Competition · AMC preparation · step 4 of 4

AMC 8 · 2024 · #14

Grade 2 geometry-2d

Archetype Path Length Comparison & Shortest Path └ Sub-archetype Shortest Route in Weighted Network

systematic-enumerationmulti-digit-arithmetic shortest-pathidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticmental-arithmetic

📏 Medium solution 💡 4 insights 📊 Diagram

📘 View easy version →

Problem

Six towns A, M, C, X, Y, Z are connected by one-way roads with given distances (km). Using only these roads (in the direction of the arrows), we want the shortest total distance from A to Z in km.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

This is a classic map problem about positions and connections, so the first move is Tool #1 — get the towns, arrows, and distances laid out in a clean picture. Then break the big question "A→ Z" into smaller subproblems: the shortest distance from A to each intermediate town (Tool #7). That lets us reuse partial sums instead of re-adding the same numbers. Finally, at the last step we list all routes into Z exhaustively (Tool #2) and pick the smallest, then check it against the answer choices with Tool #3.

1STEP 1

Draw the roads with their lengths

Organize the arrows: only A→X and A→M leave A, and only M→Z, Y→Z, C→Z enter Z, so every route ends with one final hop into Z.

A → X, A → M, M → Z, Y → Z, C → Z

2STEP 2

Find the nearest towns first

First layer: A→X = 5 (only one way in). For M, compare direct A→M = 8 with the detour A→X→M = 5+2 = 7, so the shortest to M is 7.

A→ X: 5. A→ M: min(8, 5+2) = min(8, 7) = 7

3STEP 3

Extend one layer further

Next layer, reusing those sums: A→Y = min(15, 13) = 13 and A→C = min(21, 18) = 18.

A→ Y: min(5+10, 7+6) = min(15, 13) = 13. A→ C: min(7+14, 13+5) = min(21, 18) = 18

4STEP 4

Compare the three last hops

Last hop into Z: compare via M (7+25 = 32), via Y (13+17 = 30), via C (18+10 = 28); the smallest is 28.

min(7+25, 13+17, 18+10) = min(32, 30, 28) = 28

Listing every way to reach Z and adding the two-digit pieces is a Grade 2 two-step addition exercise within 100.

The shortest distance from A to Z is the smallest of three route-totals: the shortest distance to M plus 25, the shortest distance to Y plus 17, and the shortest distance to C plus 10.

▸ Why?

Every route into Z must finish on exactly one of the three arrows that enter Z — from M, from Y, or from C — so all routes fall into three groups with none left out and none in two groups at once, and the overall shortest route lives in whichever group holds it.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

Inside one group every route ends on the very same single arrow into Z, so each route's length is its trip from A to that town plus one fixed arrow-length; since that last part is identical for the whole group, the route reaching the town in the least distance is exactly the one with the least total, which is why each group's shortest is its shortest-distance-to-the-town plus that fixed arrow.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

Each group leaves one town-total, and the overall answer is the smallest of the three; lining the three totals up and comparing them two at a time, if the least beats the middle and the middle beats the largest, then the least beats the largest too, so one value is confirmed below both others and is the shortest distance of every route.

Transitivity of orderIf the first is greater than the second and the second greater than the third, the first is greater than the third.

▸ Why?

The shortest distances to M, Y, and C plugged in here are the running totals we built up one town at a time, and reusing them is valid because adding the arrow-lengths in stages and regrouping them yields the same sum as adding the whole route's arrows at once.

Associativity of additionAdding three piles, it does not matter which two you push together first.

2.OA.A.1Make A Systematic List

5STEP 5

Match against the choices

Match 28 to the choices: it is exactly (A). The values 30 and 32 are the longer routes, and 29 and 31 never occur as a path total.

28 → (A)

Answer

Re-add the optimal route A → X → M → Y → C → Z in one shot: 5 + 2 + 6 + 5 + 10 = 28, matching the step-by-step accumulation exactly. The route starts on the cheapest arrow out of A (A→ X = 5), visits every intermediate town along the way, and ends on the cheapest arrow into Z (C→ Z = 10) — so 28 also feels right by size. The fact that C→ Z = 10 (the shortest single arrow into Z) appears in our winning route is a nice sanity check.

💡Key takeaway

This AMC 8 problem only needs Grade 2 addition within 100 and two-digit comparison you already know!

Draw the roads with their lengths
Find the nearest towns first
Extend one layer further
Compare the three last hops
Match against the choices

A parent dashboard for the family lives at sensimlab.com.

Draw the roads with their lengths

Find the nearest towns first

Extend one layer further

Compare the three last hops

Match against the choices

More like this