0
$\begingroup$

In rectangle ABCD below, points F and G lie on segment AB such that AF = FG = GB and E is the midpoint of segment DC. Also, segment AC intersects segment EF at H and segment EG at J. The area of rectangle ABCD is 70. Find the area of triangle AHF.

(Note: This question has been slightly changed from the original AMC 12 problem.)

Diagram

Work

  • Triangle AHF is similar to triangle CHE with ratio 2:3
  • Triangle AJG is similar to triangle CJE with ratio 4:3
$\endgroup$
2
  • $\begingroup$ Don't we need $J$ and $G$? or Is the $\Delta AHF$ this one? $\endgroup$ Apr 26, 2015 at 10:46
  • 1
    $\begingroup$ Did the joker do the drawing? $\endgroup$
    – Jiminion
    Nov 14, 2017 at 20:07

2 Answers 2

2
$\begingroup$

Let's say that $AB=3x$, $CD=2y$ and $BC=h$. Then, $$3xh = 2yh = 70,$$and $$x = \frac{2y}{3}.$$ The triangles $AFH$ and $HEC$ are similar and have area:

$$S_{AFH} = \frac{x h_1}{2}, S_{HEC} = \frac{y h_2}{2}$$

where $h_1$ and $h_2$ are the heights of the 2 triangles, with $h_1+h_2 = h$ and $h_1 = \frac{2h_2}{3}$. Then: $$h_1 = \frac{2h}{5}, h_2 = \frac{3h}{5},$$ and since $xh = \frac{70}{3}$, then:

$$S_{AFH} = \frac{x h}{5} = \frac{70}{15} = \frac{14}{3}$$

$\endgroup$
1
$\begingroup$

There's a "straightforward" vector solution:
Let $AB=b$, $AD=d$ the basis vectors and $[x\times y]$ be the cross product, since we're given $|[AB\times AD]|= 70$.
Vectors $AF=1/3AB=1/3b, AG=2/3AB=2/3b, AE=AD+1/2DE=AD+1/2AB=d+1/2b$.
$X$ lies on the line $YZ$ iff $AX=t\cdot AY + (1-t)\cdot AZ$, where $t$ is a real.
So, consider $AH=(1-u)\cdot 0+u\cdot (b+d)=(1-v)\cdot AE+v\cdot AF$
$u\cdot (b+d)=(1-v)\cdot (d+1/2b)+v/3\cdot b$ $u=(1-v)/2+v/3, u=(1-v) \Rightarrow u=2/5, v=3/5 \Rightarrow AH=2/5\cdot(b+d)$

$S_{\Delta AHF} = \frac{1}{2}[AH\times AF]= \frac{1}{2}[2/5\cdot(b+d)\times 1/3b]= \frac{1}{15}[(b+d)\times b] = \frac{1}{15}\left([b\times b] + [d\times b] \right)= \frac{1}{15} (0+70) = \frac{14}{3}$. More classical way:

Triangle $AHF$ is similar to triangle $CHE$ with ratio $2:3$

That is all one needs. $AH/HC=2/3$, so $AH/AC=2/5$ and the "height" of $\Delta AHF$ is $2/5AD$, while its base is $1/3AB$ , $S=\frac{1}{2}\cdot\frac{2}{5}\cdot\frac{1}{3}\cdot 70=\frac{14}{3}$

$\endgroup$
1
  • $\begingroup$ Thanks! Both answers were helpful, but this came first and so it gets the check. $\endgroup$
    – Rex
    Apr 26, 2015 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.