{"id":232,"date":"2020-01-16T14:36:58","date_gmt":"2020-01-16T14:36:58","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=232"},"modified":"2020-10-16T20:11:02","modified_gmt":"2020-10-16T20:11:02","slug":"mathcounts-exercises-13","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=232","title":{"rendered":"MATHCOUNTS Exercises – 13"},"content":{"rendered":"\n

In regular pentagon $ABCDE$, point $M$ is the midpoint of side $AE$, and segments $AC$ and $BM$ intersect at point $Z$. If $ZA = 3$, what is the value of $AB$? Express your answer in simplest radical form. Source: MATHCOUNTS 2015 State Sprint Round.<\/em> Click here for the solutions.<\/a><\/p>\n\n\n\n

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Solution 1<\/strong> Draw line $\\overline{BE}$ intersecting with line $\\overline{AC}$ at $F$. Obviously, quadrilateral $CDEF$ is a rhombus, and both $\\triangle{BAF}$ and $\\triangle{BEA}$ are similar isosceles triangles.<\/p>\n

\nLet $\\overline{AB}=\\overline{AE}=\\overline{EF}=x$, $\\overline{AF}=\\overline{BF}=y$, $$\\because \\triangle{BAF}\\sim\\triangle{BEA}$$ $$\\therefore \\dfrac{\\overline{BA}}{\\overline{BF}}=\\dfrac{\\overline{BE}}{\\overline{BA}}\\tag{1}$$ $$\\dfrac{x}{y}=\\dfrac{x+y}{x}=\\dfrac{1}{1+\\dfrac{x}{y}}$$ Solve the above equation for $\\dfrac{x}{y}$ and ignore the negative value, we have $$\\dfrac{x}{y}=\\dfrac{\\sqrt{5}+1}{2}\\tag{2}$$\n<\/p>\n

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Let $G$ be the midpoint of $\\overline{AF}$. Since $M$ is also the midpoint of $\\overline{AE}$, we have $\\overline{MF} \\parallel \\overline{BE}$, therefore $$\\triangle{AMZ}\\sim\\triangle{AEF}$$ Because $\\overline{AE}=\\overline{EF}=x$, therefore $$ \\overline{MA}=\\overline{MG}=\\dfrac{\\overline{AE}}{2}=\\dfrac{x}{2}$$ Additionally $$\\because \\triangle{MGZ}\\sim\\triangle{BFZ}$$ $$\\therefore \\dfrac{\\overline{ZG}}{\\overline{ZF}}=\\dfrac{\\overline{MA}}{\\overline{BF}}=\\dfrac{\\dfrac{x}{2}}{y}=\\dfrac{1}{2}\\cdot\\dfrac{x}{y}$$\nTherefore $$\\overline{ZG}+\\overline{ZF}=\\dfrac{1}{2}\\cdot\\dfrac{x}{y}\\cdot\\overline{ZF}+\\overline{ZF}=(\\dfrac{1}{2}\\cdot\\dfrac{x}{y}+1)\\cdot\\overline{ZF}$$\n$$\\because \\overline{ZG}+\\overline{ZF}=\\overline{GF}=\\dfrac{y}{2}$$\n$$\\therefore (\\dfrac{1}{2}\\cdot\\dfrac{x}{y}+1)\\cdot\\overline{ZF}=\\dfrac{y}{2}$$\n$$\\because \\overline{ZF}=\\overline{AF}-\\overline{AZ}=y-3$$\n$$\\therefore (\\dfrac{1}{2}\\cdot\\dfrac{x}{y}+1)\\cdot(y-3)=\\dfrac{y}{2}\\tag{5}$$\n<\/p>\n

Solve equation (2) and (5), we have $$y=\\dfrac{15-3\\sqrt{5}}{2}$$\n<\/p>\n

And finally we have the answer: $$\\overline{AB}=x=\\dfrac{x}{y}\\cdot{y}=\\dfrac{\\sqrt{5}+1}{2}\\cdot\\dfrac{15-3\\sqrt{5}}{2}=3\\sqrt{5}$$\n<\/p>\n\n\n\n

Solution 2 \u2013 Use Mass Points Method<\/strong> Follow the same process as Solution 1 to obtain the result shown in (2).<\/p>\n\n\n\n

Assign $A_{mass}=1$, because $M$ is the midpoint of side $\\overline{AE}$, we have $$E_{mass}=1, \\ \\ \\ M_{m a s s}=A_{mass}+E_{mass}=2$$ $$\\because \\dfrac{\\overline{EF}}{\\overline{BF}}=\\dfrac{x}{y}$$ $$\\therefore B_{mass}=\\dfrac{x}{y}$$ We have $$F_{mass}=E_{mass}+B_{mass}=1+\\dfrac{x}{y}\\tag{6}$$ and balanced mass at $Z$ as $$Z_{mass}=M_{mass}+B_{mass}=A_{mass}+F_{mass}=2+\\dfrac{x}{y}$$ Therefore $$A_{mass}\\cdot\\overline{AZ}=F_{mass}\\cdot\\overline{FZ}=F_{mass}\\cdot(\\overline{AF}-\\overline{AZ})$$ i.e. $$1\\cdot 3=(1+\\dfrac{x}{y})(y-3)\\tag{7}$$<\/p>\n\n\n\n

Solve equation (2) and (7), we have $$y=\\dfrac{15-3\\sqrt{5}}{2}$$<\/p>\n\n\n\n

And finally we have the answer: $$\\overline{AB}=x=\\dfrac{x}{y}\\cdot{y}=\\dfrac{\\sqrt{5}+1}{2}\\cdot\\dfrac{15-3\\sqrt{5}}{2}=3\\sqrt{5}$$<\/p>\n\n\n\n

Solution 3 \u2013 Use\u00a0Menelaus<\/em>‘\u00a0theorem<\/em><\/strong>\u00a0Follow the same process as Solution 1 to obtain the result shown in (2). <\/p>\n\n\n\n

According to Menelaus’\u00a0theorem, we have the following:<\/p>\n\n\n\n

$$\\dfrac{\\overline{EB}}{\\overline{BF}}\\cdot\\dfrac{\\overline{FZ}}{\\overline{AZ}}\\cdot\\dfrac{\\overline{AM}}{\\overline{EM}}=1$$<\/p>\n\n\n\n

Since $\\overline{FZ}=\\overline{AF}-\\overline{AZ}$, we have<\/p>\n\n\n\n

$$\\dfrac{x+y}{y}\\cdot\\dfrac{y-3}{3}\\cdot\\dfrac{\\dfrac{x}{2}}{\\dfrac{x}{2}}=1$$<\/p>\n\n\n\n

i.e. <\/p>\n\n\n\n

$$ (1+\\dfrac{x}{y})\\cdot(y-3)=3\\tag{8}$$<\/p>\n\n\n\n

Solve equation (2) and (8), we have $$y=\\dfrac{15-3\\sqrt{5}}{2}$$<\/p>\n\n\n\n

And finally we have the answer: $$\\overline{AB}=x=\\dfrac{x}{y}\\cdot{y}=\\dfrac{\\sqrt{5}+1}{2}\\cdot\\dfrac{15-3\\sqrt{5}}{2}=3\\sqrt{5}$$<\/p>\n\n\n\n<\/div>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

In regular pentagon $ABCDE$, point $M$ is the midpoint of side $AE$, and segments $AC$ and $BM$ intersect at point $Z$. If $ZA = 3$, what is the value of $AB$? Express your answer in simplest radical form. Source: MATHCOUNTS … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13,9,11],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/232"}],"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=232"}],"version-history":[{"count":43,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/232\/revisions"}],"predecessor-version":[{"id":564,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/232\/revisions\/564"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=232"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=232"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=232"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}