342. The Circumcircle Quest · Finding the center that binds all vertices

Try to break it

Drag A, B, or C. The three perpendicular bisectors always meet at one point O, and OA = OB = OC at every position — so a single circle through all three vertices is forced. Why? Any point on the perpendicular bisector of AB is equidistant from A and B; on the bisector of AC, equidistant from A and C. The unique intersection is equidistant from all three.

How you build it

Construct the circumcircle of a triangle.

• Place point A as the first vertex of the triangle.
• Place point B as the second vertex of the triangle.
• Place point C as the third vertex of the triangle.
• Draw segment AB to form one side of triangle ABC.
• Draw segment BC to form one side of triangle ABC.
• Draw segment CA to complete triangle ABC.
• Construct the perpendicular bisector of side AB.
• Construct the perpendicular bisector of side AC.
• Mark the intersection of the bisectors as point O (Circumcenter).
• Draw the circumcircle with center O and radius OA.

The proof, step by step

Prove that the constructed circle passes through all three vertices of the triangle.

• O lies on the perpendicular bisector of AB. Therefore, OA = OB.
• O lies on the perpendicular bisector of AC. Therefore, OA = OC.
• Since OA = OB and OA = OC, we have OA = OB = OC. Thus, O is equidistant from all vertices.

Worked example

In a triangle ABC, the perpendicular bisectors of the sides meet at O. If OA = 5 cm, what is the length of OB?

The circumcenter O is equidistant from all vertices of the triangle. Since OA = 5 cm, OB must also be 5 cm.

• 3 cm
• 5 cm — correct
• 10 cm
• 2.5 cm