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Difference between revisions of "1988 AHSME Problems/Problem 16"

(Created page with "==Problem== <asy> defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, B=(1,-(1/sqrt(3))), C=(-1,-(1/sqrt(3))), A=(0,(2/sqrt(3))), E=(2,-(2/sqrt(3))), F=(-2,-(2/sqrt(3))), D=...")
 
(Added a solution with explanation)
 
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==Solution==
 
==Solution==
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Let <math>\triangle ABC</math> have side length <math>s</math> and <math>\triangle A'B'C'</math> have side length <math>t</math>. Thus the altitude of <math>\triangle ABC</math> is <math>\frac{s\sqrt{3}}{2}</math>. Now observe that this altitude is made up of three parts: the distance from <math>BC</math> to <math>B'C'</math>, plus the altitude of <math>\triangle A'B'C'</math>, plus a top part which is equal to the length of the diagonal line from the bottom-left corner of <math>\triangle ABC</math> to the bottom left corner of <math>\triangle A'B'C'</math> (as an isosceles trapezium is formed with parallel sides <math>AB</math> and <math>A'B'</math>, and legs <math>AA'</math> and <math>BB'</math>). We drop a perpendicular from <math>B'</math> to <math>BC</math>, which meets <math>BC</math> at <math>D</math>. <math>\triangle BDB'</math> has angles <math>30^{\circ}</math>, <math>60^{\circ}</math>, and <math>90^{\circ}</math>, and the vertical side is that distance from <math>BC</math> to <math>B'C'</math>, which is given as <math>\frac{1}{6} \times \frac{s\sqrt{3}}{2} = \frac{s\sqrt{3}}{12}</math>, so that by simple trigonometry, the length of the diagonal line is <math>\frac{s\sqrt{3}}{12} \times \csc{30^{\circ}} = \frac{s\sqrt{3}}{6}.</math> Thus using the "altitude in three parts" idea, we get <math>\frac{s\sqrt{3}}{2} = \frac{s\sqrt{3}}{12} + \frac{t\sqrt{3}}{2} + \frac{s\sqrt{3}}{6} \implies \frac{s\sqrt{3}}{4} = \frac{t\sqrt{3}}{2} \implies t = \frac{1}{2}s.</math> Thus the sides of <math>\triangle A'B'C'</math> are half as long as <math>\triangle ABC</math>, so the area ratio is <math>(\frac{1}{2}) ^ {2} = \frac{1}{4}</math>, which is <math>\boxed{\text{C}}</math>.
  
  

Latest revision as of 13:59, 27 February 2018

Problem

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, B=(1,-(1/sqrt(3))), C=(-1,-(1/sqrt(3))), A=(0,(2/sqrt(3))), E=(2,-(2/sqrt(3))), F=(-2,-(2/sqrt(3))), D=(0,(4/sqrt(3))); draw(A--B--C--A^^D--E--F--D); label("A'", A, N); label("B'", B, SE); label("C'", C, SW); label("A", D, E); label("B", E, E); label("C", F, W); [/asy]

$ABC$ and $A'B'C'$ are equilateral triangles with parallel sides and the same center, as in the figure. The distance between side $BC$ and side $B'C'$ is $\frac{1}{6}$ the altitude of $\triangle ABC$. The ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ is

$\textbf{(A)}\ \frac{1}{36}\qquad \textbf{(B)}\ \frac{1}{6}\qquad \textbf{(C)}\ \frac{1}{4}\qquad \textbf{(D)}\ \frac{\sqrt{3}}{4}\qquad \textbf{(E)}\ \frac{9+8\sqrt{3}}{36}$

Solution

Let $\triangle ABC$ have side length $s$ and $\triangle A'B'C'$ have side length $t$. Thus the altitude of $\triangle ABC$ is $\frac{s\sqrt{3}}{2}$. Now observe that this altitude is made up of three parts: the distance from $BC$ to $B'C'$, plus the altitude of $\triangle A'B'C'$, plus a top part which is equal to the length of the diagonal line from the bottom-left corner of $\triangle ABC$ to the bottom left corner of $\triangle A'B'C'$ (as an isosceles trapezium is formed with parallel sides $AB$ and $A'B'$, and legs $AA'$ and $BB'$). We drop a perpendicular from $B'$ to $BC$, which meets $BC$ at $D$. $\triangle BDB'$ has angles $30^{\circ}$, $60^{\circ}$, and $90^{\circ}$, and the vertical side is that distance from $BC$ to $B'C'$, which is given as $\frac{1}{6} \times \frac{s\sqrt{3}}{2} = \frac{s\sqrt{3}}{12}$, so that by simple trigonometry, the length of the diagonal line is $\frac{s\sqrt{3}}{12} \times \csc{30^{\circ}} = \frac{s\sqrt{3}}{6}.$ Thus using the "altitude in three parts" idea, we get $\frac{s\sqrt{3}}{2} = \frac{s\sqrt{3}}{12} + \frac{t\sqrt{3}}{2} + \frac{s\sqrt{3}}{6} \implies \frac{s\sqrt{3}}{4} = \frac{t\sqrt{3}}{2} \implies t = \frac{1}{2}s.$ Thus the sides of $\triangle A'B'C'$ are half as long as $\triangle ABC$, so the area ratio is $(\frac{1}{2}) ^ {2} = \frac{1}{4}$, which is $\boxed{\text{C}}$.


See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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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