https://artofproblemsolving.com/wiki/index.php?title=1988_AHSME_Problems/Problem_16&feed=atom&action=history1988 AHSME Problems/Problem 16 - Revision history2024-03-28T17:38:22ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1988_AHSME_Problems/Problem_16&diff=92411&oldid=prevHapaxoromenon: Added a solution with explanation2018-02-27T17:59:27Z<p>Added a solution with explanation</p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">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>.</ins></div></td></tr>
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</table>Hapaxoromenonhttps://artofproblemsolving.com/wiki/index.php?title=1988_AHSME_Problems/Problem_16&diff=65582&oldid=prevTimneh: 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=..."2014-10-23T06:00:35Z<p>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=..."</p>
<p><b>New page</b></p><div>==Problem==<br />
<br />
<asy><br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
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)));<br />
draw(A--B--C--A^^D--E--F--D);<br />
label("A'", A, N);<br />
label("B'", B, SE);<br />
label("C'", C, SW);<br />
label("A", D, E);<br />
label("B", E, E);<br />
label("C", F, W);<br />
</asy><br />
<br />
<math>ABC</math> and <math>A'B'C'</math> are equilateral triangles with parallel sides and the same center, <br />
as in the figure. The distance between side <math>BC</math> and side <math>B'C'</math> is <math>\frac{1}{6}</math> the altitude of <math>\triangle ABC</math>. <br />
The ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math> is<br />
<br />
<math>\textbf{(A)}\ \frac{1}{36}\qquad<br />
\textbf{(B)}\ \frac{1}{6}\qquad<br />
\textbf{(C)}\ \frac{1}{4}\qquad<br />
\textbf{(D)}\ \frac{\sqrt{3}}{4}\qquad<br />
\textbf{(E)}\ \frac{9+8\sqrt{3}}{36} </math> <br />
<br />
==Solution==<br />
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== See also ==<br />
{{AHSME box|year=1988|num-b=15|num-a=17}} <br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Timneh