1996 AHSME Problems/Problem 15

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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