Constructing the orthocenter of a cyclic pentagon

Question

I’d like to construct the following thing in Georgebra:

A non-regular cyclic pentagon $ABCDE$ such that the perpendiculars from $A$ to $CD$, $B$ to $DE$... all concur in one point $H$.

I managed to prove that if $H$ exists then the pentagon formed by the midpoints of $AC,BE...$ is cyclic. Here’s how:

Let $O$ be the circumcenter of $ABCDE$ and $M$ the midpoint of $OH$ then since rays $AH$ and $AO$ are isogonal in $\triangle ACD$ so $MB_1=ME_1$ where $B_1$ and $E_1$ are the midpoints of $AC$ and $AD$. Applying this to every point of the pentagon we get $M$ is the circumcenter of the pentagon formed by the midpoints.

It turns out the converse is also true.

Answer 1

My solution is not particularly elegant, nonetheless it proves that a solution exists and might give some insight for a nicer construction.

Let the circumcircle be a unit circle centred at $O=(0,0)$ and vertex $A$ be fixed at $(1,0)$. Choose then at will circumcenter $H=(x,y)$ inside the circle. Side $CD$ is perpendicular to $AH$ and intersects ray $AH$ at some point $K$ which can be expressed as $$ K=H+t(H-A), $$ where $t$ is an unknown parameter.

From $K$ all the other vertices can be found:

• $C$ and $D$ are the intersections of the perpendicular to $AH$ passing through $K$ with the circle;
• $B$ is the second intersection of the perpendicular to $DH$ passing through $A$ with the circle;
• $E$ is the second intersection of the perpendicular to $CH$ passing through $A$ with the circle.

In general, lines $BH$ and $DE$ so constructed will not be perpendicular, but they might be for some value of $t$. To find the right $t$ I expressed the coordinates of $BCDE$ as functions of $t$ and then solved (using Mathematica) the equation expressing the perpendicularity of $BH$ and $DE$: $$ (B-H)\cdot(D-E)=0. $$ I got four real solutions, two of which represent degenerate pentagons, with some vertices being coincident. The other two solutions are fine, one of them giving a convex pentagon, the other one a self-intersecting pentagon (see figures below). These two solutions for $t$ can be expressed as follows: $$ t={\left(1-r^2\right)\over4 d^2} \left(1+r^2 \pm\sqrt{\left(1-r^2\right)^2+4}\right), $$ where $r=OH$ and $d=AH$.

Note that only square roots are involved, hence point $K$ can be constructed with straightedge and compass. Moreover, the solutions are well defined for any position of $H$ inside the circle. There is hope then, I think, that a more elegant construction can be found.

Answer 2

Not an answer but a remark that I consider as important which doesn't fit into a simple comment.

In such questions with multiple constraints, it is difficult to figure out the mutual dependencies ; the solution given by @Intelligenci Pauca has in particular the merit to show that placing first the orthocenter defines in a unique manner two solutions (convex and star-shaped).

What I would like here is to show that the $5$ orthogonality constraints are dependent, i.e., if $4$ of them are verified, the fifth one is necessarily fulfilled.

I will use for that a complex number formulation, with vertices denoted $v_0, \ k=0\cdots 4$ with cyclical convention : $v_5=v_0, v_6=v_1,$ etc.

Let $h$ denote the orthocenter. The orthogonality conditions can be written under the form :

$$\text{for} \ k=0\cdots 4, \ \text{there exists} \ \lambda_k \in \mathbb{R} \ s.t. \ h=v_k+\lambda_ki(v_{k+2}-v_{k+3}).$$

or $$\frac{h-v_k}{i(v_{k+2}-v_{k+3})} \in \mathbb{R}$$

which is equivalent to $$\frac{h-v_k}{i(v_{k+2}-v_{k+3})}=\frac{\overline{h-v_k}}{\overline{i(v_{k+2}-v_{k+3})}}$$

or equivalently, after simplification by $i$ :

$$(h-v_k)(\bar{v_{k+2}}-\bar{v_{k+3}})=-(\bar{h}-\bar{v_k})(v_{k+2}-v_{k+3})$$

$$h(\bar{v_{k+2}}-\bar{v_{k+3}})-v_k(\bar{v_{k+2}}-\bar{v_{k+3}})=-\bar{h}(v_{k+2}-v_{k+3})+\bar{v_k}(v_{k+2}-v_{k+3})\tag{2}$$

Let us set $$m:=\sum_{(cycl) \ k=0}^4v_k\overline{v_{k+2}}=v_0\bar{v_2}+v_1\bar{v_3}+v_2\bar{v_4}+v_3\bar{v_0}+v_4\bar{v_1}$$

Summing up the 5 equalities (2), we get :

$$0-m+\bar{m}=0+\bar{m}-m$$

Otherwise said, the 5 equalities (2) sum up to $0$, which is equivalent to say that any of them is a consequence of the four others.

(Maybe, I will add new elements to this "answer")