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

(Created page with "== 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--...") |
(→Solution for Problem 9) |
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label("$(1,1)$", D, NE);</asy> | label("$(1,1)$", D, NE);</asy> | ||

The vertical line that divides <math>\triangle ABC</math> into two equal regions has equation <math>x=a,</math> as shown in the diagram. | The vertical line that divides <math>\triangle ABC</math> into two equal regions has equation <math>x=a,</math> as shown in the diagram. | ||

− | The area of <math>ABC</math> is half of the height times <math>AC,</math> so because the Y coordinate of A is 1 | + | 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. |

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>). | 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>). | ||

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> | 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> |

## Latest revision as of 11:30, 19 October 2020

## Solution for Problem 9

The vertical line that divides into two equal regions has equation as shown in the diagram. The area of is half of the height times so because the Y coordinate of A is 1 and is the height, because the difference of the x coordinates between and is we have Thus the two regions must have area each.

Since has area we know that the portion of made by the points and the intersection and will be less than which is less than half of the triangle's area, or 2. Therefore is to the left of vertical line (Passing through point ).

The equation of line BC is and the vertical line intersects at the point Because the area of the portion of on the right is 2, we have or Therefore so