AMC 10 · 2007 · #6
Easy mode Grade 4A triangle ABC has sides AB=5, BC=6, and AC=7. Two bugs start at corner A at the same moment. One crawls toward B, and the other crawls toward C. Both stay on the edges of the triangle and crawl at the same speed, so they travel around it in opposite directions until they meet at a point D. How far is D from B?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A triangle has sides $AB = 5$, $BC = 6$, and $AC = 7$. Two bugs leave $A$ at the same moment and crawl around the outside of the triangle at the same speed, one going toward $B$ and the other toward $C$. They keep going until they run into each other at a point $D$. Find the distance from $B$ to $D$.
Givens: Triangle $ABC$ has $AB = 5$, $BC = 6$, and $AC = 7$.; Both bugs start at $A$ at the same time.; The bugs travel along the sides of the triangle, in opposite directions around it.; The bugs move at the same speed.
Unknowns: The length $BD$, where $D$ is the point where the two bugs meet
Understand
Restated: A triangle has sides $AB = 5$, $BC = 6$, and $AC = 7$. Two bugs leave $A$ at the same moment and crawl around the outside of the triangle at the same speed, one going toward $B$ and the other toward $C$. They keep going until they run into each other at a point $D$. Find the distance from $B$ to $D$.
Givens: Triangle $ABC$ has $AB = 5$, $BC = 6$, and $AC = 7$.; Both bugs start at $A$ at the same time.; The bugs travel along the sides of the triangle, in opposite directions around it.; The bugs move at the same speed.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #15 Organize Information in More Ways, #8 Analyze the Units, #7 Identify Subproblems
The triangle's shape does not matter here — nothing is asked about angles or area, only about distance crawled along the edge. So draw the triangle, then re-picture its border as a closed track of known total length. On a track, two walkers leaving the same spot in opposite directions at the same speed meet exactly when their two distances add up to one full lap. That turns the whole problem into two small subproblems: find the lap length, then walk half of it and see where you land. A final check with the second bug confirms both bugs really are standing on the same point.
Execute — Answer: D
3.MD.D.8 Step 1 Draw it and trace both routes
- Draw triangle $ABC$ and label $AB = 5$, $BC = 6$, $CA = 7$.
- One bug leaves $A$ heading for $B$, so its route is $A \to B \to C$.
- The other leaves $A$ heading for $C$, so its route is $A \to C \to B$.
- The two routes run around the same closed border in opposite directions.
- The total length of that border is the perimeter: $5 + 6 + 7 = 18$.
💡 The triangle is really just a loop of string 18 units long, with $A$, $B$, $C$ marked on it.
4.MD.A.2 Step 2 Straighten the border into a lap
- Cut the loop open at $A$ and lay it flat as a track of length $18$: start at $A$, mark $B$ at $5$, mark $C$ at $5 + 6 = 11$, and the far end at $18$ is $A$ again.
- The first bug moves forward along this track.
- The second bug moves backward from $18$.
- Because the speeds are equal and the clocks are the same, after crawling $t$ units the first bug sits at mark $t$ and the second at mark $18 - t$.
💡 Bending a shape does not change how far you walk along its edge, so the border can be treated as a straight ruler.
4.MD.A.2 Step 3 Meet when the laps sum to 18
- The gap between the bugs, measured the long way around from the first bug forward to the second, starts as the whole lap of $18$ and shrinks by $2$ units for every $1$ unit each bug crawls, since both are eating into it at once.
- So the gap closes exactly when the two distances together make one full lap: $t + t = 18$, giving $t = 9$.
- Any earlier time has $2t < 18$, so the loop is not yet closed and they are still apart.
- Each bug therefore crawls $9$ units, which is half the perimeter.
💡 Two walkers closing a loop from opposite sides split it fairly, so each covers half.
4.OA.A.3 Step 4 Walk 9 units and see where
- Follow the first bug $9$ units along $A \to B \to C$.
- The first $5$ units use up side $AB$ and land it exactly on $B$.
- That leaves $9 - 5 = 4$ units to crawl along $BC$.
- Since $4$ is less than $BC = 6$, the bug is still on segment $BC$ and has not reached $C$.
- So the meeting point $D$ lies on $BC$, at distance $4$ from $B$.
💡 Spend the walk one side at a time; whatever is left over after finishing a side is how far you push into the next one.
4.MD.A.2 Step 5 Confirm the second bug agrees
- The second bug must be standing on that same spot, so check it.
- Going $A \to C$ uses $7$ units, leaving $9 - 7 = 2$ units back along $CB$.
- A point $2$ from $C$ on side $BC$ is $6 - 2 = 4$ from $B$ — the very same point the first bug reached.
- Both bugs are on $BC$, four units from $B$, so the meeting point really exists and $BD = 4$, which is answer (D).
💡 Two different routes to the same spot is the proof that the spot is real, not just assumed.
3.MD.D.8 Draw triangle $ABC$ and label $AB = 5$, $BC = 6$, $CA = 7$. One bug leaves $A$ h 4.MD.A.2 Cut the loop open at $A$ and lay it flat as a track of length $18$: start at $A$ 4.MD.A.2 The gap between the bugs, measured the long way around from the first bug forwar 4.OA.A.3 Follow the first bug $9$ units along $A \to B \to C$. The first $5$ units use up 4.MD.A.2 The second bug must be standing on that same spot, so check it. Going $A \to C$ Review
Reasonableness: Three checks agree. First, $BD = 4$ is smaller than $BC = 6$, so $D$ genuinely sits on side $BC$ — an answer bigger than $6$ would have been impossible. Second, both trips measure the same: $AB + BD = 5 + 4 = 9$ and $AC + CD = 7 + 2 = 9$, exactly the equal distances that equal speeds force. Third, $9 + 9 = 18$ is the whole perimeter, used up once with nothing double-counted. Notice the numbers $5$, $6$, $7$ matter only through their sum and through $AB$: the rule is $BD = \frac{\text{perimeter}}{2} - AB = 9 - 5 = 4$, so the exact shape of the triangle is irrelevant. One trap worth naming: if the bugs kept crawling past $D$, they would meet again after $18$ units each — back at $A$, where the distance to $B$ is $5$, which is choice (E). Choice (E) is the answer to the wrong meeting, so it matters that $D$ is the first one.
Alternative: Skip the track picture and use two equations on the pieces of $BC$. Once you know $D$ lies on $BC$ (because $AB = 5 < 9 < 11 = AB + BC$), the equal-distance condition reads $AB + BD = AC + CD$, that is $5 + BD = 7 + CD$, so $BD - CD = 2$. The two pieces also make up the whole side: $BD + CD = 6$. A sum of $6$ and a difference of $2$ forces $BD = 4$ and $CD = 2$. Same answer, reached without ever mentioning the perimeter.
CCSS standards used (min grade 4)
3.MD.D.8Solve real-world problems involving perimeters of polygons (Adding the three sides to get the length of the closed path, $5 + 6 + 7 = 18$.)4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Turning equal speeds and equal times into equal distances, splitting the lap as $t + t = 18$, and re-measuring the meeting point from the second bug's side.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Spending the $9$-unit walk side by side, subtracting $9 - 5 = 4$, and comparing $4$ with $6$ to confirm the point lands on $BC$.)
⭐ Two crawlers leaving the same point in opposite directions around a loop meet once they have together covered exactly one lap, so each one has walked half the perimeter.
⭐ Two crawlers leaving the same point in opposite directions around a loop meet once they have together covered exactly one lap, so each one has walked half the perimeter.
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