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School and University Partnership for Educational Renewal in MathematicsAn NSF-funded Graduate STEM Fellows in K–12 Education Project

University of Hawai‘i Department of Mathematics

Folding a sheet of paper in three.

Overview Several methods can be used to fold a piece of paperinto exactly three equal parts. After reading aboutthese methods, students are asked to explain whythese methods work?

Grades 10th through 12thNecessaryEquipment

handouts with different methods. Sheets of paper totry methods.

necessary time 1 through 2 class periodsProposedmethod ofinstruction

Have students work in small groups. The teachergives instructions to each group.

Prior knowl-edge necessary

See the elements of synthesis

Content stan-dards ad-dressed

Trigonometry, similar triangles, sufficient conditionsfor congruent triangles

UHM Department of [email protected]



Folding a sheet of paper in three.When regular sheets of paper must be put inside a business envelope, it is folded in thirds. To do this, anempirical and rough method is used.

Here are other, more rigorous methods to fold a piece of paper in exactly three equal parts. Explain whyeach of the following methods work.




1st method:Starting with a new piece of paper, fold it twice edge-to-edge then open it again. 4 equal sections and 5parallels lines are created.

Take a second piece of paper and place it over the first as shown in the diagram. The left two creases canbe used to mark the exact locations of the thirds on the second piece of paper.




2nd method:Fold a new piece of paper twice edge-to-edge and open it again.

Fold the top of the paper down, bringing the upper right corner to touch the right crease. While foldingthe paper, be careful not to shift the other creases. The left two creases can be used to mark the exactlocations of the thirds.




3rd method:Consider the corners of the paper to have labels A, B, D, and E. Point C and F are found by folding thepaper in half. Fold the paper accordingly to create a crease along the diagonal (AD). The paper is nowcomprised of a left and right rectangle. Similarly, create creases along the diagonals BF and CE. The twointersections created from the 3 creases indicate where the paper can be folded in thirds. Note that theseintersections can be used to fold the paper into thirds the long way or the short way.




4th method:Fold a new piece of paper in half, along the short edge, and then unfold. The paper is now comprised ofan upper and lower triangle. Make creases along the diagonals of the lower triangle.

Create a new crease starting from the left corner of the paper to the intersection of these diagonals. Createa second similar crease corresponding to the right upper corner of the paper. The second should be betweenthe upper right corner and the intersection of the diagonal creases. The two endpoints of these newestcreases along the edge of the paper correspond to the two marks indicating thirds. partitioning the longerlength of the paper.

As indicated in the figure below, thirds of the short length of the paper can also be created. Just createcreases parallel to the short edge of the paper crossing the aforementioned third demarcations locatedalong the longer edge. The intersections of these newest creases and those seen in the diagram indicatethe thirds partitioning the shorter length of the paper.




5th method:Create a diagonal crease (BD). Fold the longer edge of the paper, (BC), up to the aforementioned creaseto create the point C ’ and reopen the paper.

Create a crease parallel to the shorter edge intersecting C’. Define E as the intersection of this crease andthe lower edge.

Refold the longer side BC along the diagonal BD. The point E, will intersect the diagonal BD. Denotethis intersection point E’. Re-open the paper. Creases parallel to the longer or shorter edge of the paperintersecting E’ will denote third demarcations.




Elements for synthesis

Methods 1 and 2It is sufficient, to see that each of the thirds are bases to right triangles. Each of these triangles are con-gruent due to ASA. Hence the three partitions of the paper denoted by the methods must necessarily beequal.

Method 3

Note that point I denotes the exact center of the paper.

Consider triangles ∆IHF and ∆CHB.

Because angles︷︸︸︷IHF and

︷︸︸︷BHC are vertical, they are equal. EF is parallel to BC. Due to the equality

of alternate internal angles,︷︸︸︷HIF =

︷︸︸︷HBC and

︷︸︸︷HFI =

︷︸︸︷HCB. Hence triangles ∆IHF and ∆CHB are

similar.¯IF = 1

2 CB, because I is in the vertical center of the paper.Because the two aforementioned triangles are similar,

¯IHHB

= ¯IFCB

=12CB

CB−→ ¯IH = 1

2HB

Let us note points E, F, and I as on the drawings. Because ABCD is a rectangle, I is the midpoint ofthe segment BD.12BD = ¯IB = ¯IH + HB = HB 1

2 + HBSo 1

2BD = 32HB, and HB = 1

3BD.

Consider the right triangles created in the diagram such that adjoining the hypotenuses of the threetogether, creates the diagonal (e). The topmost is indicated by points D and G. The middle is indicatedpoints G and H. The lowest is indicated by points H and B. Because the unnamed horizontal creasesintersecting H and K are parallel to the longer edges, the lower right angle of each of the three afore-mentioned triangles is equal. Because each is a right triangle, the three triangles are similar. BecauseBH = HG = GD = 1

3BD, all three triangles are congruent, using ASA. Hence, the longer side of thepaper is partitioned into thirds by the triangles’ equal bases. The shorter side of the paper is partitionedinto thirds by the triangles’ equal heights.




Method 4

Note that points I, J, and K indicate the partition of the paper into thirds, from method 4. Multiplemethods of proof are accurate.

Choice 1:

Let us consider triangles ∆CHB and ∆CGO.

These triangles are similar because︷︸︸︷HCB =

︷︸︸︷GCO.




G is the intersection of the diagonals of the rectangle ABFE, hence O is the middle of FB. Since F isthe middle of CB (due to the way the first crease was made), CO = 3

4 CBalso, note that GO = 1

2AB

Because the two triangles are equal, we have:BHOG

= CBCO←→ BH

12AB

= BC34BC←→ BH =

12AB·BC34BC

= 23AB

Hence AH = 13AB, and by symmetry, BH = 1

3AB and therefore points H and I denote partitions ofthe paper into thirds.

Choice 2:

Let us consider triangles ∆CGO and ∆PIH. These triangles are similar because GO is parallel toAB, and PI is parallel to BC. G is the intersection of the diagonals of the rectangle ABFE. Because GOis parallel to AB, O is the midpoint of BF. Due to the manner that the first crease was created, F is themiddle of CB, so CO = 3

4 CBalso, GO = 1

2ABBecause P is on the horizontal crease that bisects the paper, P I = 1

2 CBBecause the aforementioned triangles are similar:

12CB

34CB

= P ICO

= HIGO

= HI12AB

Hence, HI = 13AB

We must still show that AH = 13AB




Possibility 1:Let us consider triangles ∆CBH and ∆PIH.These triangles are similar because PI is parallel to CB.As before, P I = 1

2 CBBy what precedes HI = 1

3ABBecause the aforementioned triangles are similar:

CB12CB

= CBP I

= HBHI

= HB13AB

and therefore HB = 23AB

Hence points H and I partition the long side of the paper into thirds

Possibility 2:

Denote F as the midpoint of BC and consider triangles ∆PIH and ∆CFP .These right triangles are similar because PI is parallel to CB.As P I = FB = CF , the aforementioned triangles are congruent by ASA.Hence, ¯IB = FP = HI = 1

3AB

Hence points H and I partition the long side of the paper into thirds




Now, we will show that J partitions the shorter edge of the paper into thirds.

Consider triangles ∆EAB and ∆JHB.These triangles are similar, because JH is parallel to EABecause E is the midpoint of DA, EA = 1

2DABy previous arguments, HB = 2

3ABBecause the aforementioned triangles are similar,

¯JHEA

= HBAB←→ ¯JH

12DA

=23AB

AB←→ ¯JH = 1

3DA

Hence J partitions the shorter side of the paper into thirds




Method 5

A standard paper is 8.3 inches by 11.7 inches. Hence, we can write and equation relating BC, the longerside, and CD, the shorter side as: BC =

√2CD

Without loss of generality, suppose BC = 1. We will show that the point E ’ indicates partitions of therectangle ABCD into thirds along both the shorter and longer edges.From the equation above, CD =

√2

2

From the Pythagorean theorem, BD =√

32

Let us note the angle︷︸︸︷CBD by α. Because C’ was created by folding BC along the diagonal, 1 = BC =

¯BC ′

From trigonometry, cos(α) =√

23

sin(α) =√

33

From trigonometry, cos(α) = BE¯BC′

Hence BE = 1 · cos(α)




Because E’ was created by folding BE along the diagonal, cos(α) = BE = ¯BE′

From trigonometry, cos(α) = BJ¯BE′ = BJ

cos(α)

Hence BJ = cos2(α)Note also that sin(α) = EJ

¯BE′ = ¯ICcos(α)

Hence ¯IC = sin(α) · cos(α)Note that from previous calculations of cos(α) and sin(α)cos2(α) = 2

3

sin(α) · cos(α) =√

23

Note that because,√

23 =

√2

2 ·23

sin(α) · cos(α) = CD · 23

CI = CD · 23

¯ID = CD − CI = CD · 13

Hence point E’ indicates a third of the shorter edge of the paper.

Because 23 = 1 · 2

3cos2(α) = BC · 2

3BJ = BC · 2

3E′I = BC − BJ = 1

3BCHence point E’ indicates a third of the longer edge of the paper.


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