{"id":5393,"date":"2026-01-08T16:57:18","date_gmt":"2026-01-08T20:57:18","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=5393"},"modified":"2026-01-08T16:57:44","modified_gmt":"2026-01-08T20:57:44","slug":"geometry-challenge-2026-01-08","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=5393","title":{"rendered":"Geometry Challenge – 2026\/01\/08"},"content":{"rendered":"\n

Let $O$ be the circumcenter of $\\triangle{ABC}$, and $O$ does not lies on $AB$, $BC$ or $CA$. Let $D$, $E$, and $F$ be circumcenters of $\\triangle{OBC}$, $\\triangle{OCA}$, and $\\triangle{OAB}$, respectively. Let $G$ be the the circumcenter of $\\triangle{OEF}$. Prove that $D$, $O$, and $G$ are collinear.\ud83d\udd11<\/a><\/p>\n

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

Proof: <\/strong>Obviously, $\\triangle{ABC}$ cannot be right, otherwise, $O$ will be on its hypotenuse, and the radius of one of the circumcircles of $\\triangle{OBC}$, $\\triangle{OCA}$, and $\\triangle{OAB}$ will be infinite.<\/p>\n\n\n\n

Case 1<\/strong>: If $\\triangle{ABC}$ is acute, then $O$ is inside of $\\triangle{ABC}$. <\/p>\n\n\n\n

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

Draw line $OA$, $OB$, $OC$, $DE$, $EF$, $FD$. Additionally, $OA$ and $EF$ intersect a $P$, $OB$ and $FD$ at $Q$, and $OC$ and $DE$ at $R$. <\/p>\n\n\n\n

Draw line $OD$, $OE$, $OF$, and $OG$. And $OD$ and $BC$ intersect at $S$, $OE$ and $AC$ intersect at $T$.<\/p>\n\n\n\n

Because $OA$ is the common chord of circle centered at $E$ and $F$, $EF$ is perpendicular bisector of $OA$, so we have $OP=AP=\\dfrac{1}{2}OA$, $OP\\perp EF$. Similarly, we have $OQ=BQ=\\dfrac{1}{2}OB$, $OQ\\perp FD$, and $OR=CR=\\dfrac{1}{2}OC$, $OR\\perp DE$.<\/p>\n\n\n\n

Additionally, because $O$ is the circumcenter of $\\triangle{ABC}$, we have $OA=OB=OC$, therefore $O$ is the incenter of $\\triangle{DEF}$.<\/p>\n\n\n\n

Beacause $OP\\perp EF$ and $OQ\\perp FD$, we have $\\angle{PFQ}+\\angle{POQ}=180^\\circ$, i.e. $\\angle{PFQ}=180^\\circ-\\angle{POQ}=180^\\circ-\\angle{AOB}$.<\/p>\n\n\n\n

Because $O$ is the circumcenter of acute $\\triangle{ABC}$, we have $\\angle{AOB}=2\\angle{ACB}$. Therefore, $\\angle{PFQ}=180^\\circ-2\\angle{ACB}$<\/p>\n\n\n\n

Because $O$ is the incenter of $\\triangle{DEF}$, we have $$\\angle{EFO}=\\dfrac{1}{2}\\angle{PFO}=\\dfrac{1}{2}\\angle{PFQ}=\\dfrac{1}{2}(180^\\circ-2\\angle{ACB})=90^\\circ-\\angle{ACB}$$<\/p>\n\n\n\n

Because $O$, $E$ and $F$ are on the circle centered at $G$, $\\triangle{OGE}$ is isosceles. Therefore $$\\angle{EOG}=\\angle{OEG}=\\dfrac{1}{2}(180^\\circ-\\angle{EGO})=\\dfrac{1}{2}(180^\\circ-2\\angle{EFO})$$ $$=90^\\circ-\\angle{EFO}=90^\\circ-(90^\\circ-\\angle{ACB})=\\angle{ACB}$$<\/p>\n\n\n\n

Because $BC$ is the common chord of circles centered at $O$ and $D$, $OD\\perp BC$. Therefore $\\angle{OSC}=90^\\circ$. Similarly, $\\angle{OTC}=90^\\circ$. Therefore, $\\angle{ACB}+\\angle{SOT}=180^\\circ$. i.e. $\\angle{ACB}+\\angle{DOE}=180^\\circ$.<\/p>\n\n\n\n

Because $\\angle{ACB}=\\angle{EOG}$, we have $\\angle{EOG}+\\angle{DOE}=180^\\circ$, i.e $\\angle{DOG}=180^\\circ$. Therefore, $D$, $O$, and $G$ are collinear. <\/p>\n\n\n\n

Case 2:<\/strong> If $\\triangle{ABC}$ is obtuse, assume $\\angle{ACB}>90^\\circ$, then $O$ is outside of $\\triangle{ABC}$, on the exterior side of $AB$.<\/p>\n\n\n\n

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

Draw line $OA$, $OB$, $OC$, $DE$, $EF$, $FD$. Extend $FE$ to intersect $OA$ at $E’$. Extend $FD$ to intersect $OB$ at $D’$, and $FE$ to intersect $OA$ at $E’$. Additionally, $OC$ and $DE$ intersect at $R$.<\/p>\n\n\n\n

Draw line $OD$,$OE$, $OF$, and $OG$. <\/p>\n\n\n\n

Because $O$ is the circumcenter of $\\triangle{ABC}$, with $\\angle{ACB}>90^\\circ$, we have $$\\angle{AOB}=\\angle{AOB}+\\angle{AOC}=2\\angle{BAC}+2\\angle{ABC}=2(\\angle{BAC}+\\angle{ABC})=2(180^\\circ-\\angle{ACB})=360^\\circ-2\\angle{ACB}$$<\/p>\n\n\n\n

Because $OA$ is the common chord of circles centered at $E$ and $F$, $EF$ is perpendicular bisector of $OA$, so we have $OE’=AD’=\\dfrac{1}{2}OA$, $OE’\\perp E’F$, i.e. $OA\\perp E’F$. Similarly, $OD’=BD’=\\dfrac{1}{2}OB$, $OD’\\perp D’F$; $OR=CR=\\dfrac{1}{2}OC$, $OR\\perp DE$.<\/p>\n\n\n\n

Additionally, because $O$ is the circumcenter of $\\triangle{ABC}$, we have $OA=OB=BC$. Therefore, $O$ is the excenter opposite vertex $F$ of $\\triangle{DEF}$, with tangent points at $D’$, $E’$, and $R$. <\/p>\n\n\n\n

Therefore, $EE’$ and $ER$ are tangent to circle centered at $O$, $EE’=ER$. Because $OE’=OR$, we have $\\triangle{EOE’}\\cong\\triangle{EOR}$. Therefore $\\angle{EOE’}=\\angle{EOR}$. Similarly, we have $\\angle{DOD’}=\\angle{DOR}$<\/p>\n\n\n\n

Therefore, $$\\angle{DOE}=\\angle{DOR}+\\angle{EOR}=\\dfrac{1}{2}\\angle{E’OR}+\\dfrac{1}{2}\\angle{D’OR}=\\dfrac{1}{2}(\\angle{E’OR}+\\angle{D’OR})=\\dfrac{1}{2}\\angle{D’O’E}=\\dfrac{1}{2}{AOB}$$<\/p>\n\n\n\n

Because $OD’=OE’$, $OD’\\perp D’F$, $OE’\\perp E’F$’, we have $\\triangle{D’OF}\\cong \\triangle{E’OF}$. Thus, $\\angle{D’FO}=\\angle{E’FO}$ and $\\angle{OD’F}=\\angle{OE’F}$, i.e. $OF$ is angle bisector of both $\\angle{D’FE’}$ and $\\angle{AOB}$.<\/p>\n\n\n\n

Because $O$, $E$, and $F$ are on the circle centered at $G$, $\\triangle{EOG}$ is isosceles. Therefore $$\\angle{GOE}=\\angle{GEO}=\\dfrac{1}{2}(180^\\circ-\\angle{OGE}) =\\dfrac{1}{2}(180^\\circ-2\\angle{EFO})=90^\\circ-\\angle{EFO}=90^\\circ-\\angle{E’FO}$$<\/p>\n\n\n\n

Because $OE’\\perp E’F$’, $\\triangle{OE’F}$ is right, therefore $\\angle{E’OF}=90^\\circ-\\triangle{E’FO}$ Therefore $$\\angle{GOE}=\\angle{E’OF}$$<\/p>\n\n\n\n

Because $\\angle{E’FO}=\\angle{AOF}$, and $OF$ is angle bisector of $\\angle{AOB}$, we have $\\angle{GOE}=\\dfrac{1}{2}\\angle{AOB}$. Thus, $\\angle{DOE}=\\angle{GOE}$. Therefore, $D$, $O$, and $G$ are collinear.<\/p>\n\n\n\n

Combining Case 1 and 2, we proved that $D$, $O$, and $G$ are collinear, if $\\triangle{ABC}$ is not right.<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Let $O$ be the circumcenter of $\\triangle{ABC}$, and $O$ does not lies on $AB$, $BC$ or $CA$. Let $D$, $E$, and $F$ be circumcenters of $\\triangle{OBC}$, $\\triangle{OCA}$, and $\\triangle{OAB}$, respectively. Let $G$ be the the circumcenter of $\\triangle{OEF}$. Prove that … 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\/5393"}],"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=5393"}],"version-history":[{"count":30,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5393\/revisions"}],"predecessor-version":[{"id":5564,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5393\/revisions\/5564"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5393"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5393"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5393"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}