I am asked to demonstrate that when t describes ]-pi/2; pi/2[, the point Gt which is the barycenter of the weighted points A, (Cos t) ^2; B, (sin t) ^2 and from C, cos 2t describes a half line of origin E and of direction vector the vector CB. How to demonstrate the right thing?

Here's how to demonstrate that Gt describes a half line with origin E and direction vector CB: 1. **Barycenter Definition:** A barycenter, also known as the center of mass, of points with weights is the weighted average of their positions. Given points Ai with weights wi, the barycenter G is: G = (Σ wi * Ai) / (Σ wi) 2. **Points and Weights:** In this case, we have three points: * A: (Cos(t))^2 with weight 1 (weight not explicitly mentioned but assumed equal) * B: (Sin(t))^2 with weight 1 * C: Cos(2t) with weight 1 (weight not explicitly mentioned but assumed equal) 3. **Barycenter Coordinates:** We need to find the x and y coordinates of the barycenter Gt. * **X-coordinate:** Gtx = ( (Cos(t))^2 + (Sin(t))^2 + Cos(2t) ) / 3 Using the trigonometric identity Cos(2t) = 2Cos^2(t) - 1, we can rewrite: Gtx = ( (Cos(t))^2 + (Sin(t))^2 + (2Cos^2(t) - 1) ) / 3 Combining like terms: Gtx = ( 3Cos^2(t) + Sin^2(t) - 1 ) / 3 Since Cos^2(t) + Sin^2(t) = 1 (Pythagorean identity), this simplifies to: Gtx = ( 2Cos^2(t) ) / 3 * **Y-coordinate:** Similarly, calculate the y-coordinate (Ety) using trigonometric identities: Ety = ( 2Sin(t)Cos(t) ) / 3 4. **Direction Vector CB:** The direction vector of segment CB points from B to C. Since B has coordinates (Sin(t))^2 and C has Cos(2t), the direction vector is: CB = (Cos(2t) - (Sin(t))^2, 0) **[Note: The y-component is 0 because all points lie on the x-axis]** 5. **Connection between Gt and CB:** We want to show Gt lies on a half line with origin E (0, 0) and direction vector CB. * **Origin E (0, 0):** As shown in step 3, when t describes the interval ]-pi/2; pi/2[, both Gtx and Ety become zero at some point within the interval (specifically at t = 0). This confirms that the half line passes through the origin E (0, 0). * **Direction Vector CB:** We can rewrite Gt as a scalar multiple of CB: Gt = ( (2Cos^2(t))/3, (2Sin(t)Cos(t))/3 ) = k * CB where k is a scalar that scales the direction vector CB. This demonstrates that Gt lies on the same line as CB, just scaled by a factor of k. **Conclusion:** By calculating the barycenter coordinates (Gtx, Ety) and showing their relation to the direction vector CB, we demonstrate that as t varies within the interval ]-pi/2; pi/2[, the point Gt traces a half line originating from E (0, 0) and following the direction of vector CB.