Difference between revisions of "1996 AHSME Problems/Problem 15"
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+ | ==Problem== | ||
+ | |||
+ | Two opposite sides of a rectangle are each divided into <math>n</math> congruent segments, and the endpoints of one segment are joined to the center to form triangle <math>A</math>. The other sides are each divided into <math>m</math> congruent segments, and the endpoints of one of these segments are joined to the center to form triangle <math>B</math>. [See figure for <math>n=5, m=7</math>.] What is the ratio of the area of triangle <math>A</math> to the area of triangle <math>B</math>? | ||
+ | |||
+ | <asy> | ||
+ | int i; | ||
+ | for(i=0; i<8; i=i+1) { | ||
+ | dot((i,0)^^(i,5)); | ||
+ | } | ||
+ | for(i=1; i<5; i=i+1) { | ||
+ | dot((0,i)^^(7,i)); | ||
+ | } | ||
+ | draw(origin--(7,0)--(7,5)--(0,5)--cycle, linewidth(0.8)); | ||
+ | pair P=(3.5, 2.5); | ||
+ | draw((0,4)--P--(0,3)^^(2,0)--P--(3,0)); | ||
+ | label("$B$", (2.3,0), NE); | ||
+ | label("$A$", (0,3.7), SE); | ||
+ | </asy> | ||
+ | |||
+ | <math> \text{(A)}\ 1\qquad\text{(B)}\ m/n\qquad\text{(C)}\ n/m\qquad\text{(D)}\ 2m/n\qquad\text{(E)}\ 2n/m </math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | Place the rectangle on a coordinate grid, with diagonal vertices <math>(0,0)</math> and <math>(x,y)</math>. Each horizontal segment of the rectangle will have length <math>\frac{x}{m}</math>, while each vertical segment of the rectangle will have length <math>\frac{y}{n}</math>. | ||
+ | |||
+ | The center of this rectangle will be <math>(\frac{x}{2}, \frac{y}{2})</math>. | ||
+ | |||
+ | Triangle <math>A</math> has a base length of <math>\frac{y}{n}</math>, one of the vertical segments. It has an altitude of <math>\frac{x}{2}</math>, which is the perpendicular distance from the center of the square to the left side of the square. Thus, the area of triangle <math>A</math> is <math>\frac{1}{2}\cdot\frac{y}{n}\cdot\frac{x}{2} = \frac{xy}{4n}</math>. | ||
+ | |||
+ | Triangle <math>B</math> has a base length of <math>\frac{x}{m}</math>, one of the horizontal segments. It has an altitude of <math>\frac{y}{2}</math>, which is the perpendicular distance from the center of the square to the bottom side of the square. Thus, the area of triangle <math>B</math> is <math>\frac{1}{2}\cdot\frac{x}{m}\cdot\frac{y}{2} = \frac{xy}{4m}</math>. | ||
+ | |||
+ | The ratio of areas is <math>\frac{\frac{xy}{4n}}{\frac{xy}{4m}} = \frac{m}{n}</math>, which is answer <math>\boxed{B}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Alternately, the algebra can be made simpler if you arbitrarily assume that <math>(x,y) = (m,n)</math>, and thus that each side of the rectangle is cut into unit segments. In that case, the ratio of the areas of the two triangles <math>A</math> and <math>B</math> that have the same base length is just the ratio of the heights. Triangle <math>A</math> would have height <math>\frac{m}{2}</math>, while triangle <math>B</math>, while triangle <math>B</math> would have height <math>\frac{n}{2}</math>, giving a ratio of <math>\frac{m}{n}</math>, which is answer <math>\boxed{B}</math>. | ||
+ | |||
+ | This solution makes the extra assumption that the rectangle has dimensions <math>m</math> by <math>n</math>, instead of arbitrary <math>x</math> by <math>y</math> dimensions, and is not a formal "proof"; but since the answer is invariant, extra assumptions will not change the solution. | ||
+ | |||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=14|num-a=16}} | {{AHSME box|year=1996|num-b=14|num-a=16}} |
Revision as of 14:21, 19 August 2011
Contents
Problem
Two opposite sides of a rectangle are each divided into congruent segments, and the endpoints of one segment are joined to the center to form triangle . The other sides are each divided into congruent segments, and the endpoints of one of these segments are joined to the center to form triangle . [See figure for .] What is the ratio of the area of triangle to the area of triangle ?
Solution 1
Place the rectangle on a coordinate grid, with diagonal vertices and . Each horizontal segment of the rectangle will have length , while each vertical segment of the rectangle will have length .
The center of this rectangle will be .
Triangle has a base length of , one of the vertical segments. It has an altitude of , which is the perpendicular distance from the center of the square to the left side of the square. Thus, the area of triangle is .
Triangle has a base length of , one of the horizontal segments. It has an altitude of , which is the perpendicular distance from the center of the square to the bottom side of the square. Thus, the area of triangle is .
The ratio of areas is , which is answer .
Solution 2
Alternately, the algebra can be made simpler if you arbitrarily assume that , and thus that each side of the rectangle is cut into unit segments. In that case, the ratio of the areas of the two triangles and that have the same base length is just the ratio of the heights. Triangle would have height , while triangle , while triangle would have height , giving a ratio of , which is answer .
This solution makes the extra assumption that the rectangle has dimensions by , instead of arbitrary by dimensions, and is not a formal "proof"; but since the answer is invariant, extra assumptions will not change the solution.
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |