Paper Folding and Problem Solving

Paper Folding and Problem SolvingAuthor(s): Arsalan WaresSource: The Mathematics Teacher, Vol. 108, No. 3 (October 2014), p. 240Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/10.5951/mathteacher.108.3.0240 .

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240 MATHEMATICS TEACHER | Vol. 108, No. 3 • October 2014

MY FAVORITElesson

T H E B A C K P A G E

Problem solving is essential in mathematics. Rich problems can challenge and thrill us. This hands-on activity pro-vokes students in a secondary school geometry class to

think deeply about perimeters for several polygons as they touch, see, and manipulate a standard rectangular sheet of paper.

Designate a rectangle with corners A, B, C, and D to be p inches wide and q inches long, with p < q < 12p. Fold the paper as shown in figure 1, to locate points E and F. The diagonal AF therefore, has length 12p. Superimposing edges on creases locates points Bʹ and Dʹ.

We pose two tasks: Find the perimeters for (1) pentagon BʹXCYDʹand (2) quadrilateral AXCY.

The perimeter of pentagon BʹXCYDʹ is the sum of lengths:

BʹX + XC + CY + YDʹ + DʹBʹ = BC + CD + DE = q + p + (q – p) = 2q

Now we determine the perimeter of quadrilateral AXCY. As shown in figure 1b, �ABF is a right isosceles triangle with m∠BAF = 45o. Figures 1c and 1d illustrate that m∠BAX = m∠DAY = 22.5o, so �BAX ≅ �BʹAX is similar to �DAY ≅ �DʹAY.

In figure 1e, we see three right angles, ∠ABʹX, ∠ADʹY, and ∠FBʹX, as well as three 45o angles ∠BʹFX, ∠AFX, and ∠BʹXF. The congruent base angles assure us that �BʹFX is a right isosce-les triangle, with BʹF = BʹX. Therefore, we have

p

p

m AB X m AD Y m FB X

m B FX m AFX

m B XF m B FX

BX B X B F AF AB p p p

ABAD

BXDY

pq

p

DYDY q

AX p AY q

AX XC CY YA

AX BC BX DC DY AY

p q p p q

q

p q p

q

p q

p

2 .

2 .

90

45

90 45

2 ( 2 1) .

2 1( 2 1) .

4 2 2 and 4 2 2 .

4 2 2 2 1 2 1

4 2 2

4 2 2 4 2 2 2 2

2 2

4 2 2 2 2 .

4 2 2 ,

2

( )( )

( ) ( )

( )

( )

( ) ( )

( )( )

( ) ( )

( )

∠ ′ = ∠ ′ = ∠ ′ = °∠ ′ = ∠ = °∠ ′ = ° − ∠ ′ = °

= ′ = ′ = − = − = −

= → =−

→ = −

= − = −

+ + += + − + − +

= − + − −

+ − −

+ −

= − + − + −

+ −

= − + − +

−

Because �ABX is similar to �ADY and AD = q, we have

p

p


m B FX m AFX

m B XF m B FX


ABAD

BXDY

pq

p

DYDY q

AX p AY q

AX XC CY YA

AX BC BX DC DY AY

p q p p q

q

p q p

q

p q

p

2 .

2 .

90

45

90 45

2 ( 2 1) .

2 1( 2 1) .

4 2 2 and 4 2 2 .

4 2 2 2 1 2 1

4 2 2

4 2 2 4 2 2 2 2

2 2

4 2 2 2 2 .

4 2 2 ,

2

( )( )

( ) ( )

( )

( )

( ) ( )

( )( )

( ) ( )

( )

∠ ′ = ∠ ′ = ∠ ′ = °∠ ′ = ∠ = °∠ ′ = ° − ∠ ′ = °

= ′ = ′ = − = − = −

= → =−

→ = −

= − = −

+ + += + − + − +

= − + − −

+ − −

+ −

= − + − + −

+ −

= − + − +

−

By the Pythagorean theorem, in �ABX and �ADY , we have

p

p


m B FX m AFX

m B XF m B FX


ABAD

BXDY

pq

p

DYDY q

AX p AY q

AX XC CY YA

AX BC BX DC DY AY

p q p p q

q

p q p

q

p q

p

2 .

2 .

90

45

90 45

2 ( 2 1) .

2 1( 2 1) .

4 2 2 and 4 2 2 .

4 2 2 2 1 2 1

4 2 2

4 2 2 4 2 2 2 2

2 2

4 2 2 2 2 .

4 2 2 ,

2

( )( )

( ) ( )

( )

( )

( ) ( )

( )( )

( ) ( )

( )

∠ ′ = ∠ ′ = ∠ ′ = °∠ ′ = ∠ = °∠ ′ = ° − ∠ ′ = °

= ′ = ′ = − = − = −

= → =−

→ = −

= − = −

+ + += + − + − +

= − + − −

+ − −

+ −

= − + − + −

+ −

= − + − +

−

Thus, the perimeter of quadrilateral AXCY is

p

p


m B FX m AFX

m B XF m B FX


ABAD

BXDY

pq

p

DYDY q

AX p AY q

AX XC CY YA

AX BC BX DC DY AY

p q p p q

q

p q p

q

p q

p

2 .

2 .

90

45

90 45

2 ( 2 1) .

2 1( 2 1) .

4 2 2 and 4 2 2 .

4 2 2 2 1 2 1

4 2 2

4 2 2 4 2 2 2 2

2 2

4 2 2 2 2 .

4 2 2 ,

2

( )( )

( ) ( )

( )

( )

( ) ( )

( )( )

( ) ( )

( )

∠ ′ = ∠ ′ = ∠ ′ = °∠ ′ = ∠ = °∠ ′ = ° − ∠ ′ = °

= ′ = ′ = − = − = −

= → =−

→ = −

= − = −

+ + += + − + − +

= − + − −

+ − −

+ −

= − + − + −

+ −

= − + − +

−Readers may calculate trig ratios of the acute angles in �ABX.This problem is relatively easy to understand and model, yet

the solutions are not obvious. With the folded paper in front of them, students’ level of intellectual engagement is heightened, and the quality of communication becomes richer. Three reasons that this is one of my favorite lessons.

ARSALAN WARES, [email protected], teaches mathematics education courses at Valdosta State University in Georgia. He enjoys giving professional development workshops.

Paper Folding and Problem Solving

Arsalan Wares

Fig. 1 Folding a rectangular sheet of paper allows for calculation of areas.

p

q

D'

YB'

X F

A

C

Y

X

D'

F

A

B

D

C

B'

F

A

B

D

C

E

F

A

B

D

C

A

B C

D

The Back Page provides a forum for readers to share a favorite lesson. Lessons to be considered for publication should be submitted to mt.msubmit.net. Lessons should not exceed 600 words and are subject to abridgment.

Department editorRoger Day, [email protected], Illinois State University, Normal, IL

(a) (b) (c) (d) (e)

Copyright © 2014 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved.This material may not be copied or distributed electronically or in any other format without written permission from NCTM.



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