Difference between revisions of "1994 AHSME Problems/Problem 23"
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==Solution== | ==Solution== | ||
Let the vertices be <math>A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)</math>. It is easy to see that the line must pass through <math>CD</math>. Let the line intersect <math>CD</math> at the point <math>G=(3,3-x)</math> (i.e. the point <math>x</math> units below <math>C</math>). Since the quadrilateral <math>ABCG</math> and pentagon <math>GDEFA</math> must have the same area, we have the equation <math>3\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2</math>. This simplifies into <math>3x=2</math>, or <math>x=\frac{2}{3}</math>, so <math>G=(3,\frac{7}{3})</math>. Therefore the slope of the line is <math>\boxed{\textbf{(E)}\ \frac{7}{9}}</math> | Let the vertices be <math>A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)</math>. It is easy to see that the line must pass through <math>CD</math>. Let the line intersect <math>CD</math> at the point <math>G=(3,3-x)</math> (i.e. the point <math>x</math> units below <math>C</math>). Since the quadrilateral <math>ABCG</math> and pentagon <math>GDEFA</math> must have the same area, we have the equation <math>3\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2</math>. This simplifies into <math>3x=2</math>, or <math>x=\frac{2}{3}</math>, so <math>G=(3,\frac{7}{3})</math>. Therefore the slope of the line is <math>\boxed{\textbf{(E)}\ \frac{7}{9}}</math> | ||
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+ | == Solution 2== | ||
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+ | Consider the small rectangle between <math>x=3</math> and <math>x=5</math> with area <math>2</math>. If we exclude that area then the shape becomes a much simpler, its just a rectangle. In order to offset the exclusion of that area from the shape below the line, we must also exclude a shape with the same area above the line. We want the remainder after excluding both areas to be simple, so exclude a rectangle between <math>x=0</math> and <math>x=3</math> and <math>y=3</math> and <math>y=c</math> for some unknown <math>c</math>. For the area to be the same we need <math>(3-c) \cdot 3 = 2</math>, or <math>c=7/3</math>. After excluding our two offsetting areas, we're left with a rectangle from <math>x=0</math> to <math>x=3</math> and <math>y=0</math> to <math>y=c</math>. The area of this region is clearly bisected by its diagonal line. The slope of this line is <math>c/3 = 7/9</math> and the answer is <math>\fbox{E}</math> | ||
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+ | ==See Also== | ||
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+ | {{AHSME box|year=1994|num-b=22|num-a=24}} | ||
+ | {{MAA Notice}} |
Revision as of 03:35, 28 May 2021
Contents
Problem
In the -plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at and . The slope of the line through the origin that divides the area of this region exactly in half is
Solution
Let the vertices be . It is easy to see that the line must pass through . Let the line intersect at the point (i.e. the point units below ). Since the quadrilateral and pentagon must have the same area, we have the equation . This simplifies into , or , so . Therefore the slope of the line is
Solution 2
Consider the small rectangle between and with area . If we exclude that area then the shape becomes a much simpler, its just a rectangle. In order to offset the exclusion of that area from the shape below the line, we must also exclude a shape with the same area above the line. We want the remainder after excluding both areas to be simple, so exclude a rectangle between and and and for some unknown . For the area to be the same we need , or . After excluding our two offsetting areas, we're left with a rectangle from to and to . The area of this region is clearly bisected by its diagonal line. The slope of this line is and the answer is
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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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