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

Question

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:

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

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:

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

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...

Answer 1

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.

Answer 2

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.

Answer 3

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 $ABC$.

Answer 4

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.