In an equilateral triangle, all interior angles are equal. Since the sum of the angles in any triangle is 180^°, each interior angle measures (180^°)/(3) = 60^°. We conclude that ABC is an equilateral triangle (and equiangular).

In an equilateral triangle, all interior angles are equal. Since the sum of the angles in any triangle is 180^°, each interior angle measures (180^°)/(3) = 60^°. We conclude that ABC is an equilateral triangle (and equiangular).

Three circles are constructed such that each circle passes through the centres of the other two. Let the centres be A, B and C. Join A, B and C to form a triangle. Using GeoGebra, measure the three sides and three angles of triangle ABC. What can you conclude about the triangle?

Answer: Triangle $ABC$ is an equilateral triangle with side lengths $AB = BC = CA = r$ and angle measures $\angle A = \angle B = \angle C = 60^\circ$ .

Step-by-step solution

Step 1: Analyze the condition of mutual intersection through centres

Let the three circles have centres $A$ , $B$ , and $C$ with radii $r_A$ , $r_B$ , and $r_C$ , respectively. Since the circle with centre $A$ passes through $B$ and $C$ , the distance from $A$ to $B$ and from $A$ to $C$ must equal its radius, giving $AB = AC = r_A$ . Similarly, because the circle with centre $B$ passes through $A$ , the distance $AB = r_B$ , which implies $r_A = r_B$ . Repeating this for all pairs shows that all three circles have the identical radius $r$ .

Step 2: Determine the lengths of the sides of triangle ABC

Since all three circles have the same radius $r$ , the lengths of the segments connecting the three centres are equal: $AB = r$ , $BC = r$ , and $CA = r$ . Therefore, all three sides of $\triangle ABC$ have equal length.

Step 3: Determine the measures of the angles and conclude

In an equilateral triangle, all interior angles are equal. Since the sum of the angles in any triangle is $180^\circ$ , each interior angle measures $\frac{180^\circ}{3} = 60^\circ$ . We conclude that $\triangle ABC$ is an equilateral triangle (and equiangular).

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