AMC 8 · 2005 · #7

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

pythagorean-theoremfraction-arithmeticcoordinate-geometry identify-subproblemscoordinate-geometry ↑ Prerequisites: pythagorean-theoremfraction-arithmetic

📏 Short solution 💡 2 insights

Problem

Bill walks $\tfrac12$ mile south, then $\tfrac34$ mile east, and finally $\tfrac12$ mile south. How many miles is he, in a direct line, from his starting point?

Pick an answer.

(A)

(B)

$1\tfrac14$

(C)

$1\tfrac12$

(D)

$1\tfrac34$

(E)

AMC 8 2005 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.

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Toolkit + CCSS Solution

Understand

Restated: Bill walks $\tfrac{1}{2}$ mile south, then $\tfrac{3}{4}$ mile east, then $\tfrac{1}{2}$ mile south again. How far is his finishing point from his starting point, measured in a straight line?

Givens: First leg: $\tfrac{1}{2}$ mile south; Second leg: $\tfrac{3}{4}$ mile east; Third leg: $\tfrac{1}{2}$ mile south; Answer choices: (A) $1$, (B) $1\tfrac{1}{4}$, (C) $1\tfrac{1}{2}$, (D) $1\tfrac{3}{4}$, (E) $2$

Unknowns: The straight-line distance from start to finish

Understand

Plan

Primary tool: #1 Draw a Diagram

The problem is a path on a flat compass grid, which is the textbook signal for Tool #1 (Draw a Diagram). Sketching the three legs reveals that the two south legs stack into a single vertical segment, and the east leg sits perpendicular to it. Start and finish are then the two non-right corners of a right triangle, so the direct-line distance is the hypotenuse. Once the diagram is drawn, the Pythagorean theorem finishes the job in one line — no algebra needed beyond squaring two fractions.

Execute — Answer: B

#1 Draw a Diagram 6.NS.C.8 Step 1

Sketch the walk.
Put Bill's start at the top.
South is down and east is to the right, so the path goes down $\tfrac{1}{2}$, right $\tfrac{3}{4}$, down $\tfrac{1}{2}$.
The two south legs are along the same vertical line, separated only by the east step.

$$\text{path: } (0,0) \to (0,-\tfrac{1}{2}) \to (\tfrac{3}{4},-\tfrac{1}{2}) \to (\tfrac{3}{4},-1)$$

💡 Placing the walk on a coordinate grid is the Grade 6 "points in all four quadrants" move — south becomes negative $y$, east becomes positive $x$.

#1 Draw a Diagram 5.NF.A.1 Step 2

Close the triangle.
Draw the straight segment from the start $(0,0)$ to the finish $(\tfrac{3}{4},-1)$.
Together with the total south displacement and the east leg, this segment is the hypotenuse of a right triangle whose legs are the total southward distance and the eastward distance.

$$\text{south total} = \tfrac{1}{2} + \tfrac{1}{2} = 1, \quad \text{east total} = \tfrac{3}{4}$$

💡 Adding $\tfrac{1}{2} + \tfrac{1}{2} = 1$ to merge the two south legs is the Grade 5 "add fractions" step that the picture asks for.

#1 Draw a Diagram 8.G.B.7 Step 3

Apply the Pythagorean theorem.
With legs $1$ and $\tfrac{3}{4}$, the squared hypotenuse is $1^2 + \left(\tfrac{3}{4}\right)^2 = 1 + \tfrac{9}{16} = \tfrac{25}{16}$.
Taking the square root gives the direct-line distance.

$$d = \sqrt{1^2 + \left(\tfrac{3}{4}\right)^2} = \sqrt{\tfrac{25}{16}} = \tfrac{5}{4} = 1\tfrac{1}{4} \;\Rightarrow\; \textbf{(B)}$$

💡 On a right triangle, leg$^2$ + leg$^2$ = hypotenuse$^2$ — the Grade 8 Pythagorean theorem, applied here to legs $1$ and $\tfrac{3}{4}$.

[1] #1 6.NS.C.8 Sketch the walk. Put Bill's start at the top. South is down and east is to the r

[2] #1 5.NF.A.1 Close the triangle. Draw the straight segment from the start $(0,0)$ to the fini

[3] #1 8.G.B.7 Apply the Pythagorean theorem. With legs $1$ and $\tfrac{3}{4}$, the squared hyp

Review

Reasonableness: The straight-line distance must be less than the total walked, $\tfrac{1}{2} + \tfrac{3}{4} + \tfrac{1}{2} = 1\tfrac{3}{4}$ miles, and more than either leg of the triangle on its own, so it has to lie strictly between $1$ and $1\tfrac{3}{4}$. That rules out (A) $1$ and (E) $2$, and the diagram makes it clear the answer is only slightly more than $1$, which favors (B) $1\tfrac{1}{4}$ over (C) $1\tfrac{1}{2}$. The exact computation confirms (B). The numbers $3$-$4$-$5$ should also feel familiar: multiplying each by $\tfrac{1}{4}$ gives the triangle $\tfrac{3}{4}$-$1$-$\tfrac{5}{4}$, exactly what came out.

Alternative: Tool #9 (Try an Easier Problem): scale the walk up by $4$ to clear the fractions. The path becomes $2$ south, $3$ east, $2$ south, with legs $3$ and $4$ — a classic $3$-$4$-$5$ right triangle, hypotenuse $5$. Scaling back down by $4$ gives $\tfrac{5}{4} = 1\tfrac{1}{4}$, confirming (B).

CCSS standards used (min grade 8)

5.NF.A.1 Add and subtract fractions with unlike denominators (Combining the two south legs $\tfrac{1}{2} + \tfrac{1}{2} = 1$ so the path collapses to a single right triangle.)
6.NS.C.8 Solve real-world problems by graphing points in all four quadrants of the coordinate plane (Placing the start, the two turn points, and the finish on a coordinate grid so the right triangle is visible.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Computing the hypotenuse $\sqrt{1^2 + (\tfrac{3}{4})^2} = \tfrac{5}{4}$, the direct-line distance from start to finish.)

⭐ Two south legs separated by an east leg stack into one right triangle — and the start-to-finish line is just the hypotenuse, which here is the familiar $3$-$4$-$5$ triangle scaled down by $\tfrac{1}{4}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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