{"id":4145,"date":"2024-05-21T22:00:00","date_gmt":"2024-05-22T02:00:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=4145"},"modified":"2024-10-25T02:15:42","modified_gmt":"2024-10-25T06:15:42","slug":"circles-in-a-square-part-12","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=4145","title":{"rendered":"Circles in a Square – Part 12"},"content":{"rendered":"\n

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As shown in the figure, $ABCD$ is a square, $E$ is the mid-point of $AB$. The circle with its center at $H$ is tangent with $AD$, $AE$ and $DE$. The circle with its center at $F$ is tangent with $BC$, $BE$ and $DE$. The circle with its center at $G$ is tangent with $BC$, $CD$ and $DE$. Prove that $\\triangle{FGH}$ is a right triangle. Click here<\/a> for the proof.\n<\/p>\n\n\n\n

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Proof:<\/strong> Extend $DE$ and $CB$ so that they intersect at $K$. Draw diagonals of $ABCD$ so that they intersect at $O$. Let $M$ and $N$ are tangent points of circle $F$ on $DE$ and $AB$ respectively, we have $FM\\perp DE$ and $FN\\perp AB$. Additionally, without loss of generality, we assume that $ABCD$ is a unit square. It is obvious that $F$ is on $BD$, $G$ and $H$ are on $AC$.<\/p>\n\n\n\n

Let the radius of circle $F$, $G$, $H$ as $f$, $g$, $h$ respectively. We have ${DF}^2={FM}^2+{DM}^2$<\/p>\n\n\n\n

$DF=DB-BF=\\sqrt{2}-f\\sqrt{2}=(1-f)\\sqrt{2}$<\/p>\n\n\n\n

$DM=DE-ME=DE-NE=\\dfrac{\\sqrt{5}}{2}-(\\dfrac{1}{2}-f)=\\dfrac{\\sqrt{5}-1}{2}+f$<\/p>\n\n\n\n

We have $$((1-f)\\sqrt{2})^2=(\\dfrac{\\sqrt{5}-1}{2}+f)^2+f^2$$<\/p>\n\n\n\n

Solve the above equation, we have $$f=\\dfrac{\\sqrt{5}-1}{4}$$<\/p>\n\n\n\n

Since the perimeter of $\\triangle{DAE}$ is $\\dfrac{3+\\sqrt{5}}{2}$, and its area is $\\dfrac{1}{4}$. Therefore, $$h=\\dfrac{2\\cdot\\dfrac{1}{4}}{\\dfrac{3+\\sqrt{5}}{2}}=\\dfrac{3-\\sqrt{5}}{4}$$<\/p>\n\n\n\n

Since $\\triangle{DAE}\\sim \\triangle{KCD}$, and $CD=2{AE}$, therefore $$g=2h=\\dfrac{3-\\sqrt{5}}{2}$$<\/p>\n\n\n\n

It is easy to show that ${FO}^2={GO}\\cdot{HO}$, therefore, $\\triangle{FGH}$ is a right triangle.<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

As shown in the figure, $ABCD$ is a square, $E$ is the mid-point of $AB$. The circle with its center at $H$ is tangent with $AD$, $AE$ and $DE$. The circle with its center at $F$ is tangent with $BC$, … 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":[12,9],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4145"}],"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=4145"}],"version-history":[{"count":33,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4145\/revisions"}],"predecessor-version":[{"id":4562,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4145\/revisions\/4562"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4145"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4145"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4145"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}