AMC 8 · 2023 · #19

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference

area-trianglessimilar-figuresratio-proportion area-differenceeasier-related-problem ↑ Prerequisites: area-trianglesratio-proportion

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is $\frac23$ the side length of the larger triangle. What is the ratio of the area of one trapezoid to the area of the inner triangle?

Pick an answer.

(A)

1 : 3

(B)

3 : 8

(C)

5 : 12

(D)

7 : 16

(E)

4 : 9

AMC 8 2023 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A smaller equilateral triangle is placed inside a larger equilateral triangle so that the ring-shaped region between them splits into three congruent trapezoids. The smaller triangle's side is $\tfrac{2}{3}$ of the larger triangle's side. Find the ratio of one trapezoid's area to the inner triangle's area.

Givens: Outer figure is an equilateral triangle; inner figure is an equilateral triangle; Side length of inner triangle $= \tfrac{2}{3}\times$ side length of outer triangle; The region between the two triangles is partitioned into 3 congruent trapezoids; Answer choices: $1{:}3,\ 3{:}8,\ 5{:}12,\ 7{:}16,\ 4{:}9$

Unknowns: The ratio (one trapezoid's area) : (inner triangle's area)

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #16 Change Focus / Count the Complement

The ratio asked for is the same no matter what concrete side length we use, so we replace the abstract setup with the easiest case: outer side $= 3$ units and inner side $= 2$ units. Because both triangles are equilateral, the inner-to-outer area ratio must be $\bigl(\tfrac{2}{3}\bigr)^2 = \tfrac{4}{9}$, so we can simply name the inner area $4$ square-units and the outer area $9$ square-units — no square-root formula needed. We then split the work into two subproblems (Tool 7): (i) find the combined area of the three trapezoids, and (ii) divide by 3. The combined-trapezoid area is found by looking at the complement (Tool 16): outer minus inner. A quick diagram (Tool 1) makes the partition concrete.

Execute — Answer: C

#1 Draw a Diagram 6.G.A.1 Step 1

Sketch the two nested equilateral triangles and the three trapezoids in the ring between them.
The picture shows that the inner triangle together with the three trapezoids tiles the outer triangle, so the three trapezoid areas add up to (outer triangle area) $-$ (inner triangle area).

$$3 \cdot T \;=\; A_{\text{outer}} - A_{\text{inner}}$$

💡 When a big shape is cut into smaller shapes, the smaller areas add back to the big area.

#9 Solve an Easier Related Problem 7.G.A.1 Step 2

Replace the abstract setup with the easiest concrete case: outer side $= 3$, inner side $= 2$ (the ratio $\tfrac{2}{3}$ is preserved).
Because both triangles are equilateral, they are similar with linear scale factor $\tfrac{2}{3}$, so their areas are in ratio $\bigl(\tfrac{2}{3}\bigr)^2 = \tfrac{4}{9}$.

$$\frac{A_{\text{inner}}}{A_{\text{outer}}} \;=\; \left(\tfrac{2}{3}\right)^{2} \;=\; \tfrac{4}{9}$$

💡 Scaling every side by the same factor scales the area by the square of that factor.

#9 Solve an Easier Related Problem 6.RP.A.3 Step 3

Pick a convenient unit so the areas are whole numbers: let $A_{\text{inner}} = 4$ square-units and $A_{\text{outer}} = 9$ square-units.
This keeps the ratio $4{:}9$ exactly and avoids any square-root computation.

$$A_{\text{inner}} = 4,\qquad A_{\text{outer}} = 9$$

💡 We can pick any unit we like — only the ratio matters in the end.

#16 Change Focus / Count the Complement 5.NF.A.1 Step 4

Use the complement: the three trapezoids together fill the ring, so their combined area is the outer area minus the inner area.
Divide by $3$ to get the area of one trapezoid, since the three trapezoids are congruent.

$$3T \;=\; 9 - 4 \;=\; 5 \quad\Longrightarrow\quad T \;=\; \tfrac{5}{3}$$

💡 Looking at what's left over (outer minus inner) skips the messy trapezoid formula.

#7 Identify Subproblems 6.RP.A.3 Step 5

Form the requested ratio of one trapezoid to the inner triangle and simplify.

$$\frac{T}{A_{\text{inner}}} \;=\; \frac{5/3}{4} \;=\; \frac{5}{12} \;\Longrightarrow\; T : A_{\text{inner}} \;=\; 5 : 12$$

💡 Dividing two quantities expressed in the same units gives the ratio directly.

[1] #1 6.G.A.1 Sketch the two nested equilateral triangles and the three trapezoids in the ring

[2] #9 7.G.A.1 Replace the abstract setup with the easiest concrete case: outer side $= 3$, inn

[3] #9 6.RP.A.3 Pick a convenient unit so the areas are whole numbers: let $A_{\text{inner}} = 4

[4] #16 5.NF.A.1 Use the complement: the three trapezoids together fill the ring, so their combin

[5] #7 6.RP.A.3 Form the requested ratio of one trapezoid to the inner triangle and simplify.

Review

Reasonableness: Sanity-check the totals: one trapezoid has area $\tfrac{5}{3}$, so three trapezoids together have area $5$, and adding the inner triangle's area $4$ gives $9$, matching the outer triangle's area. The ratio $5{:}12$ is also between $\tfrac{1}{3}$ ($\approx 0.333$) and $\tfrac{4}{9}$ ($\approx 0.444$), which are the surrounding answer choices, so the magnitude is reasonable for a 'one of three trapezoids' slice. Choices (A) $1{:}3$ would mean three trapezoids equal the inner triangle (i.e. ring $=$ inner), but the ring should be larger than the inner since $9 - 4 = 5 > 4$, so (A) is too small; (E) $4{:}9$ matches the inner-to-outer ratio, not the requested one, so it is the classic distractor.

Alternative: Tool 13 (Convert to Algebra) gives the same answer symbolically: let $A_S = A_{\text{inner}}$, then $A_L = \tfrac{9}{4} A_S$ and $T = \tfrac{1}{3}(A_L - A_S) = \tfrac{1}{3}\bigl(\tfrac{9}{4} - 1\bigr) A_S = \tfrac{5}{12} A_S$. That route is just as quick once you trust the symbolic manipulation, but the concrete '4 and 9' substitution keeps every line as whole-number arithmetic — closer to a grade-5/6 student's comfort zone.

CCSS standards used (min grade 7)

5.NF.A.1 Add and subtract fractions with unlike denominators (Subtracting the inner-triangle area from the outer-triangle area ($9 - 4 = 5$) to obtain the combined area of the three trapezoids, and dividing the result by $3$.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Using the fact that the inner triangle and the three trapezoids together compose the outer triangle, so the trapezoid areas can be found from the difference instead of the trapezoid formula.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Choosing convenient unit areas $4$ and $9$ to preserve the $4{:}9$ ratio, and forming the final ratio $T : A_{\text{inner}} = 5 : 12$.)
7.G.A.1 Solve problems involving scale drawings of geometric figures (Translating the linear scale factor $\tfrac{2}{3}$ between similar equilateral triangles into the area ratio $\bigl(\tfrac{2}{3}\bigr)^2 = \tfrac{4}{9}$.)

⭐ This AMC 8 problem only needs Grade 7 'scale-drawing area ratio' (sides scaled by $k$ make areas scaled by $k^2$) you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

More like this