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# Constructing a quadrilateral inside another one

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## Abstract

Connect each vertex of a convex quadrilateral Q to the midpoint of the next (proceeding counterclockwise) side. The four connecting lines create an interior quadrilateral I. We study the ratio area(I)/area(Q). We also determine what happens to area(I)/area(Q) when the four midpoints are replaced by points which divide the sides in the ratio of rho to (1-rho) proceeding clockwise. Here rho is any fixed number satisfying 0 < rho < 1.
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
The description of Project 54 in 101 Project Ideas for the Geometer’s Sketchpad
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.)
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 (2ρ1)2
ρ2ρ+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
ρ2ρ+ 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 Mmhas 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 Mm=ρ3(ρ1)2
(ρ2+1)(ρ2ρ+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 Mmof about .034 is
achieved at the unique real zero of ρ5ρ4+ 6ρ36ρ2+ 7ρ3 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 180about point
M3, the reader can verify that we get AHGG(where Gis 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) Cand C′′ on
BC so that
Cis a distance ρBC from Cand C′′ is a distance (1)BC from Cand let
S=
CP
C′′P. Then the ratio rdeﬁned by equation (1) is maximal exactly
when Dis on S=Sint (ABC)ext (ABC ). Furthermore the set of possible
ratios is (1 ρ)3
ρ2ρ+ 1,(1 ρ)2
ρ2+ 1 #.
In the ﬁgure below, Sis 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. 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
s10
r10=ρy 0
1 + ρ(x1) 0
s10
r1ρ=10
0ρ
can easily be solved for r1and s1. The area of the interior quadrilateral is
(r1s2r2s1) + (r2s3r3s2) + (r3s4r4s3) + (r5s1r1s5)
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).
(ρ1)2
ρ4y4ρ3y43ρ5xy3+ 2ρ4xy3+ρ3xy32ρ2xy3
+2ρ5y36ρ4y3+ρ3y3+ 2ρ2y3ρy3+ρ6x2y2
ρ5x2y26ρ4x2y2+ 4ρ3x2y2ρ2x2y2ρx2y2
3ρ6xy2+ 10ρ5xy2+ 3ρ4xy213ρ3xy2+ 5
ρ2xy2+ρxy2xy2+ρ6y27ρ5y2+ 6ρ4y2+ 7
ρ3y27ρ2y2+ 2ρy2+ 2ρ5x3y2ρ4x3y3ρ3
x3y+ 2ρ2x3yρx3y3ρ6x2y5ρ5x2y+ 15ρ4
x2yρ3x2y7ρ2x2y+ 4ρx2yx2y+ 5ρ6xy
3ρ5xy 21ρ4xy + 18ρ3xy ρ2xy 3ρxy +x
y2ρ6y+ 5ρ5y+ 4ρ4y12ρ3y+ 6ρ2yρy
+ρ4x4ρ3x43ρ5x32ρ4x3+ 5ρ3x32ρ2x3
+ρ6x2+ 8ρ5x26ρ4x25ρ3x2+ 5ρ2x2ρx2
2ρ6x5ρ5x+ 12ρ4x4ρ3x2ρ2x+ρx +ρ6
4ρ4+ 4ρ3ρ2
y+ρ2xρx +x+ρ1ρy +x+ρ2ρ
ρ2y+ρx ρ+ 1
ρ2yρy +y+ρ2xρ2+ρ
Convexity means that (x, y) is constrained to the open “northeast corner” of the ﬁrst
quadrant bounded by YTX, Y ={(0, y) : y1},T={(x, 1x) : 0 x1},
X={(x, 0) : x1}. 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) =
(ρ1)2ρ5yρ+1
ρ(y(ρ1))
(ρ2y(ρ1))2((ρ2+ 1 ρ)yρ(ρ1))2.
Ash, Ash, and Ash 8
The only solution to r(y) = 0 with yYhas 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, 1x) has the following fully factored form
r(x) =
∂x r(x, 1x) =
(ρ+ 1) (ρ1)2ρ3((ρ1) xρ)xρ
ρ+1
((ρ1) xρ2)2(ρ(ρ1) 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) = (ρ1)2ρ3ρ2+ 1(x(ρ+ 1)) (x(1 ρ))
(x+ρ(ρ1))2((ρ2+ 1 ρ)x+ρ1)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. From
r(x) =
∂x rx, 1
ρx+η=
(ρ1)2ρ6ρ2+ 12ηρ+1
ρ2xρ2+ηρ ρ
ρ2+ 1
ηρ3+ρ33ρ2ηρ + 3ρ1x
ρ3+ 2ηρ2+ρηρ3η2ρ22ρ2ηρ!
(ρ+η1) ηρ2ρ+ 1ρ3ρ2+ρ1x+ρ2+ηρ ρ2
ρ3ρ2+ρ1x+ηρ3ρ3ηρ2+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 Cand C′′ on
BC so that Cis a distance (1/2) BC from C
Ash, Ash, and Ash 9
and C′′ is a distance 2BC from Cand let S=
CP
C′′P. Then the ratio rdeﬁned
by equation (1) is maximal exactly when Dis on S=Sint (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
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(1ρ)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),
Let $K$ be a convex pentagon in the plane and let $K_1$ be the pentagon bounded by the diagonals of $K$. It has been conjectured that the maximum of the ratio between the areas of $K_1$ and $K$ is reached when $K$ is an affine regular pentagon. In this paper we prove this conjecture. We also show that for polygons with at least six vertices the trivial answers are the best possible.