Where can I see the results?

$ S_{OAF}=\frac{1}{2}.AF.AO.\sin (\angle OAF)=\frac{1}{2}.\cos( A ). AC . R.\sin ( \frac{\pi}{2}-\angle C)=\frac{R}{2}.b.\cos( \angle A).\cos (\angle C) $ by the same way we get : $S_{OBF}=\frac{R}{2}. a. \cos (C) . \cos (B) $ $S_{OBD}=\frac{R}{2}. c. \cos (A) . \cos (B) $ $S_{OCD}=\frac{R}{2}. b. \cos (A) . \cos (C) $ $S_{OCE}=\frac{R}{2}. a. \cos (C) . \cos (B) $ $S_{OEA}=\frac{R}{2}. c. \cos (A) . \cos (B) $ So: $S_{OAF}=S_{OCD}$, $S_{OBF}=S_{OCE}$ and $S_{OBD}=S_{OEA}$.

if $M,N$ are midpoints of $AC$ and $AB$ then $OMAN$ is cyclic so $\angle ONM=\angle OAM=90-\angle CBA=\angle CAD=\angle HCB$($H$ is the orthocenter) similarly $\angle HBC=\angle OMN$ so $OMN\sim HBC$ and $OM/ON=HB/HC=BF/CE$(since $HBF\sim HEC$) so $OM*CE/2=ON*BF/2$ and $S_{OEC}=S_{BFO}$ similarly we get the other equalities.

omar31415 wrote: Where can I see the results? The problems were uploaded at this year's official APMO site. I don't see the results there yet.

$S_{AOE}=S_{BOD}$ and $S_{BOF}=S_{COD}$ and $S_{AOF}=S_{COD}$.

Looks like APMO#1's seem to have a love for Areas : Let, $[X]$ denote the area of the region $X$. Also, let $H$ be the orthocentre of $\bigtriangleup ABC$. Draw $OM\perp BC$ and $ON\perp AC$. Then, $\frac{[BOD]}{[AOE]}=\frac{BD\cdot OM}{AE\cdot ON}=\frac{BD\cdot AH}{AE\cdot BH}. (AH=2OM;BH=2ON)$ Now, $DHEC$ is cyclic. So, $AD\cdot AH=AC\cdot AE\Rightarrow\frac{AH}{AE}=\frac{AC}{AD};BE\cdot BH=BC\cdot BD\Rightarrow\frac{BD}{BH}=\frac{BE}{BC}$. Therefore, $\frac{BD\cdot AH}{AE\cdot BH}=\frac{BD}{BH}\cdot \frac{AH}{AE}=\frac{AC\cdot BE}{AD\cdot BC}$. Since, $\bigtriangleup ADC, \bigtriangleup BEC$ are similar,$\frac{AC}{BC}=\frac{AD}{BE}\Rightarrow\frac{AC\cdot BE}{AD\cdot BC}=1$. So, $\frac{[BOD]}{[AOE]}=\frac{AC\cdot BE}{AD\cdot BC}=1$. Therefore, $[BOD]=[AOE]$. Similarly, $[DOC]=[AOF]$, and, $[COE]=[BOF]$. And, we're done! My 1st post in AoPS..."Hello World"..

note that , $\Delta AOF=\frac{RbcosAcosC}{2}$ and $\Delta ODC=\frac{RbcosCcosA}{2}$ similarly we can get 2 other pairs

Sowmitra wrote: $(AH=2OM;BH=2ON)$ How do you know this?

Let $P$ be the second point of intersection of the line $OB$ and the circumcircle of $\triangle ABC$,we obtain that quadrilateral $ABCP$ is a parallelogram and hence $AH=CP=2OM$.Similarly we have $BH=2ON$.

Let $BC, CA, AB$ be $a, b, c$ and let $R$ be the circumradius. We will find the area of $OCD$ and the others will follow by symmetry. Notice that $DC = b\cos{C}$. We know find the height from $O$ to $BC$, so drop the perpendicular $OM$. Notice that $OM = OB\cos{BOM} = OB\cos{A} = R\cos{A}$. Hence, we have $2[OCD] = bR\cos{A}\cos{C}$. Notice, that we can find that $2[OAF] = bR\cos{C}\cos{A}$, so $[OCD] = [OAF]$. By similar reasoning, we can find that the two other pairs $OBD, OAE$ and $OCE, OBF$ have equal areas as well, so done.

We can bash this ;D, consider barycentric coordinates $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$, let $S_A = \frac{-a^2 + b^2 + c^2}{2} = bc \cdot \cos A$, etc. then $D = \left(0:S_CS_A:S_AS_B\right)$, $E = \left(S_BS_C:0:S_AS_B\right)$, $O = \Big(a^2S_A:b^2S_B:c^2S_C\Big)$ So \begin{align*}[AOE] &= [ABC]\cdot\frac{1}{(a^2S_A+b^2S_B+c^2S_C)\cdot (S_BS_C+S_AS_B)}\begin{vmatrix}1 & 0 & 0\\ a^2S_A & b^2S_B & c^2S_C\\ S_BS_C & 0 & S_AS_B\end{vmatrix}\\ &= [ABC]\cdot\frac{b^2S_AS_B^2}{(a^2S_A+b^2S_B+c^2S_C)\cdot (S_BS_C+S_AS_B)}\\ &= [ABC]\cdot\frac{b^2S_AS_B}{(a^2S_A+b^2S_B+c^2S_C)\cdot (S_C+S_A)} \\ &= [ABC] \cdot \frac{S_AS_B}{a^2S_A + b^2S_B + c^2S_C} \end{align*} \begin{align*} [BDO] &= [ABC] \cdot \frac{1}{(a^2S_A + b^2S_B + c^2S_C) \cdot (S_CS_A + S_AS_B)} \cdot \begin{vmatrix} 0 & 1 & 0\\ 0 & S_CS_A & S_AS_B \\ a^2S_A & b^2S_B & c^2S_C\\ \end{vmatrix} \\ &= [ABC] \cdot \frac{a^2S_A^2S_B}{(a^2S_A + b^2S_B + c^2S_C) \cdot (S_CS_A + S_AS_B)} \\ &= [ABC] \cdot \frac{a^2S_AS_B}{(a^2S_A + b^2S_B + c^2S_C) \cdot (S_C + S_B)} \\ &= [ABC] \cdot \frac{S_AS_B}{a^2S_A + b^2S_B + c^2S_C} \end{align*} Hence $[AOE] = [BDO]$. The two other pairs are analogous.

We will prove $[BOD]=[AOE]$. Let $Q\in BO$ such that $DQ \perp BO$ and $R\in AO$ such that $ER\perp OA$. Since $OB=OA$, it suffices to show $DQ = ER$. But $\triangle QBD \sim \triangle EBA$ and $\triangle RAE \sim \triangle DAB$. The equality follows from similarity ratios.

We will prove that $[OBD]=[OAE]$. The other equalities follow. Note $OM$, the height to $BC$, in $OBC$, is also the height to in $OBD$. Hence $\frac{1}{2} R^2\sin{2A}=\frac{1}{2} a\cdot AM$, so $AM=R^2\sin{2A}/a$. So $[OBD]=\frac{1}{2}c\cos{B}\cdot AM=\frac{cR^2\sin{2A}\cos{B}}{2a}$. Using similar reasoning, we can find $ON$, where $N$ is the foot of the perpendicular from $O$ to $AC$. $\frac{1}{2}ON\cdot b=\frac{1}{2}R^2\sin{2B}$, so $ON=R^2\sin{2B}/b$. Hence $AE=c\cos{A}$, and so $[OAE]=\frac{1}{2}ON\cdot AE=\frac{R^2c\sin{2B}\cos{A}}{2b}$. So it suffices to prove that $\frac{R^2c\sin{2B}\cos{A}}{2b}=\frac{cR^2\sin{2A}\cos{B}}{2a}$, or $\frac{\sin{A}}{a}=\frac{\sin{B}}{b}$, which is exactly the Law of Sines, so we are done. $\blacksquare$

I used the fact that the sides of the orthic triangle are perpendicular to $OA$, $OB$, $OC$. Then take the radious as the base and parts of the sides of the orthic triangle as altitudes. Prove that altitudes are equal with similar triangles (for example $AEF$ is similar to $ABC$). Thanks for reading!

We claim that $[OFB]=[OEC]$ . Then by symmetry other parts will follow .So , $[OFB]=\frac{1}{2} \cdot BF \cdot OT=\frac{1}{2} \left(acosB \right) \left(RcosC \right) = \frac{1}{2} \left(acosC \right) \left(RcosB \right) = \frac{1}{2} \cdot CE \cdot OT'=[OEC]$ . Note that $T$ is midpt. of $AB$ and $T'$ is midpt. of $AC$. $\text{Q.E.D.} \blacksquare$

I claim that $[OAF] = [OCD]$. Let $M, N, P$ be the midpoints of $BC, CA, AB$ respectively. We want to prove that $AF\cdot OP=CD\cdot OM$. Let $H$ be the orthocenter; then a well-known lemma states that $AH=2OM$. Thus it remains to prove $AF\cdot CH = CD\cdot AH$. However, note that by similar triangles $\triangle AFH\sim \triangle CDH$, $\dfrac{AF}{AH}=\dfrac{CD}{CH}\implies AF\cdot CH=CD\cdot AH$ done. This argument also works for $[OCE]=[OBF]$ and $[OAE]=[OBD]$ so the problem is finished.

Cute. We'll prove $[AFO]=[DCO]$. Let $\angle BAC = \alpha$ and $\angle ACB = \gamma$, then note that \begin{align*} [AFO] &= \frac12 \cdot |AF| \cdot R \cdot \sin{(90^{\circ}-\gamma)} \\ &= \frac12 \cdot |AF| \cdot R \cdot \frac{|CD|}{|AC|} \\ &= \frac12 \cdot |CD| \cdot R \cdot \frac{|AF|}{|AC|} \\ &= \frac12 \cdot |CD| \cdot R \cdot \sin{(90^{\circ}-\alpha)} \\ &= [DCO] \end{align*}Thus, we're done.

I claim that $[OAE] = [OBD]$. Let $X, Y$ be the intersection points of $FE$ with $AO$ and $FD$ with $BO$ respectively. We have that $\angle FDB = \angle A$ and $\angle OBC = \angle OBD = 90-A^{\circ}$. So $FD \perp OB$. Similarly $FE \perp AO$. So our claim is equivalent to $XE = YD$ since $AO = BO$. But $XE = AE*sin(90-B) = AE*\frac{BD}{AB} = BD*\frac{AE}{AB} = BD*sin(90-A) = YD$. So our claim is true and the conclusion follows.

Easy for APMO, trig bash will work well.

Barycentric coordinates were perfect for this.

Let $H$ be the orthocenter of $\Delta ABC$; since $AFHE$ is cyclic, by extended law of sines we have $\frac{HF}{HE} = \frac{\sin(\angle BAD)}{\sin(\angle CAD)}$. We also have $\angle BAD = 180^\circ - \angle B = \angle BCF = \angle OCE$ and similarly $\angle CAD = \angle OBF$ by isogonality of $O$ and $H$. Furthermore, since $BCFE$ is cyclic we have $\frac{HF}{HE} = \frac{BF}{CE}$, so thus $$\frac{\sin(\angle OCE)}{\sin(\angle OBF)} = \frac{HF}{HE} = \frac{BF}{CE} \Leftrightarrow [BFO] = \tfrac{1}{2} \cdot R \cdot BF \cdot \sin( \angle OBF) = \tfrac{1}{2} \cdot R \cdot CE \cdot \sin(\angle OCE) = [CEO]$$where $R=BO=CO$ is the circumradius of $\Delta ABC$. Similarly, we have $[BDO] = [AEO]$ and $[CDO] = [AFO]$, as desired.

Let $\angle OBC = \angle OCB = x ; \angle OCA = \angle OAC = y ; \angle OAB = \angle OBA = z$ $$\frac {[AOE]}{[BOD]} = \frac{\frac{1}{2} \cdot OA \cdot AE \cdot \sin y}{\frac{1}{2} \cdot OB \cdot BD \cdot \sin x} = \frac{AE \cdot \cos B}{BD \cdot \cos A} = \frac{AE \cdot \frac{BD}{AB}}{BD \cdot \frac{AE}{AB}} = 1$$ Thus, $[AOE] = [BOD]$ and similarly , $[DOC]=[AOF]$, and, $[COE]=[BOF]$

Quote: Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$, and let $O$ be the center of its circumcircle. Show that the segments $OA$, $OF$, $OB$, $OD$, $OC$, $OE$ dissect the triangle $ABC$ into three pairs of triangles that have equal areas. APMO 2013, Problem 1 Consider barycentric coordinates with reference triangle $ABC$. We'll show that the areas of $AOE$ and $BDO$ equal by showing that the ratio of the areas of $AOE$ and $ABC$ is symmetric in $a$ and $b$. Then we have \begin{align} A&=(1,0,0)\\ B&=(0,1,0)\\ E&=(S_{BC}:0:S_{AB}) \text{ because } AE \text{ is a cevian that passes through }H\\ O&=(a^2S_A:b^2S_B:c^2S_C)\\ \end{align}Now \begin{align*} \frac{[AOE]}{[ABC]}&=\frac{1}{(\sum {a^2S_A})\cdot (S_{BC}+S_{AB})}\begin{vmatrix}1 & 0 & 0\\ a^2S_A & b^2S_B & c^2S_C\\ S_{BC} & 0 & S_{AB}\end{vmatrix}\\ &=\frac{1}{(\sum {a^2S_A})\cdot (S_{BC}+S_{AB})}(b^2S_B^2S_A)\\ &=\frac{b^2S_{AB}}{(\sum {a^2S_A})\cdot (S_{C}+S_{A})} \end{align*}We can now remove the sum and $S_{AB}$ because both are symmetric in $a$ and $b$, so it suffices to have $ \frac{b^2}{(S_{C}+S_{A})} $ symmetric but this is trivial since this is equal to $\frac{b^2}{b^2}=1$, hence we are done. $\blacksquare$

We know that $OA=OB=OC$. So, let point $H$ be the midpoint of side $AC$, point $G$ be the midpoint of side $BC$, point $I$ be the midpoint of $AB$. Triangles $OGC$ and $AEB$ are similar, triangles $OHA$ and $BDA$ are similar, so $\frac{OG}{AE}=\frac{OC}{AB}$ and $\frac{OH}{BD}=\frac{OA}{AB} \implies BD \cdot OG=OH \cdot AE$ Triangles $OIA$ and $BEC$ are similar, triangles $FBC$ and $OHC$ are similar, so $\frac{OI}{CE}=\frac{OB}{CB}$ and $\frac{OH}{FB}=\frac{CO}{CB} \implies CE \cdot OH=OI \cdot FB$ Triangles $DAC$ and $AOI$ are similar, triangles $AFC$ and $OGC$ are similar, so $\frac{OI}{CD}=\frac{OA}{AC}$ and $\frac{OG}{AF}=\frac{OC}{AC} \implies AF \cdot OI=OG \cdot CD$. Now, $A_{OBD}=\frac{1}{2} \cdot BD \cdot OG=\frac{1}{2} \cdot AE \cdot OH=A_{OAE}$ $A_{OCD}=\frac{1}{2} \cdot CD \cdot OG=\frac{1}{2} \cdot AF \cdot OI=A_{OAF}$ $A_{OCE}=\frac{1}{2} \cdot CE \cdot OH=\frac{1}{2} \cdot OI \cdot FB=A_{OBF}$. And, that is all

Let $\angle OBC = \angle OCB = b$ and $\angle ABO =a$. It follows that: $$x = \frac{[AOF]}{[OCD]} = \frac{AO \cdot AF \cdot \frac{1}{2} \cdot \sin (a)}{OC \cdot DC \cdot \frac{1}{2} \cdot \sin (b)} = \frac{AF}{DC} \cdot \frac{\sin (a)}{\sin (b)}$$Note that since $\triangle AFH \sim \triangle CDH$, so $\frac{AF}{DC} = \frac{FH}{DH} = \frac{\frac{FH}{BH}}{\frac{DH}{BH}} = \frac{\sin (ABE)}{\sin (EBC)}$. Since $O$ is the center of the circumcircle of $ABC$ and $\angle BOC = 180 - 2b$, $\angle BAC = 90 -b$ and similarly, $\angle ACB = 90 - a$. It follows that $\angle ABE = b, \angle EBC = a$, so $\frac{AF}{DC} = \frac{\sin (ABE)}{\sin (EBC)} = \frac{\sin (b)}{\sin (a)}$. Thus $x = 1$ as desired. By symmetry, the segments $OA$, $OF$, $OB$, $OD$, $OC$, $OE$ dissect the triangle $ABC$ into three pairs of triangles that have equal areas.

I tried incredibly hard not to bash but my hand was forced. Let's look at the diagram shown below. Let $BC = a, CA = b, AB = c$ and circumradius be $R$. Note that\[[AOF] = AF \cdot \delta(O, AB) = b\cos{\angle A} \cdot R\cos{\angle C}\]\[[BOF] = BF \cdot \delta(O, AB) = a\cos{\angle B} \cdot R\cos{\angle C}\]\[[BOD] = BD \cdot \delta(O, BC) = c\cos{\angle B} \cdot R\cos{\angle A}\]\[[COD] = CD \cdot \delta(O, BC) = b\cos{\angle C} \cdot R\cos{\angle A}\]\[[COE] = CE \cdot \delta(O, AC) = a\cos{\angle C} \cdot R\cos{\angle B}\]\[[AOE] = AE \cdot \delta(O, AC) = c\cos{\angle A} \cdot R\cos{\angle B}\]hence $[AOF] = [COD]$ and $[BOF] = [COE]$ and $[BOD] = [AOE]$ as desired. $\blacksquare$ Remark: Note that $[ABC]$ represents two times the area of triangle $\triangle ABC$ in our notation. This is a terrible problem.

Wait there is a synthetic one-liner to this:

Let $H$ be the orthocenter of $ABC$ and $X, Y, Z$ be the midpoints of $BC, CA, AB$ respectively. By symmetry, it suffices to show $[AOF] = [COD]$, or $AF \cdot OZ = CD \cdot OX$. Since $AFH \sim CDH$, we know $\frac{AF}{CD} = \frac{AH}{CH}$. It's well-known that $O$ is the orthocenter of $XYZ$. Thus, the homothety from $ABC$ to $XYZ$ yields $AH = 2 \cdot OX$ and $CH = 2 \cdot OZ$. Thus, $$\frac{AF}{CD} = \frac{AH}{CH} = \frac{OX}{OZ}$$and the result easily follows. $\blacksquare$ Remark: For this problem, you literally just need to (easily) guess which triangles have equal area and "backsolve" for ratios. Woah @above's solution is really nice.

Why try synthetic and barycentric when you have a few liner trig bash which is not at all painful ?

Claim: $[AOE]=[BOD]$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20.1823433973449cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.342689704176472, xmax = 13.83965369316843, ymin = -7.423805031083167, ymax = 8.220496655616149; /* image dimensions */ draw((3.5,0.14893890675241123)--(0.62,-2.)--(-1.,-2.)--cycle, linewidth(0.4)); draw((0.62,4.22)--(3.5,0.14893890675241123)--(2.737897304214152,2.434997123006502)--cycle, linewidth(0.4)); /* draw figures */ draw((0.62,4.22)--(-1.,-2.), linewidth(0.8)); draw((-1.,-2.)--(8.,-2.), linewidth(0.8)); draw((8.,-2.)--(0.62,4.22), linewidth(0.8)); draw(circle((3.5,0.14893890675241123), 4.986776356019412), linewidth(0.8)); draw((0.62,4.22)--(0.62,-2.), linewidth(0.8)); draw((-1.,-2.)--(2.737897304214152,2.434997123006502), linewidth(0.8)); draw((-0.4282740458163086,0.19514533026083947)--(8.,-2.), linewidth(0.8)); draw((0.62,4.22)--(3.5,0.14893890675241123), linewidth(0.8)); draw((2.737897304214152,2.434997123006502)--(3.5,0.14893890675241123), linewidth(0.8)); draw((-1.,-2.)--(3.5,0.14893890675241123), linewidth(0.8)); draw((3.5,0.14893890675241123)--(0.62,-2.), linewidth(0.8)); draw((3.5,0.14893890675241123)--(8.,-2.), linewidth(0.8)); draw((3.5,0.14893890675241123)--(3.5,-2.), linewidth(0.8)); draw((3.5,0.14893890675241123)--(4.31,1.11), linewidth(0.8)); draw((3.5,0.14893890675241123)--(0.62,-2.), linewidth(0.4)); draw((0.62,-2.)--(-1.,-2.), linewidth(0.4)); draw((-1.,-2.)--(3.5,0.14893890675241123), linewidth(0.4)); draw((0.62,4.22)--(3.5,0.14893890675241123), linewidth(0.4)); draw((3.5,0.14893890675241123)--(2.737897304214152,2.434997123006502), linewidth(0.4)); draw((2.737897304214152,2.434997123006502)--(0.62,4.22), linewidth(0.4)); /* dots and labels */ dot((0.62,4.22),dotstyle); label("$A$", (0.44446917007856923,4.637832147211726), NE * labelscalefactor); dot((-1.,-2.),dotstyle); label("$B$", (-1.6056110763972935,-2.3085562607501844), NE * labelscalefactor); dot((8.,-2.),dotstyle); label("$C$", (8.505464313988515,-2.3284599524635423), NE * labelscalefactor); dot((3.5,0.14893890675241123),dotstyle); label("$O$", (2.8926232508215897,0.1794052034195543), NE * labelscalefactor); dot((0.62,-2.),dotstyle); label("$D$", (0.8823503877724428,-2.4876894861704053), NE * labelscalefactor); dot((2.737897304214152,2.434997123006502),dotstyle); label("$E$", (2.8130084839681584,2.6275592841625772), NE * labelscalefactor); dot((-0.4282740458163086,0.19514533026083947),dotstyle); label("$F$", (-0.9686929415698411,-0.11915017228081434), NE * labelscalefactor); dot((0.62,-0.07787781350482316),dotstyle); label("$H$", (1.1610020717594534,-0.0196317137140248), NE * labelscalefactor); dot((3.5,-2.),dotstyle); label("$D'$", (3.410119235368895,-2.4677857944570474), NE * labelscalefactor); dot((4.31,1.11),dotstyle); label("$E'$", (4.524725971316937,1.4134340896477446), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We have to prove $\frac{AE}{BD}=\frac{OD'}{OE'}$. Notice that $\triangle AHE\sim \triangle BHD$, $\triangle OCD'\sim \triangle CHE$ and $\triangle OCE'\sim \triangle HCD$. Which gives $$\frac{AE}{BD}=\frac{HE}{HD}=\frac{OD'}{OE'}$$as desired. $\square$ Similarly $[BOF]=[COE], [AOF]=[COD]$.

Triangles $[OBF]=[OCE]$ because they both have area $\frac12 R\cdot a \cdot \cos(C)\cdot \cos(B)$.

We will prove $ S_{OAF}$ = $ S_{OCD}$ and use same approach to prove other pairs. $ S_{OAF}$ = AO.AF/2 . Sin ∠OAF and $ S_{OCD}$ = CO/CD/2 . Sin ∠OCD so we need to prove AF/CD = Sin ∠OCD/Sin ∠OAF. Note that ACDF is cyclic so AF/CD = Sin ∠ACF/Sin ∠CAD. Note that ∠OAF = 90 - ∠C = ∠CAD and ∠OCD = 90 - ∠A = ∠ACF so Sin ∠OCD/Sin ∠OAF = Sin ∠ACF/Sin ∠CAD = AF/CD as wanted. we're Done.

We will just show that $[AOF]=[COD],$ after which the rest of the problem follows by symmetry. Let $\angle A=\alpha,\angle B=\beta,\angle C=\gamma$. Claim: $\frac{DC}{AF}=\frac{\sin(90-\gamma)}{\sin(90-\alpha)}.$ Since $AFDC$ is cyclic, $\triangle ADC$ and $\triangle AFC$ have the same circumradius, so $$\frac{DC}{AF}=\sin{\angle DAC}{\angle FCA}=\frac{\sin(90-\gamma)}{\sin(90-\alpha)}.$$ Then, $$2[AOF]=AF\cdot AO\cdot \sin\angle OAF=AF\cdot AO\cdot \sin(90-\gamma),$$$$2[COD]=CD\cdot CO\cdot \sin\angle OCD=CD\cdot CO\cdot \sin(90-\alpha)$$By our claim and the fact that $AO=CO$, we have $[AOF]=[COD]$, so we are done.

I can't believe no one did this yet. With complex numbers, we observe that $d=\frac{a+b+c-\frac{bc}{a}}{2}$ and $f=\frac{a+b+c-\frac{ab}{c}}{2}$, hence\begin{align*}[OAF] & =\frac{i}{4}\begin{vmatrix}0&0&1\\a&\frac{1}{a}&1\\\frac{a+b+c-\frac{ab}{c}}{2}&\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{c}{ab}}{2}&1\end{vmatrix} \\ & =\frac{i}{4}\begin{vmatrix}a&\frac{1}{a}\\\frac{a+b+c-\frac{ab}{c}}{2}&\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{c}{ab}}{2}\end{vmatrix} \\ & =\frac{i}{8}\left (\frac{a}{b}+\frac{a}{c}-\frac{c}{b}-\frac{b}{a}-\frac{c}{a}+\frac{b}{c}\right ) \\ & =\frac{i}{4}\begin{vmatrix}\frac{a+b+c-\frac{bc}{a}}{2}&\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}}{2}\\c&\frac{1}{c}\end{vmatrix} \\ & =\frac{i}{4}\begin{vmatrix}0&0&1\\\frac{a+b+c-\frac{bc}{a}}{2}&\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}}{2}&1\\c&\frac{1}{c}&1\end{vmatrix} \\ & =[OCD]. \end{align*}Done.

Artinya apa bang messi?

It is well known $d(O, BC) = R\cos A$. Now we have $DB = c \cos B$. Thus the areas are symmetrically $\frac{Rc\cos A \cos B}{2}$. The result follows.

Let $[X]$ denote the area of polygon $X$. We claim that\begin{align*} [AOF] &= [DOC] \\ [BOF] &= [EOC] \\ [BOD] &=[AOE]. \end{align*}We show each of the equalities above. We have $[AOF] = [DOC]$ \begin{align*} &\iff \dfrac 12 \cdot AO \cdot AF \cdot \sin FAO = \dfrac 12 \cdot CO \cdot CD \cdot \sin OCD \\ &\iff AF \cdot \sin FAO = CD \cdot \sin OCD \\ &\iff AF \cdot \sin DAC = CD \cdot \sin FCA \, \, (\text{since} \, AO, AD \, \, \text{and} \, CO,CF \, \text{are pairs of isogonals with respect to} \, \, \Delta ABC) \\ &\iff AF \cdot \dfrac{DC}{AC} = CD \cdot \dfrac{AF}{AC} \end{align*}which is obviously true, so the result follows as all steps are reversible. Similarly, we have $[BOF] = [OEC]$ \begin{align*} &\iff \dfrac 12 \cdot OB \cdot BF \cdot \sin OBF = \dfrac 12 \cdot OC \cdot EC \cdot \sin OCE \\ &\iff BF \cdot \sin EBC = EC \cdot \sin FCB \\ &\iff BF \cdot \dfrac{EC}{BC} = EC \cdot \dfrac{BF}{BC} \end{align*}giving the desired equality. Finally, we have $[BOD] = [AOE]$ \begin{align*} &\iff \dfrac 12 \cdot OB \cdot BD \cdot \sin OBD = \dfrac 12 \cdot AO \cdot AE \cdot \sin OAE \\ &\iff BD \cdot \sin EBA = AE \cdot \sin DAB \\ &\iff BD \cdot \dfrac{AE}{AB} = AE \cdot \dfrac{BD}{AB} \end{align*}and we are done. $\blacksquare$

Let $H$ be the orthocenter of $\triangle ABC$ and $B'$ and $A'$ the feet dropped from $O$ to $AC$ and $BC$ respectively. $\triangle AEH \sim \triangle BDH \implies \frac{BD}{AE} = \frac{BH}{AH} \implies BD \cdot AH = BH \cdot AE$. It's well-known that $AH=2OA'$, and similarly $BH=2OB'$ so: $BD \cdot 2OA' = 2OB' \cdot AE \implies \frac{BD \cdot OA'}{2} = \frac{OB' \cdot AE}{2}$, which are the areas of $\triangle BDO$ and $\triangle OEA$ respectively. Similarly, we can prove that $[ODC]=[OAF]$ and $[BFO]=[CEO]$ are we're done .