1984 AHSME Problems/Problem 26

Problem

In the obtuse triangle $ABC$ with $\angle C>90^\circ$ , $AM=MB$ , $MD\perp BC$ , and $EC\perp BC$ ( $D$ is on $BC$ , $E$ is on $AB$ , and $M$ is on $EB$ ). If the area of $\triangle ABC$ is $24$ , then the area of $\triangle BED$ is

$\mathrm{(A) \ }9 \qquad \mathrm{(B) \ }12 \qquad \mathrm{(C) \ } 15 \qquad \mathrm{(D) \ }18 \qquad \mathrm{(E) \ } \text{Not uniquely determined}$

Solution 1

We let side $BC$ have length $a$ , $AB$ have length $c$ , and $\angle ABC$ have angle measure $\beta$ . We then have that

$[ABC]=24=\frac{AB\cdot BC\sin{\angle ABC}}{2}=\frac{ac\sin{\beta}}{2}$

Now I shall find the lengths of $BD$ and $CE$ in terms of the defined variables. Note that $M$ is defined to be the midpoint of $AB$ , so $BM=\frac{c}{2}$ . We can then use trigonometric manipulation on triangle $BDM$ to get that $BD=\frac{c\cos{\beta}}{2}$ . We can also use trig manipulation on $BCE$ to get that $CE=a\tan{\beta}$ .

Now note that $CE$ is the height of triangle $BDE$ originating from vertex $E$ , so we have that

$[BED]=\frac{BD\cdot CE}{2}=\frac{ac\cos{\beta}\tan{\beta}}{4}=\frac{ac\sin{\beta}}{4}$

However, this is simply half the area of triangle $ABC$ , so $[BED]=12$ , which makes $\boxed{\textbf{B}}$ the correct answer.

Solution 2

It is well known that $[CMB]=12$ . But $\triangle{CBE}$ ~ $\triangle{DBM}$ . As a result $[CMB]=[BED] \implies \boxed{\textbf{B}}$ .

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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1984 AHSME Problems/Problem 26

Contents

Problem

Solution 1

Solution 2

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