AMC 8 · 1999 · #25
Grade 8 geometry-2dProblem
Points B, D, and J are midpoints of the sides of right triangle ACG. Points K, E, I are midpoints of the sides of triangle JDG, etc. If the dividing and shading process is done 100 times (the first three are shown) and AC=CG=6, then the total area of the shaded triangles is nearest
Pick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Right isosceles $\triangle ACG$ 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.
Givens: $\triangle ACG$ is right-angled at $C$ with legs $AC = CG = 6$; Each round: take a right isosceles triangle, mark midpoints of its sides, shade the corner triangle opposite vertex $G$, then recurse into the upper triangle at $G$; The process is repeated $100$ times; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Unknowns: The integer nearest the total area of all $100$ shaded triangles
Understand
Restated: Right isosceles $\triangle ACG$ 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.
Givens: $\triangle ACG$ is right-angled at $C$ with legs $AC = CG = 6$; Each round: take a right isosceles triangle, mark midpoints of its sides, shade the corner triangle opposite vertex $G$, then recurse into the upper triangle at $G$; The process is repeated $100$ times; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem
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 $\tfrac{1}{4}$. Tool #9 (Solve an Easier Related Problem) replaces "$100$ rounds" with "infinitely many rounds," because the tail $\left(\tfrac{1}{4}\right)^{100}$ is far smaller than any rounding error. Summing $\tfrac{9}{2} + \tfrac{9}{8} + \tfrac{9}{32} + \cdots$ as an unending geometric pattern is the easier problem, and its answer is the nearest integer we need.
Execute — Answer: A
7.G.B.6 Step 1 - Read the first shaded triangle off the picture.
- $\triangle ACG$ is right-angled at $C$ with legs $6$, so its area is $\tfrac{1}{2}(6)(6) = 18$.
- The first shaded triangle $\triangle BDC$ has its right angle at $C$ with legs $BC = 3$ and $CD = 3$ (each is half of a leg of $\triangle ACG$).
💡 When the right angle sits on the corner, base and height are just the two legs — a Grade 7 "area of a triangle" reading straight from the figure.
7.G.A.1 Step 2 - Repeat on the next-smaller triangle to spot the pattern.
- The recursion moves into $\triangle JDG$, a right isosceles triangle with legs $3$ (half of the original legs $6$).
- Its midpoint shading produces $\triangle KED$ with legs $\tfrac{3}{2}$.
💡 Halving every side is a Grade 7 scale-drawing move with scale factor $\tfrac{1}{2}$.
7.RP.A.2 Step 3 - Compare consecutive shaded areas to find the constant ratio.
- Side lengths halve each round, and area scales as the square of the side, so each new shaded area is $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$ of the previous.
- Check it against the numbers above.
💡 Constant area ratio $\tfrac{1}{4}$ across rounds is the Grade 7 proportional-relationship signature — every step shrinks the shaded area by the same factor.
8.EE.A.1 Step 4 - Write the running totals and watch them close in on a target.
- The shaded areas $\tfrac{9}{2}, \tfrac{9}{8}, \tfrac{9}{32}, \dots$ form a pattern where each new term is $\tfrac{1}{4}$ of the previous.
- Add the first few partial sums and look at how much room is left until the next jump.
💡 Each step closes about three-quarters of the gap to $6$, so $4.5 \to 5.625 \to 5.906 \to 5.977 \to \dots$ The remaining gap is multiplied by $\tfrac{1}{4}$ each time — that is the Grade 8 "powers of $\tfrac{1}{4}$" pattern.
8.EE.C.7 Step 5 - Use the easier related problem: replace $100$ rounds with infinitely many.
- Each round leaves $\tfrac{1}{4}$ of the current gap unfilled, so the total approaches a single limit.
- If $T$ is that limit, the sum of every term after the first equals $\tfrac{1}{4} T$ (the whole pattern, scaled by $\tfrac{1}{4}$), giving the self-similar equation $T = \tfrac{9}{2} + \tfrac{1}{4} T$.
💡 Because the pattern repeats itself at $\tfrac{1}{4}$ scale, the unknown total satisfies a Grade 8 one-step equation. Solving it gives the exact infinite-round total.
8.EE.A.1 Step 6 - Compare the $100$-round total to the infinite total.
- Stopping after $100$ rounds removes only the tail starting at term $101$, which is $\left(\tfrac{1}{4}\right)^{100}$ times $T = 6$ — a number with more than $60$ zeros after the decimal point.
- So the $100$-round total is just barely under $6$, and the nearest integer is $6$.
💡 $\left(\tfrac{1}{4}\right)^{100}$ is a Grade 8 "very small power" — far smaller than the $\tfrac{1}{2}$ rounding threshold, so the answer is the same as the infinite sum.
7.G.B.6 Read the first shaded triangle off the picture. $\triangle ACG$ is right-angled 7.G.A.1 Repeat on the next-smaller triangle to spot the pattern. The recursion moves int 7.RP.A.2 Compare consecutive shaded areas to find the constant ratio. Side lengths halve 8.EE.A.1 Write the running totals and watch them close in on a target. The shaded areas $ 8.EE.C.7 Use the easier related problem: replace $100$ rounds with infinitely many. Each 8.EE.A.1 Compare the $100$-round total to the infinite total. Stopping after $100$ rounds Review
Reasonableness: Sanity-check the limit against the picture. The first shaded triangle has area $\tfrac{9}{2} = 4.5$, 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 $6 \cdot \left(\tfrac{1}{4}\right)^{100}$, 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.
Alternative: Tool #16 (Change Focus): instead of summing the shaded triangles, look at how each round splits the current triangle. Each round divides the current right isosceles triangle into four congruent quarter-triangles; one is shaded and one (the upper-right) becomes the next round's triangle. So the shaded area in each round equals the next round's full triangle area. Summing all shaded areas equals the sum of all "current" triangle areas from round $2$ onward, which equals the area of round $2$'s triangle ($\tfrac{9}{2}$) plus the same pattern shifted by one — giving the same self-similar relation $T = \tfrac{9}{2} + \tfrac{1}{4} T$ and the same total of $6$.
CCSS standards used (min grade 8)
7.G.B.6Solve problems involving area of two-dimensional objects composed of triangles (Computing the area of the first shaded right triangle $\triangle BDC$ as $\tfrac{1}{2}(3)(3) = \tfrac{9}{2}$, and the second as $\tfrac{1}{2}\left(\tfrac{3}{2}\right)\left(\tfrac{3}{2}\right) = \tfrac{9}{8}$.)7.G.A.1Solve problems involving scale drawings of geometric figures (Recognizing that each new triangle is a scale copy of the previous one with scale factor $\tfrac{1}{2}$, so its area is $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$ of the previous area.)7.RP.A.2Recognize and represent proportional relationships between quantities (Reading the constant area ratio $\tfrac{1}{4}$ between consecutive shaded triangles as a proportional relationship — every step shrinks the shaded area by the same factor.)8.EE.A.1Know and apply the properties of integer exponents (Writing the tail $\left(\tfrac{1}{4}\right)^{100}$ as an integer power of $\tfrac{1}{4}$ and noting it is negligibly small, so $S_{100}$ rounds to the same integer as the infinite total.)8.EE.C.7Solve linear equations in one variable (Solving the self-similar equation $T = \tfrac{9}{2} + \tfrac{1}{4} T$ for the infinite total, yielding $T = 6$.)
⭐ Each round shrinks the shaded triangle to a quarter of the one before, so the totals march $4.5 \to 5.625 \to 5.906 \to 5.977 \to \dots$, 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).
⭐ Each round shrinks the shaded triangle to a quarter of the one before, so the totals march $4.5 \to 5.625 \to 5.906 \to 5.977 \to \dots$, 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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