2026 AMC 8 Problem 19

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To help you master 2026 AMC 8 Problem 19, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

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2026 AMC8 Problem 19

Solution

Solution 1

Total Distance Ratio Method

Instead of calculating times and speeds, we use the fact that distance is proportional to speed when time is constant. We set up an equation relating the distances they cover.

When two objects travel for the same amount of time, the ratio of their distances equals the ratio of their speeds: $\frac{d _{1}}{d _{2}} = \frac{v _{1}}{v _{2}}$ .

Step 1: Define the variables

Let $D$ be the distance from the entrance to the tree, and $x$ be the distance Miguel walks from the entrance before they meet.

Step 2: Determine Luna’s distance based on speeds

Since both characters travel for the same amount of time and Luna runs $5$ times as fast as Miguel, her total distance covered must be $5$ times Miguel’s distance.

d_{Luna} = 5 x

Step 3: Express Luna’s distance based on her path

Luna runs from the entrance to the tree (a distance of $D$ ) and then turns back to meet Miguel. Since Miguel has walked a distance of $x$ from the entrance, Luna’s return distance from the tree is the remaining distance between the tree and Miguel. Her total distance is the sum of these two parts.

d_{Luna} = D + (D - x)

d_{Luna} = 2 D - x

Step 4: Equate and solve for Miguel’s distance

Set the two expressions for Luna’s total distance equal to each other and solve for $x$ as a fraction of $D$ .

5 x = 2 D - x

6 x = 2 D

x = \frac{1}{3} D

Correct answer: D. 1/3
The fraction of the distance is

1/3

, which corresponds to choice D. Check: If the distance to the tree is

30

meters, Miguel walks

10

meters. Luna runs to the tree (

30

meters) and back to Miguel (

20

meters) for a total of

50

meters. The ratio of their distances is

50/10 = 5

, perfectly matching the given speed ratio.

Solution 2

Stage-by-Stage Motion Method

We break the problem into two parts: first, what happens until Luna reaches the tree, and second, what happens as they travel toward each other.

When two objects move toward each other, the distance they close is shared proportionally to their speeds. The slower object covers a fraction of the gap equal to its speed divided by their combined speed.

Step 1: Normalize the total distance and analyze Stage 1

Let the distance to the tree be $1$ unit. In the first stage, Luna runs all the way to the tree. Because she runs $5$ times as fast as Miguel, he will only cover one-fifth of the total distance in that same time.

d_{Miguel, Stage 1} = \frac{1}{5}

Step 2: Find the remaining distance between them

When Luna reaches the tree, she turns around. At this moment, the distance separating them is the total distance minus the distance Miguel has already walked.

Remaining distance = 1 - \frac{1}{5} = \frac{4}{5}

Step 3: Analyze Stage 2 as a meeting problem

They now move toward each other. Since their speeds are in a $1 : 5$ ratio, their combined rate covers $6$ parts. Miguel’s share is one-sixth of any distance they close together. We calculate the extra distance Miguel walks before they meet.

d_{Miguel, Stage 2} = \frac{1}{6} \times \frac{4}{5} = \frac{4}{30} = \frac{2}{15}

Step 4: Calculate Miguel’s total distance

Add the distance Miguel walked in Stage 1 to the distance he walked in Stage 2 to find his total fraction of the path.

Total distance = \frac{1}{5} + \frac{2}{15} = \frac{3}{15} + \frac{2}{15} = \frac{1}{3}

Correct answer: D. 1/3
The fraction is

1/3

, matching choice D. Check: This stage-by-stage calculation yields the exact same result as our single-equation distance ratio method, confirming the logic.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Constant Time Distance Ratio:

\frac{d _{1}}{d _{2}} = \frac{v _{1}}{v _{2}}

$d_{1}$ and $d_{2}$ are distances traveled, $v_{1}$ and $v_{2}$ are speeds. This applies when the travel times are identical. Miguel and Luna start moving at the same time and meet at the same time, making their total elapsed travel time identical.

Meeting Problem Principle:

When two objects move toward each other, the distance they close is shared proportionally to their speeds.

If objects move toward each other with speeds $v_{1}$ and $v_{2}$ , the first object covers $\frac{v _{1}}{v _{1} + v _{2}}$ of the initial gap. In Stage 2 of the second method, Luna and Miguel are moving toward each other, allowing us to find Miguel's traveled distance without explicitly computing the time it takes them to meet.

Key Takeaway: Remember This Strategy for Next Time

When objects start moving at the same time and stop at the same time.

Try: Try setting up an equation using the ratio of their total distances rather than finding individual times.

Why: Time is constant for both objects, so distance is strictly proportional to speed, which allows you to bypass intermediate time calculations entirely.

When an object changes direction during a motion problem.

Try: You can either model the path as a single total distance equation or break the problem into one-way stages.

Why: Both representations are physically equivalent, but accumulating total distance (Method 1) is often algebraically faster than keeping track of separated stages.

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2026 AMC 8 Problem 19

In This Page:

2026 AMC8 Problem 19