Mickey Mouse Polygon - Regular Polygons forming Mysterious Angle

Question

The figure shows a regular pentagon and two squares sharing adjacent sides of the pentagon. Some vertices are connected by straight lines as shown.

We would like to prove that angle $a$ is exactly $27°$.

The most mysterious thing of the question is that the two straight lines do NOT form any nice angle (i.e. rational) beside angle $a$ . This makes it surprisingly difficult to find the value of $a$ by basic geometry, despite the simple construction of this figure.

Yet, since the value of $a$ is indeed, again, surprising nice, my colleagues and I believe that there must be a basic geometric proof for it.

Unfortunately, we haven't been able to find such proof yet. This question has hounded us for months, so I would like to share it to you all and we are looking forward to your amazing proof.

Answer 1

Here is a solution using complex numbers geometry.

Let us "install" the given figure into a coordinate system with origin in the center of the pentagon and main vertex $A_0$ in $(1,0)$ as shown on Fig. 1, where :

$$b:=e^{2i \pi/40}\tag{1}$$

The polar angle of $b^3$ is $3 \pi/20 \ ( = 27°)$, the angular value we should obtain for $\angle BA_0C$.

Fig. 1.

Let us convert point $B$ into its complex form. Let us use for that relationship $\vec{A_2B}=R(\vec{A_2A_1})$ where $R$ is the $+\pi/2$ rotation :

$$B \leftrightarrow \beta:=\underbrace{b^{16}}_{A_2} + \underbrace{i}_R\underbrace{(b^8-b^{16})}_{A_1-A_2}$$

Or, under a different form :

$$B \longleftrightarrow \beta= b^{16} + b^{10}(b^8-b^{16}) = b^{16}+b^{18}+b^6$$

Remark : we have used various relationships such as

$$b^{10}=i, \ \ b^{20}=-1, \ \ b^{40}=1\tag{2}$$

For similar reasons, relationship $\vec{A_2C}=R^{-1}(\vec{A_2A_3})$ gives :

$$C \longleftrightarrow \gamma:=b^{16} - i(b^{24}-b^{16})$$

With simplifications :

$$C \longleftrightarrow \gamma= b^{16}+b^{14}-b^6$$

Therefore, vectors $\vec{A_0B}$ and $\vec{A_0C}$ have the complex representation :

$$\begin{cases}\vec{A_0B} \longleftrightarrow (\beta-1)\\\vec{A_0C} \longleftrightarrow (\gamma-1)\end{cases}$$

We still have to express with complex numbers that angle $(\vec{A_0B},\vec{A_0C})$ has value $3 \pi/20$ :

$$(\beta-1)b^3 \stackrel{?}{=} k(\gamma - 1)\tag{3}$$

where $k$ is a real number.

Otherwise said, we have to show that complex number

$$z:=\frac{(\beta-1)b^3}{(\gamma - 1)} \stackrel{?}{\in} \mathbb{R} \tag{4}$$

First solution : condition above can be expressed in the following way :

$$ z \stackrel{?}{=} \bar{z}\tag{5}$$

Let us give explicit expressions for $z$ and $\bar{z}$:

$$\begin{cases}z&=&\frac{(b^{16}+b^{18}+b^6-1)b^3}{(b^{16}+b^{14}-b^{6}-1)}&=&\frac{N}{D}\\ \bar{z}&=&\frac{(-b+b^{19}-b^{11}+b^{17})}{(-b^4-b^6+b^{14}-1)}&=&\frac{N'}{D'}\end{cases}$$

Remark : the expression of $\bar{z}$ is obtained by taking the conjugation of all the terms in $z$ knowing that the conjugate of $b^k$ is $b^{40-k}=-b^{20-k}$.

Relationship (4) amounts to the equivalent question for the following polynomial $$\underbrace{ND'-N'D}_{P(b)} \stackrel{?}{=} 0.\tag{6}$$

Expanding the LHS of (6):

$$P(b)=-b^{47} - b^{45}- b^{35} - b^{33} - b^{25} + b^{23} - b^{15} - b^{13} + b^7 + b^3 \stackrel{?}{=}0$$

which is indeed true due to a remarkable cancellation of all its terms ($-b^{47}$ with $b^7$, $-b^{45}$ with $- b^{25}$, etc.) (refer to formulas (2) above) $\square$

Second solution

Based on a simplification of the expression of $z$ using SAGE (a public domain Computer Algebra System) with the following code :

x = polygen(ZZ, 'x')
mp = x^16 - x^12 + x^8 -x^4 + 1
K.<b> = NumberField(mp)
z = (b^18+b^16+b^6-1)*b^3/(b^16+b^14-b^6-1)
z = z.coordinates_in_terms_of_powers() 
print(q)

where the field extension $K.<b>$ (alias $\mathbb{Q}[b]$) is classically described using the minimal polynomial of $b$, i.e., $mp(x) = x^{16} - x^{12} + x^8 - x^4 + 1$ (see remark 1 below).

Running this program gives the result :

$$z= -b^{13} + b^7$$

which is indeed a real number :

$$z=-2i b^{10}\left( \frac{b^{3}-b^{-3}}{2i}\right)=2 \sin(3 \pi/40) \color{red}{\in \mathbb{R}}$$

Remarks :

1. As $b$ is a primitive $40$th root of unity, its minimal polynomial is the so-called cyclotomic polynomial of order $40$, denoted $\Phi_{40}(x)$. Its coefficients can be obtained from the SAGE instructions below :

   from sage.rings.polynomial.cyclotomic import cyclotomic_coeffs print(cyclotomic_coeffs(40))

2. This second solution has been inspired by Steven Stadnicki's answer.

3. PM 2Ring has given a similar SAGE program that I find excellent (more concise than mine) ; it can be uploaded here. I reproduce it here:

K.<b> = CyclotomicField(40)
print(K)
print(K.defining_polynomial())
z = (b^18+b^16+b^6-1)*b^3/(b^16+b^14-b^6-1)
print(z)
q = SR(z):#type conversion to complex
print(q)
print(q.real(), q.imag())

Answer 2

A trigonometric solution is possible, with the help of a CAS (Mathematica, in my case). We can start from: $$ \cos72°={\sqrt5-1\over4},\quad \cos126°=-\sqrt{5-\sqrt5\over8},\quad \cos162°=-\sqrt{5+\sqrt5\over8},\quad \cos108°={1-\sqrt5\over4}. $$ Using the bisection formula twice, from the last equality one also gets: $$ \cos27°= \sqrt{\frac{1}{2} \left(1+\sqrt{\frac{5-\sqrt{5}}{8}}\right)}. $$ Setting the side of the pentagon to be unit, we can then find (see figure below) with the cosine rule: $$ BD={\sqrt5+1\over2},\quad GH=2\sqrt{5-\sqrt5\over8} $$ $$ BG^2=\frac{1}{2} \left(5+\sqrt{5}+\sqrt{2 \left(5+\sqrt{5}\right)}\right) $$ $$ BH^2=\frac{1}{2} \left(5+\sqrt{5}\right)+\sqrt{5+2 \sqrt{5}} $$ and finally: $$ \cos(\angle HBG)=\frac{1}{2} \sqrt{2+ \sqrt{\frac{5-\sqrt{5}}{2} }} $$ which is the same as $\cos27°$ given above.

Answer 3

A "Synthetic" Solution

Consider a rotation with center $O$ by $72^\circ$. This rotation maps the pentagon onto itself and sends the left square in the original diagram to the right square. Of the two lines in the original diagram meeting at angle $a$, call the left one $\ell_1$ and the right one $\ell_2$. The proof's thesis is equivalent to demonstrating that $\ell_2$ and the image of $\ell_1$ in our rotation (which is line $AE$ in the below diagram) meet at an angle $72^\circ-27^\circ=45^\circ$. This is equivalent to the intersection between these two lines lying on the circumcircle of the right square, which we'll call $\omega_2$. Since the line between the second intersections of $AD$ and $\ell_2$ with $\omega_2$ is parallel to side $BC$ by symmetry, this is the same as proving that lines $AD$ and $AE$ are isogonal in $\angle BAC$.

Now, consider the isogonal conjugate of point $D$ with respect to triangle $ABC$. By angle chasing, one can prove that this point is the intersection of line $CO$ and the angle bisector of $\angle CBO$. Let's call this point $W$.

By the Angle Bisector Theorem, we know that $\frac{OW}{WC} = \frac{OB}{BC}$. Since $O$ is the center of the pentagon, $OB=OE$ and $BC=CA$. Therefore, $\frac{OB}{BC} = \frac{OE}{CA}$. Notice that $OE \parallel AC$. From this follows that $W$ lies on the line segment between $E$ and $A$ since it sends one into the other via an homothety. This implies that lines $AD$ and $AE$ are isogonal with respect to $\angle BAC$, which is what we needed to prove.

Answer 4

Really grateful for recieving all of your amazing tactics to this question!

And big thanks to Mr. Bùi Văn Cường from Facebook who has provided his fabulous proof involving only basic geometry. I would like to share the proof to all of you here.

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I. Preliminary Construction

Let $P$ be a point on $CF$ such that $AC=AP$ .

$\angle ACP=45°+36°=81°$

$\angle CAP=180°-81°-81°=18°$

Reflect $B$ along $AF$ to $Q$ , such that rhombus $AQFB$ is formed.

Considering the exterior angles of pentagon $ABCDE$ and square $BCGF$ ,

$\angle ABF=72°+90°=162°$

$\angle QAB=180°-\angle ABF=18°$

Thus, $\angle QAP=\angle BAC=36°$ .

With $AQ=AB$ and $AP=AC$ , we have $\triangle QAP \cong \triangle BAC$ .

$\angle PQF=\angle AQF-\angle AQP=162°-108°=54°$

$\angle DAP=\angle EAB-\angle EAD-\angle PAB=108°-36°-18°=54°$

Together with $QF=QP$ and $AP=AD$ , we have

$$\triangle FQP \sim \triangle PAD$$

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II. Key Similar and Congruent Pairs

Claim: $\triangle ACH \sim \triangle DPF$ .

$\angle QPF=(180°-\angle PQF)\div2=63°$

$\angle DPF=\angle QPF+\angle APQ+\angle DPA=63°+36°+63°=162°$

$\angle ACH=\angle ACD+\angle DCH=72°+90°=162°$

Since $\triangle FQP \sim \triangle PAD$ (proven in I), we have

$$DP:PF=AP:QF=AC:CH$$

Thus, we have $\triangle DPF \sim \triangle ACH$ .

One shall also easily prove that $\triangle ABG \cong \triangle DCF$ .

Together with $\triangle DPF \sim \triangle ACH$ , we have

$$\angle BGA=\angle CFD=\angle CHA$$

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III. Aid of Concyclicity

In isosceles triangle $DCG$ , we have $\angle CGD=\angle CDG=9°$ .

Extend $GD$ to point $R$ such that $\angle RHD=9°$ .

$\angle GDH=\angle CDH-\angle CDG=45°-9°=36°$

$\angle DRH=\angle GDH-\angle RHD=36°-9°=27°$

Lastly, note that

$\angle AGR=45°+9°-\angle BGA=54°-\angle BGA$

$\angle AHR=45°+9°-\angle CHA=54°-\angle CHA$

Since $\angle BGA=\angle CHA$ (proven in II), we then have $\angle AGR=\angle AHR$ .

Thus, $A,G,H$ and $R$ are concyclic.

Hence, $\angle GAH=\angle GRH=27°$ .

Answer 5

This is an alternative ending to the proof in the very nice solution credited to Mr. Bùi Văn Cường.

After the step in which it is proved that angles $\angle AHC$ and $\angle AGB$ in the figure below are congruent, extend $HC$ and $GB$ to meet at $J.$

Since $\angle AHJ \cong AGJ$ and $G$ and $H$ both lie on the same side of $AJ,$ the two angles are inscribed in the same circle through $A,$ $J,$ $G,$ and $H.$ But $\angle GAH$ and $\angle GJH$ are inscribed in the same circle and intercept the same arc $GH,$ so $\angle GAH \cong GJH.$

By simple angle chasing we find that the measure of $\angle GJH$ is $27^\circ.$ Therefore the measure of $\angle GAH$ is $27^\circ.$

Answer 6

Here is a solution that requires almost no calculation.

First, consider the following tiling pattern. The white region is made up only of tiles with angles $90^\circ$, $72^\circ$, and $108^\circ$, and it is surrounded by two types of gray triangles. Remarkably, in this setting all the outer (red) edges turn out to have the same length. In other words, the blue diagonals are evenly spaced every $45^\circ$.

To see why this is the case, look at the equilateral triangle formed by joining certain diagonals of a regular dodecahedron as below. The two gray triangles on the front and top correspond exactly to the two types of gray triangles in the original figure, so we can conclude that their edge lengths are equal.

Now, if we extract the upper part of the original tiling pattern, we obtain the following configuration, where angle ABC measures $45^\circ$.

If we rotate the square and pentagon on the right by $72^\circ$ about point O, we get precisely the figure in question. Hence the desired angle is $$ 72^\circ − 45^\circ = 27^\circ. $$

Answer 7

Not an answer, but a musing upon a possible geometric solution: if we take the circumcircle of the triangle whose three vertices determine $\alpha$, then the corresponding central angle (red and dashed in the diagram below) will be $2\times27^\circ=54^\circ$; if we take circumcircles again, the new central angle (green) will be $2\times54^\circ=108^\circ$, which is the same as the angle at each vertex of the pentagon. Is there some way to argue geometrically that those angles must be equal?

Answer 8

Here are some brute-force (Mathematica-assisted) calculations from coordinates.

For a pentagon with radius $r$ (and side-length $2\sin36^\circ$) turned so that the square ears lie in the first and fourth quadrants, and taking the taking the mystery angle to be $\theta:=\angle BAC$, we can write $$\begin{align} A &= \left(\;-r(1+\cos36^\circ), r\sin36^\circ\;\right) \\ B &= \left(\;2r \cos36^\circ \sin36^\circ, \phantom{-}2r \sin^236^\circ\;\right) \\ C &= \left(\;2r \cos36^\circ \sin36^\circ, -2r \sin^236^\circ\;\right) \end{align} \tag1$$ Note that $$\sin36^\circ = \frac12\sqrt{\frac12(5 - \sqrt{5})} = \frac12\sqrt{\frac{\sqrt{5}}{\phi}}=\frac{f}{2q} \qquad \cos36^\circ = \frac14 (1 + \sqrt{5}) = \frac12\phi = \frac12q^2 \tag2$$ where $\phi$ is the Golden Ratio (ie, the positive root of $\phi^2 = \phi+1$), and $f:=\sqrt{\sqrt{5}}$ and $q:=\sqrt{\phi}$ are for convenience. For more convenience, we take $r = 2q^2/f$ so that we have $$ A = (-fq^4, q) \quad B = (q^3, f) \quad C = (q^3,-f) \tag3 $$ and calculate $$ |BC| = 2f \qquad |AB|^2 = 4 f q^3 (f q + 1) \qquad |AC|^2 = 4 f q^4 (f + q) \tag4 $$ where here and hereafter I'm manipulating expressions involving $q$ with the help of the $q$-ified Golden Ratio identity, $q^4 = q^2+1$. Continuing, the Law of Cosines tells us $$\begin{align} \cos\theta &= \frac{-|BC|^2+|AB|^2+|AC|^2}{2|AB||AC|} \tag5\\[4pt] &= \frac{4f (2q^2 + 1) (f + q)}{2\sqrt{16 f^2 q^7 (fq + 1) (f + q)}} \tag6\\[4pt] &=\frac{2q^2 + 1}{2(q^2+1)} \sqrt{\frac{f + q}{f + 1/q}} \tag7 \end{align}$$ Showing that this is equivalent to $\cos 27^\circ$ is cumbersome. Instead, we can leverage our expectation that $\theta=27^\circ$, and work on showing that $$\cos2\theta = \cos 54^\circ = \sqrt{\frac18(5-\sqrt{5})} \tag8$$ That is, $$2\cos^2\theta - 1 - \frac{f}{2q} = 0 \tag{8'}$$ Substituting from $(5)$, clearing denominators, and doing some grouping, the left-hand side becomes $$\begin{align} (q^4 - q^2 - 1) (f + 6 q - f^2 q + 2 f q^2 + 4 q^3) \;+\; q (4 - 2 f^2 + 7 q^2 - 3 f^2 q^2) \tag9 \end{align}$$ The first factor of the first term vanishes by virtue of the Golden Ratio identity; the second factor of the second term vanishes upon substituting $f$ and $q$ with their definitions. This establishes $(8)$; and, since $\theta$ is clearly acute, we may conclude $\theta = 27^\circ$, as desired. $\square$

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I notice that @intelligenti pauca's answer is a little more streamlined than mine, but still involves a bit of an algebraic slog; likewise, @Jean Marie's. I'm not sure there's a slog-less solution. Since the problem can be setup in terms of $36^\circ$, and the solution is $27^\circ$, I considered approaching this by expanding everything in terms of trigs of $9^\circ$, but it didn't seem to be getting any easier.

Answer 9

Another start of a solution without trigonometry lies in the fact that a regular icosagon can be constructed from a pentagon by alternately tiling pentagons and squares of the same side length.

The target angle (shown in red) must be tilted towards the right angle so that it intersects the 27° angle marks on the icosagon (a chord of three sides lengths)

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Here is my earlier attempt (the proof is not yet complete):

Draw the circumcircle of the triangle formed by the vertices of the angle (like an earlier answer), and multiply this five times on the circle (blue circles).

Notice the symmetric arrangement of the four cross-marked points Q, R, S and T on the circle, forming 4 right angles.

Hence the resulting inscribed angle is 135°. Divide this back by 5.

The measure of the initial angle is therefore 27°.

Answer 10

Here's a very blunt-hammer computer-aided solution via trigonometry and roots of unity. Let $\alpha=e^{2\pi i/20}$, so $\alpha^{10}=-1$, $\alpha^{20}=1$. As I noted in a comment, taking the base of the angle as the origin, the positions of the two points in the complex plane are $\alpha^4+\alpha^8+\alpha^3$ and $\alpha^4+\alpha^8+\alpha^7$. We can take the inverse of the latter mod $(\alpha^{10}+1)$ using Wolfram's PolynomialExtendedGCD, and find that (up to a constant) the result is $2+\alpha-12\alpha^2-6\alpha^3-3\alpha^4+11\alpha^5-7\alpha^6+9\alpha^7-8\alpha^8-4\alpha^9$:

Multiplying this by the first polynomial (mod $\alpha^{10}+1)$ gives (up to a constant) $2+\alpha+3\alpha^2-\alpha^3+2\alpha^4-4\alpha^5-2\alpha^6-\alpha^7+2\alpha^8+\alpha^9$; because of the way it was constructed, the argument of this number is the angle we're after. Now we can raise this to the 20th power:

The result is $7284164428710937500\alpha^9-3214836120605468750\alpha^8-16016006469726562500\alpha^7-12014865875244140625\alpha^6+12014865875244140625\alpha^4+16016006469726562500\alpha^3+3214836120605468750\alpha^2-7284164428710937500\alpha-17599964141845703125$.

Close examination(!) shows that the coefficients of $\alpha^{10-i}$ are the negative of the coefficients of $\alpha^i$; in other words, by multiplying the coefficients of degree greater than $5$ by $-\alpha^{-10}=1$ we get the result as a sum of the form $A(\alpha+\alpha^{-1})+B(\alpha^2+\alpha^{-2})+C(\alpha^3+\alpha^{-3})+D(\alpha^4+\alpha^{-4})+E$, for (very large) integers $A, B, C, D, E$. But since $\alpha$ is a root of unity, $\overline{\alpha^i}=\alpha^{-i}$ for all $i$; in other words, $\alpha^i+\alpha^{-i}$ is real for all $i$. This shows that the $20$th power of the ratio of these two values is a real number, so $20$ times the angle between them is a multiple of $\pi$. In other words, the angle in question is a multiple of $\frac\pi{20}$. From here it's easy to find bounds on the value that imply that it's $\frac{3\pi}{20}$.

Answer 11

We can use complex numbers to reduce the problem to a trigonometric identity.

Let us then assume that the pentagon is $ABCDE$ with squares on sides $CD$ and $DE$ so that the angle to be determined has vertex at $B$. Let the square on side $CD$ be $CDGF$ with $CF$ at right angles to $CD$. Also let the square on side $DE$ be $DEIH$ with $DH$ at right angles to $DE$. We assume $B$ to be origin represented by complex number $0$ and $A$ to be represented by $-1$. In what follows we use the notation $$\operatorname {Cis} \theta=\cos \theta+i\sin\theta$$ and further let $t=18^{\circ}$ so that we can express all relevant angles as integer multiples of $t$.

We then have $C=\operatorname {Cis} 4t$ and $$F=C+CF=C+C\cdot \operatorname{Cis} (-t) $$ ie $$F=\operatorname {Cis} 4t+ \operatorname {Cis} 3t$$ Next we note that $$G=F+C\cdot \operatorname {Cis} 4t$$ ie $$G= \operatorname {Cis} 4t+ \operatorname {Cis} 3t+ \operatorname {Cis} 8t$$ The length of $GH$ is $2\sin 2t$ and it is parallel to real axis and hence $$H=G-2\sin 2t=\operatorname {Cis} 4t+ \operatorname {Cis} 3t+ \operatorname {Cis} 8t-2\cos 3t$$ as $5t=90^{\circ}$.

Since $B$ is the origin the angle $GBH$ can be determined just by calculating the complex number $H/G$. One should note further that $\cos 4t+\cos 8t=-1/2$ and hence we have $$H=-\frac{1}{2}-\cos 3t+i(\sin 3t+\sin 4t+\sin 8t)$$ and $$G=-\frac{1}{2}+\cos 3t+i(\sin 3t+\sin 4t+\sin 8t)$$ Let us further note that $$\sin 4t+\sin 8t=2\sin 6t\cos 2t=2\sin 4t\sin 3t$$ which is same as $2\sin 3t\cos t$. Thus we have $$H=-\frac{1}{2}-\cos 3t+i\sin 3t(1+2\cos t)$$ and $$G=-\frac{1}{2}+\cos 3t+i\sin 3t(1+2\cos t)$$ The real part of $H/G$ is $$\frac{1}{4}-\cos^2 3t+\sin^23t(1+2\cos t) ^2$$ and it's imaginary part is $$2\sin 3t\cos 3t(1+2\cos t)=4\sin(3t/2)\cos(3t/2)\cos 3t(1+2\cos t) $$ Our job is done if we can show that $$\frac{1}{4}-\cos^2 3t+\sin^23t(1+2\cos t) ^2=4\cos^2(3t/2)\cos 3t(1+2\cos t)\tag{1} $$ because then we can write $H/G$ in polar form as $r\operatorname {Cis} (3t/2)$ and the desired angle at vertex $B$ is $3t/2=27^{\circ}$.

The left hand side of the identity $(1)$ can be written as $$\frac{1}{4}-(\cos^2 3t-\sin^23t)+4\sin^23t\cos t+4\sin^23t\cos^2t$$ which is same as $$\frac{1}{4}-\cos 6t+2(1-\cos 6t)\cos t+(1-\cos 6t)(1+\cos 2t)$$ ie $$\frac{1}{4}+\sin t+2(1+\sin t)\cos t+(1+\sin t)(1+\cos 2t)$$ The above can be simplified as $$\frac{5}{4}+\frac{3}{2}\sin t+2\cos t+\sin 2t+\frac{3}{2}\cos 2t$$ using formulas for changing product of trigonometric functions into sums of such functions.

Similarly the right hand side of $(1)$ can be simplified to $$1+\sin t+2\cos t+\sin 2t+2\cos 2t$$ and therefore all we need to prove here is that $$\frac{1}{4}+\frac{1}{2}\sin t=\frac{1}{2}\cos 2t$$ ie $$1+2\sin t=2-4\sin^2 t$$ or $$4\sin^2t+2\sin t-1=0$$ which is satisfied by $\sin t=(\sqrt{5}-1)/4$.

Answer 12

Here's another computer-assisted complex number solution, derived from comments I posted previously. I used SageMath (previously known as Sage) to do the algebra and create the diagram. My diagram uses the same point labels as Chris So's answer.

Sage can do exact calculations with symbolic quantities like $\pi$ and $\sqrt{5}$, but it can also work with number field elements. For this problem, it's convenient to use the cyclotomic field of order 40.

Following Jean Marie's answer, I use $b$ to represent the fundamental complex primitive 40th root of unity, $b = \exp(2\pi i/40)$. Multiplying a complex number by $b$ rotates it around the origin of the Argand plane by $\pi/20$ radians = $9°$.

We can create a pentagon centred at the origin of the complex plane using successive powers of $b^8$. The circumcircle of this pentagon has unit radius. The pentagon side is $b^{14} - b^6$.

We can create the attached squares by rotating appropriate pentagon sides by 90°, i.e., by multiplying by $i$ or $-i$.

We find the desired angle by division. The angle at $z_0$ between the lines $(z_0, z_1)$ and $(z_0, z_2)$ is $$\arg\left(\frac{z_2 - z_0}{z_1 - z_0}\right)$$

Here's my Sage code:

""" Draw a pentagon & squares diagram
    and calculate an angle
    CyclotomicField version
    With point labels

Written by PM 2Ring 2025.09.17
"""
def sym(seq): return map(SR, seq)
def text_pos(z):
z = SR(z)
v = vector([z.real(), z.imag()])
return v + v.normalized() * 0.125
K.<b> = CyclotomicField(40)
print(f"{b.minpoly() = }")
Pentagon
w = b^8
q = I
p = [q * w^i for i in [0 .. 4]]
pts = p.copy()
def square(p0, p1):
u = p1 - p0
s = [p1]
for i in [1 .. 3]:
u *= -I
z = s[-1] + u
s.append(z)
return s
Find side length
z = w - 1
s = sqrt(z * z.conjugate())
ss = abs(SR(z)).simplify_full()
print(f"Side length: {s}\n = {ss} ~= {abs(z)}\n")
Add some squares
s1 = square(p[4], p[0])
pts.extend(s1[1:-1])
s2 = square(p[0], p[1])
pts.extend(s2[1:-1])
pts = dict(zip("CDEABGFIH", pts))
P = Graphics()
P += line2d(sym(p + p[:1]))
P += line2d(sym(s1))
P += line2d(sym(s2))
Find the angle GAH
A, G, H = pts['A'], pts['G'], pts['H']
print(f"{A = }\n{G = }\n{H = }\n")
GA, HA = G - A, H - A
print(f"{GA = }\n{HA = }")
Q = HA / GA
print(f"{Q = }")
theta = 1/5 * arg(Q^5).simplify_full()
print(f"{theta = }, {theta * 180 / pi}°")
P += line2d(sym([G, A, H]), color="purple")
Plot the points & labels
P += points(sym(pts.values()), size=18, color="red")
for s, z in pts.items():
P += text(s, text_pos(z), color="black",
vertical_alignment="center",
horizontal_alignment="center",
fontsize=12, fontstyle="italic")
kw = dict(aspect_ratio=1, figsize=8, gridlines="minor", frame=True)
#kw = dict(aspect_ratio=1, axes=False)
P.show(**kw)
P.save("penta.png", **kw)

Output

b.minpoly() = x^16 - x^12 + x^8 - x^4 + 1
Side length: b^14 - b^6
 = 1/4*sqrt(-8*sqrt(5) + 40) ~= 1.17557050458495

A = -b^14
G = b^12 + b^10 + 1
H = b^10 + b^8 - 1
GA = b^14 + b^12 + b^10 + 1
HA = b^14 + b^10 + b^8 - 1
Q = -b^12 + b^10 + b^8 - b^6 + b^2
theta = 3/20*pi, 27°

Here's a live version, running on the SageMathCell server.

I used theta = 1/5 * arg(Q^5) because the arg function can only give symbolic values for major angles. So it can handle 135° but not 27°.

Alternatively, we can show that $Q$ is a real multiple of $b^3$. Eg,

z = SR(Q / b^3)
print(f"{z.real() = }, {z.imag() = }")

which prints

z.real() = sqrt(2) - 2*cos(9/20*pi), z.imag() = 0

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You can rotate the diagram by editing in a different value for the q variable. It's currently set to I, but try other powers of $b$, eg here's the result for $b^6$:

b.minpoly() = x^16 - x^12 + x^8 - x^4 + 1
Side length: b^14 - b^6
 = 1/4*sqrt(-8*sqrt(5) + 40) ~= 1.17557050458495

A = -b^10
G = -b^12 + 2b^8 + b^6 - b^4 + 1
H = b^12 - b^8 + b^6 + 2b^4 - 1
GA = -b^12 + b^10 + 2b^8 + b^6 - b^4 + 1
HA = b^12 + b^10 - b^8 + b^6 + 2b^4 - 1
Q = -b^12 + b^10 + b^8 - b^6 + b^2
theta = 3/20*pi, 27°

Answer 13

I am surprised that nobody presented an answer purely in vector form, since it is about as equally nice to use as the complex numbers version that everybody used. Accessing coördinates and manipulating the trigonometric functions is definitely going to look ugly and be much less elegant than a synthetic geometry proof, but I can present the kind of working that my students love me for, i.e. as elegant as I can make that, and still be clean enough to present, and yet is straightforward enough that they can hope to reproduce. I usually draw all the diagrams by hand, but it is difficult to draw online and you have also seen so many other nice ones.

Sadly, I was unable to defeat the trigonometric monster that arose, and the final result is left as a computer-aided check.

Basic Setup (Pentagon)

Let the origin $O$ be the centre of the pentagon $ABCDE$ where $AB$ and $BC$ are the sides where the squares are drawn, i.e. $B$ is the busiest intersection of all the primitive lines are. Let $\vec{OB}=(0,1)^T$; note that $\dfrac{360^\circ}5=72^\circ$ so that the other vertices of the pentagon are $(\pm\sin72^\circ,\cos72^\circ)$ and $(\pm\sin144^\circ,\cos144^\circ)$ respectively. Reducing to first quadrant, $180^\circ-144^\circ=36^\circ$ so that we can now gather all the relevant knowns into a mini data table: $$\tag1 \vec{OA}=\begin{pmatrix}-\sin72^\circ\\+\cos72^\circ\end {pmatrix}\quad \vec{OB}=\begin{pmatrix}0\\1\end {pmatrix}\quad \vec{OC}=\begin{pmatrix}+\sin72^\circ\\+\cos72^\circ\end {pmatrix}\quad \vec{OD}=\begin{pmatrix}+\sin36^\circ\\-\cos36^\circ\end {pmatrix}\quad $$

Basic Setup (Squares)

Let the Mickey mouse's left ear's vertex that goes into the original question's statement be $P$, and it is obtained by rotating $AB$ by $90^\circ$ to get $PB$ $$\tag2\vec{AB}=\vec{OB}-\vec{OA}=\begin{pmatrix}+\sin72^\circ\\+(1-\cos72^\circ)\end {pmatrix}\qquad\implies\qquad\vec{BP}=\begin{pmatrix}-(1-\cos72^\circ)\\+\sin72^\circ\end {pmatrix}$$ where we have used the right angle rotation in Cartesian coördinates trick of $x\to y$ and $y\to-x$

We will postpone finding location of $P$ since it is so obvious how to find it now. Instead, we do the same thing to find $Q$ $$\tag3\vec{BC}=\vec{OC}-\vec{OB}=\begin{pmatrix}+\sin72^\circ\\-(1-\cos72^\circ)\end {pmatrix}\qquad\implies\qquad\vec{BQ}=\begin{pmatrix}+(1-\cos72^\circ)\\+\sin72^\circ\end {pmatrix}$$

More Vector Subtraction

We are looking for the lines $DQ$ and $DP$ to find their intersecting angle. $$\tag4\vec{DQ}=\vec{OQ}-\vec{OD}=\vec{OB}+\vec{BQ}-\vec{OD}=\begin{pmatrix}1-\cos72^\circ-\sin36^\circ\\1+\sin72^\circ+\cos36^\circ\end {pmatrix}\overset{\text{def}}=\begin{pmatrix}n_-\\d\end {pmatrix}$$ $$\tag5\vec{DP}=\begin{pmatrix}-(1-\cos72^\circ+\sin36^\circ)\\1+\sin72^\circ+\cos36^\circ\end {pmatrix}\overset{\text{def}}=\begin{pmatrix}-n_+\\d\end {pmatrix}$$ where, by construction and grouping terms, I have made it so that $0<n_-<n_+<d$

Vector Product and Reduction to Trigonometry

Visual inspection to tease out the correct signs, we are thus left to compute (via 3D extension and then extracting the single non-zero component) $$\tag6a=\angle PDQ=\arg\!\sin\frac{\vec{DQ}\times\vec{DP}}{\left|\vec{DQ}\right|\left|\vec{DP}\right|}=\arg\!\sin\frac{n_-d+dn_+}{\sqrt{d^2+n_-^2}\sqrt{d^2+n_+^2}}=27^\circ$$ In the numerator, the common $d$ factorises out and the leftover terms have their $\sin36^\circ$ terms cancel out. It is an ugly mess, but it is easily checked by computer that the result is exactly $27^\circ$ as required, no approximation. This thus already constitutes a proof, even if it is an ugly trigonometric one that required computer help.

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Edit: Sadness of Futility

After a while, I managed to get a lot further than where I stopped earlier. It turns out that there is no way to simplify the above trigonometric expression, because not even computer algebra systems manage to help. In particular, my working goes as follows, where I had liberally used double angle identities, right-angle identities, and at one step there was even a triple angle identity:

$$ \begin{align} \tag7&\frac{4\sin^236^\circ(1+\sin72^\circ+\cos36^\circ)}{\sqrt{[(1+\sin72^\circ+\cos36^\circ)^2+(1-\cos72^\circ)^2+\sin^236^\circ]^2-[4\sin^336^\circ]^2}}\\ \tag8&=\frac{4\sin^236^\circ(1+\cos18^\circ+\cos36^\circ)}{\sqrt{[4+2\sin72^\circ+2\cos36^\circ+2\sin72^\circ\cos36^\circ-2\cos72^\circ]^2-[4\sin^336^\circ]^2}}\\ \tag9&=\frac{4\sin^236^\circ(\cos18^\circ+2\cos^218^\circ)}{\sqrt{[4+2\cos18^\circ+2\cos36^\circ+\cos54^\circ+\cos18^\circ-2\cos72^\circ]^2-[4\sin^336^\circ]^2}}\\ \tag{10}&=\frac{16\sin^218^\circ\cos^318^\circ(1+2\cos18^\circ)}{\sqrt{[4+3\cos18^\circ+\cos54^\circ+2\cos36^\circ-2\cos72^\circ]^2-[4\sin^336^\circ]^2}}\\ \tag{11}&=\frac{16\sin^218^\circ\cos^318^\circ(1+2\cos18^\circ)}{\sqrt{[4+2\cos18^\circ+2\cos36^\circ\cos18^\circ+4\sin54^\circ\sin18^\circ]^2-[4\sin^336^\circ]^2}}\\ \tag{12}&=\frac{16\sin^218^\circ\cos^318^\circ(1+2\cos18^\circ)}{\sqrt{[4+4\cos^318^\circ+4\cos36^\circ\sin18^\circ]^2-[4\sin^336^\circ]^2}}\\ \tag{13}&=\frac{16\sin^218^\circ\cos^318^\circ(1+2\cos18^\circ)}{\sqrt{[4+4\cos^318^\circ+\frac{\sin72^\circ}{\cos18^\circ}]^2-[4\sin^336^\circ]^2}}\\ \tag{14}&=\frac{16\sin^218^\circ\cos^318^\circ(1+2\cos18^\circ)}{\sqrt{[5+4\cos^318^\circ]^2-[4\cos^354^\circ]^2}}\\ \tag{15}&=\frac{4\sin^218^\circ(3\cos18^\circ+\cos54^\circ)(1+2\cos18^\circ)}{\sqrt{[5+3\cos18^\circ+\cos54^\circ]^2-[3\cos54^\circ+\cos162^\circ]^2}}\\ \tag{16}&=\frac{4\sin^218^\circ(3\cos18^\circ+\sin36^\circ)(1+2\cos18^\circ)}{\sqrt{[5+3\cos18^\circ+\sin36^\circ]^2-[3\sin36^\circ-\cos18^\circ]^2}}\\ \tag{17}&=\frac{4\sin^218^\circ\cos18^\circ(3+2\sin18^\circ)(1+2\cos18^\circ)}{\sqrt{[5+\cos18^\circ(3+2\sin18^\circ)]^2-\cos^218^\circ[6\sin18^\circ-1]^2}}\\ \tag{18}&=\frac{16s^2(1-s^2)(1-2s^2)[3+4s\sqrt{1-s^2}][1+2(1-2s^2)]}{\sqrt{[5+(1-2s^2)(3+4s\sqrt{1-s^2})]^2-(1-2s^2)^2[12s\sqrt{1-s^2}-1]^2}} \end {align} $$ where in this last expression, $s=\sin9^\circ$, and remarkably, the entire tower of expressions are all correct. The original aim is that Equation (18) should be reducible to $3s-4s^3$ which is none other than the triple angle identity that would produce $\sin27^\circ$. However, inserting the square of Equation (18) so as to help the computer algebra system cancel stuff out, shows that this is a futile endeavour.

As such, other than picking one of Equations (6), (14) or (16) and substituting the specific values for the trigonometric functions, so that then we can algebraically show that the result is indeed $27^\circ$ as required, I now really doubt that this direction of inquiry would nicely yield the answer. Of course, I would be grateful if someone points out a nice solution from Equation (6) to the final answer, but I am not exactly holding out hope where CAS fails.

Answer 14

This is just a hint for the solution. After determining the integer values ​​of the angles marked in the figure below, use the law of cosines to first find the length of the purple line and solve the red triangle.

[Edit] I changed the image.

Answer 15

Short solution, using complex:

Let $z$ denote a complex rotation by $18°$. Assuming sides $1$, the two vectors are $z^4+z^8+z^3$ and $z^4+z^8+z^7$ and the angle they form is the argument of

$$w:=\frac{z^{4}+z^{8}+z^{7}}{z^4+z^8+z^3}.$$

Numerically, we have $\angle w=0.4712388980385\cdots=0.15\pi$, which is no coincidence.

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It is possible to prove it analytically, by raising $w$ to the tenth power, using $z=\dfrac{\sqrt{2\sqrt5+10}}4+i\dfrac{\sqrt5-1}4$, but the computation is a little... tedious :-)

Answer 16

I have constructed the pentagon of side $2$ showed in the attached figure. Then the two necessary sides of the external squares (the other sides can be omitted).

Then the two black lines whose equations turn out to be the following: $$y=26.75435699(x-1)=m_1(x-1)\\y=-2.159208119(x-1)=m_2(x-1)$$ por lo cual $$\arctan\left(\left|\dfrac{m_1+m_2}{1-m_1m_2}\right|\right)\approx26.99095^{\circ}\approx27^{\circ}$$ (The non exact result is "natural" because of the involved decimal expressions).

Answer 17

Not an actual solution, just a picture that indicates how one can check readily the size of the angle(s) using complex numbers

$\bf{Added:}$ Another picture related to the problem

$\bf{Added:}$

Some calculations with complex numbers. We have the black dots on the circles, vertices of a regular polygon with $20$ vertices. Let their complex coordinates be $1$, $z$, $z^2$, $\ldots$, $z^{19}$, where $z= e^{\frac{2\pi i}{20}}$. Let $t$ be the coordinate of the top red point. It lies on the same (unit) circle, and the chords $z^7 z^{17}$, $z^{13} z^{19}$, $z^{16} t$ are concurrent. We conclude that the determinant is zero:

$$\det \begin{pmatrix} 1& z^7+z^{17} & z^{7} \cdot z^{17} \\ 1& z^{13} +z^{19} & z^{13} \cdot z^{19} \\ 1 &z^{16} + t& z^{16} \cdot t \end{pmatrix} = 0$$ which gives $$t = -\frac{z^{14}(z^4+z+1)}{z^4+z^3+1}$$

Now one checks that the chords $z z^7$, $z^3 z^{13}$, $z^{14} t$ are concurrent, since calculating the corresponding determinant

$$\det \begin{pmatrix} 1&z+z^7&z\cdot z^7 \\ 1& z^3 +z^{13}& z^3 \cdot z^{13} \\1 & z^{14}+t & z^{14} \cdot t \end{pmatrix} $$

with the above value of $t$ we get the factored expression with numerator $$-z^{11}(z^4-1)(z^5-1)(z^6-1)(z^8-z^6+z^4-z^2+1)(z^{12}-z^{11}+z^{10}+z^8+z^4+z^2-z+1)$$ and denominator $(z-1)(z^4+z^3+1)$. Now the last but one factor of the numerator $z^8-z^6+z^4-z^2+1$ is the cyclotomic polynomial $\Phi_{20}(z)$, hence $0$.

Note that $t=-\frac{z^{14}(z^4+z+1)}{z^4+z^3+1}$, where $z=e^{\frac{2\pi i}{20}}$, has minimal polynomial $t^8 + \frac{6}{5} t^4 + 1$ and in fact $t^4 = -(\frac{3}{5} + \frac{4}{5} i) $.

Answer 18

The side of the pentagon is 1.

First , we calculate OA with Al - Kashi to find :

OA² = 2 + 2 cos(72°)

so

OA = 2 cos(36°)

AB² = 3 + 2 cos(72°) + 2 cos(18°)

AC² = 3 + 2 cos(18°) + 2 cos(72°) + 2 cos(54°)

BC² = 2 - 2 cos(72°)

and finally we use ∆ABC to find cos(a). We know all those cosines.