equidistant 165

Equidecomposable figures

vertices lie at lattice points. Mathematicians have

proved that it is impossible to draw an equiangular lat-

tice polygon with nsides if nis a number different

from 4 or 8. (Any four-sided equiangular lattice poly-

gon is a rectangle, and any eight-sided equiangular lat-

tice polygon has eight interior angles, each equal to

135°.) The square and the octagon are the only two

regular lattice polygons.

See also C

ARTESIAN COORDINATES

.

equidecomposable Two geometric figures are said

to be equidecomposable if it is possible to dissect one

figure into a finite number of pieces that can be rear-

ranged, without overlap, to form the second figure. For

example, an equilateral triangle of side-length 1 is

equidecomposable with a square of the same area.

In the picture, ais of length and bis of

length . (The challenge to convert an equilateral

triangle into a square by dissection was a puzzle first

posed by English puzzlist Henry Ernest Dudeney in 1907.

The challenge is also known as Haberdasher’s puzzle.)

Scottish mathematician William Wallace

(1768–1843) proved that any two polygons of the

same area are equidecomposable. Mathematicians

have since proved that the result remains valid even

for figures with curved boundaries. In particular,

Hungarian mathematician Miklov Laczovich demon-

strated in 1988 that almost 1050 pieces are needed to

convert a circle into a square.

Surprisingly, the corresponding result in three

dimensions does not hold, even for simple polyhedra.

German mathematician Max Dehn (1878–1952)

proved, for instance, that a cube and a regular tetrahe-

dron of the same volume are not equidecomposable.

equidistant Two points Pand Qare said to be

equidistant from a third point Oif they are the same

distance from O. We write: |PO| = |QO|.

Given a single point Oin a plane, the set of all

points equidistant from Ois a

CIRCLE

with Oas its

center. Given two points Aand Bin a plane, the set of

all points equidistant from Aand Bis the perpendicu-

lar bisector of the line segment AB, that is, a straight

line perpendicular to AB and passing through the mid-

point of AB. (To see this, let Mbe the midpoint of the

line segment AB, and let Pbe any point in the perpen-

dicular bisector to AB. Suppose that |PM| = xand |AM|

= y= |MB|. Then, by P

YTHAGORAS

’

S THEOREM

, we

have |PA| = = |PB|, and so Pis equidistant

from Aand B. One can also use Pythagoras’s theorem

to check that any point not on this line is not equidis-

tant from those two points.)

Given three points A, B, and Cin a plane, not in a

straight line, there is just one point Pequidistant from

all three. (To see this, draw the perpendicular bisectors

of AB and BC, and let Pbe the unique point at which

they intersect. Then Pis equidistant from Aand B, and

Pis also equidistant from Band C. Consequently, Pis

the same distance from all three points.) Noting that

the points A, B, and Ccan be viewed as the vertices of

a

TRIANGLE

, this proves:

The three perpendicular bisectors of the sides

of any triangle meet at a common point P.

(This observation is used to prove that the three

ALTI

-

TUDE

s of any triangle are also

CONCURRENT

.)

Taking matters further, suppose the common dis-

tance of Pfrom each of the three points A, B, and Cis

r. It then follows that a circle of radius rcentered about

Ppasses through each of these points. This proves:

For any triangle ABC there exists a single circle

that passes through each of its vertices A, B,

and C.

This circle is called the

CIRCUMCIRCLE

of the triangle,

and the point P, the common point of intersection of

the three perpendicular bisectors of the triangle, is

called the circumcenter of the triangle.

√x2+ y2