I found the problem below in Twitter
$ABCDEFG$ is a regular heptagon. $EFHD$ is a rhombus and $HI$ is drawn perpendicular to side $AB$ find the length $HI(d)$
What I've done so far:
I can find the answer pretty easily by using trigonometric ratios. But I like to solve this using elementary geometry.
I found out that $E, H, I$ are collinear and $I$ is the midpoint of $AB$. Also, I drew lines $FD, EH, HA, HB$ but nothing helped me.
Could anyone help me to solve this?
Answers based on elementary geometry are much more appreciated.
This is a nice problem, which should have received more audience. I'll try to give solutions that do not involve trigonometry. Arguably, the solution involving complex numbers is in essence also a trigonometric solution, this is but the bridge to a more geometric solution.
The side of the given regular heptagon is normed to one, as extracted from the picture. Then it is enough to show $AH=BH=\sqrt 2$, thus getting rid of the more "complicated" point $I$, and have relations connecting only $H$ and the vertices. Equivalently, and giving up this norming, we have to show $AH/AB=BH/AB=\sqrt 2$.
First solution: We are using complex numbers, fix $z$ to be the primitive root of unity given by $z=\exp\frac {2\pi i}7=\cos\frac {2\pi}7+i\sin\frac {2\pi}7$, and we will use below no functional equation of trigonometric / exponential nature, but only the algebraic relation $$ \tag{$*$} 0 = 1 + z + z^2 + z^3 + z^4 + z^5 + z^6\ . $$ Consider the given regular heptagon with side one, and translate, rotate, and rescale it so that the vertices $A,B,C,\dots$ are mapped to their affixes, denoted by lower case corresponding letters, $a=1$, $b=z$, $c=z^2$, $\dots$ so that the parallelogram $DEFH$ is mapped to the parallelogram with vertices $d=z^3$, $e=z^4$, $f=z^5$, so that $h$, the image of $H$, is $ h = (d+f)-e=z^5-z^4+z^3$. Then the lengths $AB$ and $BH$ are $|z-1|$, and $|z^5-z^4+z^3-z|=|z^4-z^3+z^2-1|$. The quotient $\displaystyle \frac{z^4-z^3+z^2-1}{z-1}$ is $z^3+z+1$.
So let us show: $$ \sqrt 2\overset!=|z^3+z+1|\ , $$ which is a straightforward algebraic computation, using only $(*)$ and $z^7=1$: $$ \begin{aligned} |z^3+z+1|^2 &=(z^3+z+1)\overline{(z^3+z+1)}\\ &=(z^3+z+1)(z^6+z^4+1)\\ &=(z^9+z^7+z^3)\ +\ (z^7+z^5+z)\ + \ (z^6+z^4+1)\\ &=1+1+(z^6+z^5+z^4+z^3+z^2+z+1)\\ &=2\ . \end{aligned} $$ $\square$
Second solution:
(Discussion and intermediate lemmas.)
Although short, the above computation does not have a simple geometrical interpretation, which would be needed to rewrite geometrically the above "simple" algebraic solution into a "simple" geometrical one. (This was more or less the question.) We have to break into pieces, that may be easily translated. The core of the computation is $$ BH^2 = (h-b)\overline{(h-b)} = (z^5 - z^4 + z^3 - z) \underbrace{\overline{(z^5 - z^4 + z^3 - z)}}_{=(z^2 -z^3 + z^4 - z^6)} = 2\cdot (z-1) \overline{(z-1)} =2 AB^2\ . $$ One step would be to construct geometrically the factor $(h-b)/(z-1)= z(z^3+z^2+1)$ that appears in the computation. Or $z^3+z^2+1$. Or $z^3+z+1$. Or $z(z^3+z+1)=z+z^2+z^4=z+z^4+z^9$.
This is done using a couple of lemmas.
Lemma 1: Let $ABCDEFG$ be a regular pentagon with side one and center $O$. Let $H=H_E$ be the point on $OE$ so that $DEFH_E$ is a rhombus. Similarly construct $H_A,H_B,\dots,H_G$. (Or rotate $H=H_E$.) Construct the $28$-gon $\Pi_{28}$ with one vertex in $E$ and center $D$, and (also) use labels for the vertices from $\Bbb Z/28$, such that $E=2$. Then $E=2$, $H_E=4$, $H_C=8$, and $C=12$ are among its vertices. Two other vertices are $S_E=EA\cap DF=4$, and $T_C=BD\cap CG=10$.
Then (recalling alternative notations $H=H_E=6$): $$ BH_E=AH_E=AH_D=AT_C=DH_A=EH_A=EH_B=ES_C\ . $$
Proof: We have $AH=EH$ because of the $OE$-reflection symmetry. Then $AH_E=AH_D$ by $OA$-symmetry. Then $T_C$ is constructed so that $DFGT_C$ is a parallelogram. So $DT_C=FG=1=DH_E$, so $\Delta DH_ET_C$ isosceles, and its angle bisector in $D$, the line $DH_CH_BA$, is thus the side bisector of the segment $H_ET_C$. The point $A$ lies on it, so $AH_E=AT_C$. We have so far $$ BH_E=AH_E=AH_D=AT_C=\dots \text{ and } $$ reflect w.r.t. the line $OC$, so the above chain of equalities can be continued: $$ \dots = DH_B=EH_A=EH_B=ES_C\ . $$ $\square$
Comment: It is then enough to show that any of the above segments has length $\sqrt 2$. Arguably, this is in the same level of complexity as only the relation $AH_E=\sqrt 2$, but let us note the (psychological) difference. In the given initial situation, a helper construction to show a $\sqrt 2$ length segment could be done "in a natural manner" only (?) by constructing a square on a given side of length one. After having Lemma 1, we have moved to a circle $(D)$ of radius one, and related to it $\sqrt 2$ is in a natural manner a chord length in the regular $28$-gon $\Pi_{28}$, for instance the chord $29$ (or $E9$).
Lemma 2: In the situation of Lemma1, $H_A$ is the orthocenter in the triangle $\Delta$ with vertices $0$, $H_C=8$ and $C=12$.
Proof: Note that $08 \|26$, which is a direction perpendicular to the direction $FD \|GH_AC$, so $H_A$ is on the height from $C$ in $\Delta$. Furthermore, $0$, $6=H_E$, and $H_A$ are collinear, since $06\|(-4,D,10)=DB\|H_EH_A$. From $C8O\perp H_EH_A$, the point $H_A$ is also on the height from $C$ in $\Delta$, so is its orthocenter.
$\square$
Note: Unfortunately, in the given picture i could not force a proof (by simple geometric arguments) for $EH_A=E9$ (or $EH_C=E9$), so the problem was moved to one in $\Delta$, a triangle with sides chords in a regular $28$-gon inscribe in a circle with radius one. So we move to a third Lemma, given in two equivalent settings, 3a and 3b.
Lemma 3a: Let $\Pi_{28}$ be a regular $28$-gon with vertices numbered $0,1,2,\dots$ with elements from the ring $\Bbb Z/28$ of integers taken modulo $28$. Let $D$ be its center, and assume that the circle $(D)=(0123\dots)$ has radius one. Let $\Delta$ be the triangle with vertices $0,8,12$. Then the distance from its circumcenter $D$ to its orthocenter $H_A$ is $\sqrt 2$.
In fact we need for $\Delta$ only three vertices from a regular heptagon. So Lemma 3a is equivalent to the following:
Lemma 3b: Let $\Pi_7$ be the regular heptagon $ABCDEFG$ with center $O$, so that the radius of the circumscribed circle is $1=OA=OB=OC=OD=OE=OF=OG$. Let $H$ be the orthocenter of the triangle $\Delta$ with vertices $A,B,D$. Then the distance between the circumcenter $O$ and the orthocenter $H$ of $\Delta$ is $\sqrt 2$.
First proof (trigonometric): In a general triangle with sides $a,b,c$ and circumradius $R$ the distance $d$ between its circumcenter and its orthocenter is given by the formula: $$ d^2 =9R^2 - (a^2 + b^2 + c^2)\ . $$
In our special case, the circumradius is $R=1$, and the sides $a,b,c$ are $2\sin\frac\pi 7$, $2\sin\frac{2\pi} 7$, and $2\sin\frac{4\pi} 7$. Then $$ \begin{aligned} a^2+b^2+c^2 &= 4\left( \sin^2\frac{\pi}7 + \sin^2\frac{2\pi}7 + \sin^2\frac{4\pi}7 \right) \\ &= 2\left( 1-\cos\frac{2\pi}7 + 1-\cos\frac{4\pi}7 + 1-\cos\frac{8\pi}7 \right) \\ &=6 - \left( \cos\frac{2\pi}7 + \cos\frac{4\pi}7 + \cos\frac{6\pi}7 + \cos\frac{8\pi}7 + \cos\frac{10\pi}7 + \cos\frac{12\pi}7 \right) \\ &= 6-(-1)=7\ . \end{aligned} $$ Then $d^2= 9-7=2$, showing the needed relation.
$\square$
Second proof (algebraic): In essence, it is the same proof. We can and do assume that $\Delta$ has circumcenter with aff ix zero in $\Bbb C$, and vertices with affix points $z$, $z^2=z^9$, and $z^4$. Its centroid is located at $(z + z^4 + z^9)/3$, so the orthocenter is on the line $OG$ at distance three times the length of $OG$ from $O$, so $$ H= z + z^4 + z^9\ . $$ This is a Gauss sum, a Gauss period, which explains structurally the situation (from a number theoretic, rather than geometric point of view). Since $p=7$ is three modulo four, $H$ is $(-1+\sqrt {-7})/2$, so $OH$ is the modulus $\sqrt 2$ of this number.
$\square$
Third proof, a geometric proof: In essence, it is again the same proof. Let us draw a new picture:
Here $ABCDEFG$ is a regular heptagon, centered in $O$, and $\Gamma$ is so that $A\Gamma$ is a diameter. Inscribed triangles in the circle $(O)$ with $A\Gamma$ as a side have a right angle opposite to it. Let $B',C',D'$ be the projections of respectively $B,C,D$ on $A\Gamma$.
Again, we compute the distance $d$ between the center $O$ and the orthocenter $H$ of the triangle $\Delta=\Delta ABD$ using the formula $d^2 = 9R^2-(a^2+b^2+c^2)$, where $a,b,c$ are the lengths of the sides in $\Delta$ in this order, and then: $$ \begin{aligned} a^2 &= AB^2=A\Gamma\cdot AB'=2AB'\ ,\\ b^2 &= BD^2 =AC^2=A\Gamma\cdot AC'=2AC'\ ,\\ c^2 &= AD^2=A\Gamma\cdot AD'=2AD'\ ,\\ a^2+b^2+c^2 &=2AB'+2AC'+2AD'\\ &=\left|\ 2\overrightarrow{AB'}+2\overrightarrow{AC'}+2\overrightarrow{AD'}\ \right|\\ &=\left|\ \overrightarrow{AB}+ \overrightarrow{AG} +\overrightarrow{AC}+ \overrightarrow{AF} +\overrightarrow{AD}+ \overrightarrow{AE}\ \right|\\ &=\left|\ \overrightarrow{AB}+ \overrightarrow{AG} +\overrightarrow{AC}+ \overrightarrow{AF} +\overrightarrow{AD}+ \overrightarrow{AE}+ \overrightarrow{AA}\ \right|\\ &=\left|\ 7\overrightarrow{AO}\ \right|\\ &=7\ , \end{aligned} $$ using the fact that $O$ is the center of mass in $ABCDEFG$. This gives again $$ d^2 = 9R^2-(a^2+b^2+c^2) = 9-7=2\ . $$ $\square$
Conclusion: A geometric solution may work as follows:
Use Lemma 1 to move geometrically the computation from a regular heptagon with side one to a regular heptagon inscribed in a circle of radius one. The length to be computed is now the distance from the circumcenter and the orthocenter of a triangle with vertices among the vertices of the last regular heptagon. Use Lemma 2 and the geometric proof of Lemma 3 to conclude the computation. As ingredient we have used the relation $OH^2 = 9R^2-(a^2+b^2+c^2)$.