AMC 10 · 2007 · #6

Grade 4 geometry-2d
Archetype Meeting Time with Piecewise Rates └ Sub-archetype Constant-Rate Closing Speed
perimeterratepath-length-comparison identify-subproblemsphysical-representation ↑ Prerequisites: perimeterrate
📏 Medium solution 💡 2 insights
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Problem
Two bugs leave the same corner of a triangle at equal speed, crawling opposite ways around the border. They meet somewhere on the far side. Find how far the meeting point is from one corner.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
How to solve
Strategy Draw a Diagram

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.

1STEP 1

Draw it and trace both routes

The perimeter is 18, the whole track.

5 + 6 + 7 = 18
2STEP 2

Straighten the border into a lap

Straightening it out turns corners into markers.

A = 0, B = 5, C = 11, A = 18
3STEP 3

Meet when the laps sum to 18

Equal speeds mean each covers exactly half the lap.

t + t = 18 → t = 9 = 18/2
4STEP 4

Walk 9 units and see where

Walking that far lands 4 past the first corner.

9 - AB = 9 - 5 = 4, 0 ＜ 4 ＜ 6 = BC
5STEP 5

Confirm the second bug agrees

The other bug agrees on the same point, so the answer is 4, choice (B).

9 - AC = 9 - 7 = 2, BD = BC - 2 = 6 - 2 = 4
Answer
4
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 = 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.
💡Key takeaway

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.

  • Draw it and trace both routes
  • Straighten the border into a lap
  • Meet when the laps sum to 18
  • Walk 9 units and see where
  • Confirm the second bug agrees

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