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Difference between revisions of "1984 AHSME Problems/Problem 26"

(Created page with "==Problem== In the obtuse triangle <math> ABC </math> with <math> \angle C>90^\circ </math>, <math> AM=MB </math>, <math> MD\perp BC </math>, and <math> EC\perp BC </math> (<...")
 
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{{AHSME box|year=1984|num-b=25|num-a=27}}
 
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Revision as of 11:52, 5 July 2013

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

1984AHSME26.png

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 $\mathrm{(B) \ }$ the correct answer.

See Also

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