{"id":115,"date":"2020-02-01T05:06:13","date_gmt":"2020-02-01T05:06:13","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=115"},"modified":"2020-10-16T14:27:47","modified_gmt":"2020-10-16T14:27:47","slug":"mathcounts-exercises-20","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=115","title":{"rendered":"MATHCOUNTS Exercises &#8211; 20"},"content":{"rendered":"\n<p>Given an equilateral $\\triangle{ABC}$ shown below, $AB=BC=CA=1$, point $D$ and $E$ trisect side $AB$, point $F$ and $G$ trisect side $BC$, point $H$ and $I$ trisect side  $CA$. Two overlapping equilateral triangles are formed by linking various intersect points. Find the ratio of the star-shaped area covered by the overlapping equilateral triangles to the area of $\\triangle{ABC}$. <a onclick='toggle_visibility(\"mathcounts-exercises-20-sol\");'>Click here for the solutions.<\/a>\n<a href=\"wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-31-at-9.33.26-PM.png\">\n<img width=\"356\" src=\"wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-31-at-9.33.26-PM.png\" alt=\"\" class=\"aligncenter wp-image-116\" srcset=\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-31-at-9.33.26-PM.png 712w, https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-31-at-9.33.26-PM-300x253.png 300w\" sizes=\"(max-width: 712px) 100vw, 712px\" \/><\/a><\/p>\n<div id='mathcounts-exercises-20-sol' style='display:none'>\n<p><strong>Solution 1 &#8211; Use Mass Points<\/strong> Because point $D$ and $E$ trisect side $AB$, point $F$ and $G$ trisect side $BC$, we can have the following balanced mass:\n$$ A_{mass}=2, B_{mass}=1, D_{mass}=A_{mass}+B_{mass}=3$$ $$C_{mass}=1, I_{mass}=A_{mass}+C_{mass}=3$$ $$ J_{mass}=C_{mass}+D_{mass}=B_{mass}+I_{mass}=4$$\n<br\/>\nTherefore\n$$\\frac{CJ}{CD}=\\frac{D_{mass}}{J_{mass}}=\\frac{3}{4}$$\nBecause $ JK \\parallel DE $, we have $\\triangle{CJK} \\sim \\triangle{CDE}$. \nTherefore $$\\frac{JK}{DE}=\\frac{CJ}{CD}=\\frac{3}{4}$$\nBecause point $D$ and $E$ trisect side $AB$, we have $$DE=\\frac{3}{4}\\cdot DE=\\frac{3}{4}\\cdot \\frac{1}{3}=\\frac{1}{4} \\tag{1}$$\nBecause point $D$ and $E$ trisect side $AB$, point $F$ and $G$ trisect side $BC$, we can have the following balanced mass: $$A_{mass}=2, B_{mass}=1, D_{mass}=A_{mass}+B_{mass}=3$$ $$C_{mass}=2, G_{mass}=B_{mass}+C_{mass}=3$$ $$ N_{mass}=C_{mass}+D_{mass}=A_{mass}+G_{mass}=5$$<br \/>\nTherefore $$\\frac{CN}{CD}=\\frac{D_{mass}}{N_{mass}}=\\frac{3}{5}$$ Because $ NM \\parallel DE $, we have $\\triangle{CNM} \\sim \\triangle{CDE}$. Therefore $$\\frac{NM}{DE}=\\frac{NM}{CD}=\\frac{3}{5}$$ Because point $D$ and $E$ trisect side $AB$, we have $$NM=\\frac{3}{5}\\cdot DE=\\frac{3}{5}\\cdot \\frac{1}{3}=\\frac{1}{5} \\tag{2}$$ Therefore, the side length of equilateral $\\triangle{JKL}$ is $\\dfrac{1}{4}$ and the side length of equilateral $\\triangle{MNO}$ is $\\dfrac{1}{5}$. Let&#8217;s take a closer look at the two triangles as the following:<a href=\"wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-31-at-9.41.14-PM.png\"><img width=\"268\" src=\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-31-at-9.41.14-PM.png\" alt=\"\" class=\"aligncenter wp-image-117\" srcset=\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-31-at-9.41.14-PM.png 536w, https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2020-01-31-at-9.41.14-PM-300x296.png 300w\" sizes=\"(max-width: 536px) 100vw, 536px\" \/><\/a> Let $PQ=x$, $NP=y$, we have the following: $$ \\begin{array}{rcrcr}<br \/>\nx &#038; + &#038; 2y &#038; = &#038; \\frac{1}{5} \\\\\n2x &#038; + &#038; y &#038; = &#038; \\frac{1}{4}\n\\end{array} $$\nSolve the above linear equations, we have $x = \\dfrac{1}{10}$, $y=\\dfrac{1}{20}$. Therefore, the area covered by $\\triangle{JKL}$ and $\\triangle{MNO}$ is $$[\\triangle{MNO}] + 3\\cdot [\\triangle{LPQ}] = \\frac{1}{4}(\\frac{1}{5})^2\\cdot\\sqrt{3} + 3\\cdot \\frac{1}{4}\\cdot (\\frac{1}{10})^2 \\cdot \\sqrt{3} = \\frac{7}{400}\\sqrt{3}$$ Therefore, the answer to the question is $$\\frac{\\frac{7}{400}\\sqrt{3}}{[\\triangle{ABC}]}=\\frac{\\frac{7}{400}\\sqrt{3}}{\\frac{\\sqrt{3}}{4}}=\\frac{7}{100}$$<\/p>\n<p><strong>Solution 2 &#8211; Use <a href=\"https:\/\/en.wikipedia.org\/wiki\/Menelaus%27s_theorem\" target=\"_blank\" rel=\"noopener noreferrer\">Menelaus&#8217;s Theorem<\/a><\/strong> Consider $\\triangle{ACD}$ and $\\triangle{ABI}$, apply Menelaus&#8217;s theorem, we have $$\\frac{AB}{BD}\\cdot\\frac{DJ}{JC}\\cdot\\frac{DI}{DA}=1$$ Because point $D$ and $E$ trisect side $AB$, point $H$ and $I$ trisect side $CA$, and $AB=BC=CA=1$, we have $$\\frac{1}{\\frac{2}{3}}\\cdot\\frac{DJ}{JC}\\cdot\\frac{\\frac{2}{3}}{\\frac{1}{3}}=1$$ We have $$\\frac{DJ}{JC}=\\frac{1}{3}$$ Therefore $$\\frac{CJ}{CD}=\\frac{3}{4}$$<br \/>\nSimilarly, consider $\\triangle{BCD}$ and $\\triangle{ABG}$, we have $$\\frac{BA}{AD}\\cdot\\frac{DN}{NA}\\cdot\\frac{CG}{GB}=1$$  Because point $D$ and $E$ trisect side $AB$, point $F$ and $G$ trisect side $BC$, and $AB=BC=CA=1$, we have $$\\frac{1}{\\frac{1}{3}}\\cdot\\frac{DN}{NC}\\cdot\\frac{\\frac{1}{3}}{\\frac{2}{3}}=1$$ We have $$\\frac{DN}{NC}=\\frac{2}{3}$$ Therefore $$\\frac{CN}{CD}=\\frac{3}{5}$$ From here, we can calculate the length of $JK$ and $NM$ and follow the  same steps in Solution 1 to obtain the same answer.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Given an equilateral $\\triangle{ABC}$ shown below, $AB=BC=CA=1$, point $D$ and $E$ trisect side $AB$, point $F$ and $G$ trisect side $BC$, point $H$ and $I$ trisect side $CA$. Two overlapping equilateral triangles are formed by linking various intersect points. 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