Author:
myh9090
The problem: Line AB is a ladder leaning on a vertical wall. A (risk-taking) painter (P) is standing on the ladder mid-way. The ladder starts slipping down the wall (and across the floor) due to the painter's weight. What will be the path (or curve) of the painter standing at the mid-point, as the ladder slides all the way down? Explore: 1. Click on the "Show mid-point" check-box to see the painter (point P) 2. Click-and-drag point A down, to see the ladder slipping 3. Watch point P as you move point A up and down, and get a sense for the curve or path that point P makes on the plain. 4. You can visualize the path by right-clicking on the point P and selecting "Trace On" from the popup menu (to turn tracing off, right-click on P again, and de-select "Trace On") Can you guess what kind of path point P is tracing? 5. For a hint, click on the "Show hint" check-box, and move point A up and down to observe the behavior of line segment b (connecting point E to P). 6. To see more details to help you figure out the path of point P, click on the "Show more details" check-box. Observe the various shapes that are formed as they change while you move point A up and down. 7. You can click on the "Show path" check-box to see the path of point P. Formalize your observations by capturing the path of P in algebraic form.
Analyzing the path of point P: 1. Looking at triangles ABE and PBD: since P is the mid-point of AB, and angle ABE is common to both, and the triangles are both right-angle, they are therefore similar. 2. Since triangles ABE and PBD are similar, and since PB is half of AB, it follows that BD is half of BE. 3. This means that DE = BD (both are half of BE. In other words, D is the mid-point of BE) 4. This means that triangles PED and PBD are congruent (both are right-angle triangle, sharing a side PD, and another equal side (DE = BD). Since triangles PED and PBD are congruent, and since BP is fixed (half the length of AB), then PE is also fixed (at half length of AB). 5. As point A moves along the y axis, its y coordinate is given by y(A) = AB sin(ABE), and y(P) = BP sin(PBD). 6. Since triangles ABE and PBD are similar and triangles PED and PBD are congruent, y(P) = 0.5 AB sin(PED). 7. Following the same logic it can be shown that x(P) = 0.5 AB sin(PED). 8. So it follows that the x and y coordinates of point P trace a circular path (a shape of a half circle), with fixed radius (EP = 0.5 AB), and cos(PEB) and sin(PEB) respectively, as point A moves up and down. Or in a more formal notation: The path of P: 1. ∡ABE = ∡PBE, ∡AEB = ∡PDB ⇒ ΔABE ∼ ΔPBD 2. ΔABE ~ ΔPBD, AP = PB = ¹/₂ AB ⇒ DB = DE = ¹/₂ EB 3. DB = DE, PD, ∡PDE = ∡PDB ⇒ ΔPED ≅ ΔPDB 4. ΔPED ≅ ΔPDB ⇒ PE = PB = ¹/₂ AB 5. ΔPED ≅ ΔPDB ⇒ ∡PED = ∡PBD 6. X(P) = PE cos(∡PED) = ¹/₂ AB cos(∡PED), Y(P) = PE sin(∡PED) = ¹/₂ AB sin(∡PED) 7. P(¹/₂ AB cos(∡PED), ¹/₂ AB sin(∡PED)) = (R cos α, R sin α) P is moving along a circular path. QED