Subfield: Geometry


Question: Points $A_{1}, A_{2}, \ldots, A_{N}$ are equally spaced around the circumference of a circle and $N \geq 3$. Three of these points are selected at random and a triangle is formed using these points as its vertices.

Through this solution, we will use the following facts:

When an acute triangle is inscribed in a circle:

- each of the three angles of the triangle is the angle inscribed in the major arc defined by the side of the triangle by which it is subtended,
- each of the three arcs into which the circle is divided by the vertices of the triangles is less than half of the circumference of the circle, and
- it contains the centre of the circle.

Why are these facts true?

- Consider a chord of a circle which is not a diameter.

Then the angle subtended in the major arc of this circle is an acute angle and the angle subtended in the minor arc is an obtuse angle.

Now consider an acute triangle inscribed in a circle.

Since each angle of the triangle is acute, then each of the three angles is inscribed in the major arc defined by the side of the triangle by which it is subtended.

- It follows that each arc of the circle that is outside the triangle must be a minor arc, thus less than the circumference of the circle.
- Lastly, if the centre was outside the triangle, then we would be able to draw a diameter of the circle with the triangle entirely on one side of the diameter.

<img_3248>

In this case, one of the arcs of the circle cut off by one of the sides of the triangle would have to be a major arc, which cannot happen, because of the above.

Therefore, the centre is contained inside the triangle.
If $N=7$, what is the probability that the triangle is acute? (A triangle is acute if each of its three interior angles is less than $90^{\circ}$.)

Solution:
Since there are $N=7$ points from which the triangle's vertices can be chosen, there are $\left(\begin{array}{l}7 \\ 3\end{array}\right)=35$ triangles in total.

We compute the number of acute triangles.

Fix one of the vertices of such a triangle at $A_{1}$.

We construct the triangle by choosing the other two vertices in ascending subscript order. We choose the vertices by considering the arc length from the previous vertex - each of



these arc lengths must be smaller than half the total circumference of the circle.

Since there are 7 equally spaced points on the circle, we assume the circumference is 7 , so the arc length formed by each side must be at most 3 .

Since the first arc length is at most 3 , the second point can be only $A_{2}, A_{3}$ or $A_{4}$.

If the second point is $A_{2}$, then since the second and third arc lengths are each at most 3 , then the third point must be $A_{5}$. (Since the second arc length is at most 3, then the third point cannot be any further along than $A_{5}$. However, the arc length from $A_{5}$ around to $A_{1}$ is 3 , so it cannot be any closer than $A_{5}$.)

<img_4024>

If the second point is $A_{3}$, the third point must be $A_{5}$ or $A_{6}$. If the second point is $A_{4}$, the third point must be $A_{5}$ or $A_{6}$ or $A_{7}$. Therefore, there are 6 acute triangles which include $A_{1}$ as one of its vertices.

How many acute triangles are there in total?

We can repeat the above process for each of the 6 other points, giving $7 \times 6=42$ acute triangles.

But each triangle is counted three times here, as it has been counted once for each of its vertices.

Thus, there are $\frac{7 \times 6}{3}=14$ acute triangles.

Therefore, the probability that a randomly chosen triangle is acute if $\frac{14}{35}=\frac{2}{5}$.

Final answer: $\frac{2}{5}$