Sensim Math Original · sm-14
SM Original Grade 6 number-theorygeometry-2dProblem
Beatriz is restoring a stained-glass window that is built from identical unit square panes arranged in a rectangular grid. Today she discovers a single hairline crack that runs perfectly straight from one corner of the window to the diagonally opposite corner. A pane is called tinted if the crack passes through the interior of that pane (panes the crack only touches at a vertex do not count).
To test her counting method, Beatriz first practices on a tiny rehearsal frame that is 3 panes wide and 2 panes tall: a corner-to-corner crack on that small frame tints exactly 4 panes.
The real window she is restoring measures 3600 panes wide and 2400 panes tall. How many panes will the corner-to-corner crack tint?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A rectangular stained-glass window is divided into a grid of unit square panes, $3600$ panes across and $2400$ panes tall. A perfectly straight hairline crack runs from one corner of the window to the diagonally opposite corner. We need the **count of panes whose interior the crack actually passes through** — panes the crack only grazes at a vertex do not count. We are warmed up by a tiny $3\times 2$ rehearsal frame in which the corner-to-corner crack tints exactly $4$ panes.
Givens: The real window is $3600$ panes wide and $2400$ panes tall; The crack is a single straight segment from one corner of the rectangle to the diagonally opposite corner; A pane is "tinted" only when the crack enters its **interior**; grazing a vertex does not count; Rehearsal data: a $3$-wide $\times\ 2$-tall frame tints exactly $4$ panes; Answer choices: (A) 4500, (B) 4800, (C) 5400, (D) 5999, (E) 6000
Unknowns: The number of panes the crack tints on the $3600 \times 2400$ window
Understand
Restated: A rectangular stained-glass window is divided into a grid of unit square panes, $3600$ panes across and $2400$ panes tall. A perfectly straight hairline crack runs from one corner of the window to the diagonally opposite corner. We need the **count of panes whose interior the crack actually passes through** — panes the crack only grazes at a vertex do not count. We are warmed up by a tiny $3\times 2$ rehearsal frame in which the corner-to-corner crack tints exactly $4$ panes.
Givens: The real window is $3600$ panes wide and $2400$ panes tall; The crack is a single straight segment from one corner of the rectangle to the diagonally opposite corner; A pane is "tinted" only when the crack enters its **interior**; grazing a vertex does not count; Rehearsal data: a $3$-wide $\times\ 2$-tall frame tints exactly $4$ panes; Answer choices: (A) 4500, (B) 4800, (C) 5400, (D) 5999, (E) 6000
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #1 Draw a Diagram, #5 Look for a Pattern, #2 Make a Systematic List
A $3600\times 2400$ grid is far too big to count panes one by one, so we lean on Tool #9 — **replace the giant rectangle with tiny ones** ($1\times 1$, $2\times 1$, $2\times 3$, $3\times 2$, $4\times 3$, $4\times 2$, $\ldots$) and use Tool #1 to actually draw each diagonal on graph paper and tally the tinted panes. Tool #5 then turns those tallies into a rule of the form "tinted = width $+$ height $-$ (correction)", where the correction turns out to count how many interior vertices the diagonal slices through. Tool #2 lets us list those tiny cases without missing any. Finally we apply the discovered rule to the $3600\times 2400$ window — no algebra, no memorized formula needed.
Execute — Answer: B
5.G.A.2 Step 1 - Step 1: Set up a way to talk about the crack without committing to giant numbers.
- For a rectangle that is $w$ panes wide and $h$ panes tall, the corner-to-corner diagonal starts at one vertex and ends at the diagonally opposite vertex.
- We will write down the tinted-pane count $T(w,h)$ for many small $(w,h)$, look for a rule in $w$ and $h$, then plug in $w=3600,\;h=2400$.
- The rehearsal datum already gives us $T(3,2)=4$.
💡 Naming the count $T(w,h)$ and recording it for tiny rectangles is exactly the Grade 5 coordinate-plane practice of plotting a point and reading its value.
5.G.A.2 Step 2 - Step 2: Generate small cases by hand on graph paper (Tool #9 + Tool #1).
- For each tiny rectangle, draw the corner-to-corner diagonal and count how many unit squares its interior crosses.
- The results, listed in order of increasing $w+h$, are: $T(1,1)=1$, $T(2,1)=2$, $T(2,3)=4$, $T(3,2)=4$ (matches the rehearsal), $T(4,3)=6$, $T(3,4)=6$, $T(5,3)=7$, $T(4,2)=4$, $T(6,4)=8$, $T(2,2)=2$.
- Order does not matter — $T(w,h)=T(h,w)$ — which is a sanity check that we are counting correctly.
💡 Drawing each tiny rectangle, sketching the diagonal, and tallying squares is direct Grade 5 coordinate-plane work — no abstraction yet.
4.OA.C.5 Step 3 - Step 3: Use Tool #5 to look for a rule in the table.
- A first guess "$T = w+h$" fails immediately on $T(2,1)=2$ ($2+1=3$).
- The next guess "$T = w+h-1$" succeeds on every case where $w$ and $h$ share no common factor besides $1$ — for instance $(2,3),(5,3),(4,3)$ — but fails on $(4,2)$ ($4+2-1=5\neq 4$), $(6,4)$ ($6+4-1=9\neq 8$), and $(2,2)$ ($2+2-1=3\neq 2$).
- Each failure is off by exactly the amount needed to bring $w+h$ down by an *additional* shared factor.
💡 Tabling cases, testing a candidate rule, and pinpointing exactly which rows break it is the Grade 4 "generate a pattern following a rule" activity.
5.G.A.2 Step 4 - Step 4: Use Tool #1 to *see why* those special cases need an extra subtraction.
- Sketch the $(4,2)$ diagonal — it cuts straight through the interior vertex $(2,1)$.
- Sketch $(6,4)$ — the diagonal is the line $y = \tfrac{2}{3}x$, which hits the interior vertex $(3,2)$.
- Sketch $(2,2)$ — the diagonal hits the interior vertex $(1,1)$.
- Each time the diagonal lands on an interior vertex, it crosses a vertical grid line **and** a horizontal grid line at the very same instant, so it enters one fewer *new* pane than the naive count predicted.
- That is exactly the extra $-1$ correction each failing row needed.
- The corrected rule is $T(w,h) = w + h - g$, where $g - 1$ counts the interior lattice points the diagonal touches, and the extra $-1$ accounts for the starting pane.
- Equivalently, $g$ is the number of lattice points on the closed diagonal minus one.
💡 Reading off interior lattice points from a plotted segment on the coordinate plane is straight-from-the-textbook Grade 5 geometry.
6.NS.B.4 Step 5 - Step 5: Identify $g$ in number-theory terms.
- The diagonal from $(0,0)$ to $(w,h)$ is parametrized by $(tw,th)$ for $t\in[0,1]$.
- A point on this segment has integer coordinates exactly when $tw$ and $th$ are both integers — which happens for $t = 0, 1/d, 2/d, \ldots, d/d=1$, where $d$ is the **greatest common divisor** of $w$ and $h$.
- That gives $d+1$ lattice points on the closed segment, so $g = d = \gcd(w,h)$, and the rule simplifies to $$T(w,h) = w + h - \gcd(w,h).$$ Re-check with our table: $T(6,4) = 6+4-\gcd(6,4) = 10-2 = 8$ ✓, $T(4,2) = 4+2-2 = 4$ ✓, $T(2,2)=2+2-2=2$ ✓, $T(5,3) = 5+3-1 = 7$ ✓.
- The rule holds on every recorded case, including the rehearsal $T(3,2) = 3+2-1 = 4$.
💡 Identifying the spacing of lattice points along the diagonal with the greatest common divisor is the Grade 6 GCD standard in action.
6.NS.B.4 Step 6 - Step 6: Apply the rule to $w = 3600,\ h = 2400$.
- We need $\gcd(3600, 2400)$.
- Notice $3600 = 1200 \times 3$ and $2400 = 1200 \times 2$, and $\gcd(3,2) = 1$, so $\gcd(3600, 2400) = 1200 \times \gcd(3,2) = 1200$.
- Plugging in: $T(3600, 2400) = 3600 + 2400 - 1200 = 4800$.
- The crack tints $\mathbf{4800}$ panes, which is choice (B).
💡 Factoring out the shared $1200$ from $3600$ and $2400$ and using $\gcd(3,2)=1$ is a direct Grade 6 GCD computation.
5.G.A.2 Step 1: Set up a way to talk about the crack without committing to giant numbers 5.G.A.2 Step 2: Generate small cases by hand on graph paper (Tool #9 + Tool #1). For eac 4.OA.C.5 Step 3: Use Tool #5 to look for a rule in the table. A first guess "$T = w+h$" f 5.G.A.2 Step 4: Use Tool #1 to *see why* those special cases need an extra subtraction. 6.NS.B.4 Step 5: Identify $g$ in number-theory terms. The diagonal from $(0,0)$ to $(w,h) 6.NS.B.4 Step 6: Apply the rule to $w = 3600,\ h = 2400$. We need $\gcd(3600, 2400)$. Not Review
Reasonableness: The rule $T(w,h) = w + h - \gcd(w,h)$ must reproduce the rehearsal $T(3,2)=4$: $3 + 2 - \gcd(3,2) = 5 - 1 = 4$ ✓. The answer also fits the size bracket — without any lattice correction we would get $3600 + 2400 = 6000$ (choice E, a classic trap), and the biggest possible correction is $\gcd \le \min(3600, 2400) = 2400$, which would give $3600$ at the very lowest. So $4800$ sits comfortably in the legal range $[3600, 6000]$. Notice the trap choices: (E) $6000$ ignores the gcd correction entirely, (D) $5999$ uses the wrong correction of $1$ (as if the rectangle were a coprime case), and (C) $5400$ comes from mistaking $\gcd(3600, 2400)$ for $600$ instead of $1200$.
Alternative: Tool #1 alone can finish the problem by *similarity*. Draw the diagonal of a tiny $3 \times 2$ rectangle and count $T(3,2) = 4$. The big $3600 \times 2400$ window is the same shape scaled up by a factor of $1200$, so each old unit pane becomes a $1200 \times 1200$ block of new tiny panes, and inside each block the scaled diagonal tints exactly $1200$ new panes (its own little $1 \times 1$ trip). Total tinted = $4 \times 1200 = 4800$. Same answer, no formula needed.
CCSS standards used (min grade 6)
4.OA.C.5Generate a number or shape pattern following a given rule (Tabulating the tinted-pane counts $T(w,h)$ for small rectangles and testing candidate rules of the form $w + h - ?$ to see where they break.)5.G.A.2Represent real-world and mathematical problems by graphing points (Drawing the small rectangles and their corner-to-corner diagonals on the coordinate plane, then reading off the interior lattice points the diagonal passes through.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Computing $\gcd(3600, 2400) = 1200$, the correction term that captures how many lattice-point coincidences the diagonal makes.)
⭐ This AMC 8 problem only needs Grade 6 greatest common factor (GCD) you already know!
⭐ This AMC 8 problem only needs Grade 6 greatest common factor (GCD) you already know!
More like this
Same archetype — closest grade level first.