AMC 10 Step by Step

2002 AMC 10A Problem 20

P16-20~3 minPrint

Points A,B,C,D,EA,B,C,D,E and FF lie, in that order, on AF‾\overline{AF} , dividing it into five segments, each of length 1. Point GG is not on line AFAF . Point HH lies on GD‾\overline{GD} , and point JJ lies on GF‾\overline{GF} . The line segments HC‾,JE‾,\overline{HC}, \overline{JE}, and AG‾\overline{AG} are parallel. Find HC/JEHC/JE .

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Key insight

HC and JE are both parallel to AG, so each cuts off a small copy of a bigger triangle: HC is one third of AG, JE one fifth.

Solution

Draw it: A,B,C,D,E,FA, B, C, D, E, F on a horizontal line, one unit apart, and GG somewhere above. Draw GAGA, GDGD and GFGF. Both unknown segments are parallel to AGAG, so compare each of them to AGAG.

  1. Segment HCHC. It goes from CC (on ADAD) up to HH (on GDGD), parallel to AGAG. So it cuts the big triangle DAGDAG into a small triangle DCHDCH at the corner DD.
  1. Why the small triangle is a shrunken copy. Triangles DCHDCH and DAGDAG share the angle at DD, and because CHCH is parallel to AGAG, the angles at CC and AA are equal. Same angles means same shape, so every side of DCHDCH is the same fraction of the matching side of DAGDAG.

3. Which fraction? Measure along the base from DD: DC=1DC = 1 and DA=3DA = 3. So
HC=13×AG.HC = \frac{1}{3} \times AG.

4. Segment JEJE. In the same way, JEJE is parallel to AGAG and cuts triangle FAGFAG into the small triangle FEJFEJ at the corner FF. Along the base from FF: FE=1FE = 1 and FA=5FA = 5. So
JE=15×AG.JE = \frac{1}{5} \times AG.

5. Compare. To make it concrete, say AG=15AG = 15. Then HC=5HC = 5 and JE=3JE = 3, so
HCJE=53.\frac{HC}{JE} = \frac{5}{3}.
Any other length for AGAG gives the same ratio, because both segments are fractions of the same AGAG.

The answer is (D) 5/3\boxed{\textbf{(D)}\ 5/3}.

Why this works

A segment drawn parallel to one side of a triangle cuts off a smaller copy of that triangle, and the shrink factor can be read off along the side you know. Here both unknown segments are compared with the same segment AGAG, so their ratio is just the ratio of the two shrink factors. Each factor must be measured from its own corner: DD for HCHC and FF for JEJE.

Alternative approach

Use coordinates in a special case. The ratio cannot depend on where GG is, so put AA at (0,0)(0, 0), FF at (5,0)(5, 0) and GG at (0,1)(0, 1), which makes AGAG vertical. Then HCHC and JEJE are vertical too. The line from GG down to D=(3,0)D = (3, 0) has height 1−x31 - \frac{x}{3} at position xx, so at CC (x=2x = 2) the height is HC=13HC = \frac13. The line from GG to F=(5,0)F = (5, 0) has height 1−x51 - \frac{x}{5}, so at EE (x=4x = 4) the height is JE=15JE = \frac15. The ratio is 1/31/5=53\frac{1/3}{1/5} = \frac{5}{3}.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Sketch AA through FF on a horizontal line one unit apart, with GG above, and draw GDGD and GFGF.

Segment HCHC is parallel to AGAG and joins a point of GDGD to a point of ADAD, so it cuts triangle DAGDAG into a smaller similar triangle DCHDCH with the ratio of similarity
HCAG=DCDA=13.\frac{HC}{AG} = \frac{DC}{DA} = \frac{1}{3}.

Likewise JE∥AGJE \parallel AG cuts triangle FAGFAG into the similar triangle FEJFEJ, with
JEAG=FEFA=15.\frac{JE}{AG} = \frac{FE}{FA} = \frac{1}{5}.

Dividing the two relations eliminates AGAG:
HCJE=1/31/5=53.\frac{HC}{JE} = \frac{1/3}{1/5} = \frac{5}{3}.

The answer is (D) 5/3\boxed{\textbf{(D)}\ 5/3}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.