AMC 10 · 2024 · #20

Grade 10 probabilitygeometry-2d

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

geometric-probabilityindependent-eventsarea-trianglesratio-proportioncomplementary-counting convert-to-algebracomplementary-countingestimation ↑ Prerequisites: probability-basicarea-trianglesratio-proportion

📏 Long solution 💡 4 insights

Problem

In equilateral triangle △ ABC, a point P is dropped at random on side AB and a point Q is dropped at random on side AC, each spread evenly along its own side and neither depending on the other. The corner triangle △ APQ is then cut off. Decide which of the five given intervals holds the probability that △ APQ has less than half the area of △ ABC.

Pick an answer.

(A)

$\left[\frac 38, \frac 12\right]$

(B)

$\left(\frac 12, \frac 23\right]$

(C)

$\left(\frac 23, \frac 34\right]$

(D)

$\left(\frac 34, \frac 78\right]$

(E)

$\left(\frac 78, 1\right]$

How to solve

Strategy Draw a Diagram

The randomness lives in two numbers, the distances AP and AQ, so Tool #4 (Introduce a Variable) names them x and y. Tool #13 (Convert to Algebra) then turns the area comparison into the single inequality xy ＜ 1/2, which no longer mentions triangles at all. That inequality is the moment Tool #1 (Draw a Diagram) takes over: the pair (x,y) is a random point spread evenly over a unit square, so the probability is literally the area of the part of that square lying under the hyperbola xy = 1/2. Measuring that part head-on is awkward, so Tool #16 (Change Focus / Count the Complement) switches to the small corner region where xy ≥ 1/2. That corner is easy to squeeze between a square it sits inside and a triangle that sits inside it, and Tool #3 (Eliminate Possibilities) uses the squeeze to knock out every interval but one.

1STEP 1

Name the two random lengths

Both are uniform on the unit interval.

AB = AC = 1, x = AP ∈ [0,1], y = AQ ∈ [0,1], x,y independent and uniform

2STEP 2

The area ratio is just a product

Two shrinks make the ratio the product of the lengths.

[APQ]/[ABQ] = AP/AB = x, [ABQ]/[ABC] = AQ/AC = y ⟹ [APQ]/[ABC] = xy

3STEP 3

The event becomes one inequality

The whole event is the product below one half.

[△ APQ] ＜ 1/2[△ ABC] ⇔ ([△ APQ])/([△ ABC]) ＜ 1/2 ⇔ xy ＜ 1/2

4STEP 4

Draw the sample space as a unit square

The boundary is a piece of a hyperbola.

sample space = [0,1]×[0,1], P(event) = area{(x,y) : xy ＜ 1/2}, boundary y = 1/2x

5STEP 5

Switch to the bad corner

The failure region is only the top-right corner.

B = {(x,y) ∈ [0,1]² : xy ≥ 1/2}, P = 1 - area(B)

6STEP 6

Trap the bad corner between two shapes

Squeezing puts the probability between three quarters and seven eighths.

1/8 ＜ area(B) ≤ 1/4 - 1/36 = 2/9 ⟹ 7/9 ≤ P ＜ 7/8, 7/9 = 0.777… ＞ 3/4

7STEP 7

Confirm with the exact area

The integral gives about 0.847, landing in (3/4, 7/8].

P = ∫₀¹/2 1 dx + ∫₁/2¹ 1/2x dx = 1/2 + 1/2ln 2 = (1+ln 2)/2 ≈ 0.84657 → (D)

Answer

(3/4, 7/8]

The value (1+ln 2)/2 ≈ 0.84657 should feel right: the corner triangle only exceeds half the big triangle when both x and y are large, which is a small corner of the square, so the probability ought to be well above 1/2 but clearly short of 1. Two independent numerical checks agree. A deterministic 20000 × 20000 grid of cell centres over the unit square gives 0.84657357, and a Monte Carlo run of 20 million random pairs gives 0.8465616; the exact value is 0.84657359. Both sit strictly inside (3/4, 7/8], and the calculus-free sandwich 7/9 ≤ P ＜ 7/8 already excluded the neighbours on its own. The interval answer format hides two traps. Replacing the hyperbolic corner by the straight-chord triangle exactly gives 1 - 1/8 = 7/8, which is only the endpoint and would tempt a jump to (E) if the inequality direction were slipped. Guessing that halving the area means halving the probability points at (A) or (B), but the map from lengths to area is a product, not a sum, so the two halves are not symmetric. Note also that the equilateral hypothesis was never used: the same probability comes out for any triangle whatsoever.

💡Key takeaway

Two independent random picks are one random point in a square, so a probability question becomes an area question — and if the area is curved, trap it between a shape it fits inside and a shape that fits inside it.

Name the two random lengths
The area ratio is just xy
The event becomes one inequality
Draw the sample space as a unit square
Switch to the bad corner
Trap the bad corner between two shapes
Confirm with the exact area