Mathematical and Physical Journal
for High Schools
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Problem C. 824. (October 2005)

C. 824. Consider the truncated cone determined by the circumscribed circle of the base of a cube, and the inscribed circle of the top face of the cube. Find the ratio of the volume of the truncated cone to that of the cube.

(5 pont)

Deadline expired on November 15, 2005.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás: Jelölje a kocka élét a, az alaplap köré írt kör sugarát R, a fedőlapba írt körét pedig r. Ekkor:

R={\sqrt2\over2}a,

r={1\over2}a,

Vkocka=a3,

V_{csonkak\'up}={\pi\over3}\cdot a(R^2+r^2+Rr)={\pi\over3}\cdot a({2\over4}a^2+{1\over4}a^2+{\sqrt2\over4}a^2)=a^3\cdot{\pi\over12}(3+\sqrt2).

A kérdéses arány: {\pi\over12}(3+\sqrt2) : 1 = \pi(3+\sqrt2) : 12.


Statistics:

534 students sent a solution.
5 points:426 students.
4 points:25 students.
3 points:29 students.
2 points:20 students.
1 point:1 student.
0 point:23 students.
Unfair, not evaluated:10 solutions.

Problems in Mathematics of KöMaL, October 2005