AMC 10 · 2016 · #23

Grade 10 probability

Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Volume

geometric-probabilitypolygon-inequalitycomplementary-countingvolume-pyramid complementary-countingsymmetry-argument ↑ Prerequisites: polygon-inequalitygeometric-probability

📏 Medium solution 💡 3 insights

Problem

Three random lengths must build a triangle with real area. Find the probability.

Pick an answer.

(A)

$\frac16$

(B)

$\frac13$

(C)

$\frac12$

(D)

$\frac23$

(E)

$\frac56$

How to solve

Strategy Change Focus / Count the Complement

There is no finite list of outcomes here, so counting is replaced by measuring. Tool #1 and #17 turn the experiment into geometry: one draw of (x,y,z) is one point of the unit cube, and because the three numbers are uniform and independent, the probability of any event is exactly the volume of the matching region. The success region is carved out by three inequalities at once and is awkward to measure head-on. Tool #14 (Extreme Principle) shows that only the largest of the three numbers can break the triangle test, so failure is a single clean condition instead of three. That makes tool #16 (Count the Complement) the efficient route: measure the three failure blocks, which turn out to be pyramids, and subtract from the cube's volume of 1. Tool #4 keeps the bookkeeping honest by naming the three coordinates and writing the conditions as inequalities in them.

1STEP 1

Turn the draw into a point in a cube

The draw is a point in a unit cube.

(x,y,z)∈[0,1]³, volume=1, P(region)=volume of region

2STEP 2

Write the triangle test as inequalities

The test is three inequalities.

x+y ＞ z, y+z ＞ x, z+x ＞ y

3STEP 3

Only the largest side can fail

Only the largest can fail.

z=max(x,y,z) → triangle⇔ x+y ＞ z; failure⇔ z ≥ x+y

4STEP 4

Measure one failure block

One failure block is easy to measure.

slice area=z²/2; vol(B_z)=1/3·1/2 · 1=1/6

5STEP 5

Add the three failure blocks

There are three such blocks.

vol(failure)=3·1/6=1/2

6STEP 6

Subtract to get the answer

Subtracting gives 1/2, choice (C).

P(triangle)=1-1/2=1/2→(C)

Answer

1/2

A completely separate argument gives the same 1/2. Split the cube by which coordinate is largest; the piece where z is largest has volume 1/3 by symmetry. On that piece apply the map (x,y,z)↦(z-x, z-y, z). Since x,y∈[0,z], the image again has z largest, and on each horizontal slice the map is a reflection, so it preserves volume and undoes itself. It sends x+y ＞ z to (z-x)+(z-y)=2z-(x+y) ＜ z, so it pairs each triangle point with exactly one non-triangle point. Half of that piece works, and the same holds for the other two pieces, giving 1/2 overall. The size is also plausible from both sides: the answer must be less than 1 because a very small number together with two ordinary ones fails, and it must be well above 1/6 because most triples of similar-sized numbers do close up. Finally, the strict-versus-non-strict issue is harmless: the boundary z=x+y is a flat surface of zero volume, so degenerate triangles change nothing.

💡Key takeaway

Treat the three numbers as one point in a unit cube: the only way to fail is for one number to be at least the other two combined, and those three failure pyramids fill exactly half the cube.

Turn the draw into a point in a cube
Write the triangle test as inequalities
Only the largest side can fail
Measure one failure block
Add the three failure blocks
Subtract to get the answer