1982 AHSME Problems/Problem 9

Solution for Problem 9

[asy] size(350); defaultpen(fontsize(10)); pair A=origin, O=(10,0), B=(3,0), N=(0,5), C=(3,5), P=(5,0), D=(1,1), G=(9,1), F=(1,0); draw(G--A--D--cycle, linewidth(0.7)); draw(D--F, linewidth(0.6)); draw(B--C, linewidth(0.5)); draw(A--N, linewidth(0.4)); draw(A--O, linewidth(0.4)); dot(A^^B^^D^^G^^F); label("$A$", D, W); label("$B$", A, dir(0)); label("$C$", G, dir(100)); label("$E$", B, SE); label("$F$", F, SE); label("$(9,1)$", G, SE); label("$(0,0)$", A, SW); label("$8$", D--G, dir(40)); label("$y=x/9$", A--G, SE); label("$x=a$", B--N, SE); label("$(1,1)$", D, NE);[/asy] The vertical line that divides $\triangle ABC$ into two equal regions has equation $x=a,$ as shown in the diagram. The area of $ABC$ is half of the height times $AC,$ so because the Y coordinate of A is 1 and $\overline{AF}$ is the height, because the difference of the x coordinates between $A$ and $C$ is $8,$ we have $[ABC]=1/2 \cdot 8 \cdot 1 = 4.$ Thus the two regions must have area $2$ each.

Since $\triangle ABF$ has area $1/2,$ we know that the portion of $\triangle ABC$ made by the points $A,$ $B$ and the intersection $\overline{AF}$ and $\overline{BC}$ will be less than $1/2,$ which is less than half of the triangle's area, or 2. Therefore $\overline{AF}$ is to the left of vertical line $x=a$ (Passing through point $E$).

The equation of line BC is $y=x/9,$ and the vertical line $x=a$ intersects $\overline{BC}$ at the point $(a, a/9).$ Because the area of the portion of $\triangle ABC$ on the right is 2, we have \[2=1/2(1-a/9)(9-a)\] or \[(9-a)^2=36.\] Therefore $a>0$ so $a=3=x.$