Daily Math Puzzle: 2026-11-02

6 mm — This puzzle uses the principle of similar triangles. Imagine a cross-section of the setup: 1. **Identify the distances:** * The light source (L) to the organism's center (O) distance: LO = 10 mm. * The organism's radius (half its diameter): R_organism = 1 mm / 2 = 0.5 mm. * The screen is 50 mm further from the organism's center. So, the total distance from the light source (L) to the screen (S) is LS = LO + OS = 10 mm + 50 mm = 60 mm. 2. **Apply Similar Triangles:** Consider two similar right-angled triangles: * A smaller triangle formed by the light source (L), the organism's center (O), and the point on the organism's surface where the light ray is tangent (creating the shadow's edge). The 'height' of this triangle is LO (10 mm), and its 'base' is the organism's radius R_organism (0.5 mm). * A larger triangle formed by the light source (L), the center of the shadow on the screen (S), and the edge of the shadow. The 'height' of this triangle is LS (60 mm), and its 'base' is the radius of the shadow R_shadow. The ratio of base to height in similar triangles is constant: R_organism / LO = R_shadow / LS 0.5 mm / 10 mm = R_shadow / 60 mm 3. **Calculate the shadow's radius:** R_shadow = (0.5 mm / 10 mm) * 60 mm R_shadow = 0.05 * 60 mm R_shadow = 3 mm 4. **Calculate the shadow's diameter:** The question asks for the diameter of the shadow, which is twice its radius: Diameter_shadow = 2 * R_shadow = 2 * 3 mm = 6 mm.