{"id":2076,"date":"2021-08-04T13:38:35","date_gmt":"2021-08-04T17:38:35","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=2076"},"modified":"2024-10-25T12:31:20","modified_gmt":"2024-10-25T16:31:20","slug":"geometry-challenge-6-%e2%ad%90%e2%ad%90%e2%ad%90%e2%ad%90%e2%ad%90","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=2076","title":{"rendered":"Geometry Challenge – 6 \u2b50\u2b50\u2b50\u2b50\u2b50"},"content":{"rendered":"\n

In parallelogram $ABCD$, diagonal $AC$ tangents the incircle of $\\triangle{ABC}$ at $P$. Let $r_1$ and $r_2$ be the radii of incircles of $\\triangle{ADP}$ and $\\triangle{DCP}$ respectively. <\/p>\n\n\n\n

1. Prove that $\\dfrac{r_1}{r_2}=\\dfrac{AP}{CP}$ <\/p>\n\n\n\n

2. If $AD=PD$, and $\\dfrac{AD+CD}{AC}=p$, where $p>1$, prove that $\\dfrac{r_1}{r_2}=1+\\dfrac{1}{p}$.<\/p>\n\n\n\n

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

Solution<\/strong>: Draw the digram as shown. Let the incircle centers of $\\triangle{ABC}$, $\\triangle{DCP}$, and $\\triangle{ADP}$ be $E$, $F$, and $G$ respectively. <\/p>\n\n\n\n

Draw line $EP$, and $EP\\perp AC$, because $AC$ tangents circle $E$ at $P$. Let $AC$ tangent with circle $F$ and $G$ at $L$ and $M$ respectively, so $FL\\perp AC$, $GM\\perp AC$, and $r_1=GM$, $r_2=FL$. Additionally, let $AB$ and $CB$ tangent with circle $E$ at $Q$ and $R$ respectively. <\/p>\n\n\n\n

Draw line $AE$, $AG$, $CE$, $CF$, which are angle bisectors of $\\angle{BAC}$, $\\angle{CAD}$, $\\angle{ACB}$, and $\\angle{ACD}$ respectively.<\/p>\n\n\n\n

1. Proof<\/strong>: Because $ABCD$ is a parallelogram, we have $AD+AB=CD+BC$. Because $BR$ and $BQ$ tangent circle $E$, $BR=BQ$. Then $AD+AB-BQ=CD+BC-BR$. Because $AB-BQ=AQ$, $BC-BR=CR$, we have $AD+AQ=CD+CR$. Because $AQ$ and $AP$ tangent circle $E$, $AQ=AP$; $CR$ and $CP$ tangent circle $E$, $CR=CP$. Therefore $$AD+AP=CD+CP\\tag{1}$$<\/p>\n\n\n\n

Because $AM$ tangents circle $G$ at $M$, and $CL$ tangents circle $F$ at $L$, we have $AM=\\dfrac{1}{2}{AD+AP-DP}$, $CL=\\dfrac{1}{2}{CD+CP-DP}$. Because $AD+AP=CD+CP$, we have $$AM=CL\\tag{2}$$<\/p>\n\n\n\n

Because $AG$ and $CE$ are the angle bisector of $\\angle{CAD}$ and $\\angle{ACB}$ respectively, $\\angle{CAG}=\\dfrac{1}{2}\\angle{CAD}$, $\\angle{ACE}=\\dfrac{1}{2}\\angle{ACB}$. Because $ABCD$ is a parallelogram with $AD\\parallel BC$, we have $\\angle{CAD}=\\angle{ACB}$. Therefore $\\angle{CAG}=\\angle{ACE}$. As $M$ and $P$ is between $A$ and $C$, we have $\\angle{MAG}=\\angle{PCE}$. Because $M$ and $P$ are tangent points to circle $G$ and $F$ respectively, both $\\triangle{MAG}$ and $\\triangle{PCE}$ are right triangles. Therefore $\\triangle{MAG}\\sim \\triangle{PCE}$. We have $$\\dfrac{GM}{AM}=\\dfrac{EP}{CP}\\tag{3}$$ Similarly, we have $\\triangle{LCF}\\sim \\triangle{PAG}$, and $$\\dfrac{FL}{CL}=\\dfrac{EP}{AP}\\tag{4}$$ Based on $(2)$, $(3)$, and $(4)$, we have $$\\dfrac{GM}{FL}=\\dfrac{\\dfrac{GM}{AM}}{\\dfrac{FL}{CL}}=\\dfrac{\\dfrac{EP}{CP}}{\\dfrac{EP}{AP}}=\\dfrac{AP}{CP}$$ Because $r_1=GM$, $r_2=FL$, we have $$\\boxed{\\dfrac{r_1}{r_2}=\\dfrac{AP}{CP}}$$<\/p>\n\n\n\n

2. Proof<\/strong>: Let $m=AP$, $n=CP$, $a=AC=m+n$, $b=CD$, $c=AD$, $d=PD$. <\/p>\n\n\n\n

Given the assumptions that $AD=PD$, and $\\dfrac{AD+CD}{AC}=p$, we have $$c=d\\ \\ \\ \\ \\ \\ \\ \\tag{5}$$ $$\\dfrac{b+c}{m+n}=p\\tag{6}$$ Additionally, based on $(1)$, we have $$c+m=b+n\\tag{7}$$ Solving $b$ and $c$ in equation $(6)$ and $(7)$, we have $$b=\\dfrac{(p+1)m+(p-1)n}{2}\\tag{8}$$ $$c=\\dfrac{(p-1)m+(p+1)n}{2}\\tag{9}$$<\/p>\n\n\n\n

According to Stuart’s Theorem<\/a>, we have $$man+dad=bmb+cnc$$ Substituting $a$, $b$, $c$, $d$ in the above equation, we have $$mn(m+n)+(m+n)(\\dfrac{(p-1)m+(p+1)n}{2})^2$$ $$=m(\\dfrac{(p+1)m+(p-1)n}{2})^2+n(\\dfrac{(p-1)m+(p+1)n}{2})^2$$ Simplifying and factorizing the above equation, we have $$m(m+n)((p+1)n-pm)=0$$ Because $m=AP>0$ and $n=CP>0$, we have $$(p+1)n-pm=0$$ Therefore $$\\dfrac{m}{n}=\\dfrac{p+1}{p}=1+\\dfrac{1}{p}$$ Based on the result of Proof<\/strong> 1<\/strong>, we have $$\\boxed{\\dfrac{r_1}{r_2}=\\dfrac{AP}{CP}=\\dfrac{m}{n}=1+\\dfrac{1}{p}}$$<\/p>\n\n\n\n<\/div>\n\n\n\n

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

In parallelogram $ABCD$, diagonal $AC$ tangents the incircle of $\\triangle{ABC}$ at $P$. Let $r_1$ and $r_2$ be the radii of incircles of $\\triangle{ADP}$ and $\\triangle{DCP}$ respectively. 1. Prove that $\\dfrac{r_1}{r_2}=\\dfrac{AP}{CP}$ 2. If $AD=PD$, and $\\dfrac{AD+CD}{AC}=p$, where $p>1$, prove that $\\dfrac{r_1}{r_2}=1+\\dfrac{1}{p}$. … 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":[13,14,9],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2076"}],"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=2076"}],"version-history":[{"count":45,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2076\/revisions"}],"predecessor-version":[{"id":2130,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2076\/revisions\/2130"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2076"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2076"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2076"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}