Coordinatization - Mathematical Association of America · 1 From Calculus Gems, by G. Simmons, MAA,...

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8Coordinatization

In symbols one observes an advantage of discoverywhich is greatest when they express the exact natureof a thing briefly and, as it were, picture it; thenindeed the labor of thoughts is wonderfully dimin-ished.

— Gottfried Wilhelm Leibniz (1646–1716)1

In this chapter, we introduce and utilize coordinates as an analytic method for solving geometricproblems. There is some debate among scholars as to who first used this approach,2 but there is littledispute that the French mathematicians Rene Descartes (1596–1650) and Pierre Fermat (1601–1665)should be given the credit for (independently) formulating what we now call analytic geometry.

The analytic method is such a central, unifying part of mathematics that it is hard to conceiveof its nonexistence. Essentially, this method intertwines algebra and geometry in such a way thata problem in one of these areas can be converted to a problem in the other, where a solution maybe more easily obtained. The link between these two mathematical arenas is made by creating aone-to-one correspondence between numbers and points: each point on a line is represented by areal number, each point in the plane by an ordered pair of real numbers, and each point in 3-spaceby an ordered triple of real numbers. Once this correspondence is established and letters are used torepresent the real numbers, figures (sets of points) are readily identified with equations or inequalitiescontaining two or more variables. One can then study the algebraic properties of solutions to theequation f (x, y) = 0 in order to learn about the geometric properties of the figure in the plane whosepoint coordinates satisfy the equation, and vice-versa. In this way, the coordinate method serves asa “mathematical dictionary,” used for translating between geometry and algebra.

8.1 The Real Number Line

The Ruler Postulate for Euclidean geometry (Postulate 2.3), first introduced in Section 2.4, allowsus to claim the existence of a one-to-one correspondence between the set of real numbers and the set

1 From Calculus Gems, by G. Simmons, MAA, 2007.2 The Egyptians and Romans used it for surveying and the Greeks for mapmaking. Nicole Oresme graphed one variable

against another in the fourteenth century. ([18], pp. 346–347)

155

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156 8 Coordinatization

of points of an arbitrary line. The correspondence is usually defined by choosing two distinct pointson the line, and then matching one of them with the number 0, and another with a positive number(often taken to be 1). Then every point of the line becomes matched with a unique real number,and vice versa, and the line is often referred to as the real number line or the coordinate line. Theone-to-one correspondence is called the coordinatization of the line. For a given coordinatization,the number a corresponding to a point A is called the coordinate of A, and we denote this by writingA : (a).3 By the Ruler Postulate, if A : (a) and B : (b), then the distance AB is equal to |b − a|.

Because the points of the line AB correspond precisely to the real numbers, we can find a pointlocated at any desired distance from a fixed point. Furthermore, for any two distinct points, A andB, and any fixed positive constant, k, there exists a point C on

←→AB so that AC

CB= k, as we now show.

In the coordinatization of←→AB, let A : (a) and B : (b). Assume that a < b. We wish to find a point

C : (c) on←→AB such that AC

CB= |c−a|

|c−b| = k. Observe the following:

|c − a||c − b| = k ⇔

{c−a

−(c−b) = k when a < c < b

c−ac−b

= k when c < a or c > b

},

⇔{

c − a = kb − kc when a < c < b

c − a = kc − kb when c < a or c > b

},

⇔{

c = kb+ak+1 when a < c < b

c = kb−ak−1 when c < a or c > b, k �= 1

}.

Thus, unless k = 1, there are two points C dividing the segment AB into segments having ratiok : 1, one a point of internal division and the other a point of external division. (The point of externaldivision could fall to the right of B or to the left of A, as determined by the value of k.) We summarizethis result in the following proposition, accompanied by Figure 8.1.

Proposition 8.1. For points A : (a), B : (b), and C : (c), a < b, with C on←→AB and AC

CB= k ≥ 0,

c = kb − a

k − 1(when c ≤ a or c > b, k �= 1) or

c = kb + a

k + 1(otherwise).

ba

C2BC1A

kb+ak+1

kb-ak-1

FIGURE 8.1. Internal and External divisions of AB.

3 This notation can be read as an independent clause (as in “A has the coordinate a”) or as a noun (as in “A, with the

coordinate a”), depending on the context.

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8.2 Coordinates 157

(ii)(i)

Y

OX

Px

Py

XO

Y

ly

lxA: (a)

P: (xP, yP) Q: (a, b)

B: (b)

FIGURE 8.2. Coordinatization of the plane.

Of particular interest is a point C dividing AB exactly in half, that is, ACCB

= k = 1. In this case,AC = CB, C is the midpoint of AB, and c = a+b

2 .

8.2 Coordinates

To employ the full power of the coordinate method, we need to move beyond the one-dimensionalreal number line. Let be a plane, and let O, X, and Y be three non-collinear points of . Then, byPostulate 2.6, lines OX and OY are in . Each of these two lines can be coordinatized. We do thisin such a way that O is at the point corresponding to 0 on each of the lines and both X and Y havepositive coordinates. In addition, we assume that

−→OY is obtained by rotating

−→OX counterclockwise

by an angle of positive measure less than 180◦.4 Then we say that O, X, and Y define a coordinatesystem in , and we refer to it as the OXY -coordinate system. Point O is called the origin, and thecoordinate lines OX and OY are called the x-axis and the y-axis of the coordinate system OXY ,respectively.

Any such coordinate system can be used to coordinatize , as follows. Suppose that P is a point in. There is a unique line, ly , through P that is parallel to

←→OY , and a unique line, lx , through P that is

parallel to←→OX. Let Px = ←→

OX ∩ ly , and Py = ←→OY ∩ lx . Points Px and Py are called the projections

of P on the x- and y-axis, respectively. (See Figure 8.2(i).) Let xP

and yP

be the coordinates ofpoints Px and Py on the x- and y-axis, respectively, i.e., Px : (x

P) and Py : (y

P). These numbers x

P

and yP

are called the x- and y-coordinates of P in OXY , and we write P : (xP, y

P).5

From our discussion, it should be clear that once a coordinate system in the plane is chosen, eachpoint of the plane can be assigned a unique ordered pair of real numbers – its coordinates. On the otherhand, for every ordered pair of real numbers (a, b), there exists a unique point Q in the plane havingcoordinates (a, b) in the coordinate system. Indeed, point Q can be constructed as the intersectionof two lines: one passing through point A : (a) on the x-axis and parallel to the y-axis, and anotherpassing through point B : (b) on the y-axis and parallel to the x-axis. (See Figure 8.2(ii).) Therefore,as soon as a coordinate system in a plane is chosen, we obtain a one-to-one correspondence betweenthe set of points of the plane and the elements of R × R. This correspondence will justify usingexpressions like “consider the point (3, 4),” instead of “consider the point with coordinates (3, 4).”

4 This is equivalent to saying that we fix an orientation in all our coordinate systems.5 As before, this notation can be read as an independent clause (“P has the coordinates (x

P, y

P)”) or as a noun (“P , with

the coordinates (xP, y

P)”), depending on the context.

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AC':

C

A':

C: k+1

,k +1(

B

B': (xC , 0) (xB , 0)

(xA , 0)

kxB+xA kyB+yA )

FIGURE 8.3. C divides AB internally in ratio k.

We now have the tools needed to generalize the formula given in Proposition 8.1. Specifically,we are interested in finding the coordinates of a point C that divides a line segment in the plane,AB, in ratio k : 1. Let A and B be points in the plane, and assume that x

A≤ x

B. We will determine

first the coordinates (xC, y

C) of a point C that divides AB internally in ratio k.

Consider the projections of points A, C, and B on the x-axis: A′ : (xA, 0), C ′ : (x

C, 0), and

B ′ : (xB, 0), respectively, as shown in Figure 8.3. Since the lines

←→AA′,

←→CC ′, and

←→BB ′ are all parallel

to the y-axis, they are parallel to each other. By Theorem 3.13, the segments on a pair of transversalsthat are created by a set of three or more parallel lines will be proportional. That is, AC

CB= A′C ′

C ′B ′ . Ifwe want AC

CB= k, we simply need to ensure that A′C ′

C ′B ′ = k. In other words, finding xC

amounts tosimply finding the coordinate of the appropriate point C ′ on the x-axis dividing A′B ′ internally inratio k, a situation we have already encountered! We conclude that x

C= (kx

B+ x

A)/(k + 1).

Likewise, it can be shown that yC

= (kyB

+ yA)/(k + 1). Thus, the coordinates of point C on

←→AB

dividing AB internally in ratio k (k > 0) are (xC, y

C) =

(kx

B+x

A

k+1 ,ky

B+y

A

k+1

). In particular, the point

C =(

xA+x

B

2 ,y

A+y

B

2

)is the midpoint of AB. Similarly, the coordinates of point C on

←→AB dividing

AB externally in ratio k (k > 0, k �= 1) are (xC, y

C) =

(kx

B−x

A

k−1 ,ky

B−y

A

k−1

).

Theorem 8.2. For points A, B, and C with C on←→AB and

AC

CB= k ≥ 0,

(xC, y

C) =

(kx

B+ x

A

k + 1,ky

B+ y

A

k + 1

)or

(xC, y

C) =

(kx

B− x

A

k − 1,ky

B− y

A

k − 1

), k �= 1.

To create the familiar coordinatization of the plane utilized by Descartes, we make the assumptionsthat the axes OX and OY are perpendicular and that the same scale is used on both lines. The axesdivide the plane into four quadrants, with the location of a point within a quadrant or on an axisdetermined by the sign of its coordinates; this is illustrated in Figure 8.4, with x and y positive.The equality of scales on the x- and y-axes can be defined in the following way: the segment withthe end points (0, 0) and (1, 0) is congruent to the segment with the end points (0, 0) and (0, 1). A

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8.2 Coordinates 159

Quadrant 2

Quadrant 3 Quadrant 4

Quadrant 1

(-x, -y (x, -y)

(-x, y) (x, y)

y-axis

x-axis

)

FIGURE 8.4.

plane with such a coordinatization is called the Cartesian plane, and the corresponding coordinatesystem is called the Cartesian coordinate system.

Given a coordinate line and two points A : (a) and B : (b) on it, we know that the length ofAB is given by AB = |a − b|. Having a coordinate system in a plane and two points A : (x

A, y

A)

and B : (xB, y

B), we wish to likewise find AB. Here we do this for a Cartesian coordinate system

only.Consider the point C : (x

B, y

A). Points A, B, and C are vertices of a right triangle with right

angle at C (as shown in Figure 8.5). Using the Pythagorean theorem, AC2 + BC2 = AB2. Sincethe points A and C agree in the y-coordinate, and the points B and C agree in the x-coordinate,AC = |x

A− x

B| and BC = |y

A− y

B|. As |a|2 = a2 for any real number a, we obtain the following

distance formula in the plane.

Proposition 8.3. In a Cartesian coordinate system, the distance between two points A and B

is given by

AB =√

(xA

− xB)2 + (y

A− y

B)2.

A: (xA , yA)

B:

C: (xC , yC)

(xB , yB)

xB = xC

yA = yC

FIGURE 8.5.

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8.3 Equations of a Line

Recall our earlier assertion that the coordinate method provides us with a mathematical dictionary,a claim referring to the correspondence between figures in the plane and algebraic relations in x andy. Our goal in this section is to find an algebraic description of a line in the plane. As discussed inChapter 6, the set of all points satisfying a given property is called the locus for this property. Whenthe property is expressed as an equation or inequality, the locus can be portrayed as a graph of thatproperty. In this way, the graph is a visual representation of an algebraic relation.

If OXY is a coordinate system, the x-axis and all lines parallel to the x-axis are called horizontallines; the y-axis and all lines parallel to it are vertical lines. It is easy to see that any horizontal linecan be represented by an equation of the form y = b, as a point lies on the line if and only if it hascoordinates of the form (x, b), where x can assume any real value. Similarly, any vertical line canbe represented with an equation of the form x = a.

The reader is undoubtedly acquainted with the notion of the slope of a (non-vertical) line.Intuitively, the slope of a line specifies the direction in which the line rises and quantifies the“steepness” of the line. Let l be a non-vertical line. For any pair of distinct points A : (x

A, y

A) and

B : (xB, y

B) on l, we have x

A�= x

B, so we can compute the number (y

B− y

A)/(x

B− x

A). If we

compute the same ratio for any other pair of distinct points A′ and B ′ of l, the result will be the same,due to the similarity of the related triangles. (See Figure 8.6.) Therefore the ratio depends only onthe line l, and it is called the slope of l. It is often denoted by m

l, or simply m. For l = ←→

AB,

m←→AB

= yB

− yA

xB

− xA

.

Theorem 8.4. Let l be a line having slope m, and let A be a point on l. A point P lies on l ifand only if its coordinates satisfy the equation y − y

A= m(x − x

A).

Proof. Let P : (xP, y

P) be any point on l, distinct from A. Then

←→AP is the same line as l, so

m = (yP

− yA)/(x

P− x

A). Consequently, (x

P, y

P) satisfies y − y

A= m(x − x

A), as desired.

Conversely, suppose P : (xP, y

P) with x

Pand y

Psatisfying the equation y − y

A= m(x − x

A).

Hence yP

− yA

= m(xP

− xA). Let P ′ be a point on l with the same x-coordinate as P , i.e., x

P ′ = xP.

Then, using the first part of the proof, yP ′ − y

A= m(x

P ′ − xA) = m(x

P− x

A). Hence y

P− y

A=

yP ′ − y

A, so y

P= y

P ′ . This implies that P = P ′. �

m = yB-yA

xB-xA

l A:(xA , yA)

B :(xB , yB)

A'

B'

FIGURE 8.6. Slope of a line.

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8.3 Equations of a Line 161

An equation of the form y − yA

= m(x − xA) is called the point-slope form for a line. The

same line can also be described by equations presented in other forms. The point-slope formcan be rewritten as y = mx − mx

A+ y

A, or y = mx + b, where b = y

A− mx

A. The substitution

x = 0 into any equation of the form y = mx + b reveals that b is the y-coordinate of the point ofintersection of the line with the y-axis; consequently, this form is known as the slope-y-interceptform of the equation for a line. Similarly, an equation of the form y = mx + b can be rewritten as

x

(−b/m)+ y

b= 1, provided that neither m nor b is 0.

Each of these equations can be rewritten as ax + by + c = 0, which is often referred to as thestandard form, or the general form, of the equation of the line. If a = 0 and b �= 0, the graph of thisequation is a horizontal line. If a �= 0 and b = 0, it is a vertical line. If a = b = 0, then the graphof the equation is either the whole plane (if c = 0) or the empty set (if c �= 0). The forms of linearequations are summarized in the table that follows.

Name of Form FormStandard ax + by + c = 0 (at least one of a, b non-zero)

Point-Slope y − yA

= m(x − xA) (for non-vertical lines)

Slope-Intercept y = mx + b (for non-vertical lines)xy-Intercept x

a+ y

b= 1 (a �= 0, b �= 0)

One can easily observe the following.

Corollary 8.5. Lines are parallel if and only if they have the same slope (or both have noslope).

Though many interesting geometric properties can be expressed in terms of coordinates in anarbitrary coordinate system OXY , most applications use Cartesian systems. Therefore,

From now on, unless specified otherwise, a coordinate system in theplane is assumed to be Cartesian.

Suppose that l is a non-horizontal line in the coordinate plane. We define the inclination of l to bethe smallest positive measure, θ , of an angle measured counterclockwise from the positive directionof the x-axis to l.6 If l is parallel to the x-axis, then its inclination is defined to be 0◦. Observe thatthis definition restricts the value of θ so that 0◦ ≤ θ < 180◦. This gives a very useful relation: forany non-vertical line l,

ml = tan θ.

(See Figure 8.7.) Note that l is vertical if and only if θ = 90◦, in which case neither the slope nortan θ exist.

This interpretation of the slope as tangent allows us to use trigonometry when the coordinatemethod is developed. In particular, we have the following useful statement.

6 As usual, the angle is measured in degrees unless otherwise specified.

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l

xB - xA

yB - yA

tan θ = yB - yAxB - xA

A : (xA , yA )

θ

θ

B : (xB yB ),

FIGURE 8.7.

Theorem 8.6. Let α be the angle between intersecting non-vertical lines l1 and l2, havingslopes m1 and m2, respectively. Then α = 90◦ if and only if m1m2 = −1. Otherwise,

tan α =∣∣∣∣ m1 − m2

1 + m1m2

∣∣∣∣.

Proof. Let the inclinations of l1 and l2 be θ1 and θ2, respectively. (See Figure 8.8; let A, B, andC be the points of intersection shown in the figure.) Then m1 = tan θ1 and m2 = tan θ2. Since θ1 isan exterior angle to �ABC, θ1 = θ2 + m∠CAB. Recall that the measure of an angle between twointersecting lines is the measure of the smaller angle formed by them, so 0◦ ≤ α ≤ 90◦.

Suppose α �= 90◦ and m1m2 �= −1. As α = m∠CAB or α = 180◦ − m∠CAB, we always havetan α = | tan m∠CAB|. Thus, using the difference formula for the tangent function,

tan α = | tan (θ1 − θ2)| =∣∣∣∣ tan θ1 − tan θ2

1 + tan θ1 tan θ2

∣∣∣∣.Substitution yields the desired result.

θ1θ 2

l2

l1

x

y

A

BC

FIGURE 8.8.

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8.3 Equations of a Line 163

Now, α = 90◦ if and only if θ1 = θ2 + 90◦. As neither θ1 nor θ2 is a right angle, we have θ2 �= 0.Therefore,

m1 = tan θ1 = sin θ1

cos θ1= sin(θ2 + 90◦)

cos(θ2 + 90◦)= cos θ2

− sin θ2= − 1

tan θ2= −1

m2.

The implication, m1m2 = −1 ⇒ α = 90◦, is left to the reader. �

Example 29. Find the distance between the point P : (−3, 4) and the line l given byy − 1

2x − 1 = 0.

Solution: We first find the line through P that is perpendicular to l. As ml = 1/2, this line(denoted l⊥) will have slope −1/ml = −2, so its equation is given by y − 4 = −2(x + 3);equivalently, y = −2x − 2.

To find the coordinates of the point of intersection of these lines, we set −2x − 2 equalto 1

2x + 1 and find x = − 65 . Thus, y = −2

(− 65

) − 2 = 25 , and (−6/5, 2/5) is the point of

intersection.It remains to find the distance between the points (−3, 4) and (−6/5, 2/5). Using the distance

formula,

d((−3, 4), (−6/5, 2/5)

) =√(

−3 −(

−6

5

))2

+(

4 − 2

5

)2

=√(

−9

5

)2

+(

18

5

)2

=√(

9

5

)2

(12 + 22) = 9

5

√5.

�

The method of the preceding example can be used to find a general formula for the distance d(P, l)between a point P : (x

P, y

P) and a line, l, given by ax + by + c = 0. Notice first that if a = 0 and

b �= 0, then l is a horizontal line, equivalently expressed as y = −c/b. The line l⊥ through P thatis perpendicular to l is given by x = x

Pand the resulting distance from P to l is |y

P− (−c/b)|.

Similarly, if a �= 0 and b = 0, l is a vertical line, x = −c/a, and d(P, l) = |xP

− (−c/a)|.If a �= 0 and b �= 0, then ml = −a/b, and l⊥ : y − y

P= (b/a)(x − x

P). Let Q = l ∩ l⊥. Finding

its coordinates, we obtain

xQ

= b2 xP

− ac − ab yP

a2 + b2and y

Q= a2 y

P− bc − ab x

P

a2 + b2.

Note that the expression for yQ

can be obtained from the one for xQ

by interchanging a with b

and x with y, as one could anticipate. This implies

xP

− xQ

= a (a xP

+ b yP

+ c)

a2 + b2and yp − y

Q= b (a x

P+ b y

P+ c)

a2 + b2.

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Then,

d(P, l) = d(P,Q) = PQ =√

(xP

− xQ

)2 + (yP

− yQ

)2

=√

a2 (a xP

+ b yP

+ c)2

(a2 + b2)2+ b2(a x

P+ b y

P+ c)2

(a2 + b2)2

=√

(a2 + b2) (a xP

+ b yP

+ c)2

(a2 + b2)2

=√

(a xP

+ b yP

+ c)2

a2 + b2

= |a xP

+ b yP

+ c|√a2 + b2

.

Note that though the formula is obtained under the assumption that neither a nor b is zero, it alsogives the correct results if only one of them is zero. We state our finding in the following theorem.

Theorem 8.7. Let P be a point, and let l be a line with equation ax + by + c = 0. Then thedistance from P to l is given by

d(P, l) = |a xP

+ b yP

+ c|√a2 + b2

.

Note that the formula confirms our answer for Example 29:

|a xP

+ b yP

+ c|√a2 + b2

= | (−1/2) (−3) + (1)(4) + (−1)|√(−1/2)2 + 1

= 9/2√5/4

= 9√5.

8.4 Applications

. . . in some deep sense the truth about how the worldworks resides in the equations . . . to help our limitedimaginations to visualize what is going on.

— Richard Feynman (1918–1988)7

The coordinatization of the plane and the accompanying algebraic representations of geometricfigures can serve as powerful tools for solving problems of Euclidean geometry. Coordinatizationof the plane requires choosing two intersecting lines to serve as axes and, sometimes, specifying thescale. The particular selection of the axes and the scale can impact (for better or worse) the easewith which a particular problem can be solved, as the examples and theorems in this section willdemonstrate.

Theorem 8.8. The medians of any triangle are concurrent and the point of concurrency divideseach median into segments in ratio 2:1.

7 Richard Feynman: A Life in Science, John Gribbin and Mary Gribbin, Penguin Books, 1997, p. 27.

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8.4 Applications 165

P

F

E

D

A

B

C

FIGURE 8.9.

Below we consider three solutions of this problem, all using the coordinate method. Each ofthem will illustrate advantages and difficulties of a particular approach, which is typical when thecoordinate method is applied. Our first proof will be the most natural one, the second will representa substantial improvement of the first, and the third proof will be based on a completely differentapproach.

Proof 1. Consider triangle ABC with medians AD, BE, and CF . (See Figure 8.9.) Introduce anarbitrary coordinate system OXY in the plane such that no line defined by a median is vertical. (Ifone median is vertical in a given coordinate system, simply rotate the axes slightly so neither thatmedian nor any others are vertical.) Then a point-slope form of

←→AD is

y − yA

= yD

− yA

xD

− xA

(x − xA).

Being the midpoint of BC, D : (xD, y

D) =

(x

B+ x

C

2,y

B+ y

C

2

). Substituting into the equation for

←→AD and simplifying, we get

←→AD : y = y

B+ y

C− 2y

A

xB

+ xC

− 2xA

x + xBy

A− x

Ay

B+ x

Cy

A− x

Ay

C

xB

+ xC

− 2xA

. (8.1)

In a similar way, we find an equation for←→BE, but without actually redoing the computations from

scratch. The following important (and obvious) observation makes the task trivial: in order to obtainan equation for

←→BE by using (8.1), one simply needs to permute the letters A, B, and C in (8.1) so

that A becomes B, B becomes C, and C becomes A. This permutation yields

←→BE : y = y

C+ y

A− 2y

B

xC

+ xA

− 2xB

x + xCy

B− x

By

C+ x

Ay

B− x

By

A

xC

+ xA

− 2xB

. (8.2)

Similarly, if we permute the letters A, B, and C in (8.1) such that A becomes C, B becomes A, andC becomes B, we obtain an equation for

←→CF :

←→CF : y = y

A+ y

B− 2y

C

xA

+ xB

− 2xC

x + xAy

C− x

Cy

A+ x

By

C− x

Cy

B

xA

+ xB

− 2xC

. (8.3)

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Let P = ←→AD

⋂←→BE. Coordinates of P can be found by solving the system of equations (8.1) and

(8.2), which is easier said than done! After tedious algebraic manipulations, which we omit, wefinally arrive at

(xP, y

P) =

(xA

+ xB

+ xC

3,y

A+ y

B+ y

C

3

). (8.4)

Applying our first permutation to all occurrences of indices A, B, and C in the system formed by(8.1) and (8.2), we obtain the system formed by equations (8.2) and (8.3), whose solutions give thecoordinates of a point P ′ = ←→

BE⋂←→

CF . It is clear that the coordinates of P ′ can be obtained from thecoordinates of P by permuting the indices A, B, and C as before. Since the coordinates of P givenby (8.4) do not change under any permutation of {A,B,C}, P and P ′ are the same point. Hence←→AD and

←→CF intersect

←→BE at the same point, which is another way to say that all three medians are

concurrent. The statement about the medians being divided by P in proportion 2 : 1 follows directlyfrom (8.4) and Theorem 8.2. �

Proof 2. Let us again use the point labeling as in Figure 8.9. A drawback of Proof 1 is that itrequires us to determine coordinates of P given by (8.4) by solving the system formed by (8.1) and(8.2), and this requires a somewhat lengthy algebraic manipulation. The difficulty can be removedby choosing the coordinate system in a better way.

Introduce a coordinate system OXY , not necessarily Cartesian, such that E becomes the origin,A : (a, 0), and B : (0, 1). Then a < 0 and C : (−a, 0). (See Figure 8.10.)

Then D : (−a/2, 1/2), and a point-slope form of line AD is

←→AD : y = − 1

3a(x − a) = − 1

3ax + 1

3.

(This equation is, of course, what one gets from (8.1) by substituting the coordinates of A, B, andC.) Similarly (or just by replacing a with −a in the equation above) we obtain

←→CF : y = 1

3a(x + a) = 1

3ax + 1

3.

Clearly, both←→AD and

←→CF intersect

←→BE at the same point P : (0, 1/3) – their y-intercept. Again, the

second part of the theorem follows from here immediately. �

F : (a/2, 1/2)D : (-a/2, 1/2)

O = E

B : (0, 1)

C : (-a, 0)

A : (a, 0)

FIGURE 8.10.

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8.4 Applications 167

We stress that use of a not-necessarily-Cartesian system in this solution has the advantage ofreducing the number of independent coordinates. If we insist on a Cartesian system, and if nomedian of our triangle is also an altitude, then instead of one parameter a we are forced to introduceat least one more parameter. For example, keeping E as the origin and A : (a, 0), we would haveB : (x

B, 1). Or, if we wish to preserve A : (a, 0) and B : (0, 1), then x

Eor x

Cwould become a

new parameter. A similar approach will still work, but the algebraic manipulations become a littleharder.

Proof 3. We again use the point labeling of Figure 8.9. This proof will not use equations oflines, and is heavily motivated by the statement of the theorem concerning the 2 : 1 ratios andProposition 8.2.

Since D is the midpoint of BC, D has coordinates

(x

B+ x

C

2,y

B+ y

C

2

). Let P be the point on

AD such that AP/PD = 2/1. Then by Theorem 8.2, P has coordinates(xA

+ 2xD

3,y

A+ 2y

D

3

)=(x

A+ x

B+ x

C

3,y

A+ y

B+ y

C

3

).

Likewise, since E is the midpoint of AC, E has coordinates

(x

A+ x

C

2,y

A+ y

C

2

). If P ′ is the

point on BE such that BP ′/P ′E = 2/1, then P ′ has coordinates(xB

+ 2xE

3,y

B+ 2y

E

3

)=(x

B+ x

A+ x

C

3,y

B+ y

A+ y

C

3

).

Similarly, the point P ′′ on the segment CF such that CP ′′/P ′′F = 2/1 has coordinates(xC

+ xA

+ xB

3,y

C+ y

A+ y

B

3

).

That is, P , P ′, and P ′′ are all the same! This establishes both claims of the theorem. �

Note that in the last proof, the symmetry of the coordinates of P with respect to A, B, and C madeit evident – even without further calculation – that P served the same role for all three medians,which made the proof even easier. The reader is invited to compare these solutions with those usingother methods. This problem occurred as Problem 3.2.10 and it also follows from Ceva’s Theorem(Theorem 3.17); in Chapter 12, we will revisit the statement once more as Theorem 12.10.

Our second example will use an approach similar to the one of Proof 2 above. It illustrates theadvantages of a Cartesian coordinate system.

Theorem 8.9. The altitudes of any triangle are concurrent.

Proof. If our triangle is a right triangle, the proof is trivial: all altitudes meet at the vertex of theright angle. Therefore we assume that our triangle ABC has no right angle. Introduce a Cartesiancoordinate system so that A and C lie on the x-axis and B lies on the positive y-axis, and choosethe scale so that B has coordinates (0, 1). As a result, the altitude from B will lie on the y-axis withfoot at the origin. Label the remaining vertices A : (x

A, 0) and C : (x

C, 0). Since A and C are not at

the origin, xA

�= 0 and xC

�= 0. (See Figure 8.11.)

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(0,-xA xC )

C :(xC , 0)A :(xA, 0)

B :(0, 1)

O :(0, 0)

FIGURE 8.11.

l2

l 1

O

X

Y

FIGURE 8.12.

We now find the equation for the altitude through A. The slope of BC is0 − 1

xC

− 0= −1

xC

, so the

altitude through A has slope xC. The point-slope equation for the altitude is y = x

C(x − x

A). Since

the altitude from B corresponds to the vertical line x = 0, the point of intersection of the altitudesfrom A and from B has coordinates (0,−x

Ax

C).

Clearly the point of intersection of←→BO with the altitude through C will have coordinates

(0,−xC

xA); indeed, in the argument above just replace C by A and A by C. The symmetry of

the expression xCx

Awith respect to A and C proves that the altitudes through A and C intersect

←→BO

at the same point! Hence, the three altitudes are concurrent. �The reader is invited to clarify how our proof used the assumption that the coordinate system

was Cartesian. We also suggest comparing this solution with the one provided in the proof ofCorollary 3.18.

We conclude this section with an application of the coordinate method to finding a locus.

Example 30. Given a positive number k and two intersecting lines l1 and l2, find the locus ofall points whose sum of distances to the lines is k.

Solution: We chose a coordinate system OXY in such a way that the coordinate axes OX andOY are each bisectors of the angles formed by l1 and l2. Clearly, the axes OX and OY areperpendicular and the obtained coordinate system is Cartesian. (See Figure 8.12.)

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8.5 Systems of Equations, Revisited 169

Then l1 : y = mx for some m > 0, and l2 : y = −mx. A point (x, y) belongs to the locus ifand only if

d((x, y), l1) + d((x, y), l2) = k.

Using Theorem 8.7, this is equivalent to

|mx − y|√m2 + 1

+ | − mx − y|√(−m)2 + 1

= k.

To remove the clutter, let K = k√

m2 + 1. Then the equation can be rewritten as

|mx − y| + |mx + y| = K. (8.5)

It is clear that if (x, y) satisfies this equation, then so do points (−x, y), (x,−y), and (−x,−y).Therefore, the graph of the equation is symmetric with respect to both coordinate axes, andit is sufficient to construct it in the 1st quadrant only, i.e., we may assume that x ≥ 0 andy ≥ 0.

Since we are in the first quadrant, mx + y ≥ 0, and (8.5) reduces to |mx − y| + mx + y = k.We consider the two cases where mx − y ≥ 0 and mx − y ≤ 0 separately.

Let us first illustrate the general argument with the specific example where k = 1 and m = 2,so that (8.5) gives |2x − y| + |2x + y| = √

5. Then if 0 ≤ y ≤ 2x, we have that (2x − y) +(2x + y) = √

5, so x = √5/4. This means y will take on all values in the interval [0,

√5/2],

so the graph is the “vertical” segment8 with endpoints (√

5/4, 0) and (√

5/4,√

5/2).On the other hand, if 0 ≤ 2x ≤ y, then |2x − y| = −(2x − y), and we have −(2x − y) +

(2x + y) = √5, so y = √

5/2. In this case, the graph is the “horizontal” segment with endpoints(0,

√5/2) and (

√5/4,

√5/2).

We now solve the general case in a similar fashion. If 0 ≤ y ≤ mx, then

(8.5) ⇔ (mx − y) + (mx + y) = k√

m2 + 1 ⇔ x = K/(2m).

The graph of this relation in the 1st quadrant is the vertical segment

{(K/(2m), y) : 0 ≤ y ≤ K/2}.If y > mx ≥ 0, then

(8.5) ⇔ −(mx − y) + (mx + y) = K ⇔ y = K/2.

The graph of this relation in the 1st quadrant is the horizontal segment

{(x,K/2) : 0 ≤ x ≤ K/(2m)}.Therefore the locus is the rectangle shown in Figure 8.13. �

8.5 Systems of Equations, Revisited

Given a pair of distinct curves in the plane, we may wish to determine their point(s) of intersection.Such a problem can be solved with a system of equations, using algebra. Often applications of a

8 Recall that we have defined the vertical lines in a coordinate system to OXY to be those lines that are parallel to the

y-axis, OY ; the horizontal lines are those that are parallel to the x-axis, OX.

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l1

l2

FIGURE 8.13.

well-known fact or method are more exciting than the proof of the fact itself and, as a result, the“theoretical” part of the problem is not given sufficient attention. In this section, we show howinsight into the theory of the very simple idea of equivalent systems of equations can be used toprovide elegant solutions for a class of geometric problems for which students may typically choosea cumbersome algebraic approach. We begin with a simple example.

Example 31. Solve the following system of equations:

2x + 3y = 11, 4x − 2y = −2. (8.6)

Solution: Multiplying both sides of the first equation by 2 and both sides of the second by 3,then adding the results and dividing by 16, we obtain x = 1 and, subsequently, y = 3. Therefore{(1, 3)} is the solution set of the system. �

How can one be sure that the new system obtained using the above method (namely, 2x + 3y =11, x = 1) is equivalent to (8.6), i.e., has the same set of solutions as (8.6)? Using a graphicalinterpretation of (8.6), the question can be restated as “why do lines whose equations are 4x − 2y =−2 and x = 1 intersect the line whose equation is 2x + 3y = 11 at the same point?”

The method we are using above is familiar to all algebra students. Despite its importance, very fewbooks of algebra, precalculus, or even linear algebra prove that it is valid. The typical algebra studentmay have serious doubts that it requires any proof at all. Nevertheless, a thorough understandingof this method is important for understanding the solutions of later examples in this section. Thefollowing theorem legitimizes the method.

Theorem 8.10. Let F1(x, y) and F2(x, y) be two algebraic expressions. Let α and β be realnumbers, β �= 0. Then the following two systems are equivalent.

F1(x, y) = 0, F2(x, y) = 0, (8.7)

F1(x, y) = 0, αF1(x, y) + βF2(x, y) = 0. (8.8)

Proof. To prove that (8.7) and (8.8) are equivalent is to show that each solution of (8.7) is a solutionof (8.8) and vice versa. If both solution sets are empty, the statement obviously holds. If one of thesolution sets is not empty, then the following argument shows that neither is the other one, and theyare equal.

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8.5 Systems of Equations, Revisited 171

6

5

4

3

2

1

-1

-2

-3

-4

-5

-6

-7

-8

-3 -2 -1 1 2

-2F1 (x, y) + 3F2 (x, y)=0F2 (x, y)=0

F1(x, y)=0

FIGURE 8.14.

(1) If (a, b) is a solution of (8.7), then F1(a, b) = 0, F2(a, b) = 0 and, as a result, αF1(a, b) +βF2(a, b) = 0 + 0 = 0, which implies that (a, b) is also a solution of (8.8).

(2) If (a, b) satisfies (8.8) and β �= 0 (by assumption), then the relations F1(a, b) = 0 andαF1(a, b) + βF2(a, b) = 0 imply α · 0 + βF2(a, b) = 0, which is equivalent to F2(a, b) = 0,and this completes the proof.

�

In terms of the graphical interpretation of the equivalence of (8.7) and (8.8), the theorem saysthat the curves defined by the equations F2(x, y) = 0 and αF1(x, y) + βF2(x, y) = 0 intersect thegraph of the curve with the equation F1(x, y) = 0 at the same set of points. In Figure 8.14, wesee that the curve of F1(x, y) = 0 intersects the x-axis at x = −2, x = 0, and x = 2 while thecurve of F2(x, y) = 0 intersects the x-axis at x = −2, x = 0, x = 1, and x = 2. Thus, the systemF1(x, y) = 0, F2(x, y) = 0 has solutions (−2, 0), (0, 0), and (2, 0). From Theorem 8.10, then, thesystem F1(x, y) = 0, −2F1(x, y) + 3F2(x, y) = 0 will also have solutions (−2, 0), (0, 0), and (2, 0),as shown. Note that while (2/3, 0) is a solution to −2F1(x, y) + 3F2(x, y) = 0, this is not a solutionto the system.

We conclude this section with one more example. Several more impressive examples will beseen in the problem sets of future chapters, after we develop equations for different types ofcurves.

Example 32. Given an angle, ∠CBD, and a point A in its exterior, let−−→ARi, i = 1, 2, . . . , n,

intersect−→BC at a point Ci and

−→BD at a point Di . Let Qij = Qji be the intersection of the

segments CiDj and CjDi, i, j ∈ {1, 2, . . . , n} , i �= j . Prove that all points Qij are collinear.

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Qij

A

B

C

D

Ci

C

Di

Dj

j

FIGURE 8.15.

Solution: Choose a coordinate system such that B is the origin, and lines BD and BC becomethe x-axis and the y-axis respectively. (The system is not necessarily Cartesian; see Figure 8.15.)

Assume that A and−→BC lie on the same side of the line BD. Choose a scale such that point A

has coordinates (−1, a). Denote the slope of←→ACi by mi (which exists, since

←→ACi and

←→BC are not

parallel). Notice that mi �= 0, and mi �= mj for i �= j . Then an equation of←→ACi can be written

in the form: y − a = mi(x + 1). Therefore, the points Ci and Dj have coordinates (0,mi + a)and (−1 − a/mj , 0), respectively, and the coordinates of the point Qij can be found by solving

the following system of equations of the lines←−→CiDj and

←−→CjDi with

F1(x, y) = x

−1 − a/mj

+ y

mi + a− 1 = 0,

F2(x, y) = x

−1 − a/mi

+ y

mj + a− 1 = 0.

Subtracting these equations – which amounts to letting α = 1 and β = −1 in Theo-rem 8.10 – and simplifying the result, we obtain y = ax. Therefore, all points Qij lie onthe line with the equation y = ax. �

8.6 Equations of Circles

In Chapter 4, a circle was defined as the set of all points lying at a fixed distance (called the radius)from a specified point (the center of the circle). Consider any circle with center O : (h, k) and radiusr (denoted, as before, by C(O, r)). Then a point P : (x, y) is on C(O, r) if and only if d(P,O) = r .The formula for distance between two points yields the following equation for C(O, r):

(x − h)2 + (y − k)2 = r2.

Any circle can be represented with an equation of the above form, called the standard form forthe equation of a circle. By expanding and simplifying, the standard form for the equation of a circlecan be rewritten in what is called the general form:

x2 + y2 + Dx + Ey + F = 0. (8.9)

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8.6 Equations of Circles 173

Observe that the second-degree equation that describes a circle will have no xy-term. Will the graphof an equation of the form (8.9) always be a circle? Certainly not, since the solution set could beempty (such as for x2 + y2 + 1 = 0) or a single point (such as for x2 + y2 = 0).

However, if the equation x2 + y2 + Dx + Ey + F = 0 is satisfied by at least two distinct points,then its graph is indeed a circle. The center and radius can be determined by completing the squarewith respect to x and y, as we illustrate in the following example. Of course, the same is true for ageneral equation of the second degree of the form Ax2 + Cy2 + Dx + Ey + F = 0, if A = C �= 0.

Example 33. Determine the graph of 4x2 + 4y2 − 4x + 24y − 27 = 0.

Solution:

4x2 + 4y2 − 4x + 24y − 27 = 0 ⇔x2 + y2 − x + 6y = 27

4⇔(

x2 − x + 1

4

)+ (y2 + 6y + 9) = 27

4+ 1

4+ 9 ⇔(

x − 1

2

)2

+ (y + 3)2 = 16.

Thus, the equation 4x2 + 4y2 − 4x + 24y − 27 = 0 describes a circle of radius 4 with centerat the point (1/2,−3). �

Now we present several applications of the coordinate method to problems involving circles. Webegin with a “coordinate” proof of a fundamental theorem, Theorem 4.2.

Example 34. Prove that any three non-collinear points determine a unique circle.

Solution: Let A, B, and C be three non-collinear points. In order to simplify our computations,coordinatize the plane so that A : (0, 0), C : (1, 0), and B : (x

B, y

B) (with y

B�= 0), as shown in

Figure 8.16.We wish to find a unique point P : (x

P, y

P) so that d(A,P ) = d(B,P ) = d(C,P ). This

condition is equivalent to the system

d(A,P ) = d(B,P ), d(A,P ) = d(C,P ). (8.10)

Using the distance formula we get:

d(A,P ) = d(C,P ) ⇔√

x2P

+ y2P

=√

(1 − xP)2 + y2

P⇔ x

P= 1

2.

(This makes sense; a point is equidistant from A : (0, 0) and C : (1, 0) if and only if it lies onthe perpendicular bisector to AC, which is the vertical line x = 1/2.)

Using the distance formula again, with xP

= 1/2 and yB

�= 0, we get

d(A,P ) = d(B,P ) ⇔√(1

2

)2+ y2

P=√(

xB

− 1

2

)2+ (y

B− y

P)2 ⇔

yP

= x2B

− xB

+ y2B

2yB

.

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Thus, system (8.10) is equivalent to

xP

= 1

2, y

P= x2

B− x

B+ y2

B

2yB

, (8.11)

which means that the desired point P (the center of the circle) exists and is unique.

Our next two examples illustrate effective applications of Theorem 8.10.

Example 35. Find an equation of the line passing through the points of intersections of twocircles C1 and C2 given respectively by the equations

x2 + (y − 3)2 = 9 and (x − 4)2 + (y − 2)2 = 16. (8.12)

Solution: First of all we observe that the circles do intersect, because the distance between thecenters is

√17 and the radii are 3 and 4. Therefore, system (8.12) has a solution. Replacing the

second equation in (8.12) by its difference with the first one, we obtain system (8.13) which,according to Theorem 8.10, is equivalent to (8.12) (with α = 1 and β = −1 ):

x2 + (y − 3)2 = 9, 4x − y = 2. (8.13)

The line having equation 4x − y = 2 is the one we are looking for, because it crosses C1 atthe same points as C2. (See Figure 8.17.) �

Notice that we solved the problem of Example 35 without finding the points of intersectionof the circles. The latter could be done (if we wanted to) by solving the system (8.13). Manypeople solving this problem might not realize that obtaining the equation 4x − y = 2 solves theproblem.

Example 36. Let C1, C2, and C3 be three circles in the plane, such that any two of them intersectat two points. Prove that the three common chords are concurrent. (See Figure 8.18.)

(xB , yB)B:

(0, 0)A: (1, 0)C:

x = 1/2

xB2 - xB + yB

2

2yB)P:( ,1

2

FIGURE 8.16.

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8.6 Equations of Circles 175

Solution: Let Fi(x, y) denote the equation of Ci in a Cartesian coordinate system; that is,Fi(x, y) : (x − ai)2 + (y − bi)2 − (ri)2 = 0, i = 1, 2, 3. Let Lij be the line passing through thepoints of intersection of Ci and Cj . Using the same argument as in Example 35, we concludethat Fi(x, y) − Fj (x, y) = (aj − ai)x + (bj − bi)y + cij = 0 is an equation of Lij , where cij

is the constant term. The coordinates of the point of intersection of lines L12 and L13 can befound by solving the system

L12 : F1(x, y) − F2(x, y) = 0, L13 : F1(x, y) − F3(x, y) = 0. (8.14)

Replacing the second equation of (8.14) by its difference with the first one we obtain:

F1(x, y) − F2(x, y) = 0, F3(x, y) − F2(x, y) = 0, (8.15)

which, according to Theorem 8.10 (with α = 1, β = −1) is equivalent to (8.14). But the secondequation in (8.15) is an equation of the line L32 (= L23)! Therefore the lines L13 and L23 intersectthe line L12 at the same point. �

y

x

C1 C2

FIGURE 8.17.

C 2

C 3

C1L13

L12 L23

FIGURE 8.18.

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Our last example is an application of the coordinate method to finding a locus.

Example 37. Let points A and B and a positive number k be given. Find the locus of all pointsC such that CA/CB = k.Solution: Introduce a Cartesian coordinate system such that A : (0, 0) and B : (1, 0). A pointC : (x, y) is in the locus if and only if

AC

BC= k ⇔

√x2 + y2√

(x − 1)2 + y2= k ⇔ x2 + y2

(x − 1)2 + y2= k2. (8.16)

For k = 1, (8.16) is equivalent to x = 1/2, so the locus is the perpendicular bisector of AB. For0 < k �= 1, (8.16) is equivalent to each of the following.

x2(k − 1) − 2k2x + y2(k2 − 1) + k2 = 0,

x2 −( 2k2

k2 − 1

)x +

( k2

k2 − 1

)2+ y2 = k4

(k2 − 1)2− k2

(k2 − 1),(

x − k2

k2 − 1

)2+ y2 =

( k

k2 − 1

)2.

Hence the locus is the circle centered at (k2/(k2 − 1), 0) and of radius k/|k2 − 1|. It is oftenreferred to as “the circle of Apollonius.” �

Compare the solution above to the one for Problem 6.13. Isn’t this one much easier?

8.7 Problems

You just keep pushing. I made every mistake that couldbe made. But I just kept pushing.

— Rene Descartes (1596–1650)

Try to solve the following problems by using the coordinate method. Then see whether you cansolve them in other way(s), and compare the advantages of different approaches.

8.1 Prove that the diagonals of a parallelogram bisect each other.

8.2 Prove that the diagonals of a rhombus are perpendicular bisectors of each other.

8.3 Prove that in a right triangle ABC, where ∠C is the right angle, the median at C has lengthequal to half the length of side AB.

8.4 Prove that the midpoints of the sides of any convex quadrilateral are vertices of aparallelogram.

8.5 Consider the triangle with vertices A : (6, 0), B : (2, 4), and C : (1,−1).

(a) Find the length of the median at C.(b) Find the length of the altitude from C.

8.6 Given a positive number k and two intersecting lines l1 and l2, find the locus of all pointswhose difference of distances to the lines is k or −k.

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8.7 Problems 177

8.7 The midpoints of the sides AB and CD and of sides BC and DE of a convex pentagon (5-gon)ABCDE are joined by two segments. The midpoints H and K of these segments are joined.Prove that HK ‖ AE and HK = 1

4AE.

8.8 Prove that the three perpendicular bisectors to the sides of a triangle are concurrent.

8.9 Let ABCD be a parallelogram, and let F ∈ AD be such that AF = 15AD. Let E be the

intersection of BF and AC. Prove that AE = 16AC.

8.10 Let ABCD be a trapezoid with BC ‖ AD, AD = a, BC = b, a > b, and let O be theintersection of the diagonals AC and BD. A line through O that is parallel to the basesintersects lateral sides AB and CD at points E and F , respectively.

(a) Show that EO = OF .(b) Find EF .(c) Prove that the midpoints of bases AD and BC, point O, and the point of intersection of

sides AB and CD are all collinear.

8.11 Let ABCD be a trapezoid, and let P and Q be the midpoints of the bases BC and AD,respectively. Take a point M on

−→AC such that M is outside of the trapezoid. Let

←→MP and

←→MQ

intersect the lateral sides, AB and CD, at points H and K , respectively. Prove that HK isparallel to the bases.

8.12 Prove that the line joining the point of intersection of the extensions of the nonparallelsides of a trapezoid to the point of intersection of its diagonals bisects each base of thetrapezoid.

8.13 Given an equilateral triangle, �ABC, prove that the sum of three distances from any point inthe interior or on the boundary of the triangle to the sides is constant.

8.14 Prove that the area of a triangle whose vertices are (xA, y

A), (x

B, y

B), and (x

C, y

C)

equals

1

2|(x

Ay

B+ x

By

C+ x

Cy

A− x

By

A− x

Cy

B− x

Ay

C)|

= 1

2|(x

A− x

C)(y

B− y

C) − (x

B− x

C)(y

A− y

C)|.

8.15 Prove that if �ABC lies in a plane, where points A and B are fixed, then the center of thenine-point circle9 for �ABC must lie on or in the exterior of a circle centered at the midpointof AB and whose radius is equal to AB/4. Characterize those triangles for which the radiusof the nine-point circle is exactly AB/4.

8.16 (Ceva’s Theorem) Consider a triangle with vertices, A, B, and C, and points D, E, and F suchthat A − F − B, B − D − C, and C − E − A. Then AD, BE, and CF are concurrent if andonly if

AF

FB· BD

DC· CE

EA= 1.

9 The nine-point circle is a remarkable circle that can be constructed for any triangle. It contains the following nine points:

the midpoint of each side of the triangle, the foot of each altitude, and the midpoint of the segment of each altitude from its

vertex to the orthocenter.

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8.8 Supplemental Problems

S8.1 Prove that the diagonals of a rectangle are congruent.

S8.2 Each vertex of a quadrilateral is joined to the centroid of the triangle formed by the otherthree vertices. Prove that the four segments are concurrent. (This problem also appears asProblem S3.3.15.)

S8.3 Let M1, . . . , M6 be the midpoints of consecutive sides in an arbitrary hexagon (6-gon). Provethat the centroids of �M1M3M5 and �M2M4M6 coincide.

S8.4 Given �ABC, let A′ ∈ BC and B ′ ∈ AC, and let N be the point of intersection of AA′ andBB ′. Find AB ′/B ′C, if AN = 4NA′ and BN = 3NB ′.

S8.5 Let ABCD and A1B1C1D1 be parallelograms with A1B1C1D1 inscribed in ABCD in sucha way that vertices A1, B1, C1 and D1 lie on AB, BC, CD, and DA, respectively. Prove thatthe centers of these parallelograms coincide.

S8.6 Points A and B are moving along two rays that initiate at point O such that

1

OB− 1

OA= 1.

Prove that all lines AB pass through a fixed point.

S8.7 Let ABCD be a square, and let E be the midpoint of AB. Let points F ∈ BC and G ∈ CD

be chosen in such a way that EF || AG. Prove that FG is tangent to the incircle of the square.

S8.8 Given two circles in the plane such that each circle is in the exterior of another, let A1 andB1 be the points of intersection of the first circle with two tangents from its center to thesecond circle; likewise, let A2 and B2 be the points of intersection of the second circle withtwo tangents from its center to the first circle. Prove that A1B1 = A2B2. (This problem alsoappeared as Problem S4.15.)

S8.9 Fix a diameter in a circle, and let K be a point on the diameter. Let AB be a chord of thecircle that passes through K and forms a 45◦ angle with the diameter. Prove that the sumAK2 + KB2 does not depend on the position of K on the diameter. (This problem alsoappeared as Problem S4.11.)

S8.10 Two pairs of perpendicular lines in a plane intersect in such a way that four right trianglesare formed. Prove that the midpoints of the hypotenuses of these triangles are vertices of arectangle.

S8.11 Given two circles, one in the exterior of another, find the set of all points X in the plane suchthat the lengths of tangent segments from X to the circles are equal.

S8.12 Two points A1 and A2 are moving with equal constant velocity v along two intersecting linesl1 and l2, respectively. Prove that there exists a fixed point P in the plane such that at anymoment of time the distances A1P and A2P are equal.

S8.13 (Menelaus’ Theorem10) Consider �ABC. Suppose C1 ∈ AB, A1 ∈ BC, and B1 ∈ ←→AB such

that A − C − B1. Then points A1, B1, and C1 are collinear if and only if

AC1

C1B· BA1

A1C· CB1

B1A= 1.

10 Menelaus of Alexandria (c. 70–140 A.D.)

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8.8 Supplemental Problems 179

S8.14 (Newton’s Theorem) A quadrilateral is circumscribed around a circle. Prove that the center ofthe circle coincides with the midpoint of the segment joining the midpoints of the diagonalsof the quadrilateral.

S8.15 (Pappus’ Theorem11)

(a) Suppose points A1, B1, and C1 are on line l1 with A1 − B1 − C1, and points A2, B2, andC2 are on line l2 with A2 − B2 − C2. Let C3 = A1B2 ∩ A2B1, B3 = A1C2 ∩ A2C1, andA3 = B1C2 ∩ B2C1. Then points A3, B3, and C3 are collinear.

(b) Prove the same result if the six points are on a circle with

A1 − B1 − C1 − A2 − B2 − C2.

S8.16 (Bow-tie Theorem) Let AB be a chord in a circle, and C be the midpoint of AB. Let KL andMN be two chords of the circle which pass through C. Assume that K and M are on oneside of line AB. Let Q and P be the points of intersection of chords KN and ML with thechord AB, respectively. Prove that CP = CQ.

11 Pappus of Alexandria (c. 290–c. 350 B.C.)

Coordinatization - Mathematical Association of America · 1 From Calculus Gems, by G. Simmons, MAA,...

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