We can prove a huge generalization. First, since G is collinear with H and O (Euler line), it suffices to show that [AGO]+[BGO]+[CGO]=0, where [RST] is the signed area of triangle RST (two vertices form a reference axis, and if the third vertex is below this axis, the area is negative and if it is above, the area is positive). We prove much more generally that if X=(x,y,z) in barycentrics, then x[AXP]+y[BXP]+z[CXP]=0 for all points P in the plane of ABC. Let X be the origin and XP be the x-axis (reference axis) of this coordinate system. Then the vector sum xXA+yXB+zXC=(x+y+z)X=0. In particular, the y-components of these vectors, say y_a, y_b, and y_c are the components of A, B, and C repsectively, must satisfy 0=x*y_a+y*y_b+z*y_cimplying 0=x*|XP|*y_a+y*|XP|*y_b+z*|XP|*y_c=x[AXP]+y[BXP]+z[AXP], as claimed. Since G=(1,1,1), the result follows immediately. Of course, without any barycentrics or more general theory, a contestant could prove this in a completely analogous manner knowing simply that G=(A+B+C)/3.

Well, there is actually a very simple synthetic proof which only involves the line passing through G. Are we allowed to post APMO questions now, technically?

Of course we are, they are even on the Canadian MO website.

vectors OA+OB+OC=OH AHxOA+BHxOB+CHxOC =(OB+OC)xOA+(OA+OC)xOB+(OB+OC)xOA =[OBxOA+OAxOB]+[OAxOC+OCxOA]+[OBxOC+OCxOB] =0+0+0=0 so there are |OA|=|OB|+|OC| or |OB|=|OA|+|OC|or ...

We see that AH=2RcosA, <OAH=|C-B|, OA=R so S(AOH)=R ² cos A sin |C-B|. For every ordering of the angles of the triangle we transform the product into sum and we have an identity.

maybe this question can be generalized to isogonal conjugates? (i.e. if P and Q are isogonal conjugates in triangle ABC, then the area of one of APQ, BPQ and CPQ is the sum of the other two.)

vinoth_90_2004 wrote: maybe this question can be generalized to isogonal conjugates? (i.e. if P and Q are isogonal conjugates in triangle ABC, then the area of one of APQ, BPQ and CPQ is the sum of the other two.) Not really; actually, if P and Q are two arbitrary points in the plane, then the area of one of the triangles APQ, BPQ and CPQ equals the sum of the areas of the other two if and only if the line PQ passes through the centroid of triangle ABC. But, of course, not for any two isogonal conjugates P and Q, the line PQ passes through the centroid. Darij

Beautiful problem, but I'm under the impression that it's been discussed before. What we must prove is that one of the distances $d(A,OH),d(B,OH),d(C,OH)$ is the sum of the other two. Consider the homothety $\mathcal H$ centered at $G$ (the centroid of $ABC$, as usual) of ratio $-\frac 12$. It turns $OH$ into itself, and it turns $A,B,C$ into the midpoints $D,E,F$ of $BC,CA,AB$ respectively. Moreover, it shrinks all distances by a factor of $\frac 12$. Suppose $OH$ cuts the segments $AB,AC$. The distance $d(D,OH)$ is half the distance $d(A,OH)$, because $OH,D$ are the images of $OH,A$ respectively through $\mathcal H$. At the same time, we clearly have $d(D,OH)=\frac{d(B,OH)+d(C,OH)}2$. From these two facts we deduce $d(A,OH)=d(B,OH)+d(C,OH)$, as desired. Edit: Oops! Darij was here first .

Only in the acute triangle: let $a\le b\le c.$ Thus, $(AOH)=\frac {AO\cdot AH}2\cdot\sin (\widehat {OAH})=$ $R^2\cos A\sin (C-B)=-R^2\cdot \cos (B+C)\sin (C-B)=$ $\frac {R^2}2\cdot (\sin 2B-\sin 2C)$ a.s.o.

$16S\cdot [AOH]=(b^2+c^2-a^2)\cdot \left|b^2-c^2\right|$ a.s.o. Prove easily that $a\le b\le c\Longrightarrow [BOH]=[AOH]+[COH]\ .$

sorry, my English is poor. I think this is easy. wlog $B$, $C$ are on the same side of $OH$. Then we will show that $d(A,OH)=d(B,OH)+d(C,OH)$. But $G$ is on $OH$, It is obvious!

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO2004Problem2 Vo Duc Dien

very quick

are actually usable on an olympiad question...

grobber wrote: we clearly have $d(D,OH)=\frac{d(B,OH)+d(C,OH)}2$. sorry,that's seems not clear to me,can someone explain?

WLOG $AC>=BC>=AB$. We start by noting that it suffices to prove $\frac{1}{2}(AH)(AO)\sin(HOA)=\frac{1}{2}(BH)(BO)\sin(HOB)+\frac{1}{2}(CH)(CO)\sin(HOC)$, for some rotation to $A$. Note that $AO=BO=CO=R$, so that cancels out of all the terms, as does the $\frac{1}{2}$. By some angle chasing, we get that angle $HOA$=$A-\frac{2(180-2B)}{2}$, because of the isogonal conjugates. This implies angle $HOA=B-C$. In addition $AH=2R\cos(A)$ by similar triangles, and it can be similarly derived for the other two. Then, we get $2R\cos(A)\sin(B-C)=2R\cos(B)\sin(A-C)+2R\cos(C)\sin(B-A)$. Simplifying using angle addition formula for sine yields equality.

............

SOLUTION WLOG assume $ \angle A \ge \angle B \ge \angle C $ CLAIM $(AOH) + (COH) = (BOH) $ PROOF Note that ..... $ (1) $ $ (AOH) = ( AH * AO * sin\angle OAH) = (2R cos A)(R)(sin \angle OAH) = (2R cos A)(R)(sin (B-C))$ $ (2) $ $ (COH) = ( CH * CO * sin\angle OCH) = (2R cos C)(R)(sin \angle OCH)= (2R cos C)(R)(sin(A-B)) $ $ (3) $ $ (BOH) = ( BH * BO * sin\angle OBH ) =(2R cos B)(R)( sin \angle OBH) = (2R Cos B)(R)(sin (A-C)) $ We Show That $ (1) + (2) =(3).$ $ \implies 2R cos A sin(B-C) + 2R cos C sin (A-B) = 2R cos B sin (A-C) $ Which is true by simple expansion. My Solution is quite similar to tree3' s solution

Solved using hints from EGMO. WLOG assume that B and C are on the same side of the Euler line. If M is the midpoint, we claim that AD=BE+CF (note that this is preserved cyclically), where D, E, F are the projections of A, B, and C onto that line. Indeed, by sim triangles GDA and GMM' (projection of M) we have AA'=2MM'=BE+CF since MM' is the midline of right trapezoid BEFC. $\blacksquare$

Apply Cartesian coordinates with $OH$ as the real axis. The vectors $\overrightarrow{OA}$, $\overrightarrow{OB},$ and $\overrightarrow{OC}$ sum to $\overrightarrow{OH}$, so the y-coordinates of $A,B,C$ sum to 0. Thus, the absolute value of one of them is the sum of the absolute values of the other two, hence done since all three triangles have $OH$ as the base and some height away from the line.

Without loss of generality, we assume $A$ lies on the opposite side of the Euler line, shown in blue, as $B$ and $C$. (Note it is impossible for $A$, $B$, and $C$ to all be on the same side as the centroid is always inside the triangle.) We claim \[[AOH] = [BOH] + [COH].\] Let the projections of $A$, $B$, and $C$ onto the Euler line be $A_1$, $B_1$, and $C_1$, respectively. The problem reduces to proving \[AA_1 = BB_1 + CC_1.\] Denote $M$ as the midpoint of $BC$ and $M_1$ as the projection of $M$ onto the Euler line. As $MM_1$ is the midline in trapezoid $BB_1C_1C$, we have \[MM_1 = \frac{BB_1 + CC_1}{2}.\] Similar triangles $\triangle AA_1G$ and $\triangle MM_1G$, where $G$ is the centroid and on the Euler line, give \[AA_1 = \frac{AG}{GM} \cdot MM_1 = 2MM_1 = BB_1 + CC_1. \text{ } \blacksquare\] [asy][asy] size(250); pair A, B, C, H, A1, B1, C1, M, G, M1; A = dir(131); B = dir(200); C = dir(340); H = orthocenter(A, B, C); A1 = foot(A, H, 0); B1 = foot(B, H, 0); C1 = foot(C, H, 0); M = .5B + .5C; G = centroid(A, B, C); M1 = foot(M, B1, C1); markscalefactor=.005; fill(A--A1--G--cycle, mediumgreen); fill(G--M1--M--cycle, mediumgreen); draw(rightanglemark(G, A1, A)); draw(rightanglemark(B, B1, A1)); draw(rightanglemark(G, M1, M)); draw(rightanglemark(M1, C1, C)); draw(A--B--C--cycle); draw(B--B1); draw(C--C1); draw(M1--M--A--A1); draw(B1--C1, blue); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$A_1$", A1, SE); label("$B_1$", B1, NW); label("$C_1$", C1, NE); label("$M$", M, S); label("$G$", G, NE); label("$M_1$", M1, N); dot(A); dot(B); dot(C); dot(A1); dot(B1); dot(C1); dot(M); dot(G); dot(M1); [/asy][/asy]

Without loss of generality assume $[AOH] = [BOH] + [COH]$. Now proceed with complex numbers. Then the area of $[AHO]$ is given by, \begin{align*} \frac{i}{4}\begin{vmatrix}a&\frac{1}{a}&1\\ a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}&1\\ 0&0&1\end{vmatrix} = \frac{i}{4abc}\left [(a^2 - bc)(b+c)\right ] \end{align*}Similarly we can calculate the areas of $[BOH]$ and $[COH]$ and adding gives the conclusion.

I will use complex numbers to solve lets o=0 and h is reel number X,Y,Z is the foot altitude from A,B,C to OT respectively ,where t=1 WLOG AX>=BY>=CZ h=a+b+c=a'+b'+c' and x=(a+a')/2 and y=(b+b')/2 and z=(c+c')/2 AX=│a-a'│/2 BY=│b-b'│/2 CY=│c-c'│/2 because a+b+c-a'-b'-c-=0 so AX=BY+CZ ==> [AHO]=[BOH]+[COH] .

WLOG let $A$ be on the opposite side of $HO$ as $B$, and $C$. We claim that $[AOH]=[BOH]+[COH]$ Let $G$ be the centroid of $\triangle ABC$ Note that $G$ lies on $HO$ since $HO$ is the Euler line of $\triangle ABC$. Let $H_{A}$, $H_{B}$, and $H_{C}$ be the projections of $A$, $B$, and $C$ respectively onto $HO$. It suffices to show $BH_{B}+CH_{C}=AH_{A}$. Let $M$ be the midpoint of $BC$ and $H_{M}$ be its projection onto $OH$. Note that $MH_{M}$ is the midline of trapezoid $BH_{B}H_{C}C$. By centroid $2:1$ ratios we have $\tfrac{BH_{B}+ CH_{C}}{2}= MH_M=\tfrac{AH_A}{2}$ So $BH_{B}+CH_{C}=AH_{A}$, as desired.

Let $D$, $E$, and $F$ be the foots of the perpendiculars from $A$, $B$, and $C$ to $OH$, respectively. WLOG, let $A$ and $B$ lie on the same side of $OH$ and $C$ lie on the other. Then, we want to prove that $AD+BE=CF$. Since the centroid lies on $OH$, this must be true by coordinates.

WLOG $[AOH] = [BOH] + [COH]$. By Euler Line, the centroid $G$ lies on $OH$. Let the projections of $A$, $B$, and $C$ onto the Euler Line be $D$, $E$, and $F$ respectively. Then it is equivalent to show that $AD = BE + CF$. Define the midpoint of $BC$ to be $M$, and the projection of $M$ onto $OH$ being $N$. Then notice that $AD = BE + CF = 2MN$. This follows easily from noticing that $\triangle ADG \sim MNG$(by homothety) with a $2:1$ ratio, which finishes.

Use standard notation. WLOG $\angle A \ge \angle B \ge \angle C, R = 1$. From standard trigonometry formulae, $$[AOH] = \frac12 (2R \cos A) (R) (\sin(|B-C|)) = \cos A \sin( B-C).$$Similarly, $$[BOH] = \cos B \sin(A-C), [COH] = \cos C \sin (A-B).$$But now simply from expanding, $[AOH] + [COH] = [BOH]$. $\square$

darij grinberg wrote: vinoth_90_2004 wrote: maybe this question can be generalized to isogonal conjugates? (i.e. if P and Q are isogonal conjugates in triangle ABC, then the area of one of APQ, BPQ and CPQ is the sum of the other two.) Not really; actually, if P and Q are two arbitrary points in the plane, then the area of one of the triangles APQ, BPQ and CPQ equals the sum of the areas of the other two if and only if the line PQ passes through the centroid of triangle ABC. But, of course, not for any two isogonal conjugates P and Q, the line PQ passes through the centroid. Darij also if $PQ$ passes through the symmetric of one vertex in the midpoint of opposed side (this case occurs when the three areas (as signed areas) have the same sign whereas the other case occurs when one of the area has opposed sign of the others). Best regards. RH HAS.

Consider a line passing through the centroid $G$. Suppose that $A$ is the farthest from the line. Note that proving that the distance from $A$ is the sum of the distances from $B$ and $C$ to this line will suffice because $G$ lies on $\overline{OH}$. Denote the midpoint of $\overline{BC}$ as $D$, and drop atltitudes from $A$, $B$, $C$, $D$ to the line at points $A'$, $B'$, $C'$, $D'$, respectively. We have $\overline{DD'}$ as the midline of trapezoid $BB'C'C$, so \[BB'+CC' = 2DD' = AA',\] as desired. $\square$

Using signed areas, $[AOH]+[BOH]+[COH]=3[GOH]=0$.

Posting for storage. Note that if we ignore the signs of areas: $$[AOH]=\frac {AO.AH.sin(B-C)}{2}=R^2cos(A).sin(B-C)$$$$[BOH]=\frac{BO.BH.sin(C-A)}{2}=R^2cos(B)sin(C-A)$$$$[COH]=\frac{CO.CH.sin(A-B)}{2}=R^2cos(C)sin(A-B)$$Then, $[AOH]+[BOH]+[COH]$=$R^2.\sum_{cyc}cos(A)sin(B-C)=R^2.\sum_{cyc}(cos(A)sin(B)cos(C)-cos(A)sin(C)cos(B))=0$ And we are done.

So basically, prove that out of the three distances from $A,B,C$ to the Euler line, the largest one is the sum of the other two. But their average, $G$ is on the Euler line, so this is clear.