{"id":4186,"date":"2024-06-17T09:27:17","date_gmt":"2024-06-17T13:27:17","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=4186"},"modified":"2024-10-25T15:19:48","modified_gmt":"2024-10-25T19:19:48","slug":"aime-challenge-1","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=4186","title":{"rendered":"Algebra\/Geometry Challenge – 1"},"content":{"rendered":"\n

As shown in the following figure, $D$, $E$, $F$ are on the sides of $\\triangle{ABC}$, $AC$, $AB$, and $BC$ respectively. $AE=BE$, $AD=6$, $CD=7$, $BF=2$, $CF=9$. $DEFG$ is a square. The length of $AB$ can be expressed as $\\ \\ \\ a\\sqrt{\\dfrac{b}{c}}\\ \\ \\ $ in its simplest form. Find the value of $a+b+c$. Click here<\/a> for the solution.\n<\/p>\n\n\n\n

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

Solution:<\/strong> Draw line $AF$, $BD$ and $DF$. Let $S$ denote the area of a shape. We have:<\/p>\n\n\n\n

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

$S_{\\triangle{ADE}}=\\dfrac{1}{2}S_{\\triangle{ABD}}=\\dfrac{1}{2}\\cdot\\dfrac{AD}{AD+CD}S_{\\triangle{ABC}}$$ $$=\\dfrac{1}{2}\\cdot\\dfrac{6}{6+7}S_{\\triangle{ABC}}=\\dfrac{3}{13}S_{\\triangle{ABC}}$<\/p>\n\n\n\n

$S_{\\triangle{BEF}}=\\dfrac{1}{2}S_{\\triangle{BAF}}=\\dfrac{1}{2}\\cdot\\dfrac{BF}{BF+CF}S_{\\triangle{ABC}}=$$ $$\\dfrac{1}{2}\\cdot\\dfrac{1}{2+9}S_{\\triangle{ABC}}=\\dfrac{1}{11}S_{\\triangle{ABC}}$<\/p>\n\n\n\n

$S_{\\triangle{ADF}}=\\dfrac{BD}{AD+BD}S_{\\triangle{AFC}}=\\dfrac{7}{6+7}\\cdot\\dfrac{CF}{BF+CF}S_{\\triangle{ABC}}$$ $$=\\dfrac{7}{13}\\cdot\\dfrac{9}{2+9}S_{\\triangle{ABC}}=\\dfrac{63}{143}S_{\\triangle{ABC}}$<\/p>\n\n\n\n

$S_{\\triangle{DEF}}=S_{\\triangle{ABC}}-S_{\\triangle{ADE}}-S_{\\triangle{BEF}}-S_{\\triangle{ADF}}$$ $$=(1-\\dfrac{3}{13}-\\dfrac{1}{11}-\\dfrac{63}{143})S_{\\triangle{ABC}}=\\dfrac{34}{143}S_{\\triangle{ABC}}$<\/p>\n\n\n\n

$S_{CDGF}=S_{\\triangle{ADF}}-S_{\\triangle{GDF}}=S_{\\triangle{ADF}}-S_{\\triangle{DEF}}$$ $$=(\\dfrac{63}{143}-\\dfrac{34}{143})S_{\\triangle{ABC}}=\\dfrac{29}{143}S_{\\triangle{ABC}}$<\/p>\n\n\n\n

Rotate $A$ $90^\\circ$ counter-clockwise around $D$ to point $H$, and draw line $CH$, $DH$, $FH$, $GH$:<\/p>\n\n\n\n

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

Because $\\angle{ADE}+\\angle{HDE}=90^\\circ$, $\\angle{HDG}+\\angle{HDE}=90^\\circ$, $\\angle{ADE}=\\angle{HDG}$. Given $AD=HD$, $DE=DG$, $\\triangle{ADE}\\cong\\triangle{HDG}$.<\/p>\n\n\n\n

Because $\\angle{FGH}+\\angle{DGH}=90^\\circ$, $\\angle{FEB}+\\angle{DEA}=90^\\circ$, and $\\angle{DGH}=\\angle{DEA}$, we have $\\angle{FGH}=\\angle{FEB}$. <\/p>\n\n\n\n

Additionally, because $GF=EF$, $HG=AE=BE$, we have $\\triangle{BEF}\\cong\\triangle{HGF}$. Therefore $H$ is also the point when $B$ is rotated $90^\\circ$ clockwise around $F$. Both $\\triangle{CDH}$ and $\\triangle{CFH}$ are right triangles.<\/p>\n\n\n\n

$S_{CDHF}=S_{CDGF}+S_{\\triangle{HDG}}+S_{\\triangle{HGF}}=S_{CDGF}+S_{\\triangle{ADE}}+S_{\\triangle{BEF}}$$ $$=(\\dfrac{29}{143}+\\dfrac{3}{13}+\\dfrac{1}{11})S_{\\triangle{ABC}}=\\dfrac{75}{143}S_{\\triangle{ABC}}$<\/p>\n\n\n\n

Because<\/p>\n\n\n\n

$S_{CDHF}=S_{\\triangle{CDH}}+S_{\\triangle{CFH}}=\\dfrac{1}{2}CD\\cdot HD+\\dfrac{1}{2}CF\\cdot HF=\\dfrac{1}{2}(7\\cdot 6+9\\cdot 2)=30$<\/p>\n\n\n\n

Therefore <\/p>\n\n\n\n

$\\dfrac{75}{143}S_{\\triangle{ABC}}=30$<\/p>\n\n\n\n

$S_{\\triangle{ABC}}=\\dfrac{286}{5}$<\/p>\n\n\n\n

$S_{DEFG}=2\\cdot S_{\\triangle{DEF}}=2\\cdot\\dfrac{34}{143}\\cdot S_{\\triangle{ABC}}=2\\cdot\\dfrac{34}{143}\\cdot \\dfrac{286}{5}=\\dfrac{136}{5}$<\/p>\n\n\n\n

Let $x$ be the length of $AE$, $s$ the length of semi-perimeter of $\\triangle{ABC}$, we have:<\/p>\n\n\n\n

$x=\\dfrac{AB+BC+CA}{2}=\\dfrac{AE+CE+BF+CF+CD+AD}{2}$$ $$=\\dfrac{x+x+2+9+7+6}{2}=x+12$<\/p>\n\n\n\n

Per Heron’s theorem, $S_{\\triangle{ABC}}=\\sqrt{(x-12)(x+1)(x-1)(12-x)}=\\dfrac{286}{5}$<\/p>\n\n\n\n

$$(x^2-1)(x^2-144)+\\dfrac{286^2}{25}=0$$<\/p>\n\n\n\n

$$(5x^2)^2-5\\cdot 145\\cdot(5x^2)+25\\cdot 144+286^2=0$$<\/p>\n\n\n\n

$$(5x^2-148)(5x^2-577)=0$$<\/p>\n\n\n\n

$$x=\\pm 2\\sqrt{\\dfrac{37}{5}}\\ \\ \\ \\ \\ \\ \\ \\ x=\\pm\\sqrt{\\dfrac{577}{5}}$$<\/p>\n\n\n\n

Because $x=\\sqrt{\\dfrac{577}{5}}$ would result in a triangle that does not meet the description, and $x$ must be positive, therefore $x=2\\sqrt{\\dfrac{37}{5}}$ and $AE=2x=4\\sqrt{\\dfrac{37}{5}}$. We have $$a+b+c=4+37+5=\\boxed{046}$$<\/p>\n\n\n\n

Note:<\/strong> Assume $AD=a$, $CD=b$, $BF=c$, $CF=d$:<\/p>\n\n\n\n

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

Let $$t=\\dfrac{a}{a+b}+\\dfrac{c}{c+d}+\\dfrac{2bd}{(a+b)(c+d)}$$<\/p>\n\n\n\n

We have $$S_{DEFG}=\\dfrac{2-t}{2(t-1)}(a\\cdot b+c\\cdot d)$$<\/p>\n\n\n\n

$$S_{\\triangle{ABC}}=\\dfrac{a\\cdot b+c\\cdot d}{2(t-1)}$$<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

As shown in the following figure, $D$, $E$, $F$ are on the sides of $\\triangle{ABC}$, $AC$, $AB$, and $BC$ respectively. $AE=BE$, $AD=6$, $CD=7$, $BF=2$, $CF=9$. $DEFG$ is a square. The length of $AB$ can be expressed as $\\ \\ \\ … 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":[13,9],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4186"}],"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=4186"}],"version-history":[{"count":67,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4186\/revisions"}],"predecessor-version":[{"id":4628,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4186\/revisions\/4628"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4186"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4186"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4186"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}