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arXiv:0704.2716v3 [math.GM] 17 Sep 2007

Constructing a quadrilateral inside another one

J. Marshall Ash

DePaul University

Chicago, IL 60614

Michael A. Ash

Department of Economics

University of Massachusetts Amherst

Amherst, MA 01003

Peter F. Ash

Cambridge College

Cambridge, MA 02138

1 The quadrilateral ratio problem

The description of Project 54 in 101 Project Ideas for the Geometer’s Sketchpad

[Key] reads (in part):

On the Units panel of Preferences, set Scalar Precision to hundredths.

Construct a generic quadrilateral and the midpoints of the sides.

Connect each vertex to the midpoint of an opposite side in consecu-

tive order to form an inner quadrilateral. Measure the areas of the

inner and original quadrilateral and calculate the ratio of these areas.

What conjecture are you tempted to make? Change Scalar Precision

to thousandths and drag until you ﬁnd counterexamples.

A ﬁgure similar to the following accompanies the project.

Ash, Ash, and Ash 2

E

F

H

G

A

B

C

D

Let rbe the ratio of the quadrilateral areas,

r=area (EF GH )

area (ABCD). (1)

The tempting conjecture is that r= 1/5. In Theorem 1 we show that this is true in

case the original quadrilateral is a parallelogram. However, the conjecture is false in

general. Instead, the ratio can be any real number in the interval (1/6,1/5]. This

is our Corollary 3. (Since 1/6<0.17 <1/5, it is possible to ﬁnd counterexamples

even with the Sketchpad Scalar Precision set to hundredths; however, it takes very

industrious dragging to ﬁnd them.)

Suppose now that instead of a quadrilateral we had a triangle. Of course, joining

each vertex to the opposite midpoint would not yield an inner triangle, since the

three lines are medians, which are concurrent in a point. To look for an analogous

result for a triangle, we can look for points which are not midpoints, but rather

divide each side a ratio ρof the distance from one point to the next, 0 < ρ < 1. For

deﬁniteness, we assume that “next point” in this deﬁnition is based on movement in

a counterclockwise direction. We call these points ρ-points. It turns out that for a

given ρ, the ratio of the area of the inner triangle to the area of the outer triangle is

constant independent of the initial triangle and is given by (2ρ−1)2

ρ2−ρ+1 . Note that when

ρ= 1/2, this reduces to 0, which provides a convoluted proof that the medians of

a triangle are concurrent. When ρ= 1/3, the area ratio is 1/7, which the Nobel

Prize-winning physicist Richard Feynman once proved, though he was probably not

the ﬁrst to do so. The result for general ρis known, and a proof is given in [DeV],

along with the Feynman story.

Ash, Ash, and Ash 3

Inspired by this result, we will study the quadrilateral question for ρ-points. Let

ABC D be a convex quadrilateral and let N1, N2, N3,and N4be chosen so that N1is

the ρ-point of BC ,N2is the ρ-point of CD,N3is the ρ-point of DA, and N4is the

ρ-point of AB. For ﬁxed ρ(0 < ρ < 1) connect each vertex of ABC D to the ρ-point

of the next side. (Ato N1,Bto N2,Cto N3, and Dto N4.) The intersections of

the four line segments form the vertices of a convex quadrilateral EF GH .

G

H

N4

E

F

N3

N2

N1

A

B

C

D

Deﬁne the area ratio

r(ρ, ABC D) = Area (E F GH)

Area (ABCD).

Theorem 2 below states that as ABCD varies, the values of r(ρ, ABC D) ﬁll the

interval

(m, M ] := (1 −ρ)3

ρ2−ρ+ 1,(1 −ρ)2

ρ2+ 1 #(2)

and that it is possible to give an explicit characterization of the set of convex quadri-

laterals with maximal ratio M. The fact that M−mhas a maximum value of

about .034 and is usually much smaller explains the near constancy of r(ρ, ABC D)

as ABC D varies. Here are the graphs of Mand m.

Ash, Ash, and Ash 4

A more delicate look at the graph of M−m=ρ3(ρ−1)2

(ρ2+1)(ρ2−ρ+1) shows that as “con-

stant” as ris in the original ρ= 1/2 case, it is even “more constant” when ρis close

to the endpoints 0 and 1. (Actually the maximum value of M−mof about .034 is

achieved at the unique real zero of ρ5−ρ4+ 6ρ3−6ρ2+ 7ρ−3 which is about .55.)

The characterization proved in Theorem 2 below shows that not only do paral-

lelograms have maximal ratio M(ρ) for every ρ, but also they are the only quadri-

laterals that have maximal ratio M(ρ) for more than one ρ.

2 The midpoint case for parallelograms

Theorem 1 If each vertex of a parallelogram is joined to the midpoint of an opposite

side in clockwise order to form an inner quadrilateral, then the area of the inner

quadrilateral is one ﬁfth the area of the original parallelogram.

Proof. In this picture,

Ash, Ash, and Ash 5

E

H

F

G

M3

M2

M1

M4

C

D

AB

ABC D is a parallelogram, and each Miis a midpoint of the line segment it lies on.

Cut apart the ﬁgure along all lines. Then by rotating clockwise 180◦about point

M3, the reader can verify that we get AHGG′(where G′is the image of Gunder

the rotation) congruent to EF GH . Similarly, each of the triangles ABE,B CF ,

and CDG may be dissected and rearranged to form a parallelogram, each congru-

ent to EF GH . Thus, the pieces of ABCD can be rearranged into ﬁve congruent

parallelograms, one of which is EF GH , which therefore has area 1/5 the area of

ABC D.

This result is a special case of Corollary 4 below, but is included because of the

elegant and elementary nature of its proof.

3 The ﬁlling of (m, M ]and the characterization

Theorem 2 Let A, B, C, D be (counterclockwise) successive vertices of a convex

quadrilateral. Deﬁne EF GH as the inner quadrilateral formed by joining ver-

tices to ρ-points as described in Section 1. Construct point Pso that ABCP

is a parallelogram. Locate(as in the following ﬁgure) C′and C′′ on −−→

BC so that

C′is a distance ρBC from Cand C′′ is a distance (1/ρ)BC from Cand let

S=−−→

C′P∪−−→

C′′P. Then the ratio rdeﬁned by equation (1) is maximal exactly

when Dis on S∗=S∩int (∠ABC)∩ext (△ABC ). Furthermore the set of possible

ratios is (1 −ρ)3

ρ2−ρ+ 1,(1 −ρ)2

ρ2+ 1 #.

In the ﬁgure below, S∗is indicated by the thickened portions of the rays com-

posing S.

Ash, Ash, and Ash 6

S*

S*

C’’

C’

P

A

BC

Proof. Fix ρand apply an transformation that maps A, B, C, D successively to

(0,1) ,(0,0) ,(1,0) ,(x, y). Since an aﬃne transformation preserves both linear length

ratios and area ratios, it is enough to prove the theorem after the transformation has

been applied. Observe that Phas become (1,1), and the image of Shas become

a pair of perpendicular rays through (1,1) with slopes ρand −1/ρ. Here is the

situation.

((1-ρ)x,1+(ρ-1)(y-1))

(0,1-ρ)

(1+ρ(x-1),ρy)

(ρ,0)

(r4,s4)

(r3,s3)

(r2,s2)

(r1,s1)

(0,1)

(0,0)

(x,y)

(1,0)

The line from (0,0) to (x, y) divides the outer quadrilateral into two triangles, one

of area x/2 and the other of area y/2, so that its area is (x+y)/2. To ﬁnd the area

of the inner quadrilateral, we ﬁrst determine {r1, s1,...,s4}in terms of x,yand ρ

Ash, Ash, and Ash 7

by equating slopes. For example, the equations

s1−0

r1−0=ρy −0

1 + ρ(x−1) −0

s1−0

r1−ρ=1−0

0−ρ

can easily be solved for r1and s1. The area of the interior quadrilateral is

(r1s2−r2s1) + (r2s3−r3s2) + (r3s4−r4s3) + (r5s1−r1s5)

2.

This is the n= 4 case of a well-known formula for the area of an n-gon[Bra] which

can be proved by ﬁrst proving the formula for triangles and then using induction, or

by using Green’s Theorem. Some computer algebra produces this formidable and

seemingly intractable formula for r(x, y).

(ρ−1)2

ρ4y4−ρ3y4−3ρ5xy3+ 2ρ4xy3+ρ3xy3−2ρ2xy3

+2ρ5y3−6ρ4y3+ρ3y3+ 2ρ2y3−ρy3+ρ6x2y2

−ρ5x2y2−6ρ4x2y2+ 4ρ3x2y2−ρ2x2y2−ρx2y2

−3ρ6xy2+ 10ρ5xy2+ 3ρ4xy2−13ρ3xy2+ 5

∗ρ2xy2+ρxy2−xy2+ρ6y2−7ρ5y2+ 6ρ4y2+ 7

∗ρ3y2−7ρ2y2+ 2ρy2+ 2ρ5x3y−2ρ4x3y−3ρ3

∗x3y+ 2ρ2x3y−ρx3y−3ρ6x2y−5ρ5x2y+ 15ρ4

∗x2y−ρ3x2y−7ρ2x2y+ 4ρx2y−x2y+ 5ρ6xy

−3ρ5xy −21ρ4xy + 18ρ3xy −ρ2xy −3ρxy +x

∗y−2ρ6y+ 5ρ5y+ 4ρ4y−12ρ3y+ 6ρ2y−ρy

+ρ4x4−ρ3x4−3ρ5x3−2ρ4x3+ 5ρ3x3−2ρ2x3

+ρ6x2+ 8ρ5x2−6ρ4x2−5ρ3x2+ 5ρ2x2−ρx2

−2ρ6x−5ρ5x+ 12ρ4x−4ρ3x−2ρ2x+ρx +ρ6

−4ρ4+ 4ρ3−ρ2

y+ρ2x−ρx +x+ρ−1ρy +x+ρ2−ρ

∗ρ2y+ρx −ρ+ 1

∗ρ2y−ρy +y+ρ2x−ρ2+ρ

Convexity means that (x, y) is constrained to the open “northeast corner” of the ﬁrst

quadrant bounded by Y∪T∪X, Y ={(0, y) : y≥1},T={(x, 1−x) : 0 ≤x≤1},

X={(x, 0) : x≥1}. Restricting rto Y, we get a formula for r(y) = r(0, y). Taking

the derivative unexpectedly gives this simple, completely factored formula:

r′(y) =

(ρ−1)2ρ5y−ρ+1

ρ(y−(ρ−1))

(ρ2y−(ρ−1))2((ρ2+ 1 −ρ)y−ρ(ρ−1))2.

Ash, Ash, and Ash 8

The only solution to r′(y) = 0 with y∈Yhas y=ρ+1

ρ. It quickly follows that

on Y,rattains a maximum value of Mat 0,ρ+1

ρand is minimized by mat the

endpoints (0,1) and (0,∞). (By this we mean that limy→∞ r(0, y) = m.) Similarly

on T, the derivative of r(x) = r(x, 1−x) has the following fully factored form

r′(x) = ∂

∂x r(x, 1−x) =

(ρ+ 1) (ρ−1)2ρ3((ρ−1) x−ρ)x−ρ

ρ+1

((ρ−1) x−ρ2)2(ρ(ρ−1) x−(ρ2+ (1 −ρ)))2,

so that rhas minimum value mat the endpoints (0,1) and (1,0) and maximum

value Mat ρ

ρ+1 ,1

ρ+1 ; while on X,

r′(x) = ∂

∂x r(x, 0) = (ρ−1)2ρ3ρ2+ 1(x−(ρ+ 1)) (x−(1 −ρ))

(x+ρ(ρ−1))2((ρ2+ 1 −ρ)x+ρ−1)2,

so that rhas minimum value mat the endpoints (1,0) and (∞,0) and maximum

value Mat (ρ+ 1,0). Motivated by these results we now sweep the region of per-

missible values of (x, y) by line segments with y-intercept ηand slope −1/ρ. From

r′(x) = ∂

∂x rx, −1

ρx+η=

(ρ−1)2ρ6ρ2+ 12η−ρ+1

ρ2x−ρ2+ηρ −ρ

ρ2+ 1

ηρ3+ρ3−3ρ2−ηρ + 3ρ−1x

−ρ3+ 2ηρ2+ρ−ηρ3−η2ρ2−2ρ2−ηρ!

(ρ+η−1) ηρ2−ρ+ 1ρ3−ρ2+ρ−1x+ρ2+ηρ −ρ2

ρ3−ρ2+ρ−1x+ηρ3−ρ3−ηρ2+ρ2+ηρ2!

it is clear that η=ρ+1

ρproduces one arm of the image of S. Finally for all other

η,rhas mound-shaped behavior with minimum values on the coordinate axes and

reaches a maximum of Mwhere y=−1

ρx+ηintersects the other arm.

Recall that we have deﬁned ρ-points in terms of counterclockwise orientation.

Although Theorem 2 is true for clockwise orientation, we stress that the value of

rdepends, in general, on the orientation. In fact, clockwise and counterclockwise

orientations always give diﬀerent values of runless Dlies on the diagonal −−→

BP .

Setting ρ= 1/2 in Theorem 2 yields this corollary.

Corollary 3 Let A, B, C, D be (counterclockwise) successive vertices of a convex

quadrilateral. Deﬁne EF GH as the inner quadrilateral formed by joining vertices

to midpoints as described in Section 1. Construct point Pso that ABC P is a

parallelogram. Locate C′and C′′ on −−→

BC so that C′is a distance (1/2) BC from C

Ash, Ash, and Ash 9

and C′′ is a distance 2BC from Cand let S=−−→

C′P∪−−→

C′′P. Then the ratio rdeﬁned

by equation (1) is maximal exactly when Dis on S∗=S∩int (∠ABC )∩ext (△ABC).

Furthermore the set of possible ratios is

1

5,1

6.

Another corollary of Theorem 2 is the following generalization of Theorem 1

from midpoints to ρ-points.

Corollary 4 If each vertex of a parallelogram is joined to the ρ-point of an opposite

side in counterclockwise order to form an inner quadrilateral, then the area of the

inner quadrilateral is (1−ρ)2

ρ2+1 times the area of the original parallelogram.

A nice geometry exercise is to prove this corollary avoiding the calculus part of

the proof of Theorem 2. Hint: Performing the aﬃne transformation we may assume

that the original quadrilateral is the unit square. Use slope considerations to see

that the interior quadrilateral is actually a rectangle. Use length considerations to

see that it is a square of side length q(1−ρ)2

ρ2+1 .

References

[Bra] B. Braden, The Surveyor’s Area Formula, College Math. Journal 17 (1986),

326–337.

[CW] R.J. Cook and G. V. Wood, Note 88.46: Feynman’s Triangle, The Math.

Gazette 88 (2004), 299–302.

[DeV] M. De Villiers, Feedback: , Feynman’s Triangle, The Math. Gazette 89 (2005),

107. See also http://mysite.mweb.co.za/residents/profmd/feynman.pdf.

[Key] Key Curriculum Press, 101 Project Ideas for Geometer’s Sketchpad, Version

4, Key Curriculum Press, Emeryville CA, 2007.