This creative device comes from zhangyf1997, a fellow enthusiast from the Mufu Astronomy Forum.
Structure:
A and B are two fixed points, which can be considered as being connected by a rigid rod.
AC is the driving rod, rotating around point A.
BD is the driven rod, and CD is the connecting rod.
Length Data:
CD = AB = \sqrt{2}.
AC = BD = 1.
E is the midpoint of CD.
Find: The trajectory equation of point E (the black curve in the figure, which is quite interesting, isn’t it?).
To find the curve equation for this problem, I will once again use the vector method, but this time with a special approach: the “Particle Method” notation. In the article “The Winding Vector Method”, I learned something called the “Particle Method.” Of course, it is not some mysterious new thing; in my view, it is simply a convenient notation for the vector method. Specifically, the vector \vec{AB} is written as B-A (the subtraction of two points), and AB is defined as the dot product of two vectors \vec{OA} \cdot \vec{OB}. Everything else remains largely the same. During the calculation, one must remember: 1. Use clear symbols to distinguish “points” from “scalars” to avoid confusion; 2. Vectors do not have a division operation, so you cannot simply cancel out vectors from both sides of an equation.
Getting back to the point, in this problem, we establish a coordinate system with A as the origin and AB along the x-axis. We can write: B = (\sqrt{2}, 0), \quad 2E = C + D \tag{1} 1 = C^2 = (D-B)^2 = D^2 + B^2 - 2DB \tag{2} 2 = B^2 = (D-C)^2 = D^2 + C^2 - 2DC \tag{3}
From (1), we have C = 2E - D. Squaring both sides gives: 1 = 4E^2 + D^2 - 4DE \tag{4} Substituting C = 2E - D into (3) yields: 1 = D^2 - 2D(2E - D) = 3D^2 - 4ED \tag{5}
Combining (4) and (5), we obtain 2E^2 = D^2. Substituting this into (5) gives 1 = 6E^2 - 4ED, and substituting into (2) gives 2DB = 1 + 2E^2. At this point, the vector work is complete, and we must return to the coordinate system (the vectors have already helped us simplify the problem into a system of linear equations). Let E = (x, y) and D = (D_1, D_2), then we have: \begin{aligned} 6(x^2 + y^2) - 1 &= 4D_1 x + 4D_2 y \\ 1 + 2(x^2 + y^2) &= 2\sqrt{2} D_1 \end{aligned}
Solving this yields: D_1 = \frac{1 + 2(x^2 + y^2)}{2\sqrt{2}}, \quad D_2 = \frac{6(x^2 + y^2) - 1 - 4D_1 x}{4y}
From 2E^2 = D^2, we have 2(x^2 + y^2) = D_1^2 + D_2^2. Substituting the expressions for D_1 and D_2 and simplifying gives the trajectory equation. This is an extremely complex process...
2(x^2 + y^2) = \left(\frac{1 + 2(x^2 + y^2)}{2\sqrt{2}}\right)^2 + \left(\frac{6(x^2 + y^2) - 1 - 4\left(\frac{1 + 2(x^2 + y^2)}{2\sqrt{2}}\right) x}{4y}\right)^2
A relatively “simple” expanded result is: \begin{aligned} 8 x^6 &- 24 \sqrt{2} x^5 + 24 x^4 y^2 + 44 x^4 - 48 \sqrt{2} x^3 y^2 - 8 \sqrt{2} x^3 + 24 x^2 y^4 + 56 x^2 y^2 \\ &- 10 x^2 - 24 \sqrt{2} x y^4 - 8 \sqrt{2} x y^2 + 2 \sqrt{2} x + 8 y^6 + 12 y^4 - 10 y^2 + 1 = 0 \end{aligned}
I didn’t expect it to be a sixth-degree curve! However, it can be simplified slightly; this curve consists of two parts: \begin{aligned} 2y^2 + 2x^2 &= 2\sqrt{2}x + 1 \\ 4(y^2 + x^2 + 1 - \sqrt{2}x)^2 &= 8x^2 - 8\sqrt{2}x + 5 \end{aligned}
Because the calculation process is truly too complex, even for someone like me who enjoys manual calculation, I did not want to continue. Therefore, these results were completed by WolframAlpha.
If any reader can provide a simpler method, please feel free to inform BoJone.
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