{"id":129,"date":"2020-01-28T06:27:32","date_gmt":"2020-01-28T06:27:32","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=129"},"modified":"2020-10-17T18:27:41","modified_gmt":"2020-10-17T18:27:41","slug":"mathcounts-exercises-17","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=129","title":{"rendered":"MATHCOUNTS Exercises &#8211; 17"},"content":{"rendered":"\n<p>In rectangle $ABCD$, shown here, point $M$ is the midpoint of side $BC$, and point $N$ lies on $CD$ such that $DN:NC=1:4$. Segment $BN$ intersects $AM$ and $AC$ at points $R$ and $S$, respectively. If $NS:SR:RB=x:y:z$, where $x$, $y$ and $z$ are positive integers, what is the minimum possible value of $x + y + z$? <em>Source: MATHCOUNTS 2012 State Sprint Round.<\/em> <a onclick='toggle_visibility(\"mathcounts-exercises-17-sol\");'>Click here for solutions.<\/a><a href=\"wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-27-at-1.52.46-PM.png\"><img width=\"264\" src=\"wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-27-at-1.52.46-PM.png\" alt=\"\" class=\"aligncenter wp-image-130\" srcset=\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-27-at-1.52.46-PM.png 528w, https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-27-at-1.52.46-PM-300x245.png 300w\" sizes=\"(max-width: 528px) 100vw, 528px\" \/><\/a><\/p>\n<div id='mathcounts-exercises-17-sol' style='display:none'>\n<strong>Solution 1 &#8211; Using Similar Triangles<\/strong> Please visit <a href='https:\/\/artofproblemsolving.com\/videos\/mathcounts\/mc2012\/368' target=\"_blank\" rel=\"noopener noreferrer\">Art of Problem Solving<\/a> for using <em>Similar Triangles and Proportional Reasoning<\/em> to solve the problem.<\/p>\n<p><strong>Solution 2 &#8211; Using Mass Points<\/strong> This may be the simplest method found without drawing extra lines.<br\/><br\/>\n$$ \\because DN:NC=1:4 $$\n$$ \\therefore CN:AB=CN:CD=NC:(DN+NC)=4:(1+4)=4:5$$\n$$ \\because AB \\parallel CD \\ \\ \\ and \\ \\ \\angle{ASB}=\\angle{CSN} $$\n$$ \\therefore \\triangle{ASB} \\sim \\triangle{CSN}$$\n$$ \\therefore NS:BS=CS:AS=CN:AB=4:5\\tag{1}$$\nFor $\\triangle{ABC}$, let $A_{mass}=4$, $C_{mass}=5$, then, $$S_{mass}=A_{mass}+C_{mass}=4+5=9$$ Because $M$ is the midpoint of $BC$, $B_{mass}=C_{mass}=5$, and $$M_{mass}=B_{mass}+C_{mass}=5+5=10$$ Finally, we have balanced mass at $S$: $$S_{mass}=A_{mass}+M_{mass}=B_{mass}+S_{mass}=14$$ <a href=\"wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-28-at-9.47.08-AM.png\"><img width=\"288\" src=\"wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-28-at-9.47.08-AM.png\" alt=\"\" class=\"aligncenter wp-image-130\"\/><\/a>Therefore $$RB:BS=9:14\\tag{2}$$ $$SR:BS=5:14\\tag{3}$$ Combine (1), (2) and (3), we have: $$NS:SR:RB=(NS:BS):(SR:BS):(RB:BS)=(4:5):(5:14):(9:14)$$ $$= (56:70):(25:70):(45:70)=56:25:45=x:y:z$$ Therefore, $x+y+z=56+25+45=126$<\/p>\n<p><strong>Solution 3 &#8211; Using <a href='http:\/\/mathworld.wolfram.com\/CrossedLaddersTheorem.html' target=\"_blank\" rel=\"noopener noreferrer\">Crossed Ladders Theorem<\/a><\/strong> Draw line $RT$ parallel to $AB$ and intersecting $BC$ at $T$, and line $ME$ parallel to $AB$ and intersecting $BN$ at $E$.<br \/>\n<a href=\"wp-content\/uploads\/2020\/01\/Screen-Shot-2020-01-27-at-2.26.01-PM.png\"><img width=\"260\" src=\"wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-27-at-2.26.01-PM.png\" alt=\"\" class=\"aligncenter wp-image-132\" srcset=\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-27-at-2.26.01-PM.png 520w, https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-27-at-2.26.01-PM-300x249.png 300w\" sizes=\"(max-width: 520px) 100vw, 520px\" \/><\/a>$$ \\because DN:NC=1:4 $$ $$ \\therefore AB:CN=CD:CN=(DN+NC):NC=(1+4):4=5:4 \\tag{1}$$ $$ \\because AB \\parallel CD \\ \\ \\ and \\ \\ \\angle{ASB}=\\angle{CSN} $$ $$ \\therefore \\triangle{ASB} \\sim \\triangle{CSN}$$ $$ \\therefore BS:NS=AB:CN=5:4$$ $$BS:BN=BS:(BS+NS)=5:(4:5)=5:9 \\tag{2}$$ Applying <a href='https:\/\/en.wikipedia.org\/wiki\/Crossed_ladders_problem' target=\"_blank\" rel=\"noopener noreferrer\">Crossed ladders theorem<\/a>, we have $$ \\because RT \\parallel AB \\parallel EM \\parallel CD $$ $$ \\therefore \\frac{1}{RT}=\\frac{1}{AB}+\\frac{1}{EM}\\tag{3}$$ Additionally $$ \\because ME \\parallel CN $$ $$ \\therefore \\triangle{BME}\\sim\\triangle{BCD} $$ Because point $M$ is the midpoint of side $BC$, therefore $$EM:NC=BM:BC=1:2$$ Apply (1), we have $$EM=\\frac{1}{2}NC=\\frac{1}{2}\\cdot\\frac{4}{5}AB=\\frac{2}{5}AB\\tag{4} $$ Apply (4) to (3), we have $$ \\frac{1}{RT}=\\frac{1}{AB}+\\frac{1}{\\frac{2}{5}AB}=\\frac{7}{2}\\cdot\\frac{1}{AB}$$ $$\\therefore RT:AB=2:7$$ Apply (1), we have $$RT:NC=(RT:AB) \\cdot (AB:CN) = (2:7)\\cdot(5:4)=5:14$$ $$ \\because \\triangle{BTR} \\sim \\triangle{BCN} $$ $$\\therefore RB:NB=RT:NC=5:14\\tag{5}$$ $$\\therefore SR:NB=(BS-RB):NB = BS:NB-RB:NB=\\frac{5}{9}-\\frac{5}{14}=\\frac{25}{126}\\tag{6}$$ Additionally, $$NS:NB=NS:(NS+SR)=4:(4+5)=4:9\\tag{7}$$ Combine (5), (6) and (7), we have: $$NS:SR:RB=(4:9):(25:126):(5:14)=56:25:45=x:y:z$$ Therefore, we reach the same answer: $x+y+z=56+25+45=126$<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>In rectangle $ABCD$, shown here, point $M$ is the midpoint of side $BC$, and point $N$ lies on $CD$ such that $DN:NC=1:4$. Segment $BN$ intersects $AM$ and $AC$ at points $R$ and $S$, respectively. 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