https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Dingzhou&feedformat=atom AoPS Wiki - User contributions [en] 2023-12-02T09:05:23Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11&diff=26341 1983 AIME Problems/Problem 11 2008-06-07T19:12:01Z <p>Dingzhou: </p> <hr /> <div>== Problem ==<br /> The solid shown has a [[square]] base of side length &lt;math&gt;s&lt;/math&gt;. The upper edge is [[parallel]] to the base and has length &lt;math&gt;2s&lt;/math&gt;. All other edges have length &lt;math&gt;s&lt;/math&gt;. Given that &lt;math&gt;s=6\sqrt{2}&lt;/math&gt;, what is the volume of the solid?<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black;<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> D(A--B--C--D--A--E--D); D(B--F--C); D(E--F);<br /> MP(&quot;A&quot;,A);MP(&quot;B&quot;,B);MP(&quot;C&quot;,C);MP(&quot;D&quot;,D);MP(&quot;E&quot;,E,N);MP(&quot;F&quot;,F,N);<br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- Asymptote replacement for Image:1983Number11.JPG by bpms --&gt;<br /> == Solution ==<br /> First, we find the height of the figure by drawing a [[perpendicular]] from the midpoint of &lt;math&gt;AD&lt;/math&gt; to &lt;math&gt;EF&lt;/math&gt;. The [[hypotenuse]] of the triangle is the [[median]] of [[equilateral triangle]] &lt;math&gt;ADE&lt;/math&gt; one of the legs is &lt;math&gt;3\sqrt{2}&lt;/math&gt;. We apply the [[Pythagorean Theorem]] to find that the height is equal to &lt;math&gt;6&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5);<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0);<br /> D(A--B--C--D--A--E--D); D(B--F--C); D(E--F); <br /> D(B--Ba--Ca--C,dashed+d);D(A--Aa--Da--D,dashed+d);D(E--(E.x,E.y,0),dashed+l);D(F--(F.x,F.y,0),dashed+l);<br /> D(Aa--E--Da,dashed+d); D(Ba--F--Ca,dashed+d);<br /> MP(&quot;A&quot;,A);MP(&quot;B&quot;,B);MP(&quot;C&quot;,C);MP(&quot;D&quot;,D);MP(&quot;E&quot;,E,N);MP(&quot;F&quot;,F,N);MP(&quot;12\sqrt{2}&quot;,(E+F)/2,N);MP(&quot;6\sqrt{2}&quot;,(A+B)/2);MP(&quot;6&quot;,(3*s/2,s/2,3),ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Next, we complete the figure into a triangular prism, and find the area, which is &lt;math&gt;\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432&lt;/math&gt;.<br /> <br /> Now, we subtract off the two extra [[pyramid]]s that we included, whose combined area is &lt;math&gt;2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;432-144=\boxed{288}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1983|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Dingzhou