Trap e zium

Suppose $m=12a^2+8a-1$ , $n=-2a-1$ and $\phi=\frac{11\pi a}{6}$ , so $mn=-24a^3-28a^2-6a+1$ .

Therefore, the equation becomes:

$6\sin (2\theta +\phi)-6\sin \phi +mn=2m\cos (\theta +\phi )+6n\sin \theta$

$12\cos (\theta +\phi) \sin \theta +mn-2m\cos (\theta +\phi )-6n\sin \theta$ =0

$(2\cos (\theta +\phi)-n)(6\sin \theta -m)$ =0

Hence, $\cos (\theta +\phi)=\frac{n}{2}$ (I)

or $\sin \theta=\frac{m}{6}$ . (II)

We will obtain exacly 4 roots if $-1 < \frac{n}{2} < 1$ and $-1 < \frac{m}{6} < 1$ .

We notice that when draw point $(\cos \theta, \sin \theta)$ of these 4 roots, these point will be on the circle (0;1), the line connect two points of (II) will parallel to the horizontal axis but the direction of the line connect two points of (I) depends on $\phi$ . Therefore, there are two cases for these 4 points to form a trapezoid:

Case 1: $\frac{n}{2}=\frac{m}{6}$ or $\frac{n}{2}=-\frac{m}{6}$ in order to make the segment of 2 points of (I) equal to that of (II).

Solve these equation, we obtain 4 distinct values of a: $-1,\frac{-1}{6},\frac{1}{2},\frac{-2}{3}$ ,but only $-1,\frac{-1}{6},\frac{-2}{3}$ satisfy $-1 < \frac{n}{2} < 1$ and $-1 < \frac{m}{6} < 1$ . $a=-1$ is also not satisfy since it makes one root of (I) equals to one root of (II).

Case 2: $\phi=\frac{\pi}{2}+k \pi$ in order to make the line connect 2 points of (I) parallels to the horizontal axis and therefor parallels to the line connect 2 points of (II).

In order to satisfy $-1 < \frac{n}{2} < 1$ and $-1 < \frac{m}{6} < 1$ , we obtain 3 possible values of k: -2,-1,0 and therefore 3 values of a: $\frac{3}{11}, \frac{-3}{11}, \frac{-9}{11}$ .

In conclusion, there are 5 values of a satisfy the condition: $\frac{3}{11}, \frac{-3}{11}, \frac{-9}{11},\frac{-1}{6},\frac{-2}{3}$ .

Using the factor formulae, the equation factors as $\left(2\cos(\theta+\tfrac{11\pi a}{6}) + 2a+1\right)\left(6\sin\theta - (12a^2+8a-1)\right)\;=\;0$ so the solutions are $\theta_1 = \alpha \qquad \theta_2 = \pi-\alpha \\ \theta_3 = -\tfrac{11\pi a}{6} + \beta \qquad \theta_4 = -\tfrac{11\pi a}{6} - \beta$ where $\alpha \; = \; \sin^{-1}(\tfrac{12a^2+8a-1}{6}) \qquad \beta \;=\; \cos^{-1}(-\tfrac{2a+1}{2})$ For both $\alpha$ and $\beta$ to exist we need $-\frac76 \le a \le \frac12$ . If $a=\frac12$ then $\alpha=\frac{\pi}{2}$ and $\beta=\pi$ , so that $\theta_1=\theta_2$ and $\theta_3=\theta_4$ . This case does not give us four distinct points, so can be ignored. Thus $-\frac76 \le a < \frac12$ , and $-\frac{\pi}{2} < \alpha \le \frac{\pi}{2}$ , $0 < \beta < \pi$ .

The gradient of the chord between $(\cos\theta,\sin\theta)$ and $(\cos\phi,\sin\phi)$ is $-\cot\frac{\theta+\phi}{2}$ .

1. For the chords "joining" $\theta_1$ to $\theta_2$ and $\theta_3$ to $\theta_4$ to be parallel, we need $0 \;= \; \cot\tfrac{\pi}{2} \; = \; \cot(\tfrac{\theta_1+\theta_2}{2})\; = \; \cot(\tfrac{\theta_3+\theta_4}{2}) \; = \; \cot(\tfrac{-11\pi a}{6})$ and so $\tfrac{11 \pi a}{6} \,=\, (n+\tfrac{1}{2})\pi$ for some integer $n$ . This gives $a \,=\, \tfrac{3(2n+1)}{11}$ for integers $n$ . There are three possible solutions with $a = \tfrac{3}{11},-\tfrac{3}{11},-\tfrac{9}{11}$ .

2. For the chords "joining" $\theta_1$ to $\theta_3$ and $\theta_2$ to $\theta_4$ to be parallel, we need $\cot(\tfrac{\theta_1+\theta_3}{2})\; = \; \cot(\tfrac{\theta_2+\theta_4}{2})$ which breaks down to saying that $\alpha+\beta = (n+\tfrac12)\pi$ for some integer $n$ . Given the range of values of $\alpha$ and (\beta) possible, we deduce that $\alpha+\beta=\tfrac{\pi}{2}$ . This implies that $\tfrac{12a^2+8a-1}{6} \; = \; -\tfrac{2a+1}{2}$ and hence $(6a+1)(a+1) = 0$ . The value $a=-1$ gives $\theta_1=\theta_3=\tfrac{\pi}{6}$ , so may be ignored. This leaves us the solution $a=-\tfrac{1}{6}$ .

3. For the chords "joining" $\theta_1$ to $\theta_4$ and $\theta_2$ to $\theta_3$ to be parallel, we need $\cot(\tfrac{\theta_1+\theta_4}{2})\; = \; \cot(\tfrac{\theta_2+\theta_3}{2})$ which breaks down to saying that $\alpha-\beta = (n+\tfrac{1}{2})\pi$ for some integer $n$ . Given the range of values of $\alpha$ and $\beta$ , we deduce that $\alpha-\beta=-\tfrac{\pi}{2}$ . This implies that $\tfrac{12a^2+8a-1}{6} - \tfrac{2a+1}{2} \; =\; 0$ and hence $(2a-1)(3a+2) = 0$ . Since the case $a=\tfrac12$ is not permitted, we obtain just one solution $a=-\tfrac23$ .

Thus we have just five acceptable solutions, namely $a=\tfrac{3}{11}$ , $-\tfrac{3}{11}$ , $-\tfrac{9}{11}$ , $-\tfrac{1}{6}$ , and $-\tfrac{2}{3}$ .

Notice that $24a^3 + 28a^2 + 6a-1 = (12a^2+8a-1)(2a+1)$ , and that $\sin(2\theta + \frac{11\pi a}{6}) - \sin(\frac{11\pi a}{6}) = 2\cos(\theta + \frac{11\pi a}{6})\sin \theta$ . Then, as we can see, the expressions $g:=12a^2+8a-1, h:=2a+1, c:=\cos(\theta + \alpha)$ (where $\alpha=\frac{11\pi a}{6}$ ), and $d:=\sin\theta$ appear twice in the equation. This suggests a factorization may be possible. Indeed, the original equation can be written as

$6\cdot 2 cd - gh = 2g c - 6hd$ $(6d)(2c) - gh - g(2c) + h(6d) = 0$ $(6d-g)(2c+h) = 0$

Therefore, the solution set of the original equation is the union of the solution sets of the following two equations:

$6\sin\theta = g \dots \dots (1)$

$2\cos(\theta + \alpha) = -h \dots \dots (2)$

Let $z=\cos\theta + i\sin\theta$ , and $t=\cos\alpha + i\sin\alpha$ . Then the two equations becomes

$6(z-1/z)/(2i) = g$ $zt+1/(zt) = -h$

or rearrange both of them as quadratic equations:

$3z^2 - igz - 3 = 0 \dots \dots (3)$ $t^2z^2 + htz + 1 = 0 \dots \dots (4)$

Each of them has at most two solutions, and therefore there are at most four distinct solutions in total. Note that (3) has two distinct complex roots if and only if $(ig)^2 - 4(3)(-3)=-g^2+36\neq 0$ , or $g\neq \pm 6$ , and that (4) has two distinct complex roots if and only if $(ht)^2-4(t^2)=(h^2-4)t^2\neq 0$ , or $h\neq \pm 2$ .

Assume that the solution sets of (3) and (4) are $S_1 = \{z_1,z_2\}$ , $S_2=\{z_3, z_4\}$ , respectively. Then the condition of the problem can be restated as the following:

Find all real numbers $a$ such that $z_1, z_2, z_3, z_4$ are distinct complex numbers that form a trapezoid in the complex plane.

Notice that the trapezoid is inscribed in the unit circle, so it must be isosceles, and therefore there exist two pairs of points that subtends the same circular angle. Indeed (please draw a picture to see the following), if $r,s,v,w$ are four complex units that make an isosceles trapezoid and the line made by $r,s$ is parallel to the line made by $v,w$ , then there exists a complex unit $u$ such that $v/r = s/w = u$ . This fact is not affected by the orientation of the lines (Again, please draw a picture!).

Now there are two cases:

Case One. The line made by $z_1,z_2$ is parallel to the line made by $z_3,z_4$ . Thus there exists a unit $u$ such that $z_2/z_1 = z_3/z_4 = u$ , where $u\neq 1$ . So $S_1 = \{z_1, uz_1\}, S_2=\{uz_4, z_4\}$ .

(Caution: Note that we have not designate an order to the roots of the two quadratics. For example, we can only say that $p_1$ is one root of the quadratic equation (3), but we cannot say that this is "the first root". Nonetheless, $q_1$ is another root of (3). So there are two possibilities for $q_1/p_1 = p_2/q_2$ , depending on how we choose the order of the roots. However, we can still try to treat them in the same case.)

Using Vieta's theorem for equations (3) and (4), respectively, we have

$(1+u)z_1 = uz_1+z_1= ig/3$ $uz_1^2 = (uz_1)(z_1) = -1$ $(1+u)z_4 = uz_4 + z_4 = -h/t$ $uz_4^2 = (uz_4)(z_4) = 1/t^2$

Dividing the second and the fourth equations, we get $z_1^2/z_4^2 = -t^2$ , or $z_1/z_4 = p it$ , where p = 1 or -1. If $u\neq -1$ then we also divide the first and the third equations to get $z_1/z_4 = -igt/(3h)$ . Note that u=-1 is impossible, because h=0 and g=0 can't hold simultaneously. Therefore, $pit = -igt/(3h)$ $p\cdot 3h = -g$ $p \cdot 3(2a+1) = -(12a^2+8a-1)$

So we get $12a^2 + 2a - 4 = 0$ or $12a^2 -14a +2=0$ . Solving these quadratic equations, we get a = 1/2, -2/3, -1, or -1/6. These are four possible values of $a$ . When a=1/2, we get h=2 and therefore equation (4) has only one root. When a=-1, then h=-1 and g=3, then the solutions to (3) and (4) are

$z_1, z_2 = \frac{ig \pm \sqrt{-g^2+36}}{6}= \frac{3i \pm \sqrt{27}}{6}=\frac{i\pm \sqrt{3}}{2}=e^{\pi/2 \pm i \pi/3}$ $z_3, z_4 = \frac{1\pm \sqrt{3}i}{2t} = e^{\pm i \pi/3}/e^{ -i 11\pi/6} = e^{\pm i \pi/3 - i\pi/6}$

However, $\pi/2 - \pi/3 = \pi/3 - \pi/6 = \pi/6$ , which means $S_1\cap S_2$ contains $e^{i\pi/6}$ . Therefore a=1/2 and a=-1 do not satisfy the conditions.

Similarly, one can solve (3) and (4) for a=-2/3 and a=-1/6 to see that they satisfy the condition (that all solutions are distinct).

Case Two. The line made by $z_1,z_3$ is parallel to the line made by $z_2,z_4$ . Thus there exists a unit $u$ such that $z_3/z_1 = z_2/z_4 = u$ , where $u\neq 1$ . So $S_1 = \{z_1, uz_4\}, S_2=\{uz_1, z_4\}$ . Again by Vieta's Theorem,

$z_1+uz_4 = ig/3, uz_1z_4 = -1$ $uz_1+z_4 = -h/t , uz_1z_4 = 1/t^2$

Therefore $-1=1/t^2$ , and so $e^{i\pi} = -1=t^2=e^{2i\alpha}=e^{i \frac{11\pi a}{3}}$ , which is equivalent to

$\pi + 2k\pi = \frac{11\pi a}{3}$

for some integer $k$ . Then $a=\frac{3}{11}(2k+1)$ . Also, by equation (1) and (2) we know that $|g| \leq 6$ and $|h|\leq 2$ , which gives $-7/6\leq a \leq 1/2$ , so $a = -9/11, -3/11, 3/11$ . Checking the solutions to (3), (4) with these parameters, we see that they all satisfy the conditions.

In conclusion, $a=-2/3, -1/6, -9/11, -3/11, 3/11$ are all the values of $a$ satisfying the conditions.

Let $c = 2a+1$ and factor the left side of the equation, we have $\sin x = \frac{3c^2-2c-2}{6}$ and $2 \cos (x + \frac{11 \pi (c-1)}{12}) + c = 0$ . Work with the first condition, we have $-\frac{4}{3} < c< 2$ and with this value of $c$ , we get 2 points with form $(-\cos x, \sin x)$ and $(\cos x, \sin x)$ . Assume that the remain equation also have some root with such form, we find out that $c = \frac{5}{11}, \frac{7}{11}, -\frac{17}{11}$ . The other 2 values of $c$ obtaining from the case the remain equation do not have root with form $(-\cos x, \sin x)$ and $(\cos x, \sin x)$ but also satisfy that four points form a trapezoid. The total number of $c$ is 5.