{"id":1681,"date":"2021-05-23T05:22:00","date_gmt":"2021-05-23T09:22:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=1681"},"modified":"2024-10-25T12:24:36","modified_gmt":"2024-10-25T16:24:36","slug":"math-olympiad-exercise-2","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=1681","title":{"rendered":"Math Olympiad Exercise \u2013 2"},"content":{"rendered":"\n

$\\triangle{ABC}$ is a right triangle, with $\\angle{ACB}=90^\\circ$ and $\\angle{BAC}=50^\\circ$. Point $D$ and $E$ are on line $BC$ and $AC$ respectively, so that $\\angle{BAD}=\\angle{ABE}=30^\\circ$. Connect $D$ and $E$, find the value of $\\angle{ADE}$. Click here for the solution<\/a>.<\/p>\n\n\n\n

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Solution<\/strong>: Let $AD$ and $BE$ intersect at point $F$. Let $M$ and $N$ on line $AB$ so that $EM\\perp AB$, $DN\\perp AB$. Let $EM$ intersect $AD$ at $G$, $DN$ interset $BE$ at $H$. Additionally, let $GK \\parallel AB$ and intersect $DN$ at $H$, and $KP\\perp AB$.<\/p>\n\n\n\n

It is trivial to show that $\\triangle{EFG}$ and $\\triangle{DFH}$ are equilateral, $\\triangle{DEF}$ and $\\triangle{HGF}$ are congrunt. And $$\\angle{ADE}=\\angle{EHG}=\\angle{HGK}+\\angle{HKG}=\\angle{HGL}+30^\\circ \\tag{1}$$ Additionally $$\\tan{\\angle{HGL}}=\\dfrac{HL}{GL}=\\dfrac{LK\\cdot\\tan{\\angle{HKL}}}{MN}=\\dfrac{NP\\cdot\\tan{30^\\circ}}{MN}$$ $$=\\dfrac{(NB-PB)\\cdot\\tan{30^\\circ}}{MN}=\\dfrac{NB-AM}{\\sqrt{3}\\cdot MN}\\tag{2}$$<\/p>\n\n\n\n

Furthermore $$AM+MN+NB=AB$$ $$EM=AM\\cdot\\tan{50^\\circ}=(MN+NB)\\cdot\\tan{30^\\circ}=\\dfrac{MN+NB}{\\sqrt{3}}$$ $$DN=NB\\cdot\\tan{40^\\circ}=(AM+MN)\\cdot\\tan{30^\\circ}=\\dfrac{AM+MN}{\\sqrt{3}}$$<\/p>\n\n\n\n

Without loss of generality, let $AB=1$, and let $x=AM$, $y=MN$, and $z=NB$, we have $$x+y+z=1$$ $$x\\cdot\\tan{50^\\circ}=\\dfrac{y+z}{\\sqrt{3}}$$ $$z\\cdot\\tan{40^\\circ}=\\dfrac{x+y}{\\sqrt{3}}$$ Therefore $\\dfrac{1}{x}=1+\\sqrt{3}\\cdot\\tan{50^\\circ}$ and $\\dfrac{1}{z}=1+\\sqrt{3}\\cdot\\tan{40^\\circ}$<\/p>\n\n\n\n

According to $(2)$, we have $$\\tan{\\angle{HGL}}=\\dfrac{z-x}{\\sqrt{3}\\cdot y}=\\dfrac{z-x}{\\sqrt{3}\\cdot(1-x-z)}=\\dfrac{\\dfrac{1}{x}-\\dfrac{1}{z}}{\\sqrt{3}\\cdot(\\dfrac{1}{x}\\cdot\\dfrac{1}{z}-\\dfrac{1}{x}-\\dfrac{1}{z})}$$ $$=\\dfrac{(1+\\sqrt{3}\\cdot\\tan{50^\\circ})-(1+\\sqrt{3}\\cdot\\tan{40^\\circ})}{\\sqrt{3}\\cdot((1+\\sqrt{3}\\cdot\\tan{50^\\circ})\\cdot(1+\\sqrt{3}\\cdot\\tan{40^\\circ})-(1+\\sqrt{3}\\cdot\\tan{50^\\circ})-(1+\\sqrt{3}\\cdot\\tan{40^\\circ}))}$$ $$=\\dfrac{\\tan{50^\\circ}-\\tan{40^\\circ}}{3\\cdot\\tan{50^\\circ}\\cdot\\tan{40^\\circ}-1}=\\dfrac{\\tan{50^\\circ}-\\tan{40^\\circ}}{2\\cdot\\tan{50^\\circ}\\cdot\\tan{40^\\circ}-1+\\tan{50^\\circ}\\cdot\\tan{40^\\circ}}$$ $$=\\dfrac{\\tan{50^\\circ}-\\tan{40^\\circ}}{1+\\tan{50^\\circ}\\cdot\\tan{40^\\circ}}=\\tan{(50^\\circ-40^\\circ)}=\\tan{10^\\circ}$$<\/p>\n\n\n\n

Therefore, $\\angle{HGL}=10^\\circ$. Based on $(1)$, we have $\\angle{ADE}=10^\\circ+30^\\circ=\\boxed{40^\\circ}$<\/p>\n\n\n\n

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

$\\triangle{ABC}$ is a right triangle, with $\\angle{ACB}=90^\\circ$ and $\\angle{BAC}=50^\\circ$. Point $D$ and $E$ are on line $BC$ and $AC$ respectively, so that $\\angle{BAD}=\\angle{ABE}=30^\\circ$. Connect $D$ and $E$, find the value of $\\angle{ADE}$. Click here for the solution. Solution: Let $AD$ … 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":[9,15],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1681"}],"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=1681"}],"version-history":[{"count":54,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1681\/revisions"}],"predecessor-version":[{"id":4617,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1681\/revisions\/4617"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1681"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1681"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1681"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}