AMC 8 · 2007 · #22
Grade 6 geometry-2dProblem
A lemming sits at a corner of a square with side length meters. The lemming runs meters along a diagonal toward the opposite corner. It stops, makes a right turn and runs more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A lemming starts at a corner of a $10 \times 10$ square. It runs $6.2$ meters along the diagonal toward the opposite corner, makes a $90^{\circ}$ right turn, and runs $2$ more meters. From its final position, what is the average of the perpendicular distances to the four sides of the square?
Givens: Square side length is $10$ meters; The lemming travels $6.2$ meters along the diagonal, then turns $90^{\circ}$ right and travels $2$ more meters; Shortest distance from a point to a side $=$ perpendicular distance to that side; Answer choices: (A) $2$, (B) $4.5$, (C) $5$, (D) $6.2$, (E) $7$
Unknowns: The average of the four perpendicular distances from the lemming's final position to the four sides of the square
Understand
Restated: A lemming starts at a corner of a $10 \times 10$ square. It runs $6.2$ meters along the diagonal toward the opposite corner, makes a $90^{\circ}$ right turn, and runs $2$ more meters. From its final position, what is the average of the perpendicular distances to the four sides of the square?
Givens: Square side length is $10$ meters; The lemming travels $6.2$ meters along the diagonal, then turns $90^{\circ}$ right and travels $2$ more meters; Shortest distance from a point to a side $=$ perpendicular distance to that side; Answer choices: (A) $2$, (B) $4.5$, (C) $5$, (D) $6.2$, (E) $7$
Plan
Primary tool: #11 Find an Invariant
Secondary: #1 Draw a Diagram, #4 Introduce a Variable
The path detail ($6.2$ meters then a $90^{\circ}$ turn then $2$ meters) is decoy. For any point inside a $10 \times 10$ square, the two distances to a pair of opposite sides always sum to $10$ — that is the invariant. Tool #11 (Find an Invariant) catches this immediately. Tool #1 (Draw a Diagram) confirms the geometry: drop perpendiculars from the lemming's final spot to all four sides. Tool #4 (Introduce a Variable) lets us call the unknown coordinates $x$ and $y$ and watch them cancel.
Execute — Answer: C
6.G.A.3 Step 1 - Set up coordinates and draw the picture.
- Place the starting corner at the origin and the square with sides on the axes, so its corners are $(0,0)$, $(10,0)$, $(10,10)$, $(0,10)$.
- Let the lemming's final position be $(x, y)$ for some $0 < x < 10$ and $0 < y < 10$.
- Drop perpendiculars to the four sides.
💡 Coordinates turn "shortest distance to a side" into a single subtraction. Grade 6 coordinate geometry handles this directly.
6.EE.A.2 Step 2 - Name the unknown coordinates.
- We don't need the actual values of $x$ and $y$ — only their sum with their complements.
- Tool #4 lets us write the four distances symbolically.
💡 Each pair of opposite sides is exactly $10$ apart, so the two distances to that pair must add to $10$.
6.EE.A.3 Step 3 - Spot the invariant.
- Add the four distances.
- The $x$ terms cancel, the $y$ terms cancel, and the sum is always $20$ — regardless of where $(x,y)$ lands inside the square.
💡 $x$ and $-x$ cancel; $y$ and $-y$ cancel. Only the two $10$'s survive. The path numbers $6.2$ and $2$ never enter the calculation.
6.SP.B.5 Step 4 - Divide by $4$ to get the average.
- The sum is $20$, so the average of the four distances is $20 / 4 = 5$.
💡 Average = sum divided by count. The invariant sum makes the answer independent of the lemming's actual position.
6.G.A.3 Set up coordinates and draw the picture. Place the starting corner at the origin 6.EE.A.2 Name the unknown coordinates. We don't need the actual values of $x$ and $y$ — o 6.EE.A.3 Spot the invariant. Add the four distances. The $x$ terms cancel, the $y$ terms 6.SP.B.5 Divide by $4$ to get the average. The sum is $20$, so the average of the four di Review
Reasonableness: Pick any point inside the square and check. The center $(5, 5)$ has distances $5, 5, 5, 5$ averaging $5$. The point $(2, 9)$ has distances $2, 8, 9, 1$ summing to $20$ and averaging $5$. The lemming's exact final position is irrelevant — every interior point gives the same average. The path-length numbers $6.2$ and $2$ are deliberate distractors, and the answer matches choice (C).
Alternative: Tool #1 (Draw a Diagram) by pairing: pair the four distances as (left, right) and (bottom, top). Each pair adds to $10$ because opposite sides of the square are $10$ apart. Two pairs of $10$ give a total of $20$, so the average is $5$. No coordinates needed — just the picture of two opposite sides sandwiching the point.
CCSS standards used (min grade 6)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find side lengths (Placing the square on a coordinate grid so that perpendicular distances from an interior point to the sides become simple coordinate expressions.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the lemming's final position $(x, y)$ and writing the four perpendicular distances as $x$, $10 - x$, $y$, $10 - y$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Combining $x + (10 - x) + y + (10 - y)$ so the variable terms cancel, leaving the constant $20$.)6.SP.B.5Summarize numerical data sets, including reporting the number of observations and measures of center (Computing the mean of the four distances by dividing the sum $20$ by the count $4$.)
⭐ Wherever the lemming lands inside a $10 \times 10$ square, the four perpendicular distances to the sides always add to $20$, so the average is $5$. The $6.2$ and $2$ in the path are distractors.
⭐ Wherever the lemming lands inside a $10 \times 10$ square, the four perpendicular distances to the sides always add to $20$, so the average is $5$. The $6.2$ and $2$ in the path are distractors.
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