac{1}{2}\left(180^{\circ}-\theta\right) & =k \theta \\
180^{\circ} & =2 k \theta+\theta \\
\theta & =\frac{180^{\circ}}{2 k+1}
\end{aligned}
$$

Since $\theta$ is an integer, $2 k+1$ must be a divisor (an odd divisor that is at least 3 ) of 180 . The divisors of 180 are $1,2,3,4,5,6,9,10,12,15,18,20,30,36,45,60,90$, and 180, with odd divisors $1,3,5,9,15$, and 45 .

Therefore, ignoring the 1 , there are 5 possibilities for $\theta$ to be a laceable acute angle, namely $60^{\circ}, 36^{\circ}, 20^{\circ}, 12^{\circ}$, and $4^{\circ}$.