AMC 10 · 2016 · #23

Grade 10 geometry-3d

Archetype Solid Volume by Decomposition └ Sub-archetype Fraction or Difference of a Bounding Solid

absolute-valuespatial-visualizationvolume-pyramidsymmetry-argument spatial-visualizationidentify-subproblems ↑ Prerequisites: absolute-valuevolume-pyramid

📏 Medium solution 💡 3 insights

Problem

Two solids described by absolute-value inequalities overlap. Find the shared volume.

Pick an answer.

(A)

$\frac{1}{6}$

(B)

$\frac{1}{4}$

(C)

$\frac{1}{3}$

(D)

$\frac{1}{2}$

(E)

How to solve

Strategy Visualize Spatial Relationships

Two solids overlapping in space is hard to see all at once, so the move is to cut the picture into horizontal slices and handle one height at a time. Both inequalities have the same shape, |x|+|y| plus something depending only on z, so freezing z turns each of them into a flat condition of the form |x|+|y| ≤ (number) — a square standing on its corner. Two such squares share a center, so one simply sits inside the other, and the overlap is just the smaller one. That reduces the whole problem to tracking a single number, the size of the smaller square, as the height changes. Once I see how that number grows and shrinks, I can recognize the stack of squares as ordinary pyramids and finish with the pyramid volume formula instead of any calculus.

1STEP 1

See what each inequality draws

Each inequality draws a solid around a centre.

|x|+|y|+|z| ≤ 1 centered at (0,0,0), |x|+|y|+|z-1| ≤ 1 centered at (0,0,1)

2STEP 2

Cut at a fixed height

A fixed height turns each into a square.

z fixed ⟹ |x|+|y| ≤ 1-|z| and |x|+|y| ≤ 1-|z-1|

3STEP 3

Keep the smaller square

Only the smaller square survives.

k(z)=min(1-|z|, 1-|z-1|), k(z) ≥ 0 ⇔ 0 ≤ z ≤ 1, k(z)=min(z, 1-z)

4STEP 4

Stack the slices into two pyramids

Its size peaks in the middle and vanishes at the ends.

k(0)=0, k (1/2)=1/2, k(1)=0, k linear on each half

5STEP 5

Measure the shared base square

So the shape is two pyramids.

|x|+|y| ≤ k → area=(2k)(2k)/2=2k²; k=1/2 → area=2·1/4=1/2

6STEP 6

Add the two pyramid volumes

Adding gives 1/6, choice (C).

V = 2·1/3·1/2·1/2 = 2·1/12 = 1/6

Answer

1/6

Three checks agree. First, size: the whole region lives inside the box |x| ≤ 1/2, |y| ≤ 1/2, 0 ≤ z ≤ 1, whose volume is 1, so the answer must be well under 1 and choice (E) was never possible; 1/6 sits comfortably inside that bound. Second, comparison with the whole octahedron: |x|+|y|+|z| ≤ 1 is itself two pyramids with base area 2 and height 1, giving volume 2·1/3 · 2 · 1=4/3, and the overlap 1/6 is exactly 1/8 of that — consistent with the overlap being a half-size copy, since halving every length divides volume by 2³=8. Third, point tests: (0,0,1/2) satisfies both inequalities and lies inside the double pyramid, while (3/10,0,9/10) fails the first inequality (3/10+9/10=6/5 ＞ 1) and is also excluded by the slice rule, because k (9/10)=1/10 ＜ 3/10. The model and the original inequalities give the same verdict on test points.

💡Key takeaway

When two solids overlap, slice them both at the same height: each condition becomes 'stay inside this square', the overlap keeps only the smaller square, and watching that square grow and shrink tells you the shape you are measuring.

See what each inequality draws
Cut at a fixed height
Keep the smaller square
Stack the slices into two pyramids
Measure the shared base square
Add the two pyramid volumes