Why the FEET are the constraint. The mirror runs from O upward to the left, so it has plenty of glass above and none below. The head ray reflects high up the mirror and stays comfortably on it; the feet ray reflects near the bottom, and as the boy backs away that reflection point slides down towards O. The limit is when it reaches the lower edge O itself. Students who test the head first conclude “yes, he sees himself” and miss the real limit.
Image method. Reflect the boy in the mirror line: for a point P, P′ = 2(P·û)û − P with û = (−cos θ, sin θ). The feet F = (d, 0) go to
F′ = (d cos 2θ, −d sin 2θ) = (−7d/25, −24d/25) for θ = 53°.
The eye sees F by looking straight at F′, so the reflection point is where the line E–F′ crosses the mirror. Visibility is exactly the condition that E, R, F′ meet the mirror at or above O — i.e. E, O, F′ collinear is the critical case.
The numbers. The ray F → O has direction (−1, 0); reflecting it in the normal n̂ = (sin θ, cos θ) gives the outgoing direction (−cos 2θ, sin 2θ) = (7/25, 24/25), i.e. it rises 24 for every 7 across (73.7° above the horizontal, = 180° − 2θ). It reaches the eye when
144/d = 24/7 ⇒ dmax = 144 × 7/24 = 42 cm.
General result. dmax = h·tan(2θ − 90°) = −h·cot 2θ. For θ = 45° the edge ray is vertical, so dmax = 0; for θ < 45° it leans away from the boy and the feet are never visible. Steeper mirrors (θ → 90°, the ordinary wall mirror) push dmax → ∞ — which is why an upright mirror only needs to be half your height, with no distance limit at all.
On the model: everything is exact plane geometry — no approximations. The mirror is drawn 200 cm long but treated as unlimited upwards; only the lower edge at O ever matters. Distances are in cm; the boy’s head top is taken as h + 12 cm.