AMC 10 · 2002 · #22

Grade 10 geometry-2d

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

geometric-probabilitythirty-sixty-ninety-trianglearea-triangles extreme-principle ↑ Prerequisites: geometric-probabilitythirty-sixty-ninety-triangle

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Triangle ABC has its right angle at C, ∠ ABC = 60°, and hypotenuse AB = 10. A point P is picked at random inside the triangle, and the ray from B through P is extended until it hits side AC at D. Find the probability that BD exceeds 5√(2).

Pick an answer.

(A)

$\frac{2-\sqrt2}{2}$

(B)

$\frac{1}{3}$

(C)

$\frac{3-\sqrt3}{3}$

(D)

$\frac{1}{2}$

(E)

$\frac{5-\sqrt5}{5}$

How to solve

Strategy Extreme Principle

The condition BD ＞ 5√(2) is an inequality, and an inequality on a moving point is decided by its boundary case. That is Tool #14 (Extreme Principle): find the single position of D where BD equals 5√(2) exactly, and that one point splits the triangle into a winning half and a losing half. Tool #1 (Draw a Diagram) makes the split visible — the segment from B to that boundary point is the dividing line. Tool #16 (Change Focus) is the move that makes the problem finite: the question is about P, but BD depends only on where D is, so translate the condition on D back into a region for P. Tool #4 (Introduce a Variable) supplies the bridge, naming the distance CD so the length condition becomes an inequality. Tool #7 (Identify Subproblems) then leaves two easy jobs: get the side lengths of the 30-60-90 triangle, and compare two areas.

1STEP 1

Get the two legs

The 30-60-90 shape with hypotenuse 10 gives legs BC = 5 and AC = 5√(3).

BC = 10cos 60° = 5, AC = 10sin 60° = 5√(3)

2STEP 2

Rewrite BD using CD

Triangle BCD is right-angled at C, so BD² = 25 + CD² and the condition becomes CD ＞ 5.

BD² = 25 + CD² ＞ 50 ⇔ CD² ＞ 25 ⇔ CD ＞ 5

3STEP 3

Find the boundary point on AC

The point D' with CD' = 5 is the break-even spot, and it lies on AC since 5 ＜ 5√(3).

CD' = 5 → BD' = 5√(2); 5 ＜ 5√(3) = AC

4STEP 4

Turn the condition on D into a region for P

Segment BD' splits the triangle, and the ray hits beyond D' exactly when P lies in triangle ABD'.

BD ＞ 5√(2) ⇔ P ∈ △ ABD'

5STEP 5

Compare the two areas

Sharing apex B and a common line, the areas are in the ratio of the bases: AD'/AC.

[ABD']/[ABC] = AD'/AC = (5√(3)-5)/5√(3) = (√(3)-1)/√(3) = (3-√(3))/3

6STEP 6

Read off the probability

That ratio simplifies to (3-√(3))/3, about 0.423, choice (C).

P(BD ＞ 5√(2)) = (3-√(3))/3 ≈ 0.423 → (C)

Answer

(3-√3)/3

The value (3-√(3))/3≈ 0.423 sits between 0 and 1, as any probability must. It should also be a little under 1/2, and it is: the break-even point has CD' = 5, while the midpoint of AC is only 5√(3)/2≈ 4.33 from C, so D' lies past the midpoint and the winning piece ABD' is the smaller of the two. Numerically the choices are (A) 0.293, (B) 0.333, (C) 0.423, (D) 0.500, (E) 0.553, so only (C) matches. One trap is worth naming: it is tempting to measure the chance by the angle at B, since ∠ D'BC = 45° out of the full ∠ ABC = 60° would suggest 15/60 = 1/4. That is wrong, because P is spread evenly over area, not evenly over directions out of B — the wedge nearer A is longer and therefore holds more area than its angle suggests. Comparing areas, not angles, is what the phrase "randomly chosen inside" demands.

💡Key takeaway

Find the one spot where the condition just barely holds, draw the line through it, and the probability is the share of the area on the winning side.

Get the two legs
Rewrite BD using CD
Find the boundary point on AC
Turn the condition on D into a region for P
Compare the two areas
Read off the probability