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AMC 8 · 2020 · #25

Grade 8 geometry-2dalgebra
Archetype Coordinate Plane Figure Computation
area-rectanglessystems-of-equationslinear-equations-two-varspatial-visualization convert-to-algebraidentify-subproblems ↑ Prerequisites: systems-of-equationsspatial-visualization
📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Rectangles R1R_1 and R2,R_2, and squares S1, S2, S_1,\,S_2,\, and S3,S_3, shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of S2S_2 in units?

(A) 651(B) 655(C) 656(D) 662(E) 666\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666

Pick an answer.

(A)
651
(B)
655
(C)
656
(D)
662
(E)
666
View mode:

Toolkit + CCSS Solution

Understand

Restated: A big rectangle measuring $3322$ wide and $2020$ tall is tiled by three squares $S_1, S_2, S_3$ (side lengths $s_1, s_2, s_3$) and two rectangles $R_1, R_2$. From the diagram, $S_1$ sits in the upper-left, $R_1$ directly below it, $S_2$ is the middle square sharing the same vertical column as $R_1$'s right edge through $S_1$'s right edge, $S_3$ sits in the lower-right, and $R_2$ is above $S_3$. Find $s_2$.

Givens: Outer rectangle width $= 3322$; Outer rectangle height $= 2020$; $S_1, S_2, S_3$ are squares with side lengths $s_1, s_2, s_3$; $R_1$ is directly below $S_1$ and $R_2$ is directly above $S_3$; Answer choices: (A) $651$, (B) $655$, (C) $656$, (D) $662$, (E) $666$

Unknowns: The side length $s_2$ of the middle square $S_2$

Understand

Restated: A big rectangle measuring $3322$ wide and $2020$ tall is tiled by three squares $S_1, S_2, S_3$ (side lengths $s_1, s_2, s_3$) and two rectangles $R_1, R_2$. From the diagram, $S_1$ sits in the upper-left, $R_1$ directly below it, $S_2$ is the middle square sharing the same vertical column as $R_1$'s right edge through $S_1$'s right edge, $S_3$ sits in the lower-right, and $R_2$ is above $S_3$. Find $s_2$.

Givens: Outer rectangle width $= 3322$; Outer rectangle height $= 2020$; $S_1, S_2, S_3$ are squares with side lengths $s_1, s_2, s_3$; $R_1$ is directly below $S_1$ and $R_2$ is directly above $S_3$; Answer choices: (A) $651$, (B) $655$, (C) $656$, (D) $662$, (E) $666$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra

The figure is given but it pays to redraw it and label every segment with $s_1, s_2, s_3$ — that is Tool #1. Tool #7 (Identify Subproblems) splits the geometry into two independent equations: one from the horizontal direction (total width) and one from the vertical direction (height of the middle column). With two equations linking $s_1, s_2, s_3$ to the known numbers $3322$ and $2020$, Tool #13 (Convert to Algebra) lets us add or subtract the equations to eliminate $s_1$ and $s_3$ in a single move and read off $s_2$ — no need to solve for all three side lengths.

Execute — Answer: A

#1 Draw a Diagram 4.G.A.1 Step 1
  • Redraw and label.
  • Mark the side lengths $s_1, s_2, s_3$ on each square in the figure and identify which segments line up along the top edge of the big rectangle.
  • The three squares' widths sit side-by-side across the entire top, with $S_1$ on the left (above $R_1$), $S_2$ in the middle, $S_3$ on the right (above which is $R_2$).
$$\text{top edge} = s_1 + s_2 + s_3$$

💡 Labeling each segment turns a picture into a sentence: "these three widths together make the long side."

#7 Identify Subproblems 6.EE.B.6 Step 2
  • Read the width equation.
  • The labeled top edge equals the given width $3322$, giving the first equation.
$$s_1 + s_2 + s_3 = 3322 \quad (\text{Eq. 1})$$

💡 Writing the picture's width as a sum of unknowns is exactly using variables to describe a real quantity.

#7 Identify Subproblems 6.EE.B.7 Step 3
  • Read the height.
  • The left column ($S_1$ above $R_1$) has total height $2020$, so $R_1$'s height is $2020 - s_1$.
  • The right column ($S_3$ below $R_2$) has total height $2020$, so $R_2$'s height is $2020 - s_3$.
  • In the middle column, $S_2$ exactly fills the vertical gap between the top of $R_1$ (height $2020 - s_1$ from the bottom) and the bottom of $R_2$ (height $s_3$ from the bottom).
$$s_2 = (2020 - s_1) - s_3 \;\Leftrightarrow\; s_1 - s_2 + s_3 = 2020 \quad (\text{Eq. 2})$$

💡 Stacking two side columns of height $2020$ and reading what is left for the middle square turns the vertical layout into one clean equation.

#13 Convert to Algebra 8.EE.C.8 Step 4
  • Eliminate $s_1$ and $s_3$ in one stroke.
  • Subtract Eq.
  • 2 from Eq.
  • 1: the $s_1$ and $s_3$ terms cancel, and the $s_2$ terms reinforce.
$$(s_1 + s_2 + s_3) - (s_1 - s_2 + s_3) = 3322 - 2020 \;\Rightarrow\; 2 s_2 = 1302$$

💡 When two equations share most variables, subtracting them lines up the unknowns so they vanish — a classic simultaneous-equations move.

#13 Convert to Algebra 5.NBT.B.6 Step 5

Solve for $s_2$.

$$s_2 = \dfrac{1302}{2} = 651 \;\Rightarrow\; \textbf{(A)}$$

💡 Dividing a 4-digit number by $2$ is straightforward whole-number division.

[1] #1 4.G.A.1 Redraw and label. Mark the side lengths $s_1, s_2, s_3$ on each square in the fi
[2] #7 6.EE.B.6 Read the width equation. The labeled top edge equals the given width $3322$, giv
[3] #7 6.EE.B.7 Read the height. The left column ($S_1$ above $R_1$) has total height $2020$, so
[4] #13 8.EE.C.8 Eliminate $s_1$ and $s_3$ in one stroke. Subtract Eq. 2 from Eq. 1: the $s_1$ an
[5] #13 5.NBT.B.6 Solve for $s_2$.

Review

Reasonableness: The answer must lie between $0$ and $\min(3322, 2020) = 2020$, and $651$ fits comfortably. As a sanity check, since $s_1 + s_2 + s_3 = 3322$ and $s_1 + s_3 - s_2 = 2020$, adding the two equations gives $2(s_1 + s_3) = 5342$, so $s_1 + s_3 = 2671$. That leaves $s_2 = 3322 - 2671 = 651$. Matches. The figure also implies $s_1, s_3 > s_2$ (the middle square is the smallest), and any split of $2671$ into two values each above $651$ is consistent (e.g. $s_1 \approx 1369, s_3 \approx 1302$).

Alternative: Tool #3 (Eliminate Possibilities) on the choices: from Eq. 1 + Eq. 2 we know $s_1 + s_3 = 2671$, and from Eq. 1 we have $s_2 = 3322 - (s_1 + s_3) = 3322 - 2671 = 651$. Substituting each of (A)-(E) into Eq. 1 and Eq. 2 simultaneously, only $s_2 = 651$ leaves a consistent $s_1 + s_3 = 2671$ that satisfies both, confirming (A) without re-deriving the algebra.

CCSS standards used (min grade 8)

  • 4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures (Reading and labeling the line segments inside the composite rectangle so each side of every square is associated with a variable.)
  • 6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the unknown square side lengths $s_1, s_2, s_3$ and writing the top edge as the expression $s_1 + s_2 + s_3$.)
  • 6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Translating the vertical layout into the single equation $s_1 - s_2 + s_3 = 2020$ from the heights of the three columns.)
  • 8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Subtracting the two linear equations in $s_1, s_2, s_3$ to eliminate $s_1$ and $s_3$ together and isolate $2 s_2 = 1302$.)
  • 5.NBT.B.6 Find whole-number quotients with up to four-digit dividends and two-digit divisors (Dividing $1302 \div 2 = 651$ to get the final side length.)

⭐ This AMC 8 problem only needs Grade 8 simultaneous equations you already know — two pictures of the rectangle's edges become two equations, and subtracting them cancels the squares you don't care about!

⭐ This AMC 8 problem only needs Grade 8 simultaneous equations you already know — two pictures of the rectangle's edges become two equations, and subtracting them cancels the squares you don't care about!

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