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Difference between revisions of "1996 AHSME Problems/Problem 15"

(Created page with "==See also== {{AHSME box|year=1996|num-b=14|num-a=16}}")
 
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==Problem==
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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>?
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<asy>
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int i;
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for(i=0; i<8; i=i+1) {
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dot((i,0)^^(i,5));
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}
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for(i=1; i<5; i=i+1) {
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dot((0,i)^^(7,i));
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}
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draw(origin--(7,0)--(7,5)--(0,5)--cycle, linewidth(0.8));
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pair P=(3.5, 2.5);
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draw((0,4)--P--(0,3)^^(2,0)--P--(3,0));
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label("$B$", (2.3,0), NE);
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label("$A$", (0,3.7), SE);
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</asy>
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<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>
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==Solution 1==
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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>.
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The center of this rectangle will be <math>(\frac{x}{2}, \frac{y}{2})</math>.
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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>.
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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>.
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The ratio of areas is  <math>\frac{\frac{xy}{4n}}{\frac{xy}{4m}} = \frac{m}{n}</math>, which is answer <math>\boxed{B}</math>.
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==Solution 2==
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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>.
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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.
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==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

Problem

Two opposite sides of a rectangle are each divided into $n$ congruent segments, and the endpoints of one segment are joined to the center to form triangle $A$. The other sides are each divided into $m$ congruent segments, and the endpoints of one of these segments are joined to the center to form triangle $B$. [See figure for $n=5, m=7$.] What is the ratio of the area of triangle $A$ to the area of triangle $B$?

[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]

$\text{(A)}\ 1\qquad\text{(B)}\ m/n\qquad\text{(C)}\ n/m\qquad\text{(D)}\ 2m/n\qquad\text{(E)}\ 2n/m$

Solution 1

Place the rectangle on a coordinate grid, with diagonal vertices $(0,0)$ and $(x,y)$. Each horizontal segment of the rectangle will have length $\frac{x}{m}$, while each vertical segment of the rectangle will have length $\frac{y}{n}$.

The center of this rectangle will be $(\frac{x}{2}, \frac{y}{2})$.

Triangle $A$ has a base length of $\frac{y}{n}$, one of the vertical segments. It has an altitude of $\frac{x}{2}$, which is the perpendicular distance from the center of the square to the left side of the square. Thus, the area of triangle $A$ is $\frac{1}{2}\cdot\frac{y}{n}\cdot\frac{x}{2} = \frac{xy}{4n}$.

Triangle $B$ has a base length of $\frac{x}{m}$, one of the horizontal segments. It has an altitude of $\frac{y}{2}$, which is the perpendicular distance from the center of the square to the bottom side of the square. Thus, the area of triangle $B$ is $\frac{1}{2}\cdot\frac{x}{m}\cdot\frac{y}{2} = \frac{xy}{4m}$.

The ratio of areas is $\frac{\frac{xy}{4n}}{\frac{xy}{4m}} = \frac{m}{n}$, which is answer $\boxed{B}$.

Solution 2

Alternately, the algebra can be made simpler if you arbitrarily assume that $(x,y) = (m,n)$, 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 $A$ and $B$ that have the same base length is just the ratio of the heights. Triangle $A$ would have height $\frac{m}{2}$, while triangle $B$, while triangle $B$ would have height $\frac{n}{2}$, giving a ratio of $\frac{m}{n}$, which is answer $\boxed{B}$.

This solution makes the extra assumption that the rectangle has dimensions $m$ by $n$, instead of arbitrary $x$ by $y$ 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
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