Bump bump

ThE-dArK-lOrD wrote: Bump bump Hi,this problem was open before jury meeting

There is a solution in the official website, check it out.

Othre87 wrote: There is a solution in the official website, check it out. This problem was open before jury meeting ,not now.

The answer is EXISTS

Solution by Ilya Bogdanov The triangles are similar to $\triangle{XYZ}$ where $XY=1,XZ=t^3,YZ=t^2$ and $\angle{Z}=\angle{X}+2\angle{Y}$. I wonder what is the motivation for this mind-blowing solution.

Attachments:

Here's another construction that looks somewhat more human. It was found by user deep thought on Puzzling.SE. [asy][asy]size(10cm); import markers; pair X1=(0,2),X2=(0,-2),Y1=(sqrt(7),1),Y2=(-sqrt(7),1),Z1=(sqrt(7),-1),Z2=(-sqrt(7),-1); fill(X1--Y1--Z1--cycle^^X1--Y1--Z2--cycle,cyan);draw(X1--Y1--Z1--cycle^^X1--Y1--Z2--cycle,blue); draw(anglemark(Y1,Z2,X1,10)^^anglemark(Z1,X1,Y1,10),blue); draw(anglemark(X1,Y1,Z2,10,12)^^anglemark(Y1,Z1,X1,10,12),blue); D(MP("X_1\;(0,2)",X1,N)--MP("Y_1\;(\sqrt{7},1)",Y1,NE)--MP("Z_1\;(\sqrt{7},-1)",Z1,SE)--MP("X_2\;(0,-2)",X2,S)--MP("Z_2\;(-\sqrt{7},-1)",Z2,SW)--MP("Y_2\;(-\sqrt{7},1)",Y2,NW)--X1); label("$\triangle X_1Y_1Z_1\sim\triangle Z_2X_1Y_1$",(0,-3)); dot(X1^^X2^^Y1^^Y2^^Z1^^Z2); [/asy][/asy]

ThE-dArK-lOrD wrote: Solution by Ilya Bogdanov The triangles are similar to $\triangle{XYZ}$ where $XY=1,XZ=t^3,YZ=t^2$ and $\angle{Z}=\angle{X}+2\angle{Y}$. I wonder what is the motivation for this mind-blowing solution. I was trying to understand this as well and here's what I came up with. We start with a generic triangle $X_1Y_2Z_2$ (though slightly annoying, I'll follow the notation of the diagram from the official solution). Now let $X_2$ be the point such that $X_1Y_2X_2Z_2$ is a parallelogram. From this we already know that $\triangle X_1Y_2Z_2 \equiv \triangle X_2Y_2Z_2$. Next, we construct the point $Z_1$ such that $\triangle X_2Y_2Z_1 \sim \triangle Z_2Y_2X_1$ ($Z_1$ on the same side of $Z_2$ with respect to $Y_2X_2$). In other words, we consider the spiral similarity with center $Y_2$ that maps $Z_2$ to $X_2$, and define $Z_1$ as the image of $X_1$ via this transformation. Now, by definition we get that $X_2Y_2Z_1$ is similar to our original triangle $X_1Y_2Z_2$. We shall now prove that $X_1Y_2Z_1$ is similar to the original triangle as well. More precisely, we'll show that $\triangle X_1Y_2Z_1 \sim \triangle Y_2Z_2X_1$. First notice that $\angle X_1Y_2Z_1 = \angle X_1Y_2X_2 - \angle Z_1Y_2X_2 = (180^{\circ} - \angle Y_2X_1Z_2) - \angle Z_1Y_2X_2 = \angle X_1Y_2Z_2 + \angle Y_2Z_2X_1 - \angle Z_1Y_2X_2 = \angle Y_2Z_2X_1$, where in the last step we have used that $\angle X_1Y_2Z_2 = \angle Z_1Y_2X_2$, due to the similarity. So it is enough to prove that $\frac{X_1Y_2}{Y_2Z_1} = \frac{Y_2Z_2}{Z_2X_1}$. Let $x,y,z$ be the lengths of the sides opposite to $X_1, Y_2, Z_2$ in our original triangle. Now $Y_2X_2 = X_1Z_2 = y$, hence the ratio of the similarity $\triangle X_2Y_2Z_1 \sim \triangle Z_2Y_2X_1$ is $\frac{y}{x}$. Therefore, $Y_2Z_1 = \frac{y}{x} \cdot z$, and we have $\frac{X_1Y_2}{Y_2Z_1} = \frac{z}{\frac{yz}{x}} = \frac{x}{y} = \frac{Y_2Z_2}{Z_2X_1}$, as wanted. So far we have that the four triangles $X_iY_2Z_k$ are similar, and this holds without any extra requirements on the initial triangle $X_1Y_2Z_2$. If you are tired or need to go to the toilet, this is a good place to stop. The sixth point $Y_1$ is constructed such that $Y_2X_2Y_1Z_1$ is a parallelogram. Notice this implies that $Z_1Y_1Z_2X_1$ is a parallelogram as well, because it has a pair of opposite sides that are parallel and of equal length. Now it is easy to see that $\triangle X_1Y_2Z_1 \equiv \triangle Z_2X_2Y_1$ (respective sides are equal), and also $\triangle X_2Y_2Z_1 \equiv \triangle X_2Y_1Z_1$ (two halves of a parallelogram). So we have 6 similar triangles now. The remaining two triangles, $X_1Y_1Z_2$ and $X_1Y_1Z_1$ are congruent (two halves of a parallelogram). So if $X_1Y_1Z_2$ happens to be similar to the initial triangle $X_1Y_2Z_2$, then all 8 triangles $X_iY_jZ_k$ are similar and we are done. This is where the extra requirements for the initial triangle arise. Standard angle chasing yields that $\angle X_1Z_2Y_1 = \angle Y_2X_1Z_2$ if and only if $\angle X = \angle Z + 2 \angle Y$ in the starting triangle. Provided that this is the case, we further need $\frac{X_1Z_2}{Z_2Y_1} = \frac{Y_2X_1}{X_1Z_2}$, that is $\frac{y}{Z_2Y_1} = \frac{z}{y} \Rightarrow Z_2Y_1 = \frac{y^2}{z}$. Now, $Z_2Y_1 = X_1Z_1$, and because of the similarity $\triangle X_1Y_2Z_1 \sim \triangle Y_2Z_2X_1$ we have that $\frac{X_1Z_1}{X_1Y_2} = \frac{Y_2X_1}{Y_2Z_2}$, hence $X_1Z_1 = \frac{z^2}{x}$. So we need $\frac{y^2}{z} = \frac{z^2}{x}$, and if we set $x=1$ we have $y^2 = z^3$, which is the same to saying that $y=t^3$ and $z=t^2$ for some $t>0$. PS. Notice the diagram from the official solution is obviously wrong, how can it be the case that $t^3 < t^2$ AND $t^3 < t^4$??