Is this even valid? Outline: Let $A' \in BC \cap AG$ ,$H \in BC \cap DG$, and $E' \in AC \cap DG$. Menelaus on triangle $AA'C$ gives $\frac{CE'}{AE'}$$=\frac{A' H}{2\times CH}$ and likewise on triangle $ABA'$ gives $\frac{DB}{AD}=\frac{2\times A' H}{BH}.$ Now, if we prove that $D,G$, and $H$ are collinear, then $D,E$, and $G$ must be collinear which finishes the solution by proving $E=E'$, which can be shown by $\frac{AB}{AD}+\frac{AC}{AE'}=3$ and substitutions from Menelaus.

Rip didn't realize deadline was so early for my time. Went to submit and found contest ended an hour ago. Bary basically kills this. Set $\frac{AB}{AD} = m$ and $\frac{AC}{AE} = n$. Now, we know that $D = (1-\frac{1}{m}:\frac{1}{m}:0)$, and $E=(1-\frac{1}{n}:0:\frac{1}{n}$. Now, the condition for collinearity is $\begin{vmatrix} 1 & 1 & 1\\ 1-\frac{1}{m} & \frac{1}{m} & 0\\ 1-\frac{1}{n} & 0 & \frac{1}{n} \end{vmatrix}$ which is equal to $\frac{3}{mn}-\frac{1}{m}-\frac{1}{n} = \frac{3-m-n}{mn}$. Therefore, since $m+n=3$, D,E,G must be collinear.

ghu2024 wrote: Is this even valid? Outline: Let $A' \in BC \cap AG$ ,$H \in BC \cap DG$, and $E' \in AC \cap DG$. Menelaus on triangle $AA'C$ gives $\frac{CE'}{AE'}$$=\frac{A' H}{2\times CH}$ and likewise on triangle $ABA'$ gives $\frac{DB}{AD}=\frac{2\times A' H}{BH}.$ Now, if we prove that $D,G$, and $H$ are collinear, then $D,E$, and $G$ must be collinear which finishes the solution by proving $E=E'$, which can be shown by $\frac{AB}{AD}+\frac{AC}{AE'}=3$ and substitutions from Menelaus. I don't think this makes any sense...

Let $M,N $ be on $AG $ such that $\mathcal {D}M\parallel EN\parallel BC. $ Thus, we have $AM+AN=2AG$ $G $ is the midpoint of $MN. $ $\mathcal{D}MEN$ is a parallelogram $\implies \mathcal{D},G,E$ are collinear.

Try complex numbers. It works amazingly. Here was my solution:

dangerousliri wrote: Let $ABC$ be a triangle with its centroid $G$. Let $D$ and $E$ be points on segments $AB$ and $AC$, respectively, such that, $$\frac{AB}{AD}+\frac{AC}{AE}=3.$$Prove that the points $D, G$ and $E$ are collinear. Proposed by Dorlir Ahmeti, Kosovo

I have not learned complex and bary yet (seems pretty useful haha), but I came up with a synthetic solution luckily haha.

We claim the following: CLAIM: If a fixed linear combination of $\dfrac{1}{AD}$ and $\dfrac{1}{AE}$ has a fixed sum, then $DE$ always passes through a fixed point. Proof: Invert with arbitrary radius about $A$. This is then equivalent to the gliding principle. Now note that the fixed point must be $G$ by choosing $DE$ to be the $B$ and $C$ medians, so we are done.

Now i will use my old problem https://www.cut-the-knot.org/triangle/FourthIdentityInTriangle.shtml Let $A', B', C'$ be midpoints of $BC, CA, AB$ and let $DE$ cut $AA'$ at $K$ by using above theorem we have $\frac{BD}{DA}.\frac{AC'}{C'B}+\frac{CE}{EA}.\frac{AB'}{B'C}=\frac{A'K}{KA}\frac{AG}{GA'}$ $\Rightarrow \frac{BD}{DA}+\frac{CE}{EA}=2\frac{A'K}{KA}$ $\Rightarrow 3=\frac{AB}{AD}+\frac{AC}{AE}=2\frac{A'K}{KA}+2$ $\Rightarrow AK=2KA'\Rightarrow K=G$ so $D, G, E$ collinear.

See Van Aubel's Theorem. It can be proven in the same way as the first solution by ghu2024.

Huh mine was valid? I thought it was completely wrong.

Actually, I was wrong. The theorem I mentioned is not actually enough by itself to solve the problem. I got a little excited because I just learned that theorem!

I`m pretty sure I`m wrong , but ehmm... anyway...

Am I missing something or it's just the same problem ?