Geometry · G.CO.D.12

Bisecting a segment and an angle

With a compass and a straightedge that has no markings, how can you find the exact middle of a segment, or cut an angle exactly in half?

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Bisecting a segment and an angle

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When you reach for it

You have a segment \overline{AB} and you want its midpoint, or the line that crosses it at a right angle through that midpoint, its perpendicular bisector. Or you have an angle and you want the ray that splits it into two equal angles. With a ruler and a protractor you would measure and halve. A construction does it with two tools and no measuring at all:

a straightedge, which draws the line through two points and has no marks, so it cannot measure;
a compass, which draws a circle or an arc about a chosen center and holds its width, so it can carry one length to another place, as in copying a segment.

A measurement is only as good as the scale it is read from. A segment 7.3 cm long has its middle at 3.65 cm, and a ruler marked in millimeters cannot show that. A construction finds the point by a reason instead, and the reason is exact: the drawing is as good as the pencil, but the point it aims at is exactly the middle.

These two moves are building blocks for most of what follows in constructions: the center of the circle through three points, the center of the circle inside a triangle, the perpendicular through a point, and angles such as 45^\circ and 22.5^\circ.

The procedure

Bisecting a segment

The construction rests on one fact:

A point that is the same distance from A and from B lies on the perpendicular bisector of \overline{AB}.

The compass is the tool that makes equal distances. So the plan is to make two points that are each the same distance from A and from B, and join them.

Open the compass to more than half of AB. The two arcs have to meet. A crossing point C off the segment, with CA = CB = r, is the top of a triangle ACB, and two sides of a triangle add to more than the third, so r + r > AB. A width of half AB or less gives arcs that miss each other or only touch. Example 2 looks at each case.

With the point on A, swing an arc above the segment and another below it. Then, without changing the width, do the same from B. Every point of an arc about A is the distance r from A; every point of an arc about B is the distance r from B. Where an arc about A crosses an arc about B, both are true at once. Call the crossing above C and the one below D:

CA = CB = r, \quad DA = DB = r

Keeping the width the same from both ends is the whole move. It is what makes each crossing equidistant from A and B.

Draw \overleftrightarrow{CD}. It is the perpendicular bisector of \overline{AB}, and the point where it crosses \overline{AB} is the midpoint M. Two points fix a line, so no third crossing is needed.

Figure 1One compass width, four arcs. The line through the two crossings is the perpendicular bisector, and it meets \overline{AB} at the midpoint M.

Why the fact is true. Take any point P off the line \overleftrightarrow{AB} with PA = PB, and let M be the midpoint of \overline{AB}. The midpoint exists whether or not it has been drawn, so the argument may use it. Compare triangles PAM and PBM:

Statements	Reasons
PA = PB	P is equidistant from A and B
AM = BM	M is the midpoint
PM = PM	shared side
\triangle PAM \cong \triangle PBM	SSS
\angle PMA \cong \angle PMB	corresponding parts of congruent triangles

The angles \angle PMA and \angle PMB sit side by side along \overline{AB}, a linear pair, so they add to 180^\circ. Two equal angles that add to 180^\circ are 90^\circ each. So \overleftrightarrow{PM} passes through the midpoint and is perpendicular to \overline{AB}: it is the perpendicular bisector, and P lies on it. A point on \overleftrightarrow{AB} itself with PA = PB can only be M, which is also on it. This is one half of the perpendicular bisector theorem.

Figure 2If PA = PB, the triangles PAM and PBM are congruent by SSS, so the two angles at M are equal, and so each is 90^\circ.

Now use the fact twice. C is on the perpendicular bisector, D is on it, and C \neq D, so \overleftrightarrow{CD} is the perpendicular bisector. Nothing in the argument asked where C and D sit, so they may even be on the same side of the segment, as in Example 3.

A second way to see it. With one width for all four arcs, CA = CB = DA = DB, so ACBD has four equal sides: it is a rhombus. The construction is the fact that the diagonals of a rhombus are perpendicular bisectors of each other. That also says M is the midpoint of \overline{CD}, which Example 1 uses.

Bisecting an angle

The bisector of an angle is the ray from its vertex, running inside the angle, that splits it into two congruent angles. If \overrightarrow{OX} bisects \angle AOB, then

m\angle AOX = m\angle XOB = \tfrac{1}{2}\, m\angle AOB

The plan is to turn the angle into the segment problem.

With the point on the vertex O, swing one arc that crosses both sides. Call the crossings P, on \overrightarrow{OA}, and Q, on \overrightarrow{OB}. Any width will do. What matters is that one arc made both, so OP = OQ.

From P and from Q, with one width, swing arcs that cross inside the angle. The width has to be more than half of PQ, for the reason above, and it need not match the first width. Call X the crossing on the far side of \overline{PQ} from O. Then PX = QX.

Draw \overrightarrow{OX}. It bisects \angle AOB.

Figure 3Bisecting \angle AOB. The first arc makes OP = OQ, the second pair makes PX = QX, and \overrightarrow{OX} is the bisector.

Why it works. Compare triangles OPX and OQX. OP = OQ because one arc made both; PX = QX because one width made both; and OX is shared. By SSS, \triangle OPX \cong \triangle OQX, so the corresponding angles \angle POX and \angle QOX are congruent (corresponding parts of congruent triangles). Since P is on \overrightarrow{OA} and Q is on \overrightarrow{OB}, those are \angle AOX and \angle XOB.

Figure 4Three pairs of equal sides, so \triangle OPX \cong \triangle OQX by SSS, and the angles at O match.

There is a shorter way to say the same thing. O is equidistant from P and Q, and so is X. By the fact above, \overleftrightarrow{OX} is the perpendicular bisector of \overline{PQ}. Bisecting the angle is bisecting the segment \overline{PQ}, with the vertex serving as one of the two crossing points.

That also explains the instruction to take the crossing on the far side. Both crossings of the second pair of arcs lie on the perpendicular bisector of \overline{PQ}, but the near one can be O itself (it is exactly O when the second width equals the first, as in Example 5), and then it gives no new point. The far one is always a point of the bisector inside the angle.

Euclid bisects an angle in Book I, Proposition 9 of the Elements, and a segment in Proposition 10, both by building an equilateral triangle. The apex of an equilateral triangle on \overline{PQ} is a crossing of two arcs of width PQ, so the arcs used here are the same idea with a free choice of width.

Worked examples

Example 1

Where the arcs cross

Segment \overline{AB} is 10 cm long. It is bisected with the compass set to 7 cm, and the crossings C and D are on opposite sides of the segment. Find AM, CM and CD, and the perimeter of ACBD.

Solution

M is the midpoint, so AM = \tfrac{1}{2}(10) = 5 cm.

The construction makes \overline{CM} perpendicular to \overline{AB}, so triangle AMC has a right angle at M. Its hypotenuse is CA, which is the compass width, 7 cm. By the Pythagorean theorem,

\begin{aligned} CM &= \sqrt{7^2 - 5^2} = \sqrt{24} \\ &= 2\sqrt{6} \approx 4.90 \text{ cm} \end{aligned}

D is the same distance from M on the other side, because ACBD is a rhombus and its diagonals bisect each other. So

CD = 2 \cdot 2\sqrt{6} = 4\sqrt{6} \approx 9.80 \text{ cm}

The four sides of ACBD are all the compass width, so its perimeter is 4 \cdot 7 = 28 cm.

None of these numbers was needed to do the construction. The compass carried 7 cm, and everything else followed from it.

Example 2

Choosing the compass width

Segment \overline{AB} is 8 cm long. Say what happens when you try to bisect it with the compass set to 3 cm, 4 cm, 5 cm and 9 cm.

Solution

Half of AB is 4 cm, and that is the number to compare with.

3 cm: every point of an arc about A is 3 cm from A, and every point of an arc about B is 3 cm from B. A point on both would give AB \le 3 + 3 = 6, by the triangle inequality, but AB = 8. The arcs miss each other.

4 cm: a crossing point P would have PA + PB = 4 + 4 = 8 = AB. The two distances add to exactly AB only for a point on the segment, and the one with PA = PB is the midpoint. The arcs touch at M and nowhere else. That is one point, not two, so no line can be drawn, and two arcs that only touch do not show their meeting point clearly anyway.

5 cm: this works. The crossings are \displaystyle \sqrt{5^2 - 4^2} = 3 cm above and below M.

9 cm: this works too. A width larger than the whole segment is fine; the crossings are \displaystyle \sqrt{9^2 - 4^2} = \sqrt{65} \approx 8.06 cm from M. Crossings far apart fix the line more accurately than crossings close together, so a generous width is the better habit.

Example 3

Both crossings on one side

Segment \overline{AB} is 12 cm long and lies along the bottom edge of the paper, so there is no room below it. One pair of arcs is drawn with width 6.5 cm and a second pair with width 10 cm, both above the segment, crossing at C and D. How far above the segment are C and D, and why is \overleftrightarrow{CD} still the perpendicular bisector?

Solution

Each pair of arcs used one width from both ends, so

CA = CB = 6.5, \quad DA = DB = 10

Each of C and D is equidistant from A and B, so by the fact behind the construction each lies on the perpendicular bisector. They are different points, since they are different distances from A, and two different points fix a line. So \overleftrightarrow{CD} is the perpendicular bisector, even though both points are on the same side.

Figure 5Example 3. Both crossings above the segment, each made with its own width. The line through them is still the perpendicular bisector.

Half of AB is 6 cm. Each crossing is the top of a right triangle with that half as one leg and its width as hypotenuse:

\begin{aligned} CM &= \sqrt{6.5^2 - 6^2} = \sqrt{6.25} = 2.5 \text{ cm} \\ DM &= \sqrt{10^2 - 6^2} = 8 \text{ cm} \end{aligned}

So C is 2.5 cm above the segment, D is 8 cm above it, and CD = 8 - 2.5 = 5.5 cm.

What is not allowed is 6.5 cm from A and 10 cm from B for the same point. That point is closer to A than to B, so it is not on the perpendicular bisector at all.

Example 4

A segment on a coordinate grid

Segment \overline{AB} runs from A = (0, 0) to B = (8, 4), and it is bisected with the compass set to 5. Show that the crossings are C = (3, 4) and D = (5, 0), find the equation of \overleftrightarrow{CD}, and confirm that it is the perpendicular bisector.

Solution

The crossings are the points 5 from both A and B. Check each with the distance formula:

\begin{aligned} CA &= \sqrt{3^2 + 4^2} = 5 \\ CB &= \sqrt{(3 - 8)^2 + (4 - 4)^2} = 5 \\ DA &= \sqrt{5^2 + 0^2} = 5 \\ DB &= \sqrt{(5 - 8)^2 + (0 - 4)^2} = 5 \end{aligned}

Two circles of radius 5 about A and about B meet in at most two points, so these are the crossings.

The slope of \overleftrightarrow{CD} is \displaystyle \frac{0 - 4}{5 - 3} = -2. Through (3, 4):

\begin{aligned} y - 4 &= -2(x - 3) \\ y &= -2x + 10 \end{aligned}

Check. The midpoint of \overline{AB} is \displaystyle \left(\frac{0 + 8}{2}, \frac{0 + 4}{2}\right) = (4, 2), and -2(4) + 10 = 2, so the line passes through it. The slope of \overline{AB} is \displaystyle \frac{4}{8} = \frac{1}{2}, and \displaystyle (-2)\left(\frac{1}{2}\right) = -1, so the two are perpendicular. The midpoint of \overline{CD} is also (4, 2): the diagonals of the rhombus ACBD bisect each other.

Example 5

Bisecting an angle on a coordinate grid

Angle \angle AOB has its vertex at O = (0, 0), side \overrightarrow{OA} along the positive x-axis, and side \overrightarrow{OB} through (5, 12). It is bisected with width 13 for the arc about O and width 13 again for the arcs about P and Q. Find P, Q and X, and the direction of the bisector.

Solution

The arc about O crosses the x-axis at P = (13, 0). The point (5, 12) is \displaystyle \sqrt{5^2 + 12^2} = 13 from O, so the arc crosses the other side exactly there: Q = (5, 12).

The arcs of width 13 about P and Q cross where a point is 13 from both. Try X = (18, 12):

\begin{aligned} PX &= \sqrt{(18 - 13)^2 + 12^2} = 13 \\ QX &= \sqrt{(18 - 5)^2 + (12 - 12)^2} = 13 \end{aligned}

The other crossing is O itself, since OP = OQ = 13 as well. This is the case where the near crossing gives nothing new, and the far one, X, is the one to use.

The bisector is the ray from (0, 0) through (18, 12), which runs 3 across for every 2 up: it lies along y = \tfrac{2}{3}x. All four sides of OPXQ are 13, so it is a rhombus, and X is P and Q added coordinate by coordinate, (13 + 5, 0 + 12). The SSS argument guarantees that \angle AOX and \angle XOB are equal; no protractor is needed to know it.

Example 6

A 45^\circ and a 67.5^\circ angle from a right angle

\angle AOB is a right angle. Using bisection only, construct an angle of 45^\circ and an angle of 67.5^\circ.

Solution

Bisect \angle AOB with \overrightarrow{OX}. Each half is

m\angle AOX = m\angle XOB = \tfrac{1}{2}(90^\circ) = 45^\circ

Now bisect \angle XOB with \overrightarrow{OY}. Each half of it is \tfrac{1}{2}(45^\circ) = 22.5^\circ. The ray \overrightarrow{OY} is inside \angle AOB, and \overrightarrow{OX} lies between \overrightarrow{OA} and \overrightarrow{OY}, so the angle measures add:

\begin{aligned} m\angle AOY &= m\angle AOX + m\angle XOY \\ &= 45^\circ + 22.5^\circ = 67.5^\circ \end{aligned}

Check. m\angle AOY + m\angle YOB = 67.5^\circ + 22.5^\circ = 90^\circ.

The right angle itself can come from the first construction: the perpendicular bisector of any segment makes four right angles where it crosses.

Example 7

Quarters, and why not thirds

Segment \overline{AB} is 14 cm long. Mark the points that cut it into four equal parts. Then explain why bisecting, however often, never marks the point one third of the way from A to B.

Solution

Bisect \overline{AB} at M, so AM = 7 cm. Bisect \overline{AM} at K and \overline{MB} at L. Then AK = 3.5 cm, and AL = 7 + 3.5 = 10.5 cm. The points K, M and L, at 3.5, 7 and 10.5 cm from A, cut the segment into four pieces of 3.5 cm.

Each round of bisection halves the pieces. After n rounds, every marked point is a distance \displaystyle \frac{k}{2^n} \cdot AB from A for some whole number k. For the point one third of the way along, that would need

\frac{k}{2^n} = \frac{1}{3}, \quad\text{that is,}\quad 3k = 2^n

But 2^n is a product of twos and has no factor 3, while 3k does. So no number of rounds reaches it. Thirds, fifths and every other equal division need a different construction.

Boundaries

The width must be more than half the segment. At exactly half, the arcs touch only at the midpoint; at less, they do not meet (Example 2). The same holds for \overline{PQ} in the angle construction.

One width per crossing point. A crossing is equidistant from the two ends only if the same width was used from both. Different crossing points may use different widths (Example 3), but a single point may not.

Two different points. One crossing fixes nothing; the perpendicular bisector is a line, and it takes two points to fix it.

The angle needs to be at most 180^\circ. Below 180^\circ the construction runs as described. A straight angle still works: P and Q land on opposite rays, O is the midpoint of \overline{PQ}, and the bisector is the perpendicular at O, which is its own construction. For a reflex angle, bisect the ordinary angle with the same sides and take the opposite ray.

In a plane. The construction finds the points equidistant from A and B that lie in the drawing’s plane, which form a line. In space, the points equidistant from A and B form a whole plane.

Exact in principle. The reasoning is exact; a drawing is as exact as the pencil. The lengths in the examples describe where the construction lands, not steps of it: a straightedge has no marks, and nothing is measured.

Where this goes next

The first payoff is the perpendicular through a point: mark equal distances either side of the point and bisect the segment between them, which is bisecting a straight angle. Bisecting two sides of a triangle and crossing the two bisectors finds the circumcenter, the center of the circle through all three corners, which is how the circumscribed circle is built. Bisecting the angles of a triangle finds the incenter instead, because a point on an angle bisector is the same distance from both sides, which is the angle bisector distance theorem.

Both constructions are an early case of a locus: the perpendicular bisector is every point equidistant from two points, and the angle bisector is every point equidistant from two lines. And the arguments given here, SSS and the linear pair, are the pattern for justifying any construction: name the equal lengths the compass made, find the congruent triangles they form, and read off what was wanted.

Exercises

Bisecting a segment

\overline{AB} is 16 cm long and is bisected with the compass set to 10 cm, crossings C and D on opposite sides. Find AM, CM and CD.
\overline{AB} is 9 cm long. For each compass width, say whether bisecting gives two crossing points: 4 cm, 4.5 cm, 6 cm, 12 cm.
\overline{AB} is 24 cm long and is bisected with width 13 cm, crossings on opposite sides. Find CD and the perimeter of ACBD.
\overline{AB} is 16 cm long. Both crossings are above it, C made with width 10 cm and D with width 17 cm. Find CD.
\overline{AB} is 6 cm long, the crossings are on opposite sides, and CD = 8 cm. What width was the compass set to?

On a coordinate grid

A = (1, 1) and B = (7, 1), bisected with width 5. Find both crossing points and the equation of the perpendicular bisector.
A = (0, 0) and B = (4, 2), bisected with width 5. Show that the crossings are C = (0, 5) and D = (4, -3), and find the equation of \overleftrightarrow{CD}.
Is (1, 5) on the perpendicular bisector of the segment from A = (-1, 0) to B = (5, 2)?

Bisecting an angle

m\angle ABC = 118^\circ, and \overrightarrow{BD} bisects it. Find m\angle ABD.
\overrightarrow{OX} bisects \angle POQ, with m\angle POX = (3x + 7)^\circ and m\angle XOQ = (5x - 9)^\circ. Find x and m\angle POQ.
m\angle AOB = 84^\circ. \overrightarrow{OX} bisects \angle AOB, and \overrightarrow{OY} bisects \angle XOB. Find m\angle AOY.
\angle AOB has vertex O = (0, 0), side \overrightarrow{OA} along the positive x-axis and side \overrightarrow{OB} through (8, 6). It is bisected with width 10 throughout. Find P, Q and X, and the line the bisector lies along.
m\angle AOB = 60^\circ, as at a corner of an equilateral triangle. \overrightarrow{OX} bisects \angle AOB and \overrightarrow{OY} bisects \angle AOX. Find m\angle AOY and m\angle YOB.
m\angle AOB = 150^\circ, and \overrightarrow{OX} bisects it. Find m\angle AOX, and the angle that the ray opposite \overrightarrow{OX} makes with \overrightarrow{OA}.

Repeated bisection

\overline{AB} is 20 cm long. Bisect it, then bisect each half. How far from A are the three marked points?
\overline{AB} is 24 cm long. How many rounds of bisection does it take to mark a piece 3 cm long starting at A? A piece 1.5 cm long?

Reasoning

Someone sets the compass to 5 cm for the arc about A and 6 cm for the arc about B. Explain why the crossing is not on the perpendicular bisector.
In the angle construction, prove that \overleftrightarrow{OX} is the perpendicular bisector of \overline{PQ}.
Explain why the compass must be opened to more than half of AB for the two arcs to cross.
Explain why bisecting a straight angle at O gives the line perpendicular to it at O, and why that is the segment construction again.
In the angle construction the second width is set equal to the first. Prove that OPXQ is a rhombus, and say where the second crossing of the arcs about P and Q is.
Explain why bisection alone cannot cut a segment into five equal parts.

Answers

AM = 8 cm, CM = 6 cm, CD = 12 cm. \displaystyle AM = \frac{16}{2}; triangle AMC is right-angled at M with hypotenuse 10, so \displaystyle CM = \sqrt{10^2 - 8^2} = 6, and D is 6 cm on the other side.
4 cm: no crossing, since 4 + 4 = 8 < 9. 4.5 cm: the arcs only touch, at the midpoint, since 4.5 + 4.5 = 9. 6 cm and 12 cm: two crossings each, since both are more than half of 9.
CD = 10 cm, perimeter 52 cm. Half of AB is 12, so \displaystyle CM = \sqrt{13^2 - 12^2} = 5 and CD = 2 \cdot 5. The four sides of ACBD are each 13.
CD = 9 cm. Half of AB is 8. \displaystyle CM = \sqrt{10^2 - 8^2} = 6 and \displaystyle DM = \sqrt{17^2 - 8^2} = 15, both above, so CD = 15 - 6.
5 cm. \displaystyle CM = \frac{8}{2} = 4 and AM = 3, so the width is \displaystyle CA = \sqrt{3^2 + 4^2} = 5.
(4, 5) and (4, -3); the bisector is x = 4. The midpoint is (4, 1) and half of AB is 3, so the crossings are \displaystyle \sqrt{5^2 - 3^2} = 4 above and below it.
y = -2x + 5. \displaystyle CA = \sqrt{0^2 + 5^2} = 5, \displaystyle CB = \sqrt{4^2 + 3^2} = 5, \displaystyle DA = \sqrt{4^2 + 3^2} = 5, \displaystyle DB = \sqrt{0^2 + 5^2} = 5. The slope is \displaystyle \frac{-3 - 5}{4 - 0} = -2 and the line meets the y-axis at C = (0, 5). It passes through the midpoint (2, 1) of \overline{AB}, since -2(2) + 5 = 1.
No. Its distance to A is \displaystyle \sqrt{2^2 + 5^2} = \sqrt{29} and to B is \displaystyle \sqrt{4^2 + 3^2} = 5, which differ, and only equidistant points are on the bisector.
m\angle ABD = 59^\circ, half of 118^\circ.
x = 8 and m\angle POQ = 62^\circ. The halves are equal: 3x + 7 = 5x - 9 gives x = 8, and each half is 31^\circ.
m\angle AOY = 63^\circ. m\angle AOX = 42^\circ and m\angle XOY = 21^\circ, and 42^\circ + 21^\circ = 63^\circ.
P = (10, 0), Q = (8, 6), X = (18, 6); the bisector lies along y = \tfrac{1}{3}x. (8, 6) is \displaystyle \sqrt{8^2 + 6^2} = 10 from O, so it is Q. Then \displaystyle PX = \sqrt{8^2 + 6^2} = 10 and \displaystyle QX = \sqrt{10^2 + 0^2} = 10, and the line through (0, 0) and (18, 6) has slope \displaystyle \frac{6}{18} = \frac{1}{3}.
m\angle AOY = 15^\circ and m\angle YOB = 45^\circ. m\angle AOX = 30^\circ, so m\angle AOY = 15^\circ, and 60^\circ - 15^\circ = 45^\circ.
m\angle AOX = 75^\circ, and the opposite ray makes 105^\circ with \overrightarrow{OA}: the two rays form a straight angle, and 180^\circ - 75^\circ = 105^\circ. That ray bisects the reflex angle, since \displaystyle \frac{360^\circ - 150^\circ}{2} = 105^\circ too.
5 cm, 10 cm and 15 cm. The first bisection marks 10; bisecting each half marks 5 and 15.
3 rounds for 3 cm, 4 rounds for 1.5 cm. \displaystyle \frac{24}{3} = 8 = 2^3 and \displaystyle \frac{24}{1.5} = 16 = 2^4, and each round halves the piece.
The crossing is 5 cm from A and 6 cm from B, so it is closer to A. Every point of the perpendicular bisector is equidistant from A and B (fold the paper along it and A lands on B), so this point is not on it.
OP = OQ, because one arc about O made both. XP = XQ, because one width made the arcs about P and Q. So O and X are each equidistant from P and Q, and each lies on the perpendicular bisector of \overline{PQ}. They are different points, so the line through them is that bisector.
A crossing C off the segment has CA = CB = r, and A, B, C form a triangle. Two sides of a triangle add to more than the third, so 2r > AB, that is r > \tfrac{1}{2}AB. With 2r = AB the only point is the midpoint, and with 2r < AB there is none.
The arc about O crosses the two opposite rays at P and Q with OP = OQ, so O is the midpoint of \overline{PQ}. The arcs about P and Q then give a point X with XP = XQ, which is on the perpendicular bisector of \overline{PQ}. That bisector passes through the midpoint O, so \overleftrightarrow{OX} is perpendicular to the line at O; each half of the straight angle is \tfrac{1}{2}(180^\circ) = 90^\circ. It is the segment construction on \overline{PQ}, with O already known to be the midpoint.
With one width r throughout, OP = OQ = r from the first arc and PX = QX = r from the second pair, so all four sides of OPXQ equal r, and a quadrilateral with four equal sides is a rhombus. The second crossing is O: it is r from P and r from Q, so it lies on both arcs.
After n rounds of bisection every marked point is \displaystyle \frac{k}{2^n} of the way along, for a whole number k. A fifth would need \displaystyle \frac{k}{2^n} = \frac{1}{5}, that is 5k = 2^n, and 2^n has no factor 5. So no round of bisection ever marks it.