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--...") |
Tecilis459 (talk | contribs) (Add problem statement) |
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− | == | + | == Problem == |
+ | |||
+ | A vertical line divides the triangle with vertices <math>(0,0), (1,1)</math>, and <math>(9,1)</math> in the <math>xy\text{-plane}</math> into two regions of equal area. | ||
+ | The equation of the line is <math>x=</math> | ||
+ | |||
+ | <math>\textbf {(A)}\ 2.5 \qquad | ||
+ | \textbf {(B)}\ 3.0 \qquad | ||
+ | \textbf {(C)}\ 3.5 \qquad | ||
+ | \textbf {(D)}\ 4.0\qquad | ||
+ | \textbf {(E)}\ 4.5 </math> | ||
+ | |||
+ | == Solution == | ||
<asy> | <asy> | ||
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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 13:30, 16 July 2024
Problem
A vertical line divides the triangle with vertices , and
in the
into two regions of equal area.
The equation of the line is
Solution
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