Sensim Math Original · sm-7

SM Original Grade 2 geometry-2d

Inspired by AMC 8 2024 #14

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

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

📏 Medium solution 💡 4 insights

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Problem

At Cedar Bluff Summer Camp, a counselor named Hana delivers a packet of mail from the Headquarters cabin $H$ to the farthest cabin, Riverside $R$ . Six cabins sit on the hillside — $H$ , Pinewood $P$ , Lakeview $L$ , Birch $B$ , Maple $M$ , and Riverside $R$ — and Hana is only allowed to walk downhill, so each trail is one-way in the direction listed. The trail distances, measured in meters, are:

$H \to P$ : $4$ m
$H \to L$ : $9$ m
$P \to L$ : $4$ m
$P \to B$ : $11$ m
$L \to B$ : $6$ m
$L \to M$ : $12$ m
$B \to M$ : $5$ m
$B \to R$ : $18$ m
$M \to R$ : $7$ m

Walking only along the trails in their allowed direction, what is the shortest total distance, in meters, that Hana can cover going from $H$ to $R$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A summer-camp counselor, Hana, must carry mail from cabin $H$ to cabin $R$ along nine one-way (downhill-only) trails connecting six cabins $H, P, L, B, M, R$. The trail lengths in meters are fixed and listed in the problem. We must find the smallest total length of any allowed $H \to R$ trail sequence.

Givens: Six cabins: $H, P, L, B, M, R$; Nine directed trails with lengths (meters): $H\to P = 4$, $H\to L = 9$, $P\to L = 4$, $P\to B = 11$, $L\to B = 6$, $L\to M = 12$, $B\to M = 5$, $B\to R = 18$, $M\to R = 7$; Hana may only walk in the listed direction (downhill); Five answer choices: (A) 24, (B) 25, (C) 26, (D) 27, (E) 28

Unknowns: The minimum total distance, in meters, of any allowed sequence of trails from $H$ to $R$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #2 Make a Systematic List, #3 Eliminate Possibilities

The trails form a small map, so first use Tool #1 to sketch the cabins and arrows so we can see which trails leave $H$ and which trails enter $R$. Because every path from $H$ to $R$ must pass through one of the intermediate cabins $P, L, B, M$, we use Tool #7 to break the big question into the easier subproblems "shortest distance from $H$ to each intermediate cabin," growing the partial answers outward layer by layer. Tool #2 (systematic list) handles the final hop into $R$ — there are exactly two trails entering $R$, so we list both options and add. Tool #3 then matches the minimum to one of the five offered choices.

Execute — Answer: C

#1 Draw a Diagram K.G.A.1 Step 1

Sketch the cabins and arrows.
Only two trails leave $H$ directly: $H\to P = 4$ and $H\to L = 9$.
Only two trails enter $R$: $B\to R = 18$ and $M\to R = 7$.
So every valid route must start with $H\to P$ or $H\to L$ and finish with $B\to R$ or $M\to R$.
This framing is the whole solution's backbone.

$$H \xrightarrow{4} P,\ H \xrightarrow{9} L,\quad B \xrightarrow{18} R,\ M \xrightarrow{7} R$$

💡 Reading where each arrow starts and ends is the same kind of left-right-above-below position vocabulary kids learn in Kindergarten geometry.

#7 Identify Subproblems 2.NBT.B.5 Step 2

Break the journey into small pieces.
First find the shortest distance from $H$ to the two nearest cabins, $P$ and $L$.
There is only one way into $P$, so the shortest distance is $H\to P = 4$.
There are two ways into $L$: directly $H\to L = 9$, or via $P$ as $H\to P\to L = 4 + 4 = 8$.
Since $8 < 9$, the shortest distance from $H$ to $L$ is $8$.

$$d(P) = 4.\quad d(L) = \min(9,\ 4 + 4) = \min(9, 8) = 8$$

💡 Adding $4 + 4 = 8$ and comparing $8$ with $9$ is exactly Grade 2 fluent addition and comparison inside 100.

#7 Identify Subproblems 2.OA.A.1 Step 3

Push outward one more layer to $B$ and $M$.
Cabin $B$ is reached by $P\to B = 11$ or $L\to B = 6$, giving candidates $d(P) + 11 = 4 + 11 = 15$ and $d(L) + 6 = 8 + 6 = 14$, so $d(B) = 14$.
Cabin $M$ is reached by $L\to M = 12$ or $B\to M = 5$, giving $d(L) + 12 = 8 + 12 = 20$ and $d(B) + 5 = 14 + 5 = 19$, so $d(M) = 19$.
At each cabin we keep only the shorter running total.

$$d(B) = \min(4+11,\ 8+6) = \min(15, 14) = 14.\quad d(M) = \min(8+12,\ 14+5) = \min(20, 19) = 19$$

💡 Carrying two small running totals forward and keeping the smaller one is exactly the Grade 2 two-step addition word-problem move.

#2 Make a Systematic List 2.NBT.B.5 Step 4

Now use a systematic list for the final hop into $R$.
Only two trails enter $R$, so we list both: arriving via $B$ gives $d(B) + 18 = 14 + 18 = 32$, and arriving via $M$ gives $d(M) + 7 = 19 + 7 = 26$.
The smaller of $32$ and $26$ is $26$, so the shortest distance from $H$ to $R$ is $26$ m, achieved by the route $H \to P \to L \to B \to M \to R$.

$$d(R) = \min(d(B)+18,\ d(M)+7) = \min(32, 26) = 26$$

💡 Listing the last two ways into the goal and adding two two-digit numbers ($14+18$ and $19+7$) is Grade 2 fluent addition within 100.

#3 Eliminate Possibilities 1.NBT.B.3 Step 5

Compare $26$ to the five offered choices $24, 25, 26, 27, 28$.
Our value $26$ is exactly choice (C).
The neighbours $27$ and $28$ are real "near-miss" route totals — for example $H\to P\to L\to M\to R = 4+4+12+7 = 27$ matches (D), and $H\to L\to M\to R = 9+12+7 = 28$ matches (E) — but neither beats $26$.
The smaller candidates $24$ and $25$ are below the sum of the five edges along the optimal route, so they are impossible.

$$d(R) = 26 \;\Rightarrow\; \textbf{(C)}$$

💡 Comparing a handful of two-digit numbers and picking the smallest one is exactly the Grade 1 two-digit comparison skill.

[1] #1 K.G.A.1 Sketch the cabins and arrows. Only two trails leave $H$ directly: $H\to P = 4$ a

[2] #7 2.NBT.B.5 Break the journey into small pieces. First find the shortest distance from $H$ t

[3] #7 2.OA.A.1 Push outward one more layer to $B$ and $M$. Cabin $B$ is reached by $P\to B = 11

[4] #2 2.NBT.B.5 Now use a systematic list for the final hop into $R$. Only two trails enter $R$,

[5] #3 1.NBT.B.3 Compare $26$ to the five offered choices $24, 25, 26, 27, 28$. Our value $26$ is

Review

Reasonableness: Re-add the optimal route $H \to P \to L \to B \to M \to R$ in one shot: $4 + 4 + 6 + 5 + 7 = 26$, matching the layered totals exactly. The route uses the cheapest trail out of $H$ ($H\to P = 4$) and the cheapest trail into $R$ ($M\to R = 7$), and the three middle hops $4, 6, 5$ are each the small option at their respective cabin — so $26$ also looks right by feel. The two next-best totals ($27$ from $H\to L\to B\to M\to R = 9+6+5+7$, and $28$ from $H\to L\to M\to R = 9+12+7$) match choices (D) and (E), which confirms (C) is the only minimum.

Alternative: An alternative is to use Tool #2 alone and enumerate **every** simple $H\to R$ route from scratch. Counting by hand gives seven such routes — $H\!\to\! P\!\to\! L\!\to\! B\!\to\! M\!\to\! R$, $H\!\to\! P\!\to\! L\!\to\! B\!\to\! R$, $H\!\to\! P\!\to\! L\!\to\! M\!\to\! R$, $H\!\to\! P\!\to\! B\!\to\! M\!\to\! R$, $H\!\to\! P\!\to\! B\!\to\! R$, $H\!\to\! L\!\to\! B\!\to\! M\!\to\! R$, $H\!\to\! L\!\to\! B\!\to\! R$, $H\!\to\! L\!\to\! M\!\to\! R$ — and the smallest total is again $26$, but the subproblem approach we used reaches the same answer with far fewer additions because the partial sums $d(P), d(L), d(B), d(M)$ are reused.

CCSS standards used (min grade 2)

K.G.A.1 Describe positions of objects using above, below, beside, in front of (Reading the cabin map to see which trails leave $H$ and which trails enter $R$.)
1.NBT.B.3 Compare two two-digit numbers using symbols (Comparing the candidate route totals $26, 27, 28$ to pick the smallest and match it to a choice.)
2.NBT.B.5 Fluently add and subtract within 100 (Two-digit sums like $4+4=8$, $14+18=32$, and $19+7=26$ used to build and compare partial distances.)
2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Accumulating trail distances in two steps ($4+11=15$ vs. $8+6=14$, $8+12=20$ vs. $14+5=19$) and keeping the shorter running total at each cabin.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 2)

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