An Original Heptagonal Problem

I will be referring to Aditya's labelling.

Consider scalene triangle $\Delta ACE$ . The incentre of a triangle lies on the intersection of the angle bisectors. The angle bisector of $\angle ACE$ is $CB$ , as $B$ lies on the mid-arc of $AE$ , and for $\angle CEA$ it is $EL$ , by the same argument.

(This is so as $AB=BE$ and therefore the angle subtended by the arcs are of equal value. Therefore $\angle ACB = \angle BCE$ . This can be applied to the other case.)

Therefore, the intersection point of the question (point $P$ ) is the incentre of $\Delta ACE$ , and by definition the inradius must be $r$ , as $EC$ must be tangential to the incircle.

I am going to use a result that is easy enough to prove so it is fair to leave it to the readers. Refer to the first problem that appeared in INMO 1992, a very handy result useful in several places, (also in the ninth problem of the same paper!) here

$Construction$

Note in the figure that $\Delta ABC$ is one such scalene triangle. Any other scalene triangle will be congruent to it, so which triangle is worked on is irrelevant. This one eases discussion. The sides are labelled as usual with uncapitalized letters and diagonal $KB$ meets edge $CE$ upon extension at $D$ . $BF$ is the perpendicular dropped on $CD$ .

$PART$ 1:

Consider triangles $ABC$ and $DBC$ . $\angle ACB=\angle DCB$ (Angles subtended by equal chords (edges) $AB$ and $BE$ on a point on the circumference of the polygon, namely $C$ . Equal chords subtend equal angles on the circumference. In this case, each of these is in fact $\frac{\pi}{7}$ radians) $\angle ABC = \frac{4\pi}{7}$ , can be seen as a sum of angles subtended at $B$ by four equal chords. $\angle KBC=\frac{3\pi}{7}$ , can be seen as a sum of angles subtended at $B$ by three equal chords. $\implies \angle DBC=\pi-\angle KBC=\frac{4\pi}{7}$ So, $\angle ABC = \angle DBC$ . These with the common side BC imply congruency of triangles:

$\boxed{\Delta ABC \cong \Delta DBC}$ .

Let us call the altitude on side $b$ from vertex $B$ of triangle $ABC$ , $h_{b}$ . Note that, due to congruence, $BF=h_{b}$ . $PY$ is parallel to $BF$ , so triangles $CPY$ and $CBF$ are similar. This implies: $\frac{PY}{BF}=\frac{CP}{CB}$ . Consider the diagonal of the heptagon $LE$ : It is of course parallel to the diagonal $KB$ . So, $PE$ is parallel to $BD$ which means triangles $CPE$ and $CBD$ are similar. This implies: $\frac{CP}{CB}=\frac{CE}{CD}$ . Together, we get: $\frac{PY}{BF}=\frac{CE}{CD}$ . But note that, $CE$ is the same length as $AB$ (as the heptagon is regular and edges are equal), i.e. $c$ and $CD=CA=b$ due to the congruence of triangles proved above. Using notations described,

$\boxed{\frac{r}{h_{b}}=\frac{c}{b} \implies r=\frac{h_{b}\times c}{b}}$

$PART$ 2:

Now, we come to the inradius, $R$ . Note that in $\Delta ABC$ , $\angle ABC=\frac{4\pi}{7},\angle BAC=\frac{2\pi}{7},\angle BCA=\frac{\pi}{7}$

Since, $\angle B$ is twice $\angle A$ , using the result at the top,

$b^{2}=a^{2}+ac$

Since, $\angle A$ is twice $\angle C$ , using the result at the top,

$a^{2}=c^{2}+cb$

Substituting, the second relation in the first, we get, with $s$ as semiperimeter:

$b^{2}=c^{2}+cb+ac=c(a+b+c)=2cs$

Now, since inradius $R$ is given by $\frac{\Delta}{s}$ , where $\Delta$ is the area, and $\Delta$ may be written as $\frac{b\times h_{b}}{2}$ , we get, $s=\frac{b\times h_{b}}{2R}$ . Substituting,

$b^{2}=\frac{c\times b\times h_{b}}{R} \implies b=\frac{c\times h_{b}}{R}$

$\implies \boxed{R=\frac{h_{b}\times c}{b}}$ .

Hence the conclusion.