AMC 8 · 2005 · #7
Grade 8 geometry-2dPick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Sketch it with south as down, east as right: the two south legs share one vertical line with the east step between them.
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.
6.NS.C.8Draw A DiagramDraw the start-to-finish segment: the south legs add to 1 and the east leg is , giving a right triangle with that segment as hypotenuse.
Adding + = 1 to merge the two south legs is the Grade 5 "add fractions" step that the picture asks for.
5.NF.A.1Draw A DiagramBy the Pythagorean theorem on legs 1 and , the hypotenuse is the square root of — the 3-4-5 triangle scaled by .
On a right triangle, leg² + leg² = hypotenuse² — the Grade 8 Pythagorean theorem, applied here to legs 1 and .
8.G.B.7Draw A DiagramTwo 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 .