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Let $A=\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}$ and $P_0=(0,0)$. For positive arbitrary real number $a_i(i=1,2,\cdots ,6)$,

let $\begin{pmatrix} x_i\\ y_i \end{pmatrix}=A^i\begin{pmatrix} a_i\\ 0 \end{pmatrix}$ and define $\overrightarrow{P_{i-1}P_i}=(x_i,y_i)$, provide $P_6=P_0$.

and hexagon $H$ is consist of points $P_o,P_1,P_2,\cdots,P_6$ in regular sequence.

Problem 1.

Prove that $a_1-a_4=a_5-a_2=a_3-a_6$.

Problem 2.

If value provide with $\sum_{i=1}^{6}a_i=6,\ a_1-a_4=1$, find the maximum area of $H$.

Problem 3.

If value provide with $\sum_{i=1}^{6}a_i=6$, find the maximum area of $H$.

I totally can't solve these problems...

Added. I've figured out the shape of hexagon $H$, but nothing else...

enter image description here

Length of each segments are arbitrary, but angle of each segments are 120 degree and $P_0=P_6$.

Added2. Finally, I figured out problem 1. enter image description here

If you have a clear eye site, you can find the regular triangle.

So $a_1+a_2+a_3=a_3+a_4+a_5=a_5+a_6+a_1$.

We know $a_1+a_2+a_3=a_3+a_4+a_5$,

so $a_1+a_2=a_4+a_5$ and $a_1-a_4=a_5-a_2$.

and we know also $a_3+a_4+a_5=a_5+a_6+a_1$

so $a_3+a_4=a_6+a_1$ and $a_3-a_6=a_1-a_4$.

Finally we get $a_1-a_4=a_5-a_2=a_3-a_6$.

Quod Erat Demonstrandum

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Would you happen to remember what a rotation matrix looks like? Your sines and cosines? –  J. M. Sep 17 '11 at 13:48
    
You know your sines and cosines, right? Would you happen to recognize $A$'s entries? –  J. M. Sep 17 '11 at 13:51
    
The question seems unclear to me. If $a_1,\ldots,a_6$ are arbitrary, you cannot prove $a_1-a_4=a_2-a_5=a_3-a_6$. –  Martin Sleziak Sep 17 '11 at 13:56
    
@Martin stressed the fact that there seems to be no hypothesis at all on the numbers $a_i$. Say I choose $a_i=i$ for every $i\ne6$ and $a_6=23$, then these numbers DO NOT satisfy the property of your Problem 1 although my choice seems to be fully legitimate. In a word: please READ Martin's comment. –  Did Sep 18 '11 at 17:20
    
Hmmm, I see... By the way, you do not say what you know, what you tried, and so on. –  Did Sep 19 '11 at 5:37

1 Answer 1

up vote 2 down vote accepted

An easy solution to the first question uses the affixes $z_k$ of the points $P_k$, since $z_0=0$ and $z_{k+1}-z_k=a_ku^{k+1}$ with $u=\mathrm e^{\mathrm i\pi/3}$. The condition that $P_6=P_0$ reads as $\sum\limits_{k=1}^6a_ku^k=0$. Introduce the imaginary part $v$ of $u$, that is,
$$ v=\sqrt3/2. $$ Then the real and imaginary parts of the sum above are $a_1+\frac12a_2-\frac12a_3-a_4-\frac12a_5+\frac12a_6$ and $v(a_2+a_3-a_5-a_6)$ and these are both zero if and only if:

There exists $b$ such that $a_1=a_4+b$, $a_5=a_2+b$ and $a_3=a_6+b$.

Note that this implies that $a_1+a_2+a_3+a_4+a_5+a_6=2(a_2+a_4+a_6)+3b$. From now on we reduce everything to the variables $(a_2,a_4,a_6)$ and $b$.

To solve the second and third questions, first note that opposite sides of the hexagon are parallel and imagine moving points $P_3$, $P_4$ and $P_5$ in the horizontal direction, substracting some quantity $c$ to their three abscissæ. Algebraically, this amounts to adding $c$ to $a_3$ and $a_6$ and leaving the other $a_k$s and $b$ unchanged. Geometrically, this adds $c$ times the height of the hexagon to the global area.

The height of the hexagon, in other words the distance between the lines $P_5P_6$ and $P_2P_3$, is $v(a_1+a_2)=v(a_4+a_2+b)$ hence $$ A(a_2,a_4,a_6+c,b)=A(a_2,a_4,a_6,b)+cv(a_4+a_2+b), $$ where $A(x,y,z,t)$ is the area of the hexagon when $a_1=y+t$, $a_2=x$, $a_3=z+t$, $a_4=y$, $a_5=x+t$, $a_6=z$ and $b=t$. In particular, $$ A(a_2,a_4,a_6,b)=A(a_2,a_4,0,b)+va_6(a_4+a_2+b). $$ Applying the same argument to the two other directions, one gets $$ A(a_2,a_4,a_6,b)=A(0,0,0,b)+va_6(a_4+a_2+b)+va_4(a_2+b)+va_2b. $$ The hexagon parametrized by $(0,0,0,b)$ is in fact the equilateral triangle with side $b$, hence $A(0,0,0,b)=vb^2/2$. This yields finally

$$ A(a_2,a_4,a_6,b)=v(a_2a_4+a_4a_6+a_6a_2+b(a_2+a_4+a_6)+b^2/2). $$

The second question asks for the maximal value of $A$ when $b=1$ and $a_2+a_4+a_6=\frac32$. Lagrange multipliers method indicates that this happens when $a_2=a_4=a_6$, hence the maximal value is $$ A(1/2,1/2,1/2,1)=v\cdot11/4=11\sqrt3/8. $$ The third question asks for the maximal value when $a_2+a_4+a_6=3s$ is such that $2s+b=2$. For each value of $s$, $a_2a_4+a_4a_6+a_6a_2$ is maximal when $a_2=a_4=a_6=s$ hence one looks for the maximal value of $3s^2+3bs+b^2/2$. Since $b=2-2s$, this is $2+s(2-s)\le 3$ which reaches its maximal value when $s=1$ and $b=0$. Finally, the maximal value corresponds to the regular hexagon with side $1$ and is $$ A(1,1,1,0)=3v=3\sqrt3/2. $$

Edit Using your picture, one sees that $A$ is the area of the equilateral triangle with side $a_2+a_4+a_6+2b$ minus the areas of the three equilateral triangles with side $a_1=a_4+b$, $a_3=a_6+b$ and $a_5=a_2+b$. The area of an equilateral triangle with side $\ell$ being $\sqrt3\ell^2/4$, one gets directly $$ A(a_2,a_4,a_6,b)=\left((a_2+a_4+a_6+2b)^2-(a_4+b)^2-(a_2+b)^2-(a_6+b)^2\right)\sqrt3/4. $$ Expanding the squares, one falls back on the formulas above.

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@kso, Although I've figured out the shape of H, but I totally can't solve these problems is to be compared with There is another solution which use Schwarz's inequality... :-) –  Did Oct 20 '11 at 9:13

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