{"id":2399,"date":"2024-08-19T06:03:57","date_gmt":"2024-08-19T10:03:57","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=2399"},"modified":"2024-10-25T12:39:03","modified_gmt":"2024-10-25T16:39:03","slug":"a-mathcounts-problem-with-multiple-answers-%f0%9f%99%82","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=2399","title":{"rendered":"A MATHCOUNTS Problem with Multiple Answers \ud83d\ude42"},"content":{"rendered":"\n
\"\"<\/figure><\/div>\n\n\n\n

The following is MATHCOUNTS 2012 National Competition<\/em> Sprint Round Problem 22<\/em>:<\/p>\n\n\n\n

In circle $O$, shown, $OP=2$ units, $PL=8$ units, $PK=9$ units and $NK=18$ units. Points $K$, $P$ and $M$ are collinear, as are points $L$, $P$, $O$ and $N$. What is the length of segment $MN$?<\/p>\n\n\n\n

Answer 1:<\/strong> Since $\\triangle{KPN}\\sim\\triangle{LPM}$, we have $$\\dfrac{ML}{PL}=\\dfrac{NK}{PK}=\\dfrac{18}{9}=2$$ $$ML=2\\cdot PL=2\\cdot 8=16$$Since $\\triangle{LMN}$ is a right triangle, therefore $$MN=\\sqrt{LN^2-ML^2}=\\sqrt{20^2-16^2}=\\boxed{12}$$<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Answer 2: <\/strong>Draw line $KL$, $\\triangle{LKN}$ is a right triangle. $$LK=\\sqrt{LN^2-NK^2}=\\sqrt{20^2-18^2}=2\\sqrt{19}$$<\/p>\n\n\n\n

Since $\\triangle{KPL}\\sim\\triangle{NPM}$, we have: $$\\dfrac{MN}{LK}=\\dfrac{PN}{PK}=\\dfrac{12}{9}=\\dfrac{4}{3}$$ $$MN=\\dfrac{4}{3}\\cdot LK=\\dfrac{4}{3}\\cdot 2\\sqrt{19}=\\boxed{\\dfrac{8\\sqrt{19}}{3}\\approx 11.62}$$<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Answer 3:<\/strong> Since $\\triangle{KPN}\\sim\\triangle{LPM}$, we have $$\\dfrac{PM}{PL}=\\dfrac{PN}{PK}=\\dfrac{12}{9}=\\dfrac{4}{3}$$ $$PM=\\dfrac{4}{3}\\cdot PK=\\dfrac{4}{3}\\cdot 8=\\dfrac{32}{3}$$ Draw line $OM$. By the Law of Cosines, we have: $$\\cos\\angle{MOP}=\\dfrac{OP^2+OM^2-PM^2}{2\\cdot OP\\cdot OM}=\\dfrac{2^2+10^2-(\\dfrac{32}{3})^2}{2\\cdot 2\\cdot 10}=-\\dfrac{11}{45}$$<\/p>\n\n\n\n

Since $\\cos\\angle{MON}=\\cos(180^\\circ-\\angle{MOP})=-\\cos\\angle{MOP}=\\dfrac{11}{45}$, therefore $$MN=\\sqrt{OM^2+ON^2-2\\cdot OM\\cdot ON\\cos\\angle{MON}}$$ $$=\\sqrt{10^2+10^2-2\\cdot 10\\cdot 10\\cdot\\dfrac{11}{45}}=\\boxed{\\dfrac{4\\sqrt{85}}{3}\\approx 12.29}$$<\/p>\n\n\n\n

Answer 4:<\/strong> Since $\\triangle{KPN}\\sim\\triangle{LPM}$, we have $\\dfrac{PM}{PL}=\\dfrac{PN}{PK}=\\dfrac{12}{9}=\\dfrac{4}{3}$. Therefore $$PM=\\dfrac{4}{3}\\cdot PK=\\dfrac{4}{3}\\cdot 8=\\dfrac{32}{3}$$<\/p>\n\n\n\n

By the Law of of Cosines, we have: $$\\cos\\angle{KPN}=\\dfrac{PK^2+PN^2-NK^2}{2\\cdot PK\\cdot PN}=\\dfrac{9^2+12^2-18^2}{2\\cdot 9\\cdot 12}=-\\dfrac{11}{24}$$ <\/p>\n\n\n\n

Since $\\cos\\angle{MPN}=\\cos(180^\\circ-\\angle{KPN})=-\\cos\\angle{KPN}=\\dfrac{11}{24}$, therefore $$MN=\\sqrt{PN^2+PM^2-2\\cdot PN\\cdot PM\\cdot\\cos\\angle{MPN}}$$ $$=\\sqrt{12^2+(\\dfrac{32}{3})^2-2\\cdot 12\\cdot \\dfrac{32}{3}\\cdot \\dfrac{11}{24}}=\\boxed{\\dfrac{4\\sqrt{79}}{3}\\approx 11.85}$$<\/p>\n\n\n\n

Reason: <\/strong>The problem was given with incorrect length constraints, as explained by MATHCOUNTS Errata for 2010-2011 through 2014-2015<\/a> with the first two answers. In fact, if $NK=18$ units, $PK\\neq 9$ units, as shown below:<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Draw line $KL$ and line $KQ$ perpendicular to line $LN$, intersecting $LN$ at $Q$. Since $\\triangle{LKN}$ is a right triangle, we have: $$LK=\\sqrt{LN^2-NK^2}=\\sqrt{20^2-18^2}=2\\sqrt{19}$$ $$KQ=\\dfrac{NK\\cdot LK}{LN}=\\dfrac{18\\cdot 2\\sqrt{19}}{20}=\\dfrac{9\\sqrt{19}}{5}$$ $$NQ=\\sqrt{NK^2-KQ^2}=\\sqrt{18^2-(\\dfrac{9\\sqrt{19}}{5})^2}=\\dfrac{81}{5}$$ $$PQ=NQ-PN=\\dfrac{81}{5}-12=\\dfrac{21}{5}$$ Therefore $$PK=\\sqrt{KQ^2+PQ^2}=\\sqrt{(\\dfrac{9\\sqrt{19}}{5})^2+(\\dfrac{21}{5})^2}=\\dfrac{6\\sqrt{55}}{5}\\approx 8.90 \\neq 9$$ <\/p>\n\n\n\n

Because of incorrect length constraints, different calculation methods may lead to different answers $\\boxed{\ud83d\ude42}$.<\/p>\n","protected":false},"excerpt":{"rendered":"

The following is MATHCOUNTS 2012 National Competition Sprint Round Problem 22: In circle $O$, shown, $OP=2$ units, $PL=8$ units, $PK=9$ units and $NK=18$ units. Points $K$, $P$ and $M$ are collinear, as are points $L$, $P$, $O$ and $N$. What … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[7,9,11,15],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2399"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=2399"}],"version-history":[{"count":78,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2399\/revisions"}],"predecessor-version":[{"id":2486,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2399\/revisions\/2486"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2399"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2399"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2399"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}