https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Rock10916&feedformat=atom AoPS Wiki - User contributions [en] 2021-10-27T23:51:00Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=100025 2018 AMC 10B Problems/Problem 10 2019-01-03T23:00:27Z <p>Rock10916: /* Problem */</p> <hr /> <div>== Problem ==<br /> <br /> In the rectangular parallelepiped shown, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;CG&lt;/math&gt; = &lt;math&gt;2&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{FG}&lt;/math&gt;. What is the volume of the rectangular pyramid with base &lt;math&gt;BCHE&lt;/math&gt; and apex &lt;math&gt;M&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10pt));<br /> pair A =origin;<br /> pair B = (4.75,0);<br /> pair E1=(0,3);<br /> pair F = (4.75,3);<br /> pair G = (5.95,4.2);<br /> pair C = (5.95,1.2);<br /> pair D = (1.2,1.2);<br /> pair H= (1.2,4.2);<br /> pair M = ((4.75+5.95)/2,3.6);<br /> draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br /> draw(B--C);<br /> draw(F--G);<br /> draw(A--D--H--C--D,dashed);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,SW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,NW);<br /> label(&quot;$M$&quot;,M,N);<br /> dot(A);<br /> dot(B);<br /> dot(E1);<br /> dot(F);<br /> dot(G);<br /> dot(C);<br /> dot(D);<br /> dot(H);<br /> dot(M);<br /> label(&quot;3&quot;,A/2+B/2,S);<br /> label(&quot;2&quot;,C/2+G/2,E);<br /> label(&quot;1&quot;,C/2+B/2,SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the cross-sectional plane and label its area &lt;math&gt;b&lt;/math&gt;. Note that the volume of the triangular prism that encloses the pyramid is &lt;math&gt;\frac{bh}{2}=3&lt;/math&gt;, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is &lt;math&gt;\frac{bh}{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{2}&lt;/math&gt;. (AOPS12142015)<br /> <br /> ==Solution 2==<br /> We can start by finding the total volume of the parallelepiped. It is &lt;math&gt;2 \cdot 3 \cdot 1 = 6&lt;/math&gt;, because a rectangular parallelepiped is a rectangular prism.<br /> <br /> Next, we can consider the wedge-shaped section made when the plane &lt;math&gt;BCHE&lt;/math&gt; cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 3 = 3&lt;/math&gt;. Since BC is given to be &lt;math&gt;1&lt;/math&gt;, we have that FM is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Using the formula for the volume of a triangular pyramid, we have &lt;math&gt;V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}&lt;/math&gt;. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume &lt;math&gt;\frac{1}{2}&lt;/math&gt; as well.<br /> <br /> The original wedge we considered in the last step has volume &lt;math&gt;3&lt;/math&gt;, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have &lt;math&gt;3 - \frac{1}{2} \cdot 2 = 2&lt;/math&gt;. Thus, the volume of the figure we are trying to find is &lt;math&gt;2&lt;/math&gt;. This means that the correct answer choice is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Written by: Archimedes15<br /> <br /> NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says &quot;rectangular parallelepiped.&quot; If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br /> <br /> ==Solution 3==<br /> If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is &lt;math&gt;\frac{1}{3}Bh&lt;/math&gt;, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is &lt;math&gt;\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}&lt;/math&gt;. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br /> &lt;math&gt;(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= &lt;/math&gt; &lt;math&gt;\boxed{2}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Rock10916