# 2001 AMC 12 Problems/Problem 22

## 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 1

$[asy] unitsize(0.5cm); defaultpen(0.8); pair A=(0,0), B=(10,0), C=(10,7), D=(0,7), E=(C+D)/2, F=(2*A+B)/3, G=(A+2*B)/3; pair H = intersectionpoint(A--C,E--F); pair J = intersectionpoint(A--C,E--G); draw(A--B--C--D--cycle); draw(G--E--F); draw(A--C); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("E",E,N); label("F",F,S); label("G",G,S); label("H",H,SE); label("J",J,ESE); filldraw(E--H--J--cycle,lightgray,black); draw(H--D, dashed); [/asy]$ 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 $[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}$.

## Solution 3

Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are $7$ and $10$ (or any other two positive real numbers that multiply to 70). We can find $H$ and $J$ by intersecting lines, and then we calculate the area of $EHJ$ using shoelace formula. This yields $\boxed{3}$.

## Solution 4

Note that triangle $AFH$ is similar to triangle $CEH$ with ratio $\frac{2}{3}$. Similarly, triangle $AGJ$ is similar to triangle $ECJ$ with ratio $\frac{4}{3}$. Thus, if $AC = a$ then we know that $AH = \frac{2}{5}a$ and $JC = \frac{3}{7}a$ meaning $HJ = \frac{6}{35}a$ and thus the ratio of $HJ$ to $JC$ is $\frac{\frac{6}{35}}{\frac{3}{7}} = \frac{2}{5}$ which equals the ratio of the areas of $HJE$ to $JEC$. If $y = AD, x = DC$, then we know that $JEC = \text{(altitude from J to EC)} \cdot EC = \frac{3}{7}y \cdot \frac{1}{2}x \cdot \frac{1}{2}$ and since $xy = 70$ and we want to find $\frac{2}{5}$ of this, we get our answer is $\frac{2}{5} \cdot \frac{3}{7} \cdot \frac{1}{2} \cdot 70 \cdot \frac{1}{2} = \boxed{3}$. -SuperJJ

## Solution 5

$[CEF] = \frac{[ABCD]}{4} = \frac{35}{2}$

$\triangle CEH \sim \triangle AFH$, $\frac{HE}{HF} = \frac{CE}{AF} = \frac{3}{2}$, $\frac{HE}{EF} = \frac{3}{5}$

$[CEH] = \frac{HE}{EF} \cdot [CEF] = \frac{3}{5} \cdot \frac{35}{2} = \frac{21}{2}$

$\triangle CEH \sim \triangle AFH$, $\frac{AH}{HC} = \frac{AF}{CE} = \frac{2}{3}$, $\frac{AH}{AC} = \frac{2}{5}$, $\frac{CH}{AC} = \frac{3}{5}$

$\triangle CEJ \sim \triangle AGJ$, $\frac{AJ}{JC} = \frac{AG}{CE} = \frac{4}{3}$, $\frac{AJ}{AC} = \frac{4}{7}$

$\frac{HJ}{AC} = \frac{AJ}{AC} - \frac{AH}{AC} = \frac{4}{7} - \frac{2}{5} = \frac{6}{35}$

$\frac{HJ}{CH} = \frac{HJ}{AC} \cdot \frac{AC}{CH} = \frac{6}{35} \cdot \frac{5}{3} = \frac{2}{7}$

$[EHJ] = \frac{HJ}{CH} \cdot [CEH] = \frac{2}{7} \cdot \frac{21}{2} = \boxed{\textbf{(C) }3}$

## Solution 6 (Mass Points)

We need one more pair of ratios to fully define our mass point system. Let's use $\triangle AFH \sim \triangle CEH\implies EH:HF = 3:2$ and now do mass points on $\triangle AEG$:

$[asy] size(250); pair A = (0,0), B = (10,0), C = (10,7), D = (0,7); pair F = (10/3,0), G = (20/3,0), E = (5,7); pair H = intersectionpoint(A--C, E--F); pair J = intersectionpoint(A--C, E--G); filldraw(A--E--G--cycle, rgb(1,1,1)+opacity(0.3), red+2bp); draw(A--B--C--D--cycle); draw(A--C); draw(E--F); draw(E--G); draw(A--E, dashed); draw(E--B, dashed); dot("A", A, SW); dot("B", B, SE); dot("C", C, NE); dot("D", D, NW); dot("E", E, N); dot("F", F, S); dot("G", G, S); dot("H", H, ESE); dot("J", J, W); // mass point labels pair mass = A + SW; label(scale(0.8)*"3", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = F + S; label(scale(0.8)*"6", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = G + SE; label(scale(0.8)*"3", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = J + .7*ESE; label(scale(0.8)*"7", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = E + N; label(scale(0.8)*"4", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = H + .7*WNW; label(scale(0.8)*"10", mass, UnFill); draw(circle(mass, .4), linewidth(1)); [/asy]$

Now it's just a standard "Area Reduction by Ratios"™ problem going from:

$$[ABCD]\xrightarrow[]{\frac{1}{2}}[AEB]\xrightarrow[]{\frac{2}{3}}[AEG]\xrightarrow[]{\frac{3}{7}}[AEJ]\xrightarrow[]{\frac{3}{10}}[HEJ]$$

or,

$$70 \cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{7}\cdot \frac{3}{10} = \boxed{\textbf{(C) }3}$$

~ proloto