Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AMC 12 Problems/Problem 22"

(→‎See Also)
(→‎Problem)
Line 14: Line 14:
 
\text{(E) }\frac {35}{8}
 
\text{(E) }\frac {35}{8}
 
</math>
 
</math>
 +
 +
[[Category: Introductory Geometry Problems]]
  
 
== Solution ==
 
== Solution ==

Revision as of 10:36, 13 August 2014

Problem

In rectangle $ABCD$, points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$. Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$. The area of the rectangle $ABCD$ is $70$. Find the area of triangle $EHJ$.

$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$

Solution

[asy] unitsize(0.5cm); defaultpen(0.8); pair A=(0,0), B=(10,0), C=(10,7), D=(0,7), E=(C+D)/2, F=(2*A+B)/3, G=(A+2*B)/3; pair H = intersectionpoint(A--C,E--F); pair J = intersectionpoint(A--C,E--G); draw(A--B--C--D--cycle); draw(G--E--F); draw(A--C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,N); label("$F$",F,S); label("$G$",G,S); label("$H$",H,SE); label("$J$",J,ESE); filldraw(E--H--J--cycle,lightgray,black); draw(H--D, dashed); [/asy]


Solution 1

Note that the triangles $AFH$ and $CEH$ are similar, as they have the same angles. Hence $\frac {AH}{HC} = \frac{AF}{EC} = \frac 23$.

Also, triangles $AGJ$ and $CEJ$ are similar, hence $\frac {AJ}{JC} = \frac {AG}{EC} = \frac 43$.

We can now compute $[EHJ]$ as $[ACD]-[AHD]-[DEH]-[EJC]$. We have:

  • $[ACD]=\frac{[ABCD]}2 = 35$.
  • $[AHD]$ is $2/5$ of $[ACD]$, as these two triangles have the same base $AD$, and $AH$ is $2/5$ of $AC$, therefore also the height from $H$ onto $AD$ is $2/5$ of the height from $C$. Hence $[AHD]=14$.
  • $[HED]$ is $3/10$ of $[ACD]$, as the base $ED$ is $1/2$ of the base $CD$, and the height from $H$ is $3/5$ of the height from $A$. Hence $[HED]=\frac {21}2$.
  • $[JEC]$ is $3/14$ of $[ACD]$ for similar reasons, hence $[JEC]=\frac{15}2$.

Therefore $[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-\frac {21}2-\frac{15}2 = \boxed{3}$.

Solution 2

As in the previous solution, we note the similar triangles and prove that $H$ is in $2/5$ and $J$ in $4/7$ of $AC$.

We can then compute that $HJ = AC \cdot \left( \frac 47 - \frac 25 \right) = AC \cdot \frac{6}{35}$.

As $E$ is the midpoint of $CD$, the height from $E$ onto $AC$ is $1/2$ of the height from $D$ onto $AC$. Therefore we have $[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}$.

See Also

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_22&oldid=62963"