# Solve for pentagon angles/sides algebraically

Is there a way to solve for angles and side lengths of this pentagon algebraically?

We are given the equations:

$$A=90^{\circ}, 2B+C=360^{\circ}, C+E=180^{\circ}, 2a=2c=d=e \text{ and say } a=1$$

Also we need to guarantee that the angles and sides actually fit together to give a pentagon.

e.g. $$b - a \cos(A) - c \cos(B) + d \cos(B+C) + e \cos(A+E)=0$$ and

$$a \sin(A) - c \sin(B) + d \sin(B+C) - e \sin(A+E) = 0$$

Also, squaring and adding the last two equations gives a nice symmetrical equation:

$$\begin{split} 2(a b \cos (A)+b c \cos (B)+c d \cos (C)+d e \cos (D)+e a \cos (E))=a^2+b^2+c^2+d^2+e^2\\+2 (a c \cos (A+B)+b d \cos (B+C)+c e \cos (C+D)+d a \cos (D+E)+e b \cos (E+A)) \end{split}$$ which seems to be the cosine rule for pentagons or something.

Here's the solution with approximated angles. Thanks!

• The angle at D seems off by about 100 deg. Internal angle total should be 540 deg. – Narasimham Sep 27 '15 at 9:54
• @Narasimham True. I took the image from wikipedia and it should be 124.66 – Alex Sep 27 '15 at 11:43
• So you want to geometrically construct a pentagon with those angles and lengths given by Aretino? – Narasimham Sep 27 '15 at 13:35
• I was searching for a way to find the angles algebraically. All I had was a numerical approximation. – Alex Sep 27 '15 at 13:49
• Is the 4-bar mechanism is analytically known? If not, how will a pentagonal mechanism be analytically determined? – Narasimham Sep 27 '15 at 15:00

Notice first of all that $A+B+C+D+E=3\cdot180°$. Combining this with the other relations you give for the angles, we may rewrite them all as follows: $$A=90°,\quad C=360°-2B,\quad D=270°-B,\quad E=2B-180°,$$ so that all angles can be expressed in terms of $B$ alone.

Your equation $$a \sin(A) - c \sin(B) + d \sin(B+C) - e \sin(A+E) = 0$$ can be then rewritten as $$1-3\sin(B)+2\cos(2B)=0$$ where I've also inserted the values $a=c=1$ and $d=e=2$. This equation is easy to solve and gives $$\sin(B)={\sqrt{57}-3\over8}, \quad\hbox{whence}\quad B=145.338°.$$ It is then easy to find the other angles and finally to obtain $b$ from the other equation: $$b =-\cos(B)-2\sin(2B)=2.6937$$

This is the Type 14 pentagon found in 1985 by Rolf Stein. I've got it coded up at the Pentagon Tilings demonstration.

Exact placement of vertices for a 6 pentagon motif.

{{{1/8 (-3 + Sqrt[57]), Root[-1568 - 71 #1^2 + 16 #1^4 &, 2]}, {1/8 (-9 + 3 Sqrt[57]), Root[-1152 + 273 #1^2 + 16 #1^4 &, 2]}, {1, 0}, {0, 0}, {0, Root[-392 + 25 #1^2 + 4 #1^4 &, 2]}},

{{1/8 (3 - Sqrt[57]), Root[-1568 - 71 #1^2 + 16 #1^4 &, 2]}, {1/8 (9 - 3 Sqrt[57]), Root[-1152 + 273 #1^2 + 16 #1^4 &, 2]}, {-1, 0}, {0, 0}, {0, Root[-392 + 25 #1^2 + 4 #1^4 &, 2]}},

{{1/4 (1 + Sqrt[57]), Root[-32 + #1^2 + 4 #1^4 &, 1]}, {1, 0}, {1/8 (-9 + 3 Sqrt[57]), Root[-1152 + 273 #1^2 + 16 #1^4 &, 2]}, {1/8 (-1 + 3 Sqrt[57]), Root[-1152 + 273 #1^2 + 16 #1^4 &, 2]}, {1/8 (-1 + 3 Sqrt[57]), Root[-8 + #1^2 + 16 #1^4 &, 1]}},

{{1/2 (-1 + Sqrt[57]), Root[-32 + #1^2 + 4 #1^4 &, 1]}, {1/4 (-5 + 3 Sqrt[57]), 0}, {1/8 (7 + 3 Sqrt[57]), Root[-1152 + 273 #1^2 + 16 #1^4 &, 2]}, {1/8 (-1 + 3 Sqrt[57]), Root[-1152 + 273 #1^2 + 16 #1^4 &, 2]}, {1/8 (-1 + 3 Sqrt[57]), Root[-8 + #1^2 + 16 #1^4 &, 1]}},

{{1/8 (7 + 3 Sqrt[57]), Root[-1152 + 273 #1^2 + 16 #1^4 &, 2]}, {1/8 (-9 + 3 Sqrt[57]), Root[-1152 + 273 #1^2 + 16 #1^4 &, 2]}, {1/8 (-3 + Sqrt[57]), Root[-1568 - 71 #1^2 + 16 #1^4 &, 2]}, {1/4 (-3 + Sqrt[57]), Root[-72 - 15 #1^2 + #1^4 &, 2]}, {1/16 (31 + 3 Sqrt[57]), Root[-23328 + 2457 #1^2 + 64 #1^4 &, 2]}},

{{-1, 0}, {1, 0}, {1/4 (1 + Sqrt[57]), Root[-32 + #1^2 + 4 #1^4 &, 1]}, {1/8 (5 + Sqrt[57]), Root[-648 + 9 #1^2 + 16 #1^4 &, 1]}, {3/16 (-11 + Sqrt[57]), Root[-288 + 273 #1^2 + 64 #1^4 &, 1]}}}

Offset vectors are {{1/2 (1 - Sqrt[57]), Root[-72 - 15 #1^2 + #1^4 &, 2]}, {1/16 (21 + Sqrt[57]), Root[-31752 - 639 #1^2 + 64 #1^4 &, 2]}}

The method is from graphic statics of forces.

The problem of angles and lengths is to be separately solved.

Angles

Firstly assume each side unit length.

Make a table of vector angles $\theta_i$ to parallel to xaxis ( horizontal) in a CCW sense. The given figure is a closed pentagon. Given 3 angles along with interior sum ( 2* 5 -4) 90 degrees is adequate to get the angles.

Sides

Use force equilibrium treating each side as magnitude of a vector. Draw parallels to each side such that there is force equilibrium.

$$\Sigma L_x = 0, \Sigma L_y = 0.$$

ensures that the pentagon closes at specified angles.

• Please elaborate. I don't understand how you can solve for angles separately. If I completely ignore the sides then A=C=E=90 and B=D=135 is a solution, but then there are no side lengths that work for this particular choice of angles. – Alex Sep 26 '15 at 21:10
• Draw an equal sided pentagon with the above angles, extending all sides both ways. Now shift alternate position sides parallel to themselves to changing cutting points, and so the side lengths. – Narasimham Sep 26 '15 at 21:34
• There is no equal sided pentagon with the angles I posted in my last comment... – Alex Sep 27 '15 at 5:45
• Are the solutions given by Aretino not adequate for construction of the pentagon? – Narasimham Sep 27 '15 at 9:43