2002 AMC 10A Problems/Problem 20: Difference between revisions

Revision as of 19:30, 5 November 2015

Problem

Points $A,B,C,D,E$ and $F$ lie, in that order, on $\overline{AF}$ , dividing it into five segments, each of length 1. Point $G$ is not on line $AF$ . Point $H$ lies on $\overline{GD}$ , and point $J$ lies on $\overline{GF}$ . The line segments $\overline{HC}, \overline{JE},$ and $\overline{AG}$ are parallel. Find $HC/JE$ .

$\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2$

[asy] pair A,B,C,D,EE,F,G,H,J; A = (0,0); B = (0.2,0); C = 2*B; D = 3*B; EE = 4*B; F = 5*B; G = (-0.2,0.8); H = intersectionpoint(G--D,C -- (C + G)); J = intersectionpoint(G--F,EE--(EE+G)); draw(G--F--A--G--B); draw(H--C--G--D); draw(J--EE--G); label(" $A$ ",A,SW); label(" $B$ ",B,S); label(" $C$ ",C,S); label(" $D$ ",D,S); label(" $E$ ",EE,S); label(" $F$ ",F,SE); label(" $J$ ",J,NE); label(" $G$ ",G,N); label(scale(0.9)*" $H$ ",H,NE,UnFill(0.1mm)); [/asy]

Solution

Solution #1: Since $AG$ and $CH$ are parallel, triangles $GAD$ and $HCD$ are similar. Hence, $CH/AG = CD/AD = 1/3$ .

Since $AG$ and $JE$ are parallel, triangles $GAF$ and $JEF$ are similar. Hence, $EJ/AG = EF/AF = 1/5$ . Therefore, $CH/EJ = (CH/AG)/(EJ/AG) = (1/3)/(1/5) = \boxed{5/3}$ . The answer is (D).

Solution #2: As $\overline{JE}$ is parallel to $\overline{AG}$ , angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, $\triangle AGF \sim \triangle EJF$ ; hence $\frac {AG}{JE} =5$ . Similarly, $\frac {AG}{HC} = 3$ . Thus, $\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}$ .

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2002 AMC 10A Problems/Problem 20: Difference between revisions

Revision as of 19:30, 5 November 2015

Problem

Solution

See Also

@@ Line 3: / Line 3: @@
 <math>\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2</math>
+[asy]
+pair A,B,C,D,EE,F,G,H,J;
+A = (0,0);
+B = (0.2,0);
+C = 2*B;
+D = 3*B;
+EE = 4*B;
+F = 5*B;
+G = (-0.2,0.8);
+H = intersectionpoint(G--D,C -- (C + G));
+J = intersectionpoint(G--F,EE--(EE+G));
+draw(G--F--A--G--B);
+draw(H--C--G--D);
+draw(J--EE--G);
+label("<math>A</math>",A,SW);
+label("<math>B</math>",B,S);
+label("<math>C</math>",C,S);
+label("<math>D</math>",D,S);
+label("<math>E</math>",EE,S);
+label("<math>F</math>",F,SE);
+label("<math>J</math>",J,NE);
+label("<math>G</math>",G,N);
+label(scale(0.9)*"<math>H</math>",H,NE,UnFill(0.1mm));
+[/asy]
 ==Solution==