https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Aw2016&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-14T11:51:52Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_23&diff=105510 2006 AMC 10B Problems/Problem 23 2019-04-27T00:13:31Z <p>Aw2016: /* Problem */</p> <hr /> <div>== Problem ==<br /> A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?<br /> <br /> &lt;asy&gt;<br /> unitsize(1.5cm);<br /> defaultpen(.8);<br /> <br /> pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br /> pair F = intersectionpoint( A--D, B--Ep );<br /> <br /> draw( A -- B -- C -- cycle );<br /> draw( A -- D );<br /> draw( B -- Ep );<br /> filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br /> <br /> label(&quot;$7$&quot;,(1.25,0.2));<br /> label(&quot;$7$&quot;,(2.2,0.45));<br /> label(&quot;$3$&quot;,(0.45,0.35));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} &lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> Label the points in the figure as shown below, and draw the segment &lt;math&gt;CF&lt;/math&gt;. This segment divides the quadrilateral into two triangles, let their areas be &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(.8);<br /> <br /> pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br /> pair F = intersectionpoint( A--D, B--Ep );<br /> <br /> draw( A -- B -- C -- cycle );<br /> draw( A -- D );<br /> draw( B -- Ep );<br /> filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br /> <br /> label(&quot;$7$&quot;,(1.45,0.15));<br /> label(&quot;$7$&quot;,(2.2,0.45));<br /> label(&quot;$3$&quot;,(0.45,0.35));<br /> <br /> draw( C -- F, dashed );<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,Ep,NW);<br /> label(&quot;$F$&quot;,F,S);<br /> <br /> label(&quot;$x$&quot;,(1,1));<br /> label(&quot;$y$&quot;,(1.6,1));<br /> &lt;/asy&gt;<br /> <br /> Since [[triangle]]s &lt;math&gt;AFB&lt;/math&gt; and &lt;math&gt;DFB&lt;/math&gt; share an [[altitude]] from &lt;math&gt;B&lt;/math&gt; and have equal area, their bases must be equal, hence &lt;math&gt;AF=DF&lt;/math&gt;.<br /> <br /> Since triangles &lt;math&gt;AFC&lt;/math&gt; and &lt;math&gt;DFC&lt;/math&gt; share an altitude from &lt;math&gt;C&lt;/math&gt; and their respective bases are equal, their areas must be equal, hence &lt;math&gt;x+3=y&lt;/math&gt;. <br /> <br /> Since triangles &lt;math&gt;EFA&lt;/math&gt; and &lt;math&gt;BFA&lt;/math&gt; share an altitude from &lt;math&gt;A&lt;/math&gt; and their respective areas are in the ratio &lt;math&gt;3:7&lt;/math&gt;, their bases must be in the same ratio, hence &lt;math&gt;EF:FB = 3:7&lt;/math&gt;.<br /> <br /> Since triangles &lt;math&gt;EFC&lt;/math&gt; and &lt;math&gt;BFC&lt;/math&gt; share an altitude from &lt;math&gt;C&lt;/math&gt; and their respective bases are in the ratio &lt;math&gt;3:7&lt;/math&gt;, their areas must be in the same ratio, hence &lt;math&gt;x:(y+7) = 3:7&lt;/math&gt;, which gives us &lt;math&gt;7x = 3(y+7)&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;y=x+3&lt;/math&gt; into the second equation we get &lt;math&gt;7x = 3(x+10)&lt;/math&gt;, which solves to &lt;math&gt;x=\frac{15}{2}&lt;/math&gt;. Then &lt;math&gt;y=x+3 = \frac{15}{2}+3 = \frac{21}{2}&lt;/math&gt;, and the total area of the quadrilateral is &lt;math&gt;x+y = \frac{15}{2}+\frac{21}{2} = \boxed{\textbf{(D) }18}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Connect points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. Triangles &lt;math&gt;EFA&lt;/math&gt; and &lt;math&gt;FAB&lt;/math&gt; share an altitude and their areas are in the ration &lt;math&gt;3:7&lt;/math&gt;. Their bases, &lt;math&gt;EF&lt;/math&gt; and &lt;math&gt;FB&lt;/math&gt;, must be in the same &lt;math&gt;3:7&lt;/math&gt; ratio. <br /> <br /> Triangles &lt;math&gt;EFD&lt;/math&gt; and &lt;math&gt;FBD&lt;/math&gt; share an altitude and their bases are in a &lt;math&gt;3:7&lt;/math&gt; ratio. Therefore, their areas are in a &lt;math&gt;3:7&lt;/math&gt; ratio and the area of triangle &lt;math&gt;EFD&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;.<br /> <br /> Triangle &lt;math&gt;CED&lt;/math&gt; and &lt;math&gt;DEA&lt;/math&gt; share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;EA&lt;/math&gt;. The ratio is &lt;math&gt;A:(3+3) \Rightarrow A:6&lt;/math&gt; where &lt;math&gt;A&lt;/math&gt; is the area of triangle &lt;math&gt;CED&lt;/math&gt;<br /> <br /> Triangles &lt;math&gt;CEB&lt;/math&gt; and &lt;math&gt;EAB&lt;/math&gt; also share an altitude. The ratio of their areas is also equal to the ratio of bases &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;EA&lt;/math&gt;. The ratio is &lt;math&gt;(A+3+7):(3+7) \Rightarrow (A+10):10&lt;/math&gt;<br /> <br /> Because the two ratios are equal, we get the equation &lt;math&gt;\frac{A}{6} = \frac {A+10}{10} \Rightarrow 10A = 6A+60 \Rightarrow A = 15&lt;/math&gt;. We add the area of triangle &lt;math&gt;EDF&lt;/math&gt; to get that the total area of the quadrilateral is &lt;math&gt;\boxed{\textbf{(D) }18}&lt;/math&gt;.<br /> <br /> ~Zeric Hang<br /> == See also ==<br /> {{AMC10 box|year=2006|ab=B|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Aw2016 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_23&diff=105509 2006 AMC 10B Problems/Problem 23 2019-04-27T00:12:35Z <p>Aw2016: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?<br /> <br /> &lt;asy&gt;<br /> unitsize(1.5cm);<br /> defaultpen(.8);<br /> <br /> pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br /> pair F = intersectionpoint( A--D, B--Ep );<br /> <br /> draw( A -- B -- C -- cycle );<br /> draw( A -- D );<br /> draw( B -- Ep );<br /> filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br /> <br /> label(&quot;$7$&quot;,(1.25,0.2));<br /> label(&quot;$7$&quot;,(2.2,0.45));<br /> label(&quot;$3$&quot;,(0.45,0.35));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} &lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> Label the points in the figure as shown below, and draw the segment &lt;math&gt;CF&lt;/math&gt;. This segment divides the quadrilateral into two triangles, let their areas be &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(.8);<br /> <br /> pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br /> pair F = intersectionpoint( A--D, B--Ep );<br /> <br /> draw( A -- B -- C -- cycle );<br /> draw( A -- D );<br /> draw( B -- Ep );<br /> filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br /> <br /> label(&quot;$7$&quot;,(1.45,0.15));<br /> label(&quot;$7$&quot;,(2.2,0.45));<br /> label(&quot;$3$&quot;,(0.45,0.35));<br /> <br /> draw( C -- F, dashed );<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,Ep,NW);<br /> label(&quot;$F$&quot;,F,S);<br /> <br /> label(&quot;$x$&quot;,(1,1));<br /> label(&quot;$y$&quot;,(1.6,1));<br /> &lt;/asy&gt;<br /> <br /> Since [[triangle]]s &lt;math&gt;AFB&lt;/math&gt; and &lt;math&gt;DFB&lt;/math&gt; share an [[altitude]] from &lt;math&gt;B&lt;/math&gt; and have equal area, their bases must be equal, hence &lt;math&gt;AF=DF&lt;/math&gt;.<br /> <br /> Since triangles &lt;math&gt;AFC&lt;/math&gt; and &lt;math&gt;DFC&lt;/math&gt; share an altitude from &lt;math&gt;C&lt;/math&gt; and their respective bases are equal, their areas must be equal, hence &lt;math&gt;x+3=y&lt;/math&gt;. <br /> <br /> Since triangles &lt;math&gt;EFA&lt;/math&gt; and &lt;math&gt;BFA&lt;/math&gt; share an altitude from &lt;math&gt;A&lt;/math&gt; and their respective areas are in the ratio &lt;math&gt;3:7&lt;/math&gt;, their bases must be in the same ratio, hence &lt;math&gt;EF:FB = 3:7&lt;/math&gt;.<br /> <br /> Since triangles &lt;math&gt;EFC&lt;/math&gt; and &lt;math&gt;BFC&lt;/math&gt; share an altitude from &lt;math&gt;C&lt;/math&gt; and their respective bases are in the ratio &lt;math&gt;3:7&lt;/math&gt;, their areas must be in the same ratio, hence &lt;math&gt;x:(y+7) = 3:7&lt;/math&gt;, which gives us &lt;math&gt;7x = 3(y+7)&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;y=x+3&lt;/math&gt; into the second equation we get &lt;math&gt;7x = 3(x+10)&lt;/math&gt;, which solves to &lt;math&gt;x=\frac{15}{2}&lt;/math&gt;. Then &lt;math&gt;y=x+3 = \frac{15}{2}+3 = \frac{21}{2}&lt;/math&gt;, and the total area of the quadrilateral is &lt;math&gt;x+y = \frac{15}{2}+\frac{21}{2} = \boxed{\textbf{(D) }18}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Connect points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. Triangles &lt;math&gt;EFA&lt;/math&gt; and &lt;math&gt;FAB&lt;/math&gt; share an altitude and their areas are in the ration &lt;math&gt;3:7&lt;/math&gt;. Their bases, &lt;math&gt;EF&lt;/math&gt; and &lt;math&gt;FB&lt;/math&gt;, must be in the same &lt;math&gt;3:7&lt;/math&gt; ratio. <br /> <br /> Triangles &lt;math&gt;EFD&lt;/math&gt; and &lt;math&gt;FBD&lt;/math&gt; share an altitude and their bases are in a &lt;math&gt;3:7&lt;/math&gt; ratio. Therefore, their areas are in a &lt;math&gt;3:7&lt;/math&gt; ratio and the area of triangle &lt;math&gt;EFD&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;.<br /> <br /> Triangle &lt;math&gt;CED&lt;/math&gt; and &lt;math&gt;DEA&lt;/math&gt; share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;EA&lt;/math&gt;. The ratio is &lt;math&gt;A:(3+3) \Rightarrow A:6&lt;/math&gt; where &lt;math&gt;A&lt;/math&gt; is the area of triangle &lt;math&gt;CED&lt;/math&gt;<br /> <br /> Triangles &lt;math&gt;CEB&lt;/math&gt; and &lt;math&gt;EAB&lt;/math&gt; also share an altitude. The ratio of their areas is also equal to the ratio of bases &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;EA&lt;/math&gt;. The ratio is &lt;math&gt;(A+3+7):(3+7) \Rightarrow (A+10):10&lt;/math&gt;<br /> <br /> Because the two ratios are equal, we get the equation &lt;math&gt;\frac{A}{6} = \frac {A+10}{10} \Rightarrow 10A = 6A+60 \Rightarrow A = 15&lt;/math&gt;. We add the area of triangle &lt;math&gt;EDF&lt;/math&gt; to get that the total area of the quadrilateral is &lt;math&gt;\boxed{\textbf{(D) }18}&lt;/math&gt;.<br /> <br /> ~Zeric Hang<br /> == See also ==<br /> {{AMC10 box|year=2006|ab=B|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Aw2016 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_23&diff=105508 2006 AMC 10B Problems/Problem 23 2019-04-27T00:07:40Z <p>Aw2016: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?<br /> <br /> &lt;asy&gt;<br /> unitsize(1.5cm);<br /> defaultpen(.8);<br /> <br /> pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br /> pair F = intersectionpoint( A--D, B--Ep );<br /> <br /> draw( A -- B -- C -- cycle );<br /> draw( A -- D );<br /> draw( B -- Ep );<br /> filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br /> <br /> label(&quot;$7$&quot;,(1.25,0.2));<br /> label(&quot;$7$&quot;,(2.2,0.45));<br /> label(&quot;$3$&quot;,(0.45,0.35));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} &lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> Label the points in the figure as shown below, and draw the segment &lt;math&gt;CF&lt;/math&gt;. This segment divides the quadrilateral into two triangles, let their areas be &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> defaultpen(.8);<br /> <br /> pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br /> pair F = intersectionpoint( A--D, B--Ep );<br /> <br /> draw( A -- B -- C -- cycle );<br /> draw( A -- D );<br /> draw( B -- Ep );<br /> filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br /> <br /> label(&quot;$7$&quot;,(1.45,0.15));<br /> label(&quot;$7$&quot;,(2.2,0.45));<br /> label(&quot;$3$&quot;,(0.45,0.35));<br /> <br /> draw( C -- F, dashed );<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,Ep,NW);<br /> label(&quot;$F$&quot;,F,S);<br /> <br /> label(&quot;$x$&quot;,(1,1));<br /> label(&quot;$y$&quot;,(1.6,1));<br /> &lt;/asy&gt;<br /> <br /> https://latex.artofproblemsolving.com/a/3/7/a373a9412edc8a46f24525404560b3b355922171.png<br /> Since [[triangle]]s &lt;math&gt;AFB&lt;/math&gt; and &lt;math&gt;DFB&lt;/math&gt; share an [[altitude]] from &lt;math&gt;B&lt;/math&gt; and have equal area, their bases must be equal, hence &lt;math&gt;AF=DF&lt;/math&gt;.<br /> <br /> Since triangles &lt;math&gt;AFC&lt;/math&gt; and &lt;math&gt;DFC&lt;/math&gt; share an altitude from &lt;math&gt;C&lt;/math&gt; and their respective bases are equal, their areas must be equal, hence &lt;math&gt;x+3=y&lt;/math&gt;. <br /> <br /> Since triangles &lt;math&gt;EFA&lt;/math&gt; and &lt;math&gt;BFA&lt;/math&gt; share an altitude from &lt;math&gt;A&lt;/math&gt; and their respective areas are in the ratio &lt;math&gt;3:7&lt;/math&gt;, their bases must be in the same ratio, hence &lt;math&gt;EF:FB = 3:7&lt;/math&gt;.<br /> <br /> Since triangles &lt;math&gt;EFC&lt;/math&gt; and &lt;math&gt;BFC&lt;/math&gt; share an altitude from &lt;math&gt;C&lt;/math&gt; and their respective bases are in the ratio &lt;math&gt;3:7&lt;/math&gt;, their areas must be in the same ratio, hence &lt;math&gt;x:(y+7) = 3:7&lt;/math&gt;, which gives us &lt;math&gt;7x = 3(y+7)&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;y=x+3&lt;/math&gt; into the second equation we get &lt;math&gt;7x = 3(x+10)&lt;/math&gt;, which solves to &lt;math&gt;x=\frac{15}{2}&lt;/math&gt;. Then &lt;math&gt;y=x+3 = \frac{15}{2}+3 = \frac{21}{2}&lt;/math&gt;, and the total area of the quadrilateral is &lt;math&gt;x+y = \frac{15}{2}+\frac{21}{2} = \boxed{\textbf{(D) }18}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Connect points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. Triangles &lt;math&gt;EFA&lt;/math&gt; and &lt;math&gt;FAB&lt;/math&gt; share an altitude and their areas are in the ration &lt;math&gt;3:7&lt;/math&gt;. Their bases, &lt;math&gt;EF&lt;/math&gt; and &lt;math&gt;FB&lt;/math&gt;, must be in the same &lt;math&gt;3:7&lt;/math&gt; ratio. <br /> <br /> Triangles &lt;math&gt;EFD&lt;/math&gt; and &lt;math&gt;FBD&lt;/math&gt; share an altitude and their bases are in a &lt;math&gt;3:7&lt;/math&gt; ratio. Therefore, their areas are in a &lt;math&gt;3:7&lt;/math&gt; ratio and the area of triangle &lt;math&gt;EFD&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;.<br /> <br /> Triangle &lt;math&gt;CED&lt;/math&gt; and &lt;math&gt;DEA&lt;/math&gt; share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;EA&lt;/math&gt;. The ratio is &lt;math&gt;A:(3+3) \Rightarrow A:6&lt;/math&gt; where &lt;math&gt;A&lt;/math&gt; is the area of triangle &lt;math&gt;CED&lt;/math&gt;<br /> <br /> Triangles &lt;math&gt;CEB&lt;/math&gt; and &lt;math&gt;EAB&lt;/math&gt; also share an altitude. The ratio of their areas is also equal to the ratio of bases &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;EA&lt;/math&gt;. The ratio is &lt;math&gt;(A+3+7):(3+7) \Rightarrow (A+10):10&lt;/math&gt;<br /> <br /> Because the two ratios are equal, we get the equation &lt;math&gt;\frac{A}{6} = \frac {A+10}{10} \Rightarrow 10A = 6A+60 \Rightarrow A = 15&lt;/math&gt;. We add the area of triangle &lt;math&gt;EDF&lt;/math&gt; to get that the total area of the quadrilateral is &lt;math&gt;\boxed{\textbf{(D) }18}&lt;/math&gt;.<br /> <br /> ~Zeric Hang<br /> == See also ==<br /> {{AMC10 box|year=2006|ab=B|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Aw2016 https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_29&diff=104715 1982 AHSME Problems/Problem 29 2019-03-19T23:55:33Z <p>Aw2016: Created page with &quot;Answer is 69&quot;</p> <hr /> <div>Answer is 69</div> Aw2016 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_12&diff=87453 2015 AMC 8 Problems/Problem 12 2017-09-10T22:35:33Z <p>Aw2016: </p> <hr /> <div>How many pairs of parallel edges, such as &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{GH}&lt;/math&gt; or &lt;math&gt;\overline{EH}&lt;/math&gt; and &lt;math&gt;\overline{FG}&lt;/math&gt;, does a cube have?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }6 \quad\textbf{(B) }12 \quad\textbf{(C) } 18 \quad\textbf{(D) } 24 \quad \textbf{(E) } 36&lt;/math&gt;<br /> &lt;asy&gt; import three;<br /> currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */<br /> draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle);<br /> draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1));<br /> draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); <br /> draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle);<br /> label(&quot;$D$&quot;,(0,0,0),S);<br /> label(&quot;$A$&quot;,(0,0,1),N);<br /> label(&quot;$H$&quot;,(0,1,0),S);<br /> label(&quot;$E$&quot;,(0,1,1),N);<br /> label(&quot;$C$&quot;,(1,0,0),S);<br /> label(&quot;$B$&quot;,(1,0,1),N);<br /> label(&quot;$G$&quot;,(1,1,0),S);<br /> label(&quot;$F$&quot;,(1,1,1),N);<br /> &lt;/asy&gt;<br /> ==Solution 1==<br /> We first count the number of pairs of parallel lines that are in the same direction as &lt;math&gt;\overline{AB}&lt;/math&gt;. The pairs of parallel lines are &lt;math&gt;\overline{AB}\text{ and }\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{CD}\text{ and }\overline{GH}&lt;/math&gt;, &lt;math&gt;\overline{AB}\text{ and }\overline{CD}&lt;/math&gt;, &lt;math&gt;\overline{EF}\text{ and }\overline{GH}&lt;/math&gt;, &lt;math&gt;\overline{AB}\text{ and }\overline{GH}&lt;/math&gt;, and &lt;math&gt;\overline{CD}\text{ and }\overline{EF}&lt;/math&gt;. These are &lt;math&gt;6&lt;/math&gt; pairs total. We can do the same for the lines in the same direction as &lt;math&gt;\overline{AE}&lt;/math&gt; and &lt;math&gt;\overline{AD}&lt;/math&gt;. This means there are &lt;math&gt;6\cdot 3=\boxed{\textbf{(C) } 18}&lt;/math&gt; total pairs of parallel lines.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Aw2016