Dissections of regular polygons

An hexagon into an equilateral triangle

The hexagon area is  6×a/2×a√3/2 = 3a²√3/2.
Naming c the side of the equivallent triangle :
 c/2×c√3/2 = √3/4 c² = 3√3/2 a²
That is c² = 6a² or c/2 = a√3/2 × √2 which gives a construction of c/2.
Draw HM = HF and DN = DI then the perpendicular bisector of IN intersects FM in P.
FM = IN = c/2, and triangle INP is equilateral. (PP' = IF/√2 etc...)
The target triangle is constructed by copying IT = IP then the parallel TR to PN. Its internal cutting lines are drawn by MU = MA, MW = MF and finally UV = UW.

The piece ICBST is symmetric of IODNP through I. Piece MUW is la symmetric of MAF through M. Triangle UVW is equilateral (isosceles by construction and angle WUV = 60°) and equal to triangle COD. As FM is half the side of triangle and MW = FM by construction, FW = PR and the piece WRSV is get by translation of FPNE, and vector FW.

Another older construction, not the best as using 6 pieces :
AN = AI, then IN = AI√2 = AF√6/2 is half the side of triangle.
Construct HM = HC in the same way (that is CM = IN), then P on the perpendicular bisector of CM.
The altitude of triangle CMP is distance between the lines IN and MC at 45° from IC, hence this altitude is IC/√2 = 3a/(2√2) = (√3/2).(a√6/2) = MC√3/2.
Triangle MCP is then equilateral : MP = PC = c/2
Draw CF = CP and EF // MC to complete the target triangle.
IEL symmetric of IAN and MGS symmetric of MLP
Pieces AFI and FIE are equal to pieces CDH and HKC, finally CFGK is le symmetric of CPNB

Hexagon to square

The first dissection here is the one by E. Lucas (1891)
The hexagon AB'C'D'EF is cut by a diameter to get a parallelogram ACDF.
The side s of the equivallent square is the mean proportionnal between the length of this parallelogram and its altitude : s² = FI.FD
Draw FH = FI, the perpendicular to FD from H intersects the circle with diameter FD in M. FM is the side of the equivallent square.
Complete the square FMNP. Draw FK = UC (or PK // CD)
The triangles UWD and FHM are similar (angles from perpendicular lines) and because UW = FH, they are equal and UD = FM = FP.
Triangle FKP can then be obtained by translation from UCD.
KP // CD // AF and equal, triangles PKV and FAG are then equal from translation.
Finally PV = FG gives VN = GM and triangles VND and GMU are equal.

Important note : here also, avoid approximations. H and K are different points and FG is not parallel to EB.
FK = UC = WD – AI = HM – AI. As FM² = FH.FD = 3a²√3/2 = HM² + HF², HM² = 3a²√3/2 – 3a²/4 that is FK ≈ FH × 0,9924... here also about the thickness of the line.
As for angle α = MFH, cos(α) = FH/FM gives a ≈ 57° 30'. More than 2° apart !

Sorry, your browser is not Java compliant. Another dissection, allowing to vary the exact shape of the pieces (M draggable) :

The side of the equivallent square is constructed as above : AK = AH
Then L, intersection of perpendicular in K to AD and circle centered in O with radius OA = 3/2 a (diameter = 3a then).
A circle centered in A with radius AL intersects CE in P.
Choose any line MN // CE, not too close from CE so that point I be between A and P, then draw the perpendiculars MI and NJ to AP.
Copy JT = AP and complete the square TJRS.
MI + NJ doesn't depend on line MN because MIX, NJX and AOP are similar MI/MX = NJ/NX = MI + NJ/MN = AO/AP and as MN.AO = AP², MI + NJ = AP. Then NR = MI.
Because AG/JR = MX/MI = AP/AO, AG = MN = CE = PW then EW = CP, the piece NEWR is then equal to piece MCPI.
TAGS is equal to JPWR, with its two parts.
AFG equals to CDE, and finally GFNR is equal to ABMI.

We can choose MN freely, anywhere as above,
or so that I = P, or so that F = N and M = B, which results into different dissection patterns.

A pentagon into a triangle
Let 2c the side of the equivallent triangle. Its area is  1/2.2c.2c√3/2 = c²√3 = h.MN.
Draw AN' = MN and circle with diameter AN'. From A as center and with the same radius, an arc of circle intersects it at S, AS intersects MN in S'. AS' = h/√3.
Copy AH = AS', the perpendicular in H to AB intersects the circle with diameter AN' in T : AT² = AH.AN' = h.MN/√3 = c². The circle centered in A with radius AT intersects DE in X, farthest from E. AX = c. Draw the parallels FG to AX from M, WV through N and IJ through midpoint O of MN.
Draw the perpendicular OP to FG from O. Complete the drawing by constructing PIQ and PJR.
Important note : Lines AX, FG, IJ and QR are NOT parallel to BD
Calculating gives AX ≈ h × 1.04794... and BD ≈ h × 1.05146...
IJ = c by construction. Triangles PIJ and PQR are similar QR/IJ = PQ/PI = MO/MN = 1/2, QR = 2c.
Triangles POM and YIJ are similar, OP/IY = OM/IJ, that is OP = OM.IY/IJ = (MN/2).h/c = c√3/2. Triangle PIJ has angle in I equal to 60°, is then equilateral, PI = PJ = c. Triangle PQR is then the target equilateral triangle.
AFM and EGM symmetric through center M are equal, triangles GPJ and VRJ symmetric from center J are also eqal. Triangles FIP and WIQ are symmetric through center I and finally triangles WCN and VDN are symmetric through center N. All pieces of the pentagon then fit in the triangle PQR.

Another dissection of pentagon into triangle :

Let M the midpoint of CD, IJ the parallel to AE from M.
Copy DU = DM, V midpoint of EU, the perpendicular to DE from V, and DF with angle FDH 60° (arcs centered in D and H with radius DH intersect on line DF)
The circle centered in F with radius FV intersects AB in P. Draw DT // VP.
The perpendicular to DT from its midpoint intersects PV in Q.
Extend TR = TQ and DS = DQ.

DV = (ED + DU)/2 = MN/2. FDU = 30° then FV = MN/2√3 and VW = MN/√3 that is VP² = VX.VW = h.MN/√3 = c half side of the equivallent triangle.
The triangles VDY and AVX are similar VD/AV = DY/VX that is DY = VD × VX / AV = MN/2 × h/c = c√3/2.
Triangle DQT is then equilateral, QT = QD = c. Triangle QRS double of this one is therfore the equivallent triangle. And more RS // TD.
Triangle DMJ is symmetric of CMI through center M. Angle MJD = 2π/5 and MDJ = π/5, hence DMJ = 2π/5, triangle DMJ is isosceles and DM = DJ.
QVD is the symmetric of SLD through center D, PTQ is symmetric of KTR.
Finally JL = DL – DJ = DV – DU = UV = EV, AE // IJ, PV // KL and the pieces APVE and IKLJ are then equal.

Beware : PV is not parallel to AE.

A pentagon into a square
Dissect a regular pentagon to build a square.

The "cap" (BCD) of pentagon is cut to get a trapezoid ACDE. A strip from equivallent squares is then set so as its edges go through the midpoints M and N of the trapezoid's sides.

Copy BF = MN and BI = BH = h, altitude of trapezoid.
The perpendicular in I to BF intersects the circle with diameter BF in J. BJ² = BI.BF = h.MN gives the side of the equivalent square.
The perpendicular to BJ in B intersects DE in P. Draw the perpendiculars to BJ from M and N and the parallel to BJ from P. This constructs the "square" GSQR, remains to prove it is really a square and all its sides are equal, up to now it is just a rectangle.

Triangles BJI and PBH are similar, and even equal as BI = BH, hence BP = BJ and SG = QR = BP by construction.
Triangles BJI and MNT are similar, BJ/MN = BI/NT that is BJ.NT = MN.BI = h.MN = BJ²
Hence NT = BJ and QS = GR = NT = BJ. It is the equivallent square.
Triangles MAK and MEU are equal, symmetric through M, UK = PB = GS hence UG = KS and triangles UGP and KSB are equal. UGP is made from the two pieces UME and MGPE.
Important note : G and M are different, MGPE is a true quadrilateral, even if its side MG is quite small.
Triangles NWC and NVD are equal from symmety through N.
GP = SB hence PR = BQ and triangles BWQ and PVR are equal.
All the pieces of pentagon fit then in the square.

Lucas gave the dissection of pentagon into square with 7 pieces. This one has then only a historical interest.

The "cap" BCD of pentagon is cut to get the trapezoid ACDE.
The parallel IJ to AE from midpoint M of CD gives the equivallent parallelogram AJIE.
Copy EU = EH, altitude of this parallelogram.
The perpendicular to EI in U intersects the circle with diameter EI in P. IP intersects AB in L. Complete the square EPQR.
EP² = EU.ED gives the side of the equivallent square, mean proportionnal between base EI and altitude EU = EH.
Triangles EPU and ILK are similar, and because EU = IK, they are equal, hence IL = EP = PQ.
Let J'R the parallel to JI in R. Triangles EJ'R and LJI are similar, and equal because ER = IL
As IL = PQ, PL = IQ. The pieces ALPE are J'IQR are then equal, all pieces fit.

The polygons with 7, 9, 11 sides are not constructible with compass and straightedge, and won't be handled here.
Stay then about dissections in triangles and squares and be cool with the dissection of octogon into a square.

Octogon into square

Let M, N, I, J the midpoints of opposite sides of octogon.
We have to construct a square with side the side of octogon, and going through M,N,I,J.
That is to construct two parallel lines MA and NC at distance = c, side of octogon.
The circle with diameter MN and circle with radius c intersect in P.
MP is the searched MA line, and the four other immediately result.

The proof is from translation of the pieces, obvious angle conditions, symmetries and equal segments.

This dissection comes from a tiling of the plane with octogons + squares with side c, overlayed with a tiling from equivallent squares + squares with side c.

Beware : P is not the midpoint of a side ! hence MP is not parallel to the diagonal of octogon.

Octogon into a triangle

Compared, the dissection of octogon into a triangle is much more complicated.

This dissection is done by first dissecting the octogon into a periodic strip. The condition ZT = TE = EY gives ET/ED = HA/HK and construction of this dissection through a dilation centered in O, intersection of HE and KD.
The area of octogon is then just the half a complete period (2 octogons) in this strip. That is d×h.
We can then construct the side of equivallent triangle.
The side s of this triangle is deduced from  s²√3/4 = d×h that is  (s/2)² = d.h/√3
Line KM from the vertex of the square circumscribed to octogon, at 30° of vertical, gives LM = h/√3. A circle with diameter d intersects the vertical MS in S then LS is half the side of the equivallent triangle.
Let U the midpoint of AH.
Construct the equilateral equilateral PWU, then the parallel NV to UW from P.

We specifically care of points which are NOT at midpoints of segments. Specifically W is not at midpoint of FG : WG < WF, hence NV and UW are not parallel to GH.
This implies that the piece NCXV is not a trapezoid.
But V is really the midpoint of XT, and NV = UW.
The fastest way to prove it, is by folding the pieces on the tiling, considering that the triangle so constructed is equilateral and has same area as octogon, by construction.

Decagon to square

The following dissection results into "weird" shapes, but it is the minimum (7 pieces). Discovered by Gavin Theobald.

Let I the midpoint of AB = a, the decagon area is 20× that of triangle OIB. 20 copies of this triangle can then cover a rectangle with sides BD et BH = 5×BI. The perpendicular in H to BD intersects the circle with diameter BD in M. BM is the mean proportionnal between BH and BD, and is then the side of the equivallent square.

Draw FG = FE on radius OF and GL parallel to EF, L being the intersection of this parallel with BC.
The circle centered in B with radius BM intersects line GL in P. Draw CN perpendicular to BP.
Draw FV = LP. The circles centered in V with radius PN and centered in G with radius CN intersect in W, that is VW = PN and GW = CN.
Draw EU with angle UEO to radius EO equal to angle NBD = α
Copy AR = GU and draw the perpendicular RS to BP.

To prove the covering of the square, after considering angles, segments equal to side of decagon and equal segments by construction, remains to prove that LC = FG = a.
BF _|_ GL intersects GL in X. BF = 2a(sin(π/5) + cos(π/10)), FX = a cos(π/10) hence BX = a(2sin(π/5) + cos(π/10)).
BC = a(1 + 2cos(π/5)) and BX = BL sin(π/5) = CL sin(π/5) + BC sin(π/5) = CL sin(π/5) + a sin(π/5) + 2a cos(π/5)sinπ/5) and because 2cos(π/5)sin(π/5) = sin(2π/5) = –cos(π/10), CL equals a.

Note : the perpendicular in H does NOT go through C.
The projection of BC = a(1 + 2cos(π/5)) on BD is a(1 + 2cos(π/5))cos(π/10) ≈ a×2.489... ≠ 5a/2

Dodecagon to square

Yet another "easy" dissection !
No comments...

The dissections of decagon and dodecagon to triangle are "hideous*" hence we won't say a word about them, and care only of easier dissections.
* the difficulty being the weird dissection of the n-gon into a strip !
We can find on Gavin Theobald web site many other dissections. Including n-gons to p-gons with n, p > 4.

Stay a while with quite easy dissections, and dissect a pentagon to hexagon :

Pentagon to hexagon

From an already seen dissection of pentagon into a parallelogram and dissection of hexagon into a strip. The method gives, by overlaying the strips, the wanted dissection (H. Lindgren 1964).
The hexagon area is xy = x²√3/2 we have then dh = x²√3/2. That is x² = d × 2h/√3
We have to construct, given the pentagon, the side x of the rectangle equivallent to hexagon.

Construct EI = h/√3 hence EJ = 2h/√3, then x² = EI.EV gives the construction of x : The perpendicular in J intersects the circle with diameter EV in K. EP = EK = x
Complete the cap EPN of hexagon, and the hexagon piece NPXYZ. NZ _|_ EP and YZ goes through V (strips have equal area etc.) avoid the effective drawing of  X,Y : the perpendicular NZ to EP from N intersects ED in W. Finally WVZ is copied in SPT to give the last cut ST (PS = VW and ST _|_ EP).

7 pieces suffice then, which is remarquable for a n-gon to p-gon.

Another easy dissection of n-gon to p-gon is to dissect the hexagon into a dodecagon with just 6 pieces. Adding an equilateral triangle to get a tiling hexagon + triangle overlayed to a tiling dodecagon + triangles (as for octogon into square above).
Unfortunately, one of the pieces is very small. The trick is then to use curved cuts for instance (in magenta).
The method to get this construction is to consider the locus of T and choose T so that the added triangles are all equal.

Angle ATB being 120°, T is on a circle centered in I, intersect point of Ax1 and By1.
The angle bisector of ATB goes through J, midpoint of external arc AB, and intersection point of perpendicular bisector Cx2 of AB with this circle.
For triangles TUV and CMN to be equal, and equal to the added triangle, distance from C to JT must be twice the altitude h of this triangle.
We then construct point H, intersection of circle with diameter JC and circle centered in C with radius 2h. JH intersects the locus of T in the searched T.
The sequel is obvious.
We get then in the neighbourhood of N a very small triangular piece.
This piece is replaced by a larger one by replacing the triangles in A, B, C by curved triangles : CMN is rotation through center C of triangle Cx8y4, the circle arc is then centered in C with radius Cy4 = Cx8 = CN = CM.

Note : z5 is on the hexagon side, from translation of B by vector z2y5> (tiling period)

Users of a calculation software could check that :
- JT doesn't go through y1
- AT is not a radius (doesn't go through dodecagon center, neither through x6)
- UV is not parallel to dodecagon axis x2x8


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