AMC 10 · 2002 · #20
Grade 8 geometry-2dPoints A,B,C,D,E and F lie, in that order, on AF, dividing it into five segments, each of length 1. Point G is not on line AF. Point H lies on GD, and point J lies on GF. The line segments HC,JE, and AG are parallel. Find HC/JE.
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Six points $A,B,C,D,E,F$ sit in order on segment $\overline{AF}$, splitting it into five pieces each of length $1$. A point $G$ lies off the line. $H$ is a point on $\overline{GD}$ and $J$ is a point on $\overline{GF}$, chosen so that the three segments $\overline{HC}$, $\overline{JE}$, and $\overline{AG}$ are all parallel. Find the ratio $HC/JE$.
Givens: $A,B,C,D,E,F$ lie in order on $\overline{AF}$; The five pieces $AB,BC,CD,DE,EF$ each have length $1$; $G$ is not on line $AF$; $H$ is on $\overline{GD}$ and $J$ is on $\overline{GF}$; $\overline{HC}\parallel\overline{JE}\parallel\overline{AG}$; Answer choices: (A) $5/4$, (B) $4/3$, (C) $3/2$, (D) $5/3$, (E) $2$
Unknowns: The ratio $HC/JE$
Understand
Restated: Six points $A,B,C,D,E,F$ sit in order on segment $\overline{AF}$, splitting it into five pieces each of length $1$. A point $G$ lies off the line. $H$ is a point on $\overline{GD}$ and $J$ is a point on $\overline{GF}$, chosen so that the three segments $\overline{HC}$, $\overline{JE}$, and $\overline{AG}$ are all parallel. Find the ratio $HC/JE$.
Givens: $A,B,C,D,E,F$ lie in order on $\overline{AF}$; The five pieces $AB,BC,CD,DE,EF$ each have length $1$; $G$ is not on line $AF$; $H$ is on $\overline{GD}$ and $J$ is on $\overline{GF}$; $\overline{HC}\parallel\overline{JE}\parallel\overline{AG}$; Answer choices: (A) $5/4$, (B) $4/3$, (C) $3/2$, (D) $5/3$, (E) $2$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
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.
Execute — Answer: D
6.RP.A.1 Step 1 Draw the figure and read the distances
- Put the points on a line with each gap equal to $1$: let $A,B,C,D,E,F$ sit at positions $0,1,2,3,4,5$.
- Mark $G$ somewhere off the line and draw $\overline{AG}$.
- Draw $\overline{HC}$ from $C$ up to $H$ on $\overline{GD}$, and $\overline{JE}$ from $E$ up to $J$ on $\overline{GF}$, both parallel to $\overline{AG}$.
- Now read the along-the-line distances straight off the spacing: from $D$ (at $3$) the piece to $C$ (at $2$) is $DC=1$ while the whole stretch to $A$ (at $0$) is $DA=3$; from $F$ (at $5$) the piece to $E$ (at $4$) is $FE=1$ while the whole stretch to $A$ is $FA=5$.
- So the two ratios that will matter are $DC:DA=1:3$ and $FE:FA=1:5$.
💡 Equal unit gaps let you count distances right off the line instead of measuring.
8.G.A.5 Step 2 First parallel gives a similar triangle
- Look at triangle $GAD$: its vertices are $G$, $A$, and $D$, and $C$ lies on side $AD$ while $H$ lies on side $GD$.
- Since $\overline{HC}\parallel\overline{AG}$, the parallel lines make equal corresponding angles at $D$ and along the sides, so triangle $DHC$ has the same shape as triangle $DGA$ (angle-angle).
- In similar triangles matching sides are in the same ratio, and $HC$ matches $AG$, so $HC/AG$ equals the ratio of the sides they sit on, $DC/DA=1/3$.
- Writing $AG$ as a symbol, $HC=\tfrac{1}{3}\,AG$.
💡 A cut parallel to one side of a triangle makes a shrunk copy, so its lengths scale by the base ratio.
8.G.A.5 Step 3 Second parallel gives another similar triangle
- Do the same with the bigger triangle $GAF$: here $E$ lies on side $AF$ and $J$ lies on side $GF$.
- Since $\overline{JE}\parallel\overline{AG}$, triangle $FJE$ has the same shape as triangle $FGA$ (again angle-angle from the parallel lines).
- The side $JE$ matches $AG$, so $JE/AG$ equals the base ratio $FE/FA=1/5$.
- Carrying the same symbol $AG$, $JE=\tfrac{1}{5}\,AG$.
💡 The same parallel-cut trick works on the larger triangle, just with its own base ratio.
7.RP.A.2 Step 4 Divide the two lengths
- Both lengths are now written with the same symbol $AG$, so form the asked ratio and let $AG$ cancel: $HC/JE=\left(\tfrac{1}{3}AG\right)\big/\left(\tfrac{1}{5}AG\right)=\tfrac{1}{3}\cdot\tfrac{5}{1}=\tfrac{5}{3}$.
- The unknown length $AG$ divides out, so the answer does not depend on where $G$ is.
- The ratio $HC/JE=5/3$, which is choice (D).
💡 When both quantities carry the same unknown factor, dividing wipes it out and leaves a pure number.
6.RP.A.1 Put the points on a line with each gap equal to $1$: let $A,B,C,D,E,F$ sit at po 8.G.A.5 Look at triangle $GAD$: its vertices are $G$, $A$, and $D$, and $C$ lies on side 8.G.A.5 Do the same with the bigger triangle $GAF$: here $E$ lies on side $AF$ and $J$ l 7.RP.A.2 Both lengths are now written with the same symbol $AG$, so form the asked ratio Review
Reasonableness: 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 $5/3\approx1.67$ is exactly $(1/3)\div(1/5)=FA/DA\cdot(DC/FE)=5/3$; equivalently $HC/JE=\frac{DC/DA}{FE/FA}=\frac{FA}{DA}=\frac{5}{3}$ 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 $HC\parallel AG$ (vertical) sits at $x=2$, giving $H=(2,1)$ so $HC=1$; similarly $J=(4,\tfrac{3}{5})$ so $JE=\tfrac{3}{5}$, and $HC/JE=1/(3/5)=5/3$. Everything agrees, and $5/3$ is choice (D).
Alternative: Skip the symbol and pick a convenient position for $G$, say $G=(0,3)$, then compute $H$ and $J$ directly from the parallel condition as in the check above: $HC=1$ and $JE=3/5$, so $HC/JE=5/3$. Because both similar-triangle ratios are independent of $G$, any choice of $G$ gives the same $5/3$, which is why fixing coordinates is legal here.
CCSS standards used (min grade 8)
6.RP.A.1Understand the concept of a ratio and use ratio language to describe a relationship between two quantities (Reading the base ratios $DC:DA=1:3$ and $FE:FA=1:5$ off the equally spaced line.)8.G.A.5Use informal arguments to establish facts about the angle relationships in figures, including the angle-angle criterion for similarity of triangles (Using $HC\parallel AG$ and $JE\parallel AG$ to get $\triangle DHC\sim\triangle DGA$ and $\triangle FJE\sim\triangle FGA$ by angle-angle.)7.RP.A.2Recognize and represent proportional relationships between quantities (Turning similarity into the proportions $HC=\tfrac13 AG$, $JE=\tfrac15 AG$ and dividing to get $HC/JE=5/3$.)
⭐ 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 $5/3$.
⭐ 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 $5/3$.
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