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Given the regular heptagon, prove the four point in circle.

enter image description here

Here's my attempt to construct an isosceles trapezoid, but I do not know how to do next:

enter image description here

Blue
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lu654312
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    Hello, my English is not very good. I know that there are ways to prove that four points are on a circle, such as proving that the angles are equal, Ptolemy's Theorem, etc. In this case, I want to prove the five-point common circle by proving another isosceles trapezoid – lu654312 Apr 02 '22 at 04:44
  • Unfortunately, my first comment was a little optimistic; the question's a bit trickier than I thought. Still, it's useful for you to tell what you know, so that answerers don't spend time explaining things you already understand or using techniques you haven't seen yet. It would also help to know the source of the problem. If it's a textbook exercise, what topics were covered in the chapter that seem relevant? If it's an online challenge or contest, can you say something about the intended level of the audience? The more context you can provide, the better. Cheers! – Blue Apr 02 '22 at 04:52
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    In fact, this is just an exercise set under the theme of four-point in circle. I have mastered many methods to prove the four-point in circle, but I found that I could not prove it in this problem. I even tried to calculate the side length and Angle by using trigonometric functions to get the solution, but I knew that such a calculation would be too much, and it was not the method that the question hoped to see – lu654312 Apr 02 '22 at 05:22

2 Answers2

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Label points as in the picture.

Note that $\angle AGH = \dfrac{2\pi}{7} = \angle GHA$, hence $AH=AG$. Also, $\angle ICB = \dfrac{3\pi}{7} = \angle BIC$, therefore $BI=BC$. So $AH=AG=BC=BI$. Moreover $AE=BE$ and $\angle DBE = \dfrac{\pi}{7} = \angle DAE$. It follows from SAS that $\triangle HAE = \triangle IBE$. In particular $$\angle DHE = \pi - \angle EHA = \pi - \angle EIB = \angle DIE$$ from which it follows that $H, E, D, I$ are concyclic.

timon92
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enter image description here

Let $a=\frac{2 \pi}{7}.$ Let us consider the heptagon $ABCDEFG$ with resp. coordinates $(\cos ka, \sin ka), k=0,1,\cdots 6$.

Let us denote by $W$ the center of the circumscribed circle to $(B,C,K)$. It is situated at the intersection of the two perpendicular bissectors of $BC$ and $BK$.

The perpendicular bissector of $BC$ passes through $O$ and $F$, due to the symmetry of the figure wrt line $FJ$. Therefore $$\vec{OW}=p\vec{OJ}$$ which is equivalent to the fact that, for a certain $p$:

$$(W_x,W_y)=p(\cos 3a/2,\sin 3a/2)\tag{1}$$

(because polar angle of $\vec{OJ}$ is $3a/2$).

Triangle $KBD$ is an homothetic image of $ADE$ because they have the same angles. It is therefore an isoceles triangle too with corresponding parallel bases $BK$ and $DE$ ; therefore parallel altitudes, altitude $DH$ being the perpendicular bissector of $BK$, we can conclude that $W$ and $D$ have the same ordinate $\sin(3a)$. Using (1), we can conclude that

$$(W_x,W_y)=2 \cos(3a/2)(\cos 3a/2,\sin 3a/2).\tag{2}$$

Let $H=AE \cap BF$. Due to the symmetry of the figure wrt line $GI$, we have, for a certain $q$::

$$\vec{OH}=q\vec{OG} \iff (H_x,H_y)=q(\cos(6a),\sin(6a))=q(\cos a, -\sin a)$$

The value of $q$ is easily found to be $q=2 \cos a -1$.

Now, if we turn to complex number representation setting $b:=a/2=\pi/7$, we have:

$$W=2 \cos(b)e^{ib}, \ \ H=(2 \cos(a)-1)e^{i5b}, \ \ B=e^{i2b}$$

it remains to check that the following squared distances are equal:

$$WB^2=WH^2$$

Meaning that:

$$|2\cos(b)e^{ib}-(2 \cos(2b)-1)e^{-2ib}|^2=|2\cos(b)e^{ib}-e^{2ib}|^2$$

which is true (verification using general trigonometry formulas and particular relationships specific to angle $a=\frac{2 \pi}{7}$ ; I can give more details).

Reamrk: a very nice article here on the heptagon.

Jean Marie
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