{"id":3931,"date":"2024-04-04T16:35:00","date_gmt":"2024-04-04T20:35:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=3931"},"modified":"2024-10-25T02:29:08","modified_gmt":"2024-10-25T06:29:08","slug":"geometry-challenge-16","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=3931","title":{"rendered":"Geometry Challenge – 16"},"content":{"rendered":"\n

$O$ is an interior point of regular hexgon $ABCDEF$. <\/p>\n\n\n\n

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

Prove that $[\\triangle{OEF}]=2[\\triangle{OAB}]+2[\\triangle{OCD}]-3[\\triangle{OBC}]$<\/p>

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

\n\n\n\n

Proof 1<\/strong>: Without loss of generality, assume that $ABCDEF$ is unit regular hexgon with its center at $(0,0)$, $A$ at $(-\\dfrac{1}{2},-\\dfrac{\\sqrt{3}}{2})$, $B$ at $(\\dfrac{1}{2},-\\dfrac{\\sqrt{3}}{2})$, $C$ at $(1,0)$, etc. Then the liner equations for line $AB$, $BC$, $CD$, and $EF$ are:<\/p>\n\n\n\n

$y+\\dfrac{\\sqrt{3}}{2}=0$<\/p>\n\n\n\n

$\\sqrt{3}x-y-\\sqrt{3}=0$<\/p>\n\n\n\n

$\\sqrt{3}x+y-\\sqrt{3}=0$<\/p>\n\n\n\n

$\\sqrt{3}x-y+\\sqrt{3}=0$<\/p>\n\n\n\n

Let $O$ at $(a, b)$. Denote $O(XY)$ as the distance from $O$ to line $XY$. The distances from $O$ to $AB$, $BC$, $CD$ and $EF$ are:<\/p>\n\n\n\n

$O(AB)=\\dfrac{\\sqrt{3}}{2}+b$<\/p>\n\n\n\n

$O(BC)=-\\dfrac{\\sqrt{3}a-b-\\sqrt{3}}{\\sqrt{(\\sqrt{3})^2+1^2}}=\\dfrac{b+(1-a)\\sqrt{3}}{2}$<\/p>\n\n\n\n

$O(CD)=-\\dfrac{\\sqrt{3}a+b-\\sqrt{3}}{\\sqrt{(\\sqrt{3})^2+1^2}}=\\dfrac{(1-a)\\sqrt{3}-b}{2}$<\/p>\n\n\n\n

$O(EF)=\\dfrac{\\sqrt{3}a-b+\\sqrt{3}}{\\sqrt{(\\sqrt{3})^2+1^2}}=\\dfrac{(1+a)\\sqrt{3}-b}{2}$<\/p>\n\n\n\n

Therefore: $2\\cdot O(AB)+2\\cdot O(CD)-3\\cdot O(BC)$<\/p>\n\n\n\n

$=2\\cdot(\\dfrac{\\sqrt{3}}{2}+b)+2\\cdot\\dfrac{(1-a)\\sqrt{3}-b}{2}-3\\cdot\\dfrac{b+(1-a)\\sqrt{3}}{2}=\\dfrac{(1+a)\\sqrt{3}-b}{2}=O(EF)$<\/p>\n\n\n\n

which implies $[\\triangle{OEF}]=2[\\triangle{OAB}]+2[\\triangle{OCD}]-3[\\triangle{OBC}]$.<\/p>\n\n\n\n

Proof 2:<\/strong> Without loss of generality, assume that $ABCDEF$ is unit regular hexgon.<\/p>\n\n\n\n

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

Let point $K$, $L$, $M$, and $N$ on $EF$, $AB$, $BC$, and $CD$ respectively so that $OK\\perp EF$, $OL\\perp AB$, $OM\\perp BC$, and $ON\\perp CD$. <\/p>\n\n\n\n

Let point $P$ and $Q$ on $OL$ and $ON$ respectively so that $MP\\perp OL$ and $MQ\\perp ON$.<\/p>\n\n\n\n

Let point $R$ and $S$ on $MP$ and $MQ$ respectively so that $BR\\perp MP$ and $CS\\perp MQ$. Therefore both $BLPR$ and $CNQS$ are rectangles.<\/p>\n\n\n\n

$\\angle{LOM}=540^\\circ-\\angle{OLB}-\\angle{OMB}-\\angle{ABC}=360^\\circ-90^\\circ-90^\\circ-120^\\circ=60^\\circ$<\/p>\n\n\n\n

Therefore, $OP=\\dfrac{1}{2}OM$ and $PL=OL-OP=OL-\\dfrac{1}{2}OM$.<\/p>\n\n\n\n

Because $MP\\perp OL$, $OL\\perp AB$, we have $MP\\parallel AB$, and <\/p>\n\n\n\n

$\\angle{RMB}=180^\\circ-\\angle{ABC}=180^\\circ-120^\\circ=60^\\circ$<\/p>\n\n\n\n

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

$MB=\\dfrac{2}{\\sqrt{3}}RB=\\dfrac{2\\sqrt{3}}{2}PL=\\dfrac{2\\sqrt{3}}{2}(OL-\\dfrac{1}{2}OM)$<\/p>\n\n\n\n

Similarly, we have<\/p>\n\n\n\n

$MC=\\dfrac{2\\sqrt{3}}{3}(ON-\\dfrac{1}{2}OM)$<\/p>\n\n\n\n

Because $MB+MC=BC=1$, we have $$\\dfrac{2\\sqrt{3}}{3}(OL-\\dfrac{1}{2}OM)+\\dfrac{2\\sqrt{3}}{3}(ON-\\dfrac{1}{2}OM)=1\\tag{1}$$<\/p>\n\n\n\n

Simplifying $(1)$, we have $$OL+ON-OM=\\dfrac{\\sqrt{3}}{2} \\tag{2}$$<\/p>\n\n\n\n

Because $BC\\parallel EF$, $OK\\perp EF$ and $OM\\perp BC$, $M$, $O$ and $M$ are co-linear and $$OK+OM=KM=\\sqrt{3}$$ <\/p>\n\n\n\n

Applying $(2)$, We have $$OK=\\sqrt{3}-OM=2(OL+ON-OM)-OM=2\\cdot OL+2\\cdot ON-3\\cdot OM$$<\/p>\n\n\n\n

which implies $[\\triangle{OEF}]=2[\\triangle{OAB}]+2[\\triangle{OCD}]-3[\\triangle{OBC}]$.<\/p>\n\n\n\n

Proof 3:<\/strong> Without loss of generality, assume that $ABCDEF$ is unit regular hexgon.<\/p>\n\n\n\n

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

Let $GL$ pass $O$ and $GL\\perp AB$, $KM$ pass $O$ and $KM\\perp BC$, $HN$ pass $O$ and $HN\\perp CD$. As $ABCDEF$ is a regular hexgon, $AB\\parallel DE$, $BC\\parallel EF$, and $CD\\parallel FA$. Therefore $GL\\perp DE$, $KM\\perp EF$, and $HN\\perp FA$.<\/p>\n\n\n\n

Extend $AB$, $CD$, and $EF$ to form equilateral $\\triangle{TUV}$, with its side length as $3$. Because $OK\\perp VT$, $OL\\perp TU$, and $ON\\perp UV$, we have $$OK+OL+ON=\\dfrac{3\\sqrt{3}}{2}\\tag{1}$$<\/p>\n\n\n\n

as $OK$, $OL$ and $ON$ are heights of $\\triangle{OEF}$, $\\triangle{OAB}$ and $\\triangle{OCD}$ respectively and their sum is equal to the height of equilateral $\\triangle{TUV}$. Similarly, $$OG+OH+OM=\\dfrac{3\\sqrt{3}}{2}\\tag{2}$$<\/p>\n\n\n\n

Additionally, $$OK+OM=\\sqrt{3}\\tag{3}$$ $$OG+OL=\\sqrt{3}\\tag{4}$$ $$ OH+ON=\\sqrt{3}\\tag{5}$$<\/p>\n\n\n\n

Subtracting both side $(3)$ from $(1)$, we have $$OL+ON-OM=\\dfrac{\\sqrt{3}}{2}\\tag{6}$$<\/p>\n\n\n\n

Because $(3)$ and $(6)$, we have $$OK=\\sqrt{3}-OM=2\\cdot(OL+ON-OM)-OM=2\\cdot OL+2\\cdot ON-3\\cdot OM$$<\/p>\n\n\n\n

which implies $[\\triangle{OEF}]=2[\\triangle{OAB}]+2[\\triangle{OCD}]-3[\\triangle{OBC}]$.<\/p>\n\n\n\n

Proof 4<\/strong>: Extend $AB$, $CD$, and $EF$ to form equilateral $\\triangle{TUV}$, with its side length $3\\cdot AB$. <\/p>\n\n\n\n

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

Draw line $OT$, $OU$, and $OV$. Therefore:<\/p>\n\n\n\n

$[\\triangle{OTU}]=[\\triangle{OTA}]+[\\triangle{OAB}]+[\\triangle{OBU}]=3[\\triangle{OAB}]$<\/p>\n\n\n\n

$[\\triangle{OUV}]=[\\triangle{OUC}]+[\\triangle{OCD}]+[\\triangle{ODV}]=3[\\triangle{OCD}]$<\/p>\n\n\n\n

$[\\triangle{OVT}]=[\\triangle{OVE}]+[\\triangle{OEF}]+[\\triangle{OFT}]=3[\\triangle{OEF}]$<\/p>\n\n\n\n

Let the $[\\triangle{FTA}=\\triangle{BUC}=\\triangle{DVE}=S$. The area of hexgon $ABCDEF$ is $6S$, and $[\\triangle{TUV}]=9S$. Therefore<\/p>\n\n\n\n

$$[\\triangle{OTU}]+[\\triangle{OUV}]+[\\triangle{OVT}]=9S$$<\/p>\n\n\n\n

$$3[\\triangle{OAB}]+3[\\triangle{OCD}]+3[\\triangle{OEF}]=9S$$<\/p>\n\n\n\n

$$[\\triangle{OAB}]+[\\triangle{OCD}]+[\\triangle{OEF}]=3S\\tag{1}$$<\/p>\n\n\n\n

Because $BC\\parallel EF$, therefore $$[\\triangle{OBC}]+[\\triangle{OEF}]=2S\\tag{2}$$<\/p>\n\n\n\n

Replace $S$ in $(1)$ with $(2)$, we have:<\/p>\n\n\n\n

$$2([\\triangle{OAB}]+[\\triangle{OCD}]+[\\triangle{OEF}])=3([\\triangle{OBC}]+[\\triangle{OEF}])$$<\/p>\n\n\n\n

We have $[\\triangle{OEF}]=2[\\triangle{OAB}]+2[\\triangle{OCD}]-3[\\triangle{OBC}]$.<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

$O$ is an interior point of regular hexgon $ABCDEF$. Prove that $[\\triangle{OEF}]=2[\\triangle{OAB}]+2[\\triangle{OCD}]-3[\\triangle{OBC}]$ Click here for the proof. Proof 1: Without loss of generality, assume that $ABCDEF$ is unit regular hexgon with its center at $(0,0)$, $A$ at $(-\\dfrac{1}{2},-\\dfrac{\\sqrt{3}}{2})$, $B$ at … 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\/3931"}],"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=3931"}],"version-history":[{"count":68,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3931\/revisions"}],"predecessor-version":[{"id":4567,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/3931\/revisions\/4567"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=3931"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=3931"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=3931"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}