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Compass and straightedge

Geometry · level 5, lesson 1 of 6

A question to start

With nothing but a compass and a ruler that has no marks on it, you can cut any angle exactly in half. You can cut a line exactly in half, draw a perfect right angle, and build a perfect equilateral triangle. No measuring, no numbers, and the result is exact, not "close enough".

So here is the puzzle the ancient Greeks set themselves: with the same two tools, can you cut any angle into three equal parts? People tried for more than two thousand years. In 1837 a young French mathematician, Pierre Wantzel, proved that it cannot be done in general. Not "nobody has found the way yet", but "there is no way". Halving is easy; thirding is impossible. By the end of this lesson you will be able to do the easy part with your own hands, and you will see why each step works.

The idea

The two tools and the rules of the game

A straightedge is a ruler with no marks. You may use it only to draw a straight line through two points you already have. You may not use it to measure.

A compass draws circles. You put its point on one point (the centre) and open it so the pencil sits on another point; the gap between point and pencil is the radius. Then you swing it round. Every point on the circle you draw is exactly that radius away from the centre. That is the whole secret of compass constructions:

A compass is a machine for making equal lengths.

A construction is a recipe that uses only these two tools. You may draw a line through two points, a circle with a given centre through a given point, and mark the points where lines and circles cross. Nothing else. You do not need to draw whole circles: a short piece of a circle, called an arc, is enough, as long as it is long enough to cross what it needs to cross.

Why play with such limited tools? Because a construction is exact. A protractor can be off by half a degree; a ruler can be off by half a millimetre. A construction, done with perfect tools, is perfect, and the reason it works can be proved. That is why Euclid, about 2300 years ago, built the whole of his geometry book, the Elements, on these two tools.

Copying a segment

A segment is a piece of a straight line with two ends, such as \(AB\). To copy it onto another line:

  1. Put the compass point on \(A\) and open it until the pencil touches \(B\).
  2. Without changing the opening, move the point to the start \(P\) of your new line and draw an arc crossing it.
  3. Call the crossing point \(Q\). Then \(PQ\) is exactly as long as \(AB\).

Why it works: the compass kept the same radius, and every point on its arc is that radius from the centre.

Bisecting a segment: the perpendicular bisector

To bisect means to cut into two equal parts. The line that cuts segment \(AB\) in half and meets it at a right angle is called the perpendicular bisector of \(AB\). ("Perpendicular" means "at right angles".)

  1. Open the compass to more than half of \(AB\).
  2. With the point on \(A\), draw an arc above \(AB\) and an arc below it.
  3. With the same opening and the point on \(B\), draw arcs that cross the first two. Call the crossings \(P\) (above) and \(Q\) (below).
  4. Draw the straight line through \(P\) and \(Q\). It crosses \(AB\) at \(M\), the midpoint (the middle point).
ABPQM

Why it works. Look at point \(P\). It sits on the arc drawn from \(A\), so \(PA\) equals the compass opening. It also sits on the arc drawn from \(B\) with the same opening, so \(PB\) is the same length. So \(PA = PB\): \(P\) is the same distance from \(A\) as from \(B\). The same is true of \(Q\). The four points \(A, P, B, Q\) make a shape with four equal sides, a rhombus (the dashed lines show two of its sides). The diagonals of a rhombus always cut each other in half at right angles: fold the rhombus along \(PQ\) and \(A\) lands exactly on \(B\), because the two halves match side for side. So the fold line \(PQ\) cuts \(AB\) in half, and since folding sends the angle on one side of \(M\) onto the angle on the other, those two angles are equal and add to \(180°\), so each is \(90°\).

There is a beautiful fact hiding here, worth remembering on its own:

The perpendicular bisector of \(AB\) is exactly the set of all points that are the same distance from \(A\) and from \(B\).

Every point on the line is equally far from both ends, and every point equally far from both ends is on the line. You will use this again and again: for finding the centre of a circle, for reflections, and for the "fair meeting point" puzzles in the examples below.

Bisecting an angle

An angle is made by two rays (half-lines) from one point, the vertex. The angle bisector is the ray that cuts the angle into two equal angles.

  1. With the compass point on the vertex \(V\), draw an arc crossing both arms. Call the crossings \(P\) and \(Q\).
  2. With the point on \(P\), draw an arc inside the angle. With the same opening and the point on \(Q\), draw an arc that crosses it. Call the crossing \(R\).
  3. Draw the ray from \(V\) through \(R\). That is the bisector.
VPQR

Why it works. Join \(P\) to \(R\) and \(Q\) to \(R\) (dashed). Now compare the two triangles \(VPR\) and \(VQR\):

  • \(VP = VQ\), because both came from the first arc centred on \(V\);
  • \(PR = QR\), because both came from arcs with the same opening;
  • \(VR\) is shared by both triangles.

The two triangles have all three sides equal, one for one. Two triangles with the same three sides are the same triangle, just possibly flipped over (you will see in lesson 5 that this rule is called SSS). So their angles match too, and in particular angle \(PVR\) equals angle \(QVR\). The ray \(VR\) splits the angle into two equal parts. The shape \(VPRQ\) is a kite, and the bisector is its line of symmetry.

Copying an angle

To copy angle \(V\) onto a new ray starting at \(V'\): draw an arc from \(V\) crossing both arms at \(P\) and \(Q\); draw the same arc from \(V'\), crossing the new ray at \(P'\); set the compass to the distance \(PQ\), put its point on \(P'\) and mark \(Q'\) on the new arc. The ray \(V'Q'\) makes the same angle. The reason is the same as before: triangles \(VPQ\) and \(V'P'Q'\) have the same three sides.

Building special angles

The very first construction in Euclid's Elements is the equilateral triangle (all three sides equal). Draw a circle centred on \(A\) through \(B\), and a circle centred on \(B\) through \(A\). Where they cross is \(C\). Then \(CA = AB\) (same circle) and \(CB = BA\) (same circle), so all three sides are equal and every angle is \(60°\).

From there, everything else follows:

  • 60°: an angle of the equilateral triangle.
  • 30°: bisect 60°. 15°: bisect 30°.
  • 90°: the perpendicular bisector makes one. 45°: bisect 90°.
  • 120°: two 60° angles side by side. 75° is 60° + 15°, and so on.

What the tools cannot do

You can halve any angle, and halve again, as often as you like. But trisecting (cutting into three) a general angle, like 60°, is impossible with compass and straightedge; that is Wantzel's 1837 theorem. Two other old problems are also impossible: making a square with the same area as a given circle, and making a cube with double the volume of a given cube. The proofs use algebra, not geometry: every length you can construct turns out to be built from whole numbers using only \(+\), \(-\), \(\times\), \(\div\) and square roots, and the lengths those problems need are not of that kind. A 20° angle (a third of 60°) would need a cube root.

Some angles can be trisected: 90° can (just build 30°), and so can 180° (build 60°). The theorem says no single recipe works for every angle.

Worked examples

1. Find the midpoint of a 9 cm plank with no ruler.

  1. Put the compass point on one end \(A\) and open it well past halfway, say about 6 cm (the exact amount does not matter, as long as it is more than half).
  2. Draw arcs above and below the plank. Keep the opening and repeat from the other end \(B\).
  3. Join the two crossings. Where that line meets the plank is the midpoint, exactly 4.5 cm from each end.

If the opening were less than half (say 4 cm), the two arcs would not reach each other and there would be no crossings. That is why step 1 says "more than half".

2. Construct a 45° angle.

  1. Draw a line and mark a point \(A\) on it. Build a right angle at \(A\): draw a circle centred on \(A\) cutting the line at \(B\) and \(C\), then construct the perpendicular bisector of \(BC\). It passes through \(A\) (because \(A\) is the same distance from \(B\) and \(C\)), so it meets the line at \(A\) at 90°.
  2. Bisect that right angle. Each half is \(90° \div 2 = 45°\).

3. A fair place for a water tank. Three houses stand at \(A\), \(B\) and \(C\), not in a line. Where should a tank go so that it is the same distance from all three?

  1. Every point the same distance from \(A\) and \(B\) lies on the perpendicular bisector of \(AB\). Construct it.
  2. Every point the same distance from \(B\) and \(C\) lies on the perpendicular bisector of \(BC\). Construct it.
  3. Where the two lines cross, the point \(O\) is as far from \(A\) as from \(B\), and as far from \(B\) as from \(C\). So it is equally far from all three. Put the tank there.

A bonus: since \(O\) is equally far from \(A\) and \(C\) too, it must also lie on the perpendicular bisector of \(AC\). The three bisectors always meet at one point. And a circle centred on \(O\) through \(A\) passes through \(B\) and \(C\) as well: this is how to draw the circle through any three points, called the circumcircle.

4. Construct a 75° angle. 75 = 60 + 15. Build a 60° angle from an equilateral triangle. On the other side of one arm, build another 60° angle, bisect it to get 30°, and bisect again to get 15°. The two angles side by side make \(60° + 15° = 75°\).

Try it

  1. Why must the compass opening be more than half of \(AB\) when you bisect a segment?
  2. Describe how to construct a 30° angle.
  3. Which of these angles can you construct by starting from 60° and 90° and bisecting or adding: 22.5°, 105°, 135°, 20°?
  4. A point \(P\) is 7 cm from \(A\) and 7 cm from \(B\). Where must \(P\) lie?
  5. You construct the perpendicular bisectors of two sides of a triangle and they cross at \(O\). \(O\) is 5 cm from one corner. How far is it from each of the other two corners?
  6. A circular plate has broken; you have only a curved piece of its edge. Explain how to find where the centre of the plate was.
  7. Harder: an angle is bisected, then each half is bisected again, then each of those halves is bisected again. The smallest angles are 6° each. What was the original angle, and how many small angles are there?
Answers
  1. If the opening is half or less, the arc from \(A\) and the arc from \(B\) do not reach past the middle, so they never cross (or just touch at one point on \(AB\)), and you get no points \(P\) and \(Q\) to join.
  2. Construct an equilateral triangle (circles centred on each end of a segment, through the other end); one of its angles is 60°. Bisect that angle: \(60° \div 2 = 30°\).
  3. 22.5° yes: bisect 90° to get 45°, then bisect again. 105° yes: \(90° + 15°\), where 15° comes from bisecting 60° twice. 135° yes: \(90° + 45°\). 20° no: it is a third of 60°, and needs a trisection, which Wantzel showed cannot be done with these tools.
  4. \(P\) is the same distance from \(A\) and \(B\), so it lies on the perpendicular bisector of \(AB\). (There are two such points 7 cm from each end, one on each side of \(AB\), as long as \(AB\) is shorter than 14 cm.)
  5. 5 cm from each. \(O\) is on the bisector of the first side, so it is equally far from those two corners; it is on the bisector of the second side, so it is equally far from those two. Since the sides share a corner, all three distances are equal.
  6. Mark three points \(A\), \(B\), \(C\) on the curved edge. Construct the perpendicular bisectors of \(AB\) and \(BC\). The centre is the same distance from every point on the rim, so it lies on both bisectors; it is where they cross.
  7. Three rounds of bisecting cut the angle into \(2 \times 2 \times 2 = 8\) equal parts. Each is 6°, so the original angle was \(8 \times 6° = 48°\).

Common mistake

Changing the compass opening halfway through. In the perpendicular bisector and the angle bisector, the arcs from the two ends must use the same opening. If you adjust the compass between them, \(P\) is no longer equally far from both ends, and the "why it works" falls apart: the line you draw will miss the midpoint. Set the opening once and do not touch it until both arcs are drawn.

Measuring "just to be sure". Using the marks on a ruler, or a protractor, is not a construction. It may give a good picture, but it is not exact and it is not what the question asks. Use the straightedge only to join points.

Where it leads