# Prove that the area of an inscribed hexagon is twice the area of a triangle

Let $$\Delta ABC$$ be an acute triangle and denote the circumscribed circle of $$\Delta ABC$$ $$\Gamma$$ with midpoint $$O$$. Let $$A_1, B_1, C_1$$ be the points on $$\Gamma$$ where the lines $$AO, BO, CO$$ intersect $$\Gamma$$. Show that the area of the hexagon $$AC_1BA_1CB_1$$ is twice the area of $$\Delta ABC$$.

My first approach was to realize that the hexagon $$AC_1BA_1CB_1$$ contains $$\Delta ABC$$, which means that we have to prove that $$|\Delta ABC_1| + |\Delta BA_1C| + |\Delta CB_1A| = |\Delta ABC|$$. This could equivalently be written as:

$$$$\frac{|\Delta ABC_1| + |\Delta BA_1C| + |\Delta CB_1A|}{|\Delta ABC|} = 1.$$$$

But since the three "outer" triangles each share a side with $$\Delta ABC$$, the ratio between the areas of $$\Delta ABC$$ and the respective "outer" triangle" will be $$\frac{h_i}{H_i}$$, where $$h_i$$ is the height of the outer triangle and $$H_i$$ is the height of $$\Delta ABC$$ (where both heights are perpendicular to the shared side). This means that we can write the above equation as:

$$$$\frac{h_1}{H_1} + \frac{h_2}{H_2} + \frac{h_3}{H_3} = 1$$$$

Now, label the points where the lines $$AO, BO, CO$$ intersect the opposite side of $$\Delta ABC$$ $$P, R, Q$$ respectively ($$P$$ on $$AB$$, $$R$$ on $$AC$$, $$Q$$ on $$AB$$). Since we have angles in the same segment of $$\Gamma$$, we get the following similar triangles: $$\Delta ABC_1\sim\Delta PBC, \Delta CQA_1\sim\Delta ABQ, \Delta ARB_1\sim\Delta RBC$$, which means that we can write:

$$\frac{h_1}{H_1} = \frac{|AP|}{|BP|} \\ \frac{h_2}{H_2} = \frac{|CQ|}{|BQ|} \\ \frac{h_3}{H_3} = \frac{|AR|}{|CR|}$$

Which means that we have to prove that the sum of these (above) ratios equals $$1$$. But from here, I can't seem to make much progress...

• Hint: Connect the orthocenter of the triangle to the vertices of the triangle.
– Blue
Mar 4, 2023 at 13:10
• @Blue what do you mean? the orthocenter of $\Delta ABC$ (where the heights of $\Delta ABC$ meet) are already connected to the vertices of $\Delta ABC$, since each height goes from one vertex (perpendicular) to the opposite side...? Mar 4, 2023 at 14:19
• @MartinWestin Yes. That is very likely what Blue means. What observations can you make about $HA_1$? Try drawing an accurate picture. Mar 4, 2023 at 14:32
• @CalvinLin I see what Blue means. But I am not sure what it has to do with $HA_1$.
– ACB
Mar 5, 2023 at 6:11
• I have taken the liberty to change "circumscribed" into "inscribed" in your title. Mar 8, 2023 at 14:27

I found the full solution myself as well:

Drawing out the heights of $$\Delta ABC$$ and labeling their intersection point $$I$$, we get the following:

Since $$\Delta C'AC$$ is a triangle with one point on $$\Gamma$$ and the base as the diameter of $$\Gamma$$, we have $$\angle C'AC = 90°$$, and since $$BP$$ is a height in $$\Delta ABC$$, $$\angle BPA = 90°$$. Hence, $$\angle C'AC + \angle BPA = 180°$$, which is true if and only if $$AC' || BP$$. In the same manner, we find that $$C'B || AQ$$, and hence $$AIBC'$$ is a parallelogram, and since $$AB$$ is a diagonal in this parallelogram, it is true that $$\Delta ABC_1 \cong \Delta ABI \implies |\Delta ABC'| = |\Delta ABI|$$. In the same way, we can find the following parallel lines:

$$BI || CA', CI || BA', AI || CB', CI || AB'$$. This gives two more parallelograms: $$BICA'$$ and $$AICB'$$. Hence, $$|\Delta ABC| = |\Delta ABI| + |\Delta BCI| + |\Delta ACI| = |\Delta ABC'| + |\Delta BCA'| + |\Delta ACB'|$$, which was to be proved.

• (Generally use H for Orthocenter, I for Incircle. I'm going to use H instead of I in my comments). The angles at H match up with the angles at $AC'B$, which is what motivates looking at H (as opposed to just something magical that is pulled out of a hat). Mar 5, 2023 at 13:26

Here is a solution using complex numbers geometry.

Have a look at this picture and its associated complex numbers (we have assumed that the circumradius is $$1$$ WLOG); for example, $$A'$$, opposite of $$A$$ is $$-e^{ia}=e^{i(a+\pi)}$$ :

The area $$\frak{A}$$ of hexagon $$AC'BA'CB'$$ is the sum of areas of triangles $$AOA',C'OB$$, etc. where each area is simple to compute ; for example :

$$area(AOA')=\frac12 OA.OA'.\sin(\angle AOC')=\frac12 \sin((c+\pi)-a)$$

Therefore :

$$\frak{A}=\begin{cases} \frac12 \sin((c+\pi)-a)+\\ \frac12 \sin(b-(c+\pi))+\\ \frac12 \sin((a+\pi)-b)+\\ \frac12 \sin(c-(a+\pi))+\\ \frac12 \sin((b+\pi)-c)+\\ \frac12 \sin(a-(b+\pi)) \end{cases}=\begin{cases} -\frac12 \sin(c-a)+\\ -\frac12 \sin(b-c)+\\ -\frac12 \sin(a-b)+\\ -\frac12 \sin(c-a)+\\ -\frac12 \sin(b-c)+\\ -\frac12 \sin(a-b) \end{cases}=\begin{cases} -\sin(c-a)+\\ -\sin(b-c)+\\ -\sin(a-b) \end{cases}=\begin{cases} \sin(a-c)+\\ \sin(c-b)+\\ \sin(b-a) \end{cases}$$

which is the sum of triangles $$OAB, \ OBC, \ OCA$$ i.e., the area of triangle $$ABC$$.

• well it seems a bit over the top for such a simple geometry question... but still an elegant solution i must say Mar 7, 2023 at 11:44
• Thanks for your answer. Well, its interest IMHO is that it involves very few concepts, followed by a mechanical verification that can be given to a machine. It may not look "glamorous" compared to pure geometry methods... but is in the spirit of automated theorem proving which is progressing year after year. Mar 7, 2023 at 14:32
• yes of course, good point Mar 7, 2023 at 17:53

Let's assume $$R$$ is the circumradius of $$\triangle ABC$$. Then, we have:

$$S_{\triangle ABC}=\frac{1}{2} |AC||AB| \sin A=\frac{1}{2} \times 2R \sin B \times 2R\sin C \times \sin A \\ = 2R^2 \sin A\sin B \sin C.$$

Now, let's compute the area of $$\triangle A_1BC$$. We have:

$$S_{\triangle A_1BC}=\frac{1}{2}|A_1B||A_1C| \sin A =\frac{1}{2} \times 2R \cos C \times 2R \cos B \times \sin A \\= 2R^2 \cos C \cos B \sin A.$$ Similarly,

$$S_{\triangle B_1AC}=2R^2 \cos A \cos C \sin B \\ S_{\triangle C_1AB}=2R^2 \cos A \cos B \sin C.$$

Thus, as you observed, we need to show that:

$$\cos C \cos B \sin A+\cos A \cos C \sin B+\cos A \cos B \sin C =\sin A\sin B \sin C,$$

while $$\sin A=\sin(B+C)$$ and $$\cos A= - \cos (B+C).$$

So, we must show that:

$$\cos C \cos B \sin(B+C)- \cos (B+C) \cos C \sin B- \cos (B+C)\cos B \sin C \\= \sin(B+C)\sin B \sin C \\ \iff \sin(B+C) (\cos C \cos B - \sin B \sin C )= \cos (B+C) (\cos C \sin B + \cos B \sin C),$$

which obviously holds.

We are done.

• Nice! I didn't think you could use trig to solve this question, but good job Mar 5, 2023 at 9:35

Here is a generalization of this issue, with a surprisingly simple answer.

Let us select, in triangle $$T=ABC$$, acute or not, a point $$I$$ in its interior (see "Restriction on the choice of $$I$$" at the bottom of this answer).

Let triangle $$T'=A'B'C'$$ be obtained from $$T$$ by point-symmetry with respect to $$I$$ (said otherwise by the homothety centered in $$I$$ with ratio $$-1$$).

Result : The area of hexagon $$AC'BA'CB'A$$ is (still) twice the area of triangle $$T$$.

Proof by barycentric coordinates (b.c.) with respect to triangle $$T$$.

Let $$(a,b,c)$$ be the b.c. of $$I$$. The b.c. of the different points are easily obtained and can be seen of figure 1. (please, check that all of them have a sum equal to $$1$$).

Fig. 1.

The hexagon area being twice the area of quadrilateral $$CA'BC'$$, it remains then to show that $$area(CA'BC)=area(T)$$ (we take the area of triangle $$T$$ as reference area).

$$area(CA'BC)=area(CIA')+area(A'IB)+area(BIC')$$

$$=\begin{vmatrix}0&a&(2a-1)\\0&b&2b\\1&c&2c\end{vmatrix}+\begin{vmatrix}(2a-1)&a&0\\2b&b&1\\2c&c&0\end{vmatrix}+\begin{vmatrix}0&a&2a\\1&b&2b\\0&c&(2c-1)\end{vmatrix}$$

$$=b+c+a=1$$

as desired.

Some particular cases :

• If $$I=O$$, the circumcenter, it is the original question.

• If $$I=A_1$$, midpoint of $$BC$$, some points aggregate, turning the hexagon into a parallelogram whose area is blattantly twice the area of triangle $$T$$.

• If $$I=G$$, the centroid of triangle $$T$$, I don't resist the pleasure to give a specific proof (see figure 2 below). In fact, I leave it like this, a "proof without words" !

Fig. 2.

"Restriction on the choice of $$I$$" : in fact, for the hexagon having ... 6 vertices (!), it is necessary that points $$A',B',C'$$ have a negative b.c., i.e., that we have simulatneously $$a \le 1/2, b \le 1/2, c \le 1/2$$ which means that $$I$$ must be chosen inside the mid-triangle $$A_1B_1C_1$$ where this points are the midpoints of the triangle sides. This will always be the case for centroid $$G$$, but will necessitate to have an acute triangle for the circumcenter.

• Very nice generalization. I knew of a proof by comparing areas essentially like yours, showing that $[CA'I] = [CIA]$ due to the same base length, which you showed by calculating the determinant equals to $b$. Mar 10, 2023 at 14:33