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Difference between revisions of "1982 AHSME Problems/Problem 9"

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label("$x=a$", B--N, SE);
 
label("$x=a$", B--N, SE);
 
label("$(1,1)$", D, NE);</asy>
 
label("$(1,1)$", D, NE);</asy>
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The vertical line that divides <math>\triangle ABC</math> into two equal regions has equation <math>x=a,</math> as shown in the diagram.
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The vertical line that divides <math>\triangle ABC</math> into two equal regions has equation <math>x=a</math>, as shown in the diagram.
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The area of <math>ABC</math> is half of the height times <math>AC,</math> so because the Y coordinate of A is 1 and <math>\overline{AF}</math> is the height, because the difference of the x coordinates between <math>A</math> and <math>C</math> is <math>8,</math> we have <math>[ABC]=1/2 \cdot 8 \cdot 1 = 4.</math> Thus the two regions must have area <math>2</math> each.
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The area of <math>ABC</math> is <math>\frac{1}{2}\cdot AC\cdot AF = \frac{1}{2} \cdot 8 \cdot 1 = 4</math>, so the two regions must each have area <math>2</math> .
  
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Since <math>\triangle ABF</math> has area <math>1/2,</math> we know that the portion of <math>\triangle ABC</math> made by the points <math>A,</math> <math>B</math> and the intersection <math>\overline{AF}</math> and <math>\overline{BC}</math> will be less than <math>1/2,</math> which is less than half of the triangle's area, or 2. Therefore <math>\overline{AF}</math> is to the left of vertical line <math>x=a</math> (Passing through point <math>E</math>).
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Since <math>\triangle ABF</math> has area <math>\frac{1}{2}</math>, the portion of <math>\triangle ABC</math> to the left of <math>AF</math> will be less than <math>\frac{1}{2}</math> and therefore less than <math>2</math>. So <math>\overline{AF}</math> is to the left of vertical line <math>x=a</math> (Passing through point <math>E</math>).
  
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The equation of line BC is <math>y=x/9,</math> and the vertical line <math>x=a</math> intersects <math>\overline{BC}</math> at the point <math>(a, a/9).</math> Because the area of the portion of <math>\triangle ABC</math> on the right is 2, we have <cmath>2=1/2(1-a/9)(9-a)</cmath> or <cmath>(9-a)^2=36.</cmath> Therefore <math>a>0</math> so <math>a=3=x.</math>
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The equation of line BC is <math>y=\frac{1}{9}x</math> and the vertical line <math>x=a</math> intersects <math>\overline{BC}</math> at the point <math>\left(a, \frac{a}{9}\right)</math> .  Because the area of the portion of <math>\triangle ABC</math> on the right is <math>2</math>, we have <cmath>2=\frac{1}{2}\left(1-\frac{a}{9}\right)(9-a)</cmath> which simplifies to <cmath>(9-a)^2=36</cmath>
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This equation has solutions <math>a = 3</math> and <math>a = 15</math>, but <math>1 < a < 9</math> so <math>a = \boxed{(\mathbf{B}) 3.0}</math>.
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==See Also==
 +
{{AHSME box|year=1982|num-b=8|num-a=10}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 22:55, 29 June 2025

Problem

A vertical line divides the triangle with vertices $(0,0), (1,1)$, and $(9,1)$ in the $xy\text{-plane}$ into two regions of equal area. The equation of the line is $x=$

$\textbf {(A)}\ 2.5 \qquad \textbf {(B)}\ 3.0 \qquad \textbf {(C)}\ 3.5 \qquad \textbf {(D)}\ 4.0\qquad \textbf {(E)}\ 4.5$

Solution

[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 $\frac{1}{2}\cdot AC\cdot AF = \frac{1}{2} \cdot 8 \cdot 1 = 4$, so the two regions must each have area $2$ .

Since $\triangle ABF$ has area $\frac{1}{2}$, the portion of $\triangle ABC$ to the left of $AF$ will be less than $\frac{1}{2}$ and therefore less than $2$. So $\overline{AF}$ is to the left of vertical line $x=a$ (Passing through point $E$).

The equation of line BC is $y=\frac{1}{9}x$ and the vertical line $x=a$ intersects $\overline{BC}$ at the point $\left(a, \frac{a}{9}\right)$ . Because the area of the portion of $\triangle ABC$ on the right is $2$, we have \[2=\frac{1}{2}\left(1-\frac{a}{9}\right)(9-a)\] which simplifies to \[(9-a)^2=36\]

This equation has solutions $a = 3$ and $a = 15$, but $1 < a < 9$ so $a = \boxed{(\mathbf{B}) 3.0}$.

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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