Spiral of right triangles

Since $\triangle A_{i-1}A_{i}O$ has two sides equal and one right angle at $\angle OA_{i-1}A_{i}$ , it must be a $45-45-90$ triangle with $OA_{i-1}$ as a leg and hypotenuse $OA_i$ , as well as both $\angle A_{i-1}A_{i}O$ and (more importantly) $\angle A_{i-1}OA_{i}$ as $45^\circ$ .

This means that between each point $A_i$ and $A_{i-1}$ , we move $45^\circ$ in a set direction, without moving back the other direction at any point since that would make $O$ and some $A_{i-1}$ , $A_{i+1}$ collinear, which isn't allowed.

In addition, since $OA_{i-1}$ is a leg of a $45-45-90$ or $1-1-\sqrt{2}$ triangle with hypotenuse $OA_i$ , we have $\frac{OA_i}{OA_{i-1}} = \sqrt{2}$ , so the distance from each successive point to the origin is $\sqrt{2}$ that of the point before it.

Since $A_{i}$ is $A_{i-1}$ rotated $45^\circ$ in a set direction and farther away from the origin, we know that the direction of $A_{12}$ is the direction of $A_{1}$ rotated $12\times 45^\circ = 540^\circ\Rightarrow 180^\circ$ , which is along the +x axis, no matter the direction of rotation. Knowing that $|OA_0| = 1$ and $|OA_{i}| = |OA_{i-1}|\times \sqrt{2}$ for all $i\ge 1$ , we find that $|OA_{12}| = 1\times (\sqrt{2})^{12} = 64$ . The only point that is $64$ away from the origin and along the +x axis is $(64, 0)$ , which has x coordinate $\boxed{64}$ .

You can take $A_1=(-1,1)$ and the following points would follow in the clockwise direction or take $A_1=(-1,-1)$ and the points would follow in the anticlockwise direction. Anyway, both the $A_{12}$ points would coincide. Let's take $A_1=(-1,-1)$ . The two possible co-ordinates for $A_2$ are $(0,-2) or (-2,0)$ by the first two conditions, but we can't take $(-2,0)$ due to $A_0, A_2$ and O not being collinear(third condition). Similarly when we proceed the points would follow in the anticlockwise direction while making sure it fulfils all three conditions. Every alternate point would fall on either the x or y axis. Thus when we proceed taking $A_1=(-1,1)$ , the points would be:- $A_2=(0,-2) A_3=(2,-2) A_4=(4,0) A_5=(4,4) A_6=(0,8) A_7=(-8,8) A_8=(-16,0) A_9=(-16,16) A_{10}=(0,-32) A_{11}=(32,-32),$ finally arriving at $A_{12}=(64,0)$

Note:- We can skip this process by simply observing the co-ordinates $A_0=(-1,0) A_4=(4,0) A_8=(-16,0)$ thus $A_{12}=(64,0)$ as after $A_0$ every fourth point will lie on the X axis (but on the opposite side) with the abcissa multiplied by 4.

Since |A {i-1}A {i}|=|A {i-1}O| and the angle ∠OA {i−1} A{i} is a right angle, the triangle A {i-1}A {i}O is an isosceles right triangle with the right angle at A {i-1}. From above we derive that ∠OA{i}Ox = ∠OA{i-1}Ox + 45 degrees and |OA {i}|= \sqrt{2}|OA {i-1}| Thus with A {12}, we obtain ∠OA{12}Ox: -180+45\times12 = 360 degrees (A {12} lies on Ox). So x {A {12}} = |OA {12}| = 1\times(sqrt2)^12=64

Note that for all right angle triangle A_{i-1}OA_{i} , the hypothenuse A_{i-1}A_{i} is \sqrt{2} times the base. Since the distance fro (-1,0) to (0,0) is 1, the length of A_1A_0 is \sqrt{2}. Note that for A_n where n is multiple of 4, the point A_n must lies on the x -axis. Thus, A_{12} must lies on the positive x -axis. Thus, since the hypothenuse of the right triangle is \sqrt{2} times bigger than the previous one, the length of OA_{12} is \sqrt{2}^12, which is 64.

Take the first triangle OA 0A 1.Here,\angle OA 0A 1 = 90 ^ \circ and sides OA 0 and A 1A 0 adjacent to the right angle are both equal.Hence,\angle OA 1A 0 and \angle A 1OA 0 are both equal to 45 ^ \circ (angle sum property of a triangle ).Therefore,the hypotenuse to the first triangle will be inclined at an angle of 45 ^ \circ to the base OA 0.The next hypotenuse will be inclined at 45 ^ \circ w.r.t. OA 0.Thus,after one rotation,the hypotenuse formed is rotated by 45 ^ \circ .This will continue for each subsequent construction.Hence,after 12 rotations the final hypotenuse formed will have been rotated by an angle 12 \times {45 ^ \circ} = 540 ^ \circ .Hypotenuses of the triangles constructed coincide with the positive y-axis after a rotation of 90 ^ \circ or coincides with the positive x-axis after a rotation of 180 ^ \circ ,the hypotenuse OA {12} will again coincide with the positive x-axis after a rotation of 540 ^ \circ .Hence,the coordinates of the point A {12} are (x,0).It is given that the length of OA 0 ,one of the equal sides in the first triangle we have to construct using it as base,is 1 unit.By Pythagoras theorem,length of hypotenuse of first triangle = \sqrt{ 1 ^ 2 + 1 ^ 2 } = \sqrt{2} units.This hypotenuse will act as one of the equal sides of the next triangle to be constructed as well as its base.Length of hypotenuse of second triangle = \sqrt{ {\sqrt{2}} ^ 2 + {\sqrt{2}} ^ 2 } = 2 units.The length of the next hypotenuse will be 2\sqrt{2} units.We see that the lengths of the consecutive hypotenuses form a G.P. with first term,a = \sqrt{2} units and common ratio,r = \sqrt{2} .Thus,the length of the hypotenuse after 12 rotations = 12^{th} term of the G.P. = ar^{12-1} = ar^{11} = \sqrt{2} \times {\sqrt{2}}^{11} = 64 .As this hypotenuse OA {12} coincides with the positive x-axis,the length OA {12} is the x-coordinate of A_{12} = 64 .

I.

Basically, it can be noticed from the construction that the angle between the segments $OA_{i-1}$ and $OA_{i}$ is $45^\circ$ . Therefore, using the positive x-axis as the reference line, the angle subtended by $OA_{i}$ is angle subtended by $OA_{i-1}$ less $45^\circ$ . From this sequence, a general iteration equation can be derived:

$\angle OA_{i} = \angle OA_{i-1} - (45^\circ)(1)$ $\angle OA_{i} = \angle OA_{i-2} - (45^\circ)(2)$ $\angle OA_{i} = \angle OA_{i-3} - (45^\circ)(3)$ . . . $\angle OA_{i} = \angle OA_{i-n} - (45^\circ)(n)$

let n = i

$\angle OA_{i} = \angle OA_{0} - (45^\circ)(i)$ $= 180^\circ - (45^\circ)(i)$

II.

Again from the construction triangle $OA_{i-1}A_{i}$ is A 45-45-90 right triangle with $OA_{i}$ as the hypotenuse. Therefore:

$|OA_{i}| = |OA_{i-1}| times {\sqrt{2}}^{1}$ $|OA_{i}| = |OA_{i-2}| times {\sqrt{2}}^{2}$ $|OA_{i}| = |OA_{i-3}| times {\sqrt{2}}^{3}$ . . . $|OA_{i}| = |OA_{i-m}| times {\sqrt{2}}^{m}$

Let m=i

$|OA_{i}| = |OA_{0}| times {\sqrt{2}}^{i}$ $= {\sqrt{2}}^{i}$

III.

To get the x-coordinate of point A_{i}:

$x_{A_{i}} = |OA_{i}|\cos \angle OA_{i} ={\sqrt{2}}^{i}\cos \180 - 45i$

Simplifying by trigonometric identity:

for i= 12

$x_{A_{i}}={\sqrt{2}}^{12}\cos \180 - 45\times12$
= 64

Since $\angle OA_{i-1}A_i$ is a right angle and $|A_{i-1}A_i| = |A_{i-1}O|$ , $OA_{i-1}A_i$ is a right isosceles triangle, and $\angle A_{i-1}OA_i = \frac{\pi}{4}$ . More generally, angle $\angle A_{i}OA_j = \frac{(i-j)\pi}{4}$ .

Since $O, A_{i-1},$ and $A_{i+1}$ are not collinear, the angles point in the same direction. Then $\angle A_{12}OA_0 = 3\pi$ , implying the point $A_{12}$ lies on the positive $x$ -axis.

In a right isosceles triangle, the length of the hypotenuse is $\sqrt{2}$ times the length of the respective sides. Since $A_iO$ is the hypotenuse of a triangle with side $A_{i-1}O$ , we have $|A_i| = \sqrt{2}|A_{i-1}|$ . Since $|A_0O| = 1$ , we have $|A_iO| = 2^{i/2}$ . Thus, $A_{12}$ has length $2^{6} = 64$ , so it has $x$ -coordinate $64$ .

Note: There are 2 different series of points that can be constructed, depending on if the rotation is clockwise, or anti-clockwise.

By drawing, the coordinates of A0 is (-1,0), A1 is (-1,1), A3 is (2,2), A5 is (4,-4), A7 is (-8,-8).

Notice that the odd-numbered points runs clockwise, and the absolute value of the x-coordinates double every time.

Following this pattern, A12 lies in the second quadrant, and the absolute value of the x-coordinate is 64. Hence the x-coordinate will be 64