AMC 10 · 2010 · #7
Grade 8 geometry-2dCrystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?
Pick an answer.
AMC 10 2010 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: Crystal runs a four-leg loop: $1$ mile due north, then $1$ mile northeast, then $1$ mile southeast, then a straight line back to the start. Find the length of that final straight leg.
Givens: She starts at a point and runs due north for $1$ mile.; She then runs northeast for $1$ mile (a $45^\circ$ turn toward the east).; She then runs southeast for $1$ mile.; The fourth leg is a straight line back to the starting point.; Answer choices: (A) $1$, (B) $\sqrt{2}$, (C) $\sqrt{3}$, (D) $2$, (E) $2\sqrt{2}$.
Unknowns: The length of the final straight leg — the distance from her position after three legs back to where she started.
Understand
Restated: Crystal runs a four-leg loop: $1$ mile due north, then $1$ mile northeast, then $1$ mile southeast, then a straight line back to the start. Find the length of that final straight leg.
Givens: She starts at a point and runs due north for $1$ mile.; She then runs northeast for $1$ mile (a $45^\circ$ turn toward the east).; She then runs southeast for $1$ mile.; The fourth leg is a straight line back to the starting point.; Answer choices: (A) $1$, (B) $\sqrt{2}$, (C) $\sqrt{3}$, (D) $2$, (E) $2\sqrt{2}$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #17 Visualize Spatial Relationships
The problem is all positions and paths, so the natural tool is #1 (Draw a Diagram): drop everything onto a coordinate grid with east as $+x$ and north as $+y$. The slanted miles are hard to add directly, so use Tool #7 (Identify Subproblems) to break each $45^\circ$ mile into an east piece and a north piece using a 45-45-90 triangle. Add the pieces to find the endpoint, then read off a right triangle and use the Pythagorean theorem for the straight distance home.
Execute — Answer: C
6.NS.C.8 Step 1 Set up coordinates
- Put the start at the origin.
- Let east be the positive $x$-direction and north the positive $y$-direction.
- Now every leg is just a change in $x$ (east) and $y$ (north) that you can add.
💡 A grid turns compass directions into simple $x$ and $y$ moves you can add up.
6.NS.C.8 Step 2 The north leg
- The first mile goes due north, so it only changes the $y$-coordinate.
- After this leg she is at $(0,1)$.
💡 Moving straight north changes only the north coordinate.
8.G.B.7 Step 3 Split the diagonal legs
- Northeast and southeast each make a $45^\circ$ angle with east.
- A $1$-mile step at $45^\circ$ is the hypotenuse of a 45-45-90 triangle, so its east piece and its north piece each have length $\tfrac{1}{\sqrt{2}} = \tfrac{\sqrt{2}}{2}$.
- Northeast goes east and north; southeast goes east and south.
💡 A tilted step is really two straight steps of equal size — one east and one up or down — because the 45° triangle is isosceles.
6.NS.C.8 Step 4 Add the pieces
- Add the three displacement pieces.
- The north parts of the two diagonals are $+\tfrac{\sqrt{2}}{2}$ and $-\tfrac{\sqrt{2}}{2}$, which cancel.
- The east parts add to $\sqrt{2}$.
- Her north coordinate stays at the $1$ from the first leg.
- So after three legs she is at $(\sqrt{2},\ 1)$.
💡 The two diagonals lean opposite ways vertically, so their up and down parts cancel and only the eastward push survives.
8.G.B.8 Step 5 Distance back to start
- The final leg is the straight-line distance from $(\sqrt{2},\ 1)$ back to $(0,0)$.
- That segment is the hypotenuse of a right triangle with legs $\sqrt{2}$ and $1$.
- By the Pythagorean theorem the distance is $\sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}$, which is choice (C).
💡 The straight path home is the hypotenuse of a right triangle whose two legs you can read straight off the grid.
6.NS.C.8 Put the start at the origin. Let east be the positive $x$-direction and north th 6.NS.C.8 The first mile goes due north, so it only changes the $y$-coordinate. After this 8.G.B.7 Northeast and southeast each make a $45^\circ$ angle with east. A $1$-mile step 6.NS.C.8 Add the three displacement pieces. The north parts of the two diagonals are $+\t 8.G.B.8 The final leg is the straight-line distance from $(\sqrt{2},\ 1)$ back to $(0,0) Review
Reasonableness: Check the size: $\sqrt{3} \approx 1.73$ miles, which sits between $1$ and $2$ — sensible for a loop built from $1$-mile legs, not tiny and not huge. Re-trace the path: north to $(0,1)$; northeast adds $\left(\tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{2}}{2}\right)$ giving $\left(\tfrac{\sqrt{2}}{2},\ 1+\tfrac{\sqrt{2}}{2}\right)$; southeast adds $\left(\tfrac{\sqrt{2}}{2}, -\tfrac{\sqrt{2}}{2}\right)$ giving $(\sqrt{2},\ 1)$. Distance $\sqrt{2+1} = \sqrt{3}$. Rule out neighbors: $\sqrt{2}$ would need the endpoint's squared distance to be $2$, but $(\sqrt{2})^2 + 1^2 = 3 \ne 2$; the value $2$ would need the squared legs to sum to $4$. So (C) is locked in.
Alternative: Tool #17 (Visualize Spatial Relationships): the northeast and southeast legs together form an isosceles right triangle opening toward the east — two $1$-mile sides meeting at a $90^\circ$ corner. Their combined effect is a single horizontal jump of $\sqrt{2}$ miles east with zero net vertical change. Stack that on the $1$-mile north leg and you immediately see a right triangle with legs $1$ and $\sqrt{2}$, hypotenuse $\sqrt{3}$ — no coordinates required.
CCSS standards used (min grade 8)
6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Placing the start at the origin and tracking each leg of the run as a change in the $x$ (east) and $y$ (north) coordinates.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Splitting each $45^\circ$ mile into equal east and north legs of length $\tfrac{\sqrt{2}}{2}$ using the 45-45-90 triangle.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Computing the final straight-line distance from $(\sqrt{2}, 1)$ back to the origin $(0,0)$.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Checking that $\sqrt{3} \approx 1.73$ sits sensibly between the $1$-mile and $\sqrt{2}$-mile lengths.)
⭐ Put the run on a grid and split each slanted mile into an east part and a north part — the two diagonal north parts cancel, leaving a plain right triangle with legs $\sqrt{2}$ and $1$, so the walk home is $\sqrt{3}$ miles.
⭐ Put the run on a grid and split each slanted mile into an east part and a north part — the two diagonal north parts cancel, leaving a plain right triangle with legs $\sqrt{2}$ and $1$, so the walk home is $\sqrt{3}$ miles.
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