Study Notes: Constructions

Overview

Constructions is a hands-on geometry topic that tests your ability to use only a compass and straight-edge (unmarked ruler) to draw precise geometric figures. In SOF IMO, you won't physically draw constructions during the exam, but you must understand the **logical steps**, the **order of operations**, and the **geometric principles** behind each construction. Questions typically ask you to identify the correct sequence of steps, recognize which construction method applies, or determine properties of constructed figures.

This topic bridges pure geometry with practical problem-solving. Mastering constructions strengthens your understanding of angle bisectors, perpendicular bisectors, triangle properties, and tangent-circle relationships—concepts that appear across multiple IMO sections. Focus on memorizing the step sequences and understanding *why* each step works, not just mechanical reproduction.

The most common constructions tested are: bisecting angles and line segments, constructing perpendiculars, drawing triangles from given data (SSS, SAS, ASA, RHS criteria), and drawing tangents to circles from external points. Expect 2–4 questions in the exam, often in the Achievers Section as multi-step reasoning problems.

Key Concepts

• **Compass and straight-edge only**: All constructions must be done without measuring angles or lengths. The compass transfers distances; the straight-edge connects points. No protractor or marked ruler allowed.

• **Arc intersections create key points**: Most constructions rely on drawing arcs from two different centers. Where arcs intersect defines new points with specific geometric properties (equidistant from centers, for example).

• **Angle bisector divides an angle into two equal parts**: Constructed by drawing equal arcs from the angle vertex, then arcs from where those arcs cut the arms, creating an intersection that lies on the bisector.

• **Perpendicular bisector of a segment**: Every point on this line is equidistant from the segment's endpoints. Constructed by arcs of equal radius from both endpoints; their intersections determine the perpendicular bisector.

• **Triangle construction uses congruence criteria**: SSS (three sides), SAS (two sides and included angle), ASA (two angles and included side), RHS (right angle, hypotenuse, one side). Each criterion has a specific construction sequence.

• **Tangent from external point**: A tangent is perpendicular to the radius at the point of contact. Constructing tangents requires finding the perpendicular from the external point to the radius, often using a semicircle on the line joining the point to the center.

• **60° and 30° angles**: Easily constructed using equilateral triangle properties (compass set to radius = side length) and bisecting 60° to get 30°.

• **90° angle (perpendicular)**: Constructed by replicating the perpendicular bisector method at a point on a line or from a point to a line.

Formulas / Key Facts

• **Angle bisector construction**: Arcs from vertex with same radius → arcs from intersection points on arms with same radius → line through vertex and final intersection is the bisector.

• **Perpendicular bisector of AB**: Arcs centered at A and B, radius > ½AB → two intersection points above and below AB → line through these points is perpendicular bisector.

• **60° angle**: Draw arc from point O on line → same radius arc from intersection point → connect O to second intersection = 60°.

• **Perpendicular at point P on line l**: Arcs centered at P cutting l at two points → arcs from those points (larger radius) intersecting on opposite side → line through P and intersection is perpendicular.

• **Perpendicular from point P to line l**: Arcs from P cutting l at two points → equal radius arcs from those points intersecting on opposite side of P → line through P and intersection is perpendicular.

• **Triangle SSS construction**: Draw base BC → arc centered at B, radius = AB → arc centered at C, radius = AC → intersection point is A.

• **Triangle SAS construction**: Draw base BC → construct given angle at B → mark length AB on the angle ray → join A to C.

• **Tangent from external point P to circle (center O, radius r)**: Join PO → construct perpendicular bisector of PO to find midpoint M → draw circle centered at M, radius = MO (or MP) → intersections with original circle are tangent points.

• **Two tangents from external point are equal in length**: If PT₁ and PT₂ are tangents from P, then PT₁ = PT₂.

Worked Examples

**Example 1: Construct the perpendicular bisector of segment AB = 6 cm**

*Step 1*: Draw line segment AB of any length (exam questions won't require exact measurement).

*Step 2*: Open compass to radius greater than half of AB (visual estimate: more than 3 cm).

*Step 3*: Place compass point at A, draw arc above and below AB.

*Step 4*: Without changing compass width, place point at B, draw arcs intersecting previous arcs at points P (above) and Q (below).

*Step 5*: Draw line through P and Q using straight-edge. This line PQ is the perpendicular bisector of AB.

*Why it works*: Points P and Q are equidistant from A and B (radius was same), so they lie on the perpendicular bisector by definition.

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**Example 2: Construct a triangle ABC where AB = 5 cm, BC = 6 cm, AC = 4 cm (SSS)**

*Step 1*: Draw base BC = 6 cm (longest side often chosen as base).

*Step 2*: Open compass to 5 cm, place point at B, draw arc above BC.

*Step 3*: Open compass to 4 cm, place point at C, draw arc intersecting previous arc at point A.

*Step 4*: Join A to B and A to C with straight-edge. Triangle ABC is constructed.

*Why it works*: Point A is exactly 5 cm from B and 4 cm from C, satisfying all three side conditions.

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**Example 3: Construct a 30° angle**

*Step 1*: Draw a ray OX.

*Step 2*: Construct 60° first: arc from O cutting OX at A, same radius arc from A cutting first arc at B, join OB → ∠XOB = 60°.

*Step 3*: Bisect ∠XOB: equal arcs from O cutting OX and OB, then equal arcs from those points, intersection gives point C. Join OC.

*Step 4*: ∠XOC = 30° (half of 60°).

Common Mistakes

• **Changing compass width mid-construction** → Arcs won't intersect properly or will create wrong distances. **Fix**: Set compass once per step and maintain that radius for all arcs in that step.

• **Using too small a radius for perpendicular bisector** → Arcs won't intersect if radius ≤ ½AB. **Fix**: Always open compass wider than half the segment length (visually estimate 60–70% of the segment).

• **Forgetting to bisect first to get 30°, 45°, or other derived angles** → Students try to construct 30° directly. **Fix**: 30° = bisect 60°; 45° = bisect 90°; 15° = bisect 30°. Build complex angles from basic ones.

• **Drawing tangent without constructing the perpendicular** → Tangent must be perpendicular to radius at point of contact; guessing creates errors. **Fix**: Use the circle-on-PO method (semicircle with diameter PO) to find tangent points precisely.

• **Mixing up ASA and AAS in triangle construction** → ASA uses included angle; AAS doesn't. **Fix**: For ASA, construct the angle first, then mark sides on its arms. For AAS, draw one side, construct both angles at endpoints.

Quick Reference

• Angle bisector: equal arcs from vertex → equal arcs from arm intersections → join vertex to final intersection.
• Perpendicular bisector: arcs from both endpoints (radius > ½length) → join arc intersections.
• SSS triangle: draw base → arcs from both ends with given radii → join intersection to base ends.
• 60° angle: equilateral triangle property (compass = side length) gives 60° at each vertex.
• Tangent from external point P to circle O: draw PO → perpendicular bisector of PO → circle on midpoint → intersections are tangent points.
• Tangents from external point are equal: PT₁ = PT₂ (use in calculations and proofs).