AMC 8 · 1999 · #25

Grade 8 geometry-2d

Archetype Geometric Series Share └ Sub-archetype Nested-Similar-Figure Area Sum

area-trianglessequences-geometricsimilar-figurespattern-recognitionfraction-arithmetic pattern-recognitionidentify-subproblemsarea-difference ↑ Prerequisites: area-trianglesfraction-arithmeticsequences-geometric

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Right isosceles △ ACG has legs AC = CG = 6. Midpoints of its sides create a midpoint triangle, leaving three small triangles — one is shaded. The same midpoint-shade move is repeated 100 times, each time inside the upper-right corner triangle (the new small triangle that shares vertex G). Find the integer nearest the total shaded area.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

The picture already shows three shaded triangles, so Tool #1 (Draw a Diagram) lets us read their sizes straight off the figure. Tool #5 (Look for a Pattern) is the natural fit: each new triangle is a half-scale copy of the previous one, so each shaded area is one-quarter of the one before — a clean geometric pattern with ratio $\frac{1}{4}$ . Tool #9 (Solve an Easier Related Problem) replaces "100 rounds" with "infinitely many rounds," because the tail ( $\frac{1}{4}$ )¹⁰⁰ is far smaller than any rounding error. Summing $\frac{9}{2}$ + $\frac{9}{8}$ + $\frac{9}{32}$ + … as an unending geometric pattern is the easier problem, and its answer is the nearest integer we need.

1STEP 1

△ ACG has area 18; the first shaded △ BDC is right-angled at C with legs 3, so its area is $\frac{9}{2}$ .

[△ BDC] =

\frac{1}{2}

(3)(3) =

\frac{9}{2}

2STEP 2

Recurse into △ JDG (legs 3); its midpoint shading gives △ KED with legs $\frac{3}{2}$ , so area $\frac{9}{8}$ .

[△ KED] =

\frac{1}{2}

(

\frac{3}{2}

)(

\frac{3}{2}

) =

\frac{9}{8}

3STEP 3

Compare consecutive areas: ( $\frac{9}{8}$ )/( $\frac{9}{2}$ ) = $\frac{1}{4}$ , so each shaded area is a quarter of the one before — a constant ratio.

(

\frac{9}{8}

)/(

\frac{9}{2}

) =

\frac{2}{8}

\frac{1}{4}

4STEP 4

The areas $\frac{9}{2}$ , $\frac{9}{8}$ , $\frac{9}{32}$ , … each a quarter of the last, so partial sums 4.5, 5.625, 5.906, 5.977, … close in on a limit.

\frac{9}{2}

= 4.5, +

\frac{9}{8}

= 5.625, +

\frac{9}{32}

≈ 5.906, +

\frac{9}{128}

≈ 5.977

5STEP 5

Replace 100 rounds by infinitely many: the pattern repeats at $\frac{1}{4}$ scale, so the total obeys T = $\frac{9}{2}$ + ( $\frac{1}{4}$ )T, giving T = 6.

T -

\frac{1}{4}

T =

\frac{9}{2}

→

\frac{3}{4}

T =

\frac{9}{2}

→ T = 6

6STEP 6

Stopping at 100 rounds drops only a tail of 6·( $\frac{1}{4}$ )¹⁰⁰, far below $\frac{1}{2}$ , so the total is just under 6 and the nearest integer is 6.

S₁00 = 6 - 6 · (

\frac{1}{4}

)¹⁰⁰ ≈ 6 → (A)

Answer

Sanity-check the limit against the picture. The first shaded triangle has area

\frac{9}{2}

= 4.5, already most of the way to 6. Adding the second pushes the total to 5.625, the third to about 5.906, the fourth to about 5.977 — every step closes three-quarters of the remaining gap to 6, so the total can never reach (let alone pass) 6. After 100 rounds the gap is 6 · (

\frac{1}{4}

)¹⁰⁰, which is essentially zero. The total is just under 6, and the nearest integer is 6, matching choice (A). The size also makes sense: the whole big triangle has area 18, and the shaded triangles tile a long thin staircase toward G, so a total of about a third of 18 is geometrically plausible.

💡Key takeaway

Each round shrinks the shaded triangle to a quarter of the one before, so the totals march 4.5 → 5.625 → 5.906 → 5.977 → …, closing three-quarters of the gap to 6 every time but never quite reaching it. After 100 rounds the gap is microscopic, so the total area is essentially 6 — answer (A).

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