Let A0=(−1,0), and let O be the origin (0,0). For each integer i≥1, we construct the point Ai so that |Ai−1Ai|=|Ai−1O| and the angle ∠OAi−1Ai is a right angle. If O,Ai−1,Ai+1 are not collinear for any value of i, what is the x-coordinate of A12?
To find the x-coordinate of A12, we need to first understand the construction of the points A0, A1, A2, ... , A12.
Starting with A0 = (-1, 0), we know that |A0O| = |A1A0| and the angle ∠OA0A1 is a right angle. This means that A1 lies on a circle centered at O with radius |A0O|.
To find the coordinates of A1, we can use the Pythagorean theorem. Since A1 lies on the circle centered at O, it lies on the unit circle (circle with radius 1). So, we can visualize A1 to be (cos θ, sin θ) on the unit circle.
Since the angle ∠OA0A1 is a right angle, it means that ∠OA0A1 = π/2 radians or 90 degrees. Therefore, θ = π/2.
Using the values of cos θ and sin θ for θ = π/2, we get A1 = (0, 1).
Now, let's move on to finding the coordinates of A2. Again, A2 lies on the unit circle, so we can visualize A2 to be (cos θ, sin θ) on the unit circle.
The construction tells us that |A1O| = |A2A1| and ∠OA1A2 is a right angle. Since A1 = (0, 1) lies on the unit circle, |A1O| = 1.
Now, we need to find the angle ∠OA1A2. Since A1 = (0, 1), it means that the line joining O and A1 is vertical. The angle ∠OA1A2 is a right angle, so the line joining A1 and A2 must be horizontal.
Since A1 = (0, 1), A2 = (x, 1), where x is the x-coordinate of A2.
Given that ∠OA1A2 is a right angle, we can use the distance formula to find the value of x:
|A1O|^2 + |OA2|^2 = |A1A2|^2
1^2 + x^2 = |A1A2|^2
1 + x^2 = |A1A2|^2
But, |A1A2| = |A1O| = 1
So, 1 + x^2 = 1^2
x^2 = 1^2 - 1
x^2 = 0
x = 0
Therefore, A2 = (0, 1).
Following similar steps, we can calculate the coordinates for A3, A4, ... , A12. Continuing with the process, we find that A12 = (0, 1).
So, the x-coordinate of A12 is 0.