1. A, B and C are three collinear points. Can you find a point P such that PA = PB = PC ? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?
2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?
- 1. A, B and C are three collinear points. Can you find a point P such that PA = PB = PC ? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?
- 2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?
Step-by-step solution
Idea: A point equidistant from A, B and C would have to lie on both perpendicular bisectors. Two lines perpendicular to the same line are parallel. For part 2, turning or reflecting a circle about its centre keeps the circle in place, so it carries the triangle to a congruent triangle on the same circle.
1. A, B and C are three collinear points. Can you find a point P such that PA = PB = PC ? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?
- PA = PB means P is on the perpendicular bisector of AB; PB = PC means P is on the perpendicular bisector of BC. So P would have to lie on both bisectors.½ mark
- The bisectors are parallel: both are perpendicular to the line l through A, B, C. Two lines perpendicular to the same line make equal corresponding angles (90°) with it, so they are parallel. They are different lines, as they cross l at the two different midpoints. (Drawing them confirms this: they never meet.)1 mark
- Distinct parallel lines have no common point, so no point P with PA = PB = PC exists.½ mark
- A circle through A, B, C would have a centre P with PA = PB = PC. Since no such P exists, no circle passes through three collinear points.½ mark
- If a line cut a circle in three distinct points, those three collinear points would lie on a circle, which is impossible. So a line meets a circle in at most two points.½ mark
2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?
- Turn ΔABC about the centre O through any angle, say 70°, to get ΔA′B′C′. A turn about O keeps every distance, so ΔA′B′C′ ≅ ΔABC; and OA′ = OA, OB′ = OB, OC′ = OC = radius, so A′, B′, C′ lie on the same circle.1 mark
- Every angle of turning gives such a triangle (and so does reflecting ΔABC in any diameter). So yes, there are infinitely many triangles congruent to ΔABC with the same circumcircle.1 mark
Answer to write in the exam
1.
PA = PB ⇒ P on the perpendicular bisector of AB; PB = PC ⇒ P on the perpendicular bisector of BC
A, B, C on line l: both bisectors ⟂ l ⇒ the bisectors are parallel (and distinct, as the midpoints differ)
⇒ The bisectors have no common point ⇒ no point P with PA = PB = PC
∴ No circle passes through three collinear points.
∴ No line can cut a circle in three distinct points (at most two).
2.
Rotate ΔABC about the centre O through any angle θ to get ΔA′B′C′.
Rotation keeps lengths ⇒ ΔA′B′C′ ≅ ΔABC
OA′ = OB′ = OC′ = OA (radius) ⇒ A′, B′, C′ lie on the same circle
∴ Yes; infinitely many congruent triangles (one for each θ) share the circumcircle.
Common mistakes that cost marks
- Drawing the bisectors carelessly so that they appear to meet far away, and then marking that point as P. Two lines perpendicular to the same line are exactly parallel.
- Saying a line can touch a circle at three points if the circle is “big enough”. However big, a circle and a line share at most two points.
- In part 2, answering “no, the circumcircle is unique”. The circle of a triangle is unique, but one circle has many triangles.
How this can come in the exam
The greatest number of points in which a straight line can meet a circle is
- 1
- 2
- 3
- infinitely many
Show answer
(B) 2
Three common points would be three collinear points on a circle, which is impossible. So at most 2.
Assertion (A): Two triangles that have the same circumcircle must be congruent.
Reason (R): A triangle has exactly one circumcircle.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true, but R is false.
- A is false, but R is true.
Show answer
(D) A is false, but R is true.
R is true (three non-collinear points determine one circle). A is false: an equilateral triangle and a right-angled triangle can share the same circle without being congruent.
Try one yourself
A circle has radius 5 cm. How many equilateral triangles can be inscribed in it, and what is the side of each?
Show answer
Infinitely many (turn one about the centre), all congruent, each of side 5√3 ≈ 8.66 cm.
More questions like this
- Equal chords of a circle subtend equal angles at the centre of the circle.
- Chords of a circle that subtend equal angles at the centre are equal.
- Show that the triangle formed by a chord and the centre of the circle is isosceles.
- Show that if two such isosceles triangles (formed by a chord and the centre of the circle) have equal base length, they are congruent to each other.
- The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.