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Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Prove that the area of one of the triangles $AOH, BOH,$ and $COH$ is equal to the sum of the areas of the other two. enter image description here In this figure, we want to prove $[AOH]+ [BOH]= [COH]$. Where $[.] $ is the area of a given shape.

Let $A',B'$ and $C'$ be the othogonal projections of the points $A,B,C$, respectively onto the line $OH$. Now note that $$[AOH]+ [BOH]= [COH]\iff $$ $$AA'\cdot OH+BB'\cdot OH=CC'\cdot OH$$ $$\iff AA'+BB'=CC'$$

Any idea how to show this? By the way, I know this problem already exists here but I don't want to solve it using analytical techniques (Vectors, coordinates).

PNT
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  • E.g. $A'$ is the orthogonal projection of $A$ onto what line? – coffeemath May 24 '22 at 19:36
  • onto the line $OH$ @coffeemath – PNT May 24 '22 at 19:37
  • O yeah I see that in the diagram now. – coffeemath May 24 '22 at 19:50
  • On the third line you use a double implication error, which would hold provided it happened that $OA'=OH$ since one could cancel that from the two sides of the equation. Is it somehow the case that $OA'=OH,$ or on the other hand what is needed to show that double implication? – coffeemath May 24 '22 at 20:20
  • I'm sorry that this was a mistake I meant to write $OH $ – PNT May 24 '22 at 20:46
  • Btw, what you want to prove , i.e $AA'+BB'=CC'$, is true for any line through the centroid not just the Euler line. (See Serbia 2017, Opstinsko IIA) – ZNatox May 24 '22 at 22:08
  • How would you prove it @ZNatox – PNT May 24 '22 at 22:12
  • Normally, I'd proceed with barycentric coordinates, but you don't want to use coordinates. I can try to find a synthetic solution. (Note that this is problem 16 from the book "A beautiful journey through Olympiad geometry", so you can find the solution on AOPS) – ZNatox May 24 '22 at 22:34

2 Answers2

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Let $(d)$ be any line through the centroid $G$.

Let $A'$ be the feet of the perpendicular from $A$ to $(d)$, $(AC)\cap (d)=A'',$ $ \ (BC)\cap(d)=B''$, and $M$ be the midpoint of side BC.

Define $B', C'$ similarly.

Wlog, $A$ and $B$ lie on the same side of $(d)$

Note, by Thales's Theorem:

$$\frac{BB'}{CC'}=\frac{BB''}{B''C}$$ $$\frac{AA'}{CC'}=\frac{AA''}{A''C}$$

Thus, what we want is equivalent to:

$$\frac{AA''}{CA''}+\frac{BB''}{CB''}=1$$

which we obtain directly by applying Menelaus's theorem, in triangle $\triangle ACM$ and collinearity of $G, A''$ and $B''$.

ZNatox
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  • Good proof I just have on question how could you think that this fact is true for any line passing through $G$ – PNT May 25 '22 at 11:44
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    As I've mentioned earlier, I've already seen this problem in Serbia 2017, Opstinsko IIA. But even without seeing the problem beforehand, using barycentric coordinates, the only thing we use about the Euler line, is that the sum of its coefficients is zero, which is equivalent to saying that the line passes through the centroid. – ZNatox May 25 '22 at 18:36
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Here is a possibility to proceed geometrically. We isolate the Euler line as the main ingredient in the proof, then use the fact that the centroid is a point on it. Picture of the situation:

math stackexchange euler line problem 4457924

Here, $G,H,O$ are respectively the centroid, the orthocenter, and the circumcenter of $\Delta ABC$. Thle points $G,H,O$ are on a line, the Euler line of the triangle. For each vertex $X$ among $A,B,C$ i have denoted

  • by $X_G$ the intersection of $XG$ with the opposite side,
  • by $X_H$ the intersection of $XH$ with the opposite side,
  • by $X_E$ the intersection of the Euler line with the opposite side,
  • by $X'$ the projection of $X$ on the Euler line, as in the question.

In the picture the vertex $A$ is chosen so that $A_E$ is not on the opposite side (seen as a segment). We apply the theorem of Menelaus in $\Delta ABA_G$, $\Delta ACA_G$ w.r.t. the Euler line: $$ \begin{aligned} 1 &= \frac{C_EB}{C_EA}\cdot \frac{GA}{GA_G}\cdot \frac{A_EA_G}{A_EB} = \frac{B'B}{A'A}\cdot 2\cdot \frac{A_EA_G}{A_EB} \ , \\ 1 &= \frac{B_EC}{B_EA}\cdot \frac{GA}{GA_G}\cdot \frac{A_EA_G}{A_EC} = \frac{C'C}{A'A}\cdot 2\cdot \frac{A_EA_G}{A_EC} \ , \end{aligned} $$ and from here: $$ \frac{B'B}{A'A} + \frac{C'C}{A'A} = \frac 12\left( \frac{A_EB}{A_EA_G}+ \frac{A_EC}{A_EA_G} \right) =1 \ , $$ since $A_G$ is the mid point of $BC$.

dan_fulea
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