{"id":2573,"date":"2022-04-04T03:59:00","date_gmt":"2022-04-04T07:59:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=2573"},"modified":"2024-10-25T12:30:03","modified_gmt":"2024-10-25T16:30:03","slug":"geometry-challenge-12","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=2573","title":{"rendered":"Geometry Challenge – 12"},"content":{"rendered":"\n

Acute $\\triangle{ABC}$ is inscribed inside circle centered at $O$. $P$ is on $BC$ and $AP\\perp BC$, and $\\angle{ACB}>\\angle{ABC}$. Prove the following:<\/p>\n\n\n\n

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  1. $\\angle{BAC}+\\angle{OBC}=90^\\circ$<\/li>
  2. $\\angle{OAP}=\\angle{ACB}-\\angle{ABC}$<\/li>
  3. If $\\angle{ACB}-\\angle{ABC}\\ge 30^\\circ$, and $MB=MC$, then $MP\\gt CP$<\/li>
  4. If $\\angle{ACB}-\\angle{ABC}\\ge 30^\\circ$, then $\\angle{BAC}+\\angle{POC}<90^\\circ$<\/li><\/ol>\n\n\n\n

    Click here<\/a> for the solutions.\n<\/p>\n

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    Proof<\/strong> 1: Because $O$ is the center of the circle, $\\triangle{OBC}$ is isosceles. and $\\angle{OBC}=\\angle{OCB}$. As $\\angle{BOC}=2\\angle{BAC}$, and $\\angle{BOC} +\\angle{OBC}+\\angle{OCB}=180^\\circ$, we have $$2\\angle{BAC}+\\angle{OBC}+\\angle{OBC}=180^\\circ$$ Therefore $\\boxed{\\angle{BAC}+\\angle{OBC}=90^\\circ}$<\/p>\n\n\n\n

    Proof <\/strong>2: Because $O$ is the center of the circle, $\\triangle{OAC}$ is isosceles. Therefore $$\\angle{OAC}=\\dfrac{1}{2}(180^\\circ-\\angle{AOC})$$ Because $\\angle{AOC}=2\\angle{ABC}$, we have $$\\angle{OAC}=\\dfrac{1}{2}(180^\\circ-2\\angle{ABC})=90^\\circ-\\angle{ABC}$$ Because $AP\\perp BC$, $\\triangle{APC}$ is right, $\\angle{CAP}=90^\\circ-\\angle{ACB}$. Therefore $$\\angle{OAP}=\\angle{OAC}-\\angle{CAP}=(90^\\circ-\\angle{ABC})-(90^\\circ-\\angle{ACB})=\\boxed{\\angle{ACB}-\\angle{ABC}}$$<\/p>\n\n\n\n

    Proof<\/strong> 3: Let $ON\\perp AP$ and $OA=OB=OC=r$.<\/p>\n\n\n\n

    Since $\\triangle{AON}$ is right, and $\\angle{AOP}=\\angle{ACB}-\\angle{ABC}\\ge 30^\\circ$, $$ON=sin(\\angle{AOP})\\cdot OA\\ge sin(30^\\circ)\\cdot r=\\dfrac{r}{2}$$ Since $MB=MC$, and $\\triangle{BOC}$ is isosceles, $OM\\perp BC$, $OMPN$ is a rectangle. Therefore, $MP=ON\\ge \\dfrac{r}{2}$.<\/p>\n\n\n\n

    Since $\\triangle{ABC}$ is acute, $O$ is inside $\\triangle{ABC}$, and $BC<2r$. Therefore $MB=MC=\\dfrac{BC}{2}<r$. Because $MB=MP+CP$, $MP+CP<r$. <\/p>\n\n\n\n

    Because $MP\\ge\\dfrac{r}{2}$, $CP<r-MP\\le r-\\dfrac{r}{2}=\\dfrac{r}{2}$. Since $MP\\ge\\dfrac{r}{2}$, $CP<\\dfrac{r}{2}$, $\\boxed{MP>CP}$<\/p>\n\n\n\n

    Proof <\/strong>4: Since $\\triangle{MOP}$ is right, and $OM\\perp BC$, $OP>MP>CP$. Therefore $\\angle{POC}<\\angle{OCB}$. <\/p>\n\n\n\n

    Since $\\angle{BAC}+\\angle{OCB}=\\angle{BAC}+\\angle{OBC}=90^\\circ$, Therefore <\/p>\n\n\n\n

    $$\\boxed{\\angle{BAC}+\\angle{POC}<\\angle{BAC}+\\angle{OCB}=90^\\circ}$$<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

    Acute $\\triangle{ABC}$ is inscribed inside circle centered at $O$. $P$ is on $BC$ and $AP\\perp BC$, and $\\angle{ACB}>\\angle{ABC}$. Prove the following: $\\angle{BAC}+\\angle{OBC}=90^\\circ$ $\\angle{OAP}=\\angle{ACB}-\\angle{ABC}$ If $\\angle{ACB}-\\angle{ABC}\\ge 30^\\circ$, and $MB=MC$, then $MP\\gt CP$ If $\\angle{ACB}-\\angle{ABC}\\ge 30^\\circ$, then $\\angle{BAC}+\\angle{POC}<90^\\circ$ Click here for the … 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],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2573"}],"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=2573"}],"version-history":[{"count":40,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2573\/revisions"}],"predecessor-version":[{"id":2617,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2573\/revisions\/2617"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2573"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2573"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2573"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}