Question 3

A right circular cone is cut into three parts of volume $$v_1, v_2, v_3$$ from the vertex respectively, by giving two cuts parallel to the base, which trisect the altitude of the cone. Then $$v_1 \colon v_2 \colon v_3 =$$

Solution

The cuts create cones similar to the whole cone with linear ratios $$\frac{1}{3}$$, $$\frac{2}{3}$$ and $$1$$, so the cumulative volumes from the vertex are in the ratio $$1^3 \colon 2^3 \colon 3^3 = 1 \colon 8 \colon 27$$. Subtracting successive cumulative volumes gives $$v_1 = 1$$, $$v_2 = 8 - 1 = 7$$ and $$v_3 = 27 - 8 = 19$$. Hence $$v_1 \colon v_2 \colon v_3 = 1 \colon 7 \colon 19$$.

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