A challenging problem of platonic solid by HCR

Geometry Level 5

A tiny sphere with radius 5 mm is resting in one of 12 vertices (corners) of a sufficiently large icosahedron (hollow thin shell type). Find out the minimum gap (in mm) between sphere & one of the five edges meeting at that vertex (corner).

Details and Assumptions :

  • Icosahedron is hollow thin shell & sufficiently large having a tiny sphere resting in one of its 12 vertices.

  • Sphere touches all five equilateral triangular faces meeting at that vertex.

  • Sphere doesn't touch any of five edges meeting at that vertex (corner) of icosahedron.


The answer is 0.352331345.

Harish Chandra Rajpoot by Harish Chandra Rajpoot , 30, India

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1 solution

Ujjwal Rane
Jul 14, 2016

Imgur Imgur Let us focus on any five equilateral triangular faces meeting at a vertex to form a 'pyramidal cup' in which the 5 unit radius sphere (ice cream scoop? :-) is placed.

a = Side of Base

R = Circumcircle Radius of Base = a 2 × sin ⁡ 36 ° \frac{a}{2 \times \sin 36°}

r = Incircle Radius of Base = a 2 × tan ⁡ 36 ° \frac{a}{2 \times \tan 36°}

h = Altitude of Triangular Face = 3 a 2 \frac{\sqrt{3} a}{2}

z = depth of the cup = 5 h r = 5 3 tan ⁡ 36 ° \frac {5h}{r} = 5 \sqrt{3} \tan 36°

D = Distance of an Edge from the Center of the Sphere = z R a = 5 3 2 cos ⁡ 36 ° = 10 3 5 + 1 \frac{zR}{a} = \frac{5 \sqrt{3}}{2 \cos 36°} = \frac{10 \sqrt{3}}{\sqrt{5}+1}

Δ = D − 5 = 5.352331346596 − 5 = 0.352331346596 \Delta = D - 5 = 5.352331346596 - 5 = 0.352331346596

@Harish Chandra Rajpoot @Niranjan Khanderia

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