AMC 10 · 2002 · #20

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
similar-trianglesratio-proportion identify-subproblems ↑ Prerequisites: similar-triangles
📏 Medium solution 💡 2 insights
Problem
Six points A,B,C,D,E,F sit in order on segment AF, splitting it into five pieces each of length 1. A point G lies off the line. H is a point on GD and J is a point on GF, chosen so that the three segments HC, JE, and AG are all parallel. Find the ratio HC/JE.

Pick an answer.

(A)
5/4
(B)
4/3
(C)
3/2
(D)
5/3
(E)
2

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Nothing in this problem can be seen without a picture, so tool #1 (Draw a Diagram) comes first: put the six points on a line, mark G above it, and draw AG plus the two parallels HC and JE. Once drawn, the figure splits into two look-alike triangle pairs, so tool #7 (Identify Subproblems) handles them one at a time: HC sits inside triangle GAD and JE sits inside triangle GAF. A segment parallel to a side of a triangle cuts off a smaller similar triangle, which turns each unknown length into a simple fraction of the shared segment AG. Because AG never gets a number, tool #4 (Introduce a Variable) lets us carry it as a symbol; when we finally form HC/JE it cancels, so its actual length never matters.

1STEP 1

Draw the figure and read the distances

Put A through F at 0,1,2,3,4,5 with G off the line; the spacing hands you DC=1, DA=3, FE=1, FA=5.

A,B,C,D,E,F = 0,1,2,3,4,5; DC=1, DA=3, FE=1, FA=5
2STEP 2

First parallel gives a similar triangle

Since HC ∥ AG, triangle DHC is a scaled copy of DGA, so HC/AG=DC/DA=1/3 — that is, HC=1/3 AG.

△ DHC ∼ △ DGA → HC/AG=DC/DA=1/3 → HC=1/3AG
3STEP 3

Second parallel gives another similar triangle

The same cut in the bigger triangle GAF gives JE/AG=FE/FA=1/5, so JE=1/5 AG.

△ FJE ∼ △ FGA → JE/AG=FE/FA=1/5 → JE=1/5AG
4STEP 4

Divide the two lengths

Divide: HC/JE=(1/3 AG)/(1/5 AG)=5/3 — AG cancels, so where G sits never matters. Choice (D).

HC/JE=1/3AG/1/5AG=1/3·5/1=5/3 → (D)
Answer
5/3
The two triangles share vertex-line AG, and HC comes from the base ratio 1/3 while JE comes from 1/5. Since 1/3 ＞ 1/5, HC should be the longer segment, so HC/JE must be greater than 1 — which rules out nothing yet but confirms the ratio is ＞ 1. Its value 5/3≈1.67 is exactly (1/3)÷(1/5)=FA/DA·(DC/FE)=5/3; equivalently HC/JE=(DC/DA)/(FE/FA)=FA/DA=5/3 because DC=FE=1. A quick coordinate check with, say, G=(0,3) gives C=(2,0),D=(3,0),F=(5,0); the line GD is y=-x+3 scaled, and the point H with HC ∥ AG (vertical) sits at x=2, giving H=(2,1) so HC=1; similarly J=(4,3/5) so JE=3/5, and HC/JE=1/(3/5)=5/3. Everything agrees, and 5/3 is choice (D).
💡Key takeaway

A line drawn parallel to a triangle's side makes a shrunken copy, so each length is just a fraction of AG — and dividing the two fractions makes AG vanish, leaving 5/3.

  • Draw the figure and read the distances
  • First parallel gives a similar triangle
  • Second parallel gives another similar triangle
  • Divide the two lengths

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