# Triangle in hexagon

In a regular hexagon ABCDEF is the midpoint (G)of the sides FE  and S
intersection of lines AC and GB.


(a) What is the relationship shared point of straight line with GB?

AB=a
BC=b

1. BS = αBG = α(-2a+ 2/3b)
2. BS = -a + ΨAC= -a + Ψ(a+b)
3. BS=-4/7a + 3/7b

(b) What proportion of the area of ​​the regular hexagon ABCDEF a surface area of triangle ABS?

How can i calculate ABS triangle and ABCDEF hexagon ? Thanks

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How should we know if is your answer? What kind of answer are you looking for? – Phira Nov 21 '12 at 20:36
I've removed the 'linear algebra' tag. May I recommend that you read the description of a tag before you select it? Also, your first question is pretty much unintelligible. Are you trying to find the ratio BS:BG? – Daniel Littlewood Nov 21 '12 at 20:38
The problem statement is unclear and does not match the picture. In the picture, G is the midpoint of DF, and the hexagon is ABCEFD. In part A, are you being asked for the relationship between two things? Finally, a regular hexagon has equal sides and angles, so is composed of 6 equilateral triangles and can be parametrized by $e^{in\pi/6}$ or $(\cos\frac{in\pi}6,\sin\frac{in\pi}6)$ for $0\le n<6$. – bgins Nov 21 '12 at 20:46
"linear algebra" is an appropriate tag. Question (a) is about vectors: With $\mathbf{a} := \vec{AB}$ and $\mathbf{b} := \vec{BC}$, what is $\vec{BS}$ in terms of $\mathbf{a}$ and $\mathbf{b}$? Options 1 and 2 describe approaches to the answer: (1) $\vec{BS} = \alpha \vec{BG}$, where we can write $\vec{BG} = -2\mathbf{a} + 3/2\mathbf{b}$ (OP has typo) and $\alpha$ happens to be $2/7$. (2) $\vec{BS}$ decomposes as $-\vec{AB} + \phi \vec{AC} = - \mathbf{a} + \phi(\mathbf{a}+\mathbf{b})$, where $\phi$ happens to be $3/7$. So, from either of these, (3) $\vec{BS} = -4/7\mathbf{a} + 3/7\mathbf{b}$. – Blue Nov 22 '12 at 1:24

Expanding my comment to an answer ...

I'll re-state the question.

(a) With vectors $\mathbf{a} := \vec{AB}$ and $\mathbf{b} := \vec{BC}$, how do we write $\vec{BS}$ in terms of $\mathbf{a}$ and $\mathbf{b}$?

OP lists three relations. I believe the the first two are approaches to finding the final, third, relation.

1. $\vec{BS} = \alpha \vec{BG}$, where we can write $\vec{BG} = -2\mathbf{a} + \frac{3}{2}\mathbf{b}$ (OP has typo) and scalar $\alpha$ happens to be $\frac{2}{7}$ (as explained below).

2. $\vec{BS}$ decomposes as $-\vec{AB} + \phi \vec{AC} = - \mathbf{a} + \phi(\mathbf{a}+\mathbf{b})$, where scalar $\phi$ happens to be $3/7$.

3. So, from either of the above, $$\vec{BS} = \frac{2}{7}\left(-2\mathbf{a}+\frac{3}{2}\mathbf{b}\right) = -\mathbf{a} + \frac{3}{7}\left(\mathbf{a}+\mathbf{b}\right) = -\frac{4}{7}\mathbf{a} +\frac{3}{7}\mathbf{b}$$

To see that $\alpha = 2/7$ in (1), tilt the diagram and construct segment $GP$ parallel to segment $DA$, with $P$ on the perpendicular dropped from $B$ to $DA$. Decompose the bottom half of the hexagon into equilateral triangles, and note that vertical lines through the vertices break these into congruent 30-60-90 triangles.

As $G$ is the midpoint of a side of these triangles, and $GP$ is parallel to their short legs, $GP$ cuts through midpoints all the way across, and is cut into 7 congruent segments. Note that $BS : BG = PQ : PG = 2 : 7$, so that $\vec{BS} = \frac{2}{7}\vec{BG}$.

To determine that $\phi=3/7$ in (2) ... well ... I just worked backwards from (3), but there's probably a nice geometric explanation here, too.

Moving on ...

(b) What is the proportion of the area of the hexagon to the area of $ABS$?

Note that $\triangle ABR$, as one of those 30-60-90 triangles, makes up one-twelfth of the hexagon. Calling the hexagon's area $H$, we can write

$$\frac{H}{\triangle ABS} = \frac{12 \triangle ABR}{\triangle ABS} = \frac{12 \cdot \frac{1}{2} |BR| |AR|}{\frac{1}{2}|BR||AS|} = 12 \frac{|AR|}{|AS|} = 12\frac{\frac{1}{2}|AC|}{|AS|} = 6\frac{|AC|}{|AS|} = 6\frac{1}{\phi} = 14$$

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One can use the geometry. However, the point of the sketch below is to show that the problem can be solved mechanically.

We could let the sides of the regular hexagon be, say, $a$. However, it is easier, at least for typing, to choose a specific length. Then we can if necessary scale our answer. Sides $1$ are not a bad idea, but sides $4$ are more convenient, fewer fractions for a while.

Let the $x$-axis be along $AB$, wjth $A=(-2,0)$ and $B=(2,0)$. The coordinates of $C$ are now easy to find. They are $(2+4\cos(60^\circ), 4\sin(60^\circ))$, that is, $(4, 2\sqrt{3})$.

Now that we know the coordinates of $A$ and $C$, we can find the equation of line $AC$.

Do a similar calculation for the other side. The coordinates of $D$ are, by symmetry, $(-4, 2\sqrt{3})$. To get to $G$, we add $2\cos(60^\circ)$ to the $x$-coordinate, and $2\sin(60^\circ)$ to the $y$-coordinate. Thus $G$ has coordinates $(-3,3\sqrt{3})$.

Now that we know the coordinates of $B$ and $G$, we can find the equation of line $BG$.

To find the coordinates of $S$, find where our two lines with known equations meet.

Now that we know the coordinates of $S$, we can find answers to the questions. The only one that is clear is the asked ratio between the area of $\triangle ABS$ and the area of the hexagon. Each area can be computed. For example, the area of $\triangle ABS$ is one-half of $4$ times the $y$-coordinate of $S$.

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