AMC 10 · 2016 · #21

Grade 10 geometry-2d
coordinate-geometryrecursive-sequencetelescoping-sumarea-triangles coordinate-geometrytelescoping-sum ↑ Prerequisites: coordinate-geometryarea-triangles
📏 Long solution 💡 4 insights
Problem
A construction repeats forever, each round making a smaller triangle. Add all their areas.

Pick an answer.

(A)
$\frac{1}{6}$
(B)
$\frac{1}{4}$
(C)
$\frac{1}{3}$
(D)
$\frac{1}{2}$
(E)
1
How to solve
Strategy Convert to Algebra

An endless picture cannot be drawn all the way, so the picture has to become a formula. Tool #1 (Draw a Diagram) puts the square on coordinate axes with D at the origin, which turns "drop a perpendicular to CD" into "keep the x-coordinate, set y to 0" and turns diagonal BD into the line y = x. Tool #4 (Introduce a Variable) then handles all the rounds at once: call the current foot Q_i = (q, 0) and do the work one time, in terms of q. Tool #13 (Convert to Algebra) converts one full round into a single rule that sends q to the next value, and tool #5 (Look for a Pattern) reads a closed form out of that rule. Tool #7 (Identify Subproblems) splits the goal into two easy pieces, the area of one triangle and then the sum. Tool #15 (Organize Information in More Ways) rewrites each area as a difference of two fractions so the pieces cancel, and tool #9 (Solve an Easier Related Problem) finishes by summing the first N terms and watching what is left over shrink.

1STEP 1

Put the square on coordinates

Coordinates make one round computable.

D=(0,0), C=(1,0), B=(1,1), A=(0,1), BD: y = x, Q₁=(1/2,0)
2STEP 2

Do one round with a letter

One letter carries the whole round.

Q_i=(q,0), AQ_i: y = 1 - x/q
3STEP 3

Cross the diagonal to find P_i

The crossing point comes out cleanly.

x = 1 - x/q ⟹ x = q/(q+1), P_i = (q/(q+1),q/(q+1)), Q_i+1 = (q/(q+1),0)
4STEP 4

Read the pattern of the feet

The feet follow a simple pattern.

q_i+1 = q_i/(q_i+1), q₁ = 1/2 ⟹ DQ_i = q_i = 1/(i+1)
5STEP 5

Area of one triangle

Each area is a product of neighbours.

[△ DQ_iP_i] = 1/2 · DQ_i · DQ_i+1 = 1/2·1/(i+1)·1/(i+2) = 1/(2(i+1)(i+2))
6STEP 6

Rewrite each area as a difference

It rewrites as a difference.

1/(i+1) - 1/(i+2) = (i+2)-(i+1)/(i+1)(i+2) = 1/(i+1)(i+2) ⟹ [△ DQ_iP_i] = 1/2(1/(i+1) - 1/(i+2))
7STEP 7

Add up and watch it collapse

The sum collapses to 1/4, choice (B).

Σ_i=1^N1/2(1/(i+1)-1/(i+2)) = 1/2(1/2-1/(N+2)) → 1/4 = (B)
Answer
1/4
Check the running totals by hand. The first five areas are 1/12, 1/24, 1/40, 1/60, 1/84, which add to 5/28 ≈ 0.179, and the closed form agrees: 1/2(1/2-1/7) = 5/28. That single computation kills three choices at once. Every partial sum is strictly below 1/4, so (C) 1/3, (D) 1/2 and (E) 1 are impossible, while the total already exceeds (A) 1/6 ≈ 0.167 after only five triangles. Only 1/4 survives. A geometric check agrees too: each triangle DQ_iP_i sits in the wedge between DC and the diagonal, whose total area is 1/2, and the triangles pile up on a shrinking sliver near D, so a modest total like 1/4 is the right size.
💡Key takeaway

When each piece of an endless sum can be written as one fraction minus the next, almost everything cancels and only the very first fraction is left.

  • Put the square on coordinates
  • Do one round with a letter
  • Cross the diagonal to find P_i
  • Read the pattern of the feet
  • Area of one triangle
  • Rewrite each area as a difference
  • Add up and watch it collapse