Isosceles triangles $A_3A_1O_2$ and $A_1A_2O_3$ are constructed on the sides of a triangle $A_1A_2A_3$ as the bases, outside the triangle. Let $O_1$ be a point outside $\Delta A_1A_2A_3$ such that $\angle O_1A_3A_2 =\frac 12\angle A_1O_3A_2$ and $\angle O_1A_2A_3 =\frac 12\angle A_1O_2A_3$. Prove that $A_1O_1\perp O_2O_3$, and if $T$ is the projection of $O_1$ onto $A_2A_3$, then $\frac{A_1O_1}{O_2O_3} = 2\frac{O_1T}{A_2A_3}$.
Problem
Source: 17-th Iranian Mathematical Olympiad 1999/2000
Tags: geometry, power of a point, radical axis, geometry proposed
lasha
19.04.2010 12:59
Denote $\angle{A_3O_2A_1}=x$ and $\angle{A_1O_3A_2}=y$. Take point $X$ different from $A_{3}$ on a line $O_1A_3$ so that $X$ belongs to the circle with center at $O_2$ and radii equal to $O_2A_3=O_2A_1$.(Call this circle $S_1$). Take point $Y$ on a line $A_2O_1$ in a similar way.(Call the second circle $S_2$). Then, $\angle{O_1XA_1}=x$ and $\angle{A_1YO_1}=y$, while $\angle{A_3O_1A_2}=180-x-y$ from triangle $A_3O_1A_2$. It means that points $X, A_1, Y$ are collinear. Assume the point $B$ belongs to $A_1O_1$ and circle $S_1$ as well(different from $A_1$). Than, $\angle{A_3BO_1}=\angle{A_3XA_1}=x=\angle{A_3A_2O_1}$. It means that $A_3BA_2O_1$ is cyclic and $\angle{O_1BA_2}=\angle{O_1A_3A_2}=y=\angle{A_1YA_2}$. The last equality means that $B$ belongs to the circle $S_2$ as well. So, line $A_1B$ is the common chord of circles $S_1$ and $S_2$. So, $O_2O_3$ is perpendicular to $A_1B$, the same as $A_1O_1$, q . e . d . Denote by $M$ the projection of $O_1$ into the line $XY$. Easy to notice the similarity of triangles $O_1A_3A_2$ and $O_1YX$. As $O_1T$ and $O_1M$ are respective altitudes, $\frac{O_1T}{A_2A_3}=\frac{O_1M}{XY}$. It means we have to prove $\frac{A_1O_1}{O_1M}=\frac{2O_2O_3}{XY}$. $(*)$. Note the similiraty of triangles $O_2BO_3$ and $XBY$. (as $\angle{BXY}=\angle{BO_2O_3}$ and $\angle{BYX}=\angle{BO_3O_2}$. ) So, $\frac{2O_2O_3}{XY}=\frac{2O_2B}{XB}=\frac{1}{sin\angle{XA_1B}}$, while $\frac{A_1O_1}{O_1M}=\frac{1}{sin{\angle{XA_1B}}}$. Hence, we have proved $(*)$, which completes our proof.
sayantanchakraborty
14.04.2014 08:27
lasha wrote:
Denote $\angle{A_3O_2A_1}=x$ and $\angle{A_1O_3A_2}=y$. Take point $X$ different from $A_{3}$ on a line $O_1A_3$ so that $X$ belongs to the circle with center at $O_2$ and radii equal to $O_2A_3=O_2A_1$.(Call this circle $S_1$). Take point $Y$ on a line $A_2O_1$ in a similar way.(Call the second circle $S_2$). Then, $\angle{O_1XA_1}=x$ and $\angle{A_1YO_1}=y$, while $\angle{A_3O_1A_2}=180-x-y$ from triangle $A_3O_1A_2$. It means that points $X, A_1, Y$ are collinear. Assume the point $B$ belongs to $A_1O_1$ and circle $S_1$ as well(different from $A_1$). Than, $\angle{A_3BO_1}=\angle{A_3XA_1}=x=\angle{A_3A_2O_1}$. It means that $A_3BA_2O_1$ is cyclic and $\angle{O_1BA_2}=\angle{O_1A_3A_2}=y=\angle{A_1YA_2}$. The last equality means that $B$ belongs to the circle $S_2$ as well. So, line $A_1B$ is the common chord of circles $S_1$ and $S_2$. So, $O_2O_3$ is perpendicular to $A_1B$, the same as $A_1O_1$, q . e . d .
Denote by $M$ the projection of $O_1$ into the line $XY$. Easy to notice the similarity of triangles $O_1A_3A_2$ and $O_1YX$. As $O_1T$ and $O_1M$ are respective altitudes, $\frac{O_1T}{A_2A_3}=\frac{O_1M}{XY}$. It means we have to prove $\frac{A_1O_1}{O_1M}=\frac{2O_2O_3}{XY}$. $(*)$. Note the similiraty of triangles $O_2BO_3$ and $XBY$. (as $\angle{BXY}=\angle{BO_2O_3}$ and $\angle{BYX}=\angle{BO_3O_2}$. ) So, $\frac{2O_2O_3}{XY}=\frac{2O_2B}{XB}=\frac{1}{sin\angle{XA_1B}}$, while $\frac{A_1O_1}{O_1M}=\frac{1}{sin{\angle{XA_1B}}}$. Hence, we have proved $(*)$, which completes our proof.
Very nice and elegant solution indeed,but with few errors.It will be $\angle{A_1O_3A_2}=2y$ and $\angle{A_1O_2A_3=2x}$.You can also prove that $A_1,B,O_1$ are collinear in a more simple way:$\angle{O_1A_2A_3}=\angle{O_1XY}=x \Rightarrow$ points $A_2,A_3,X,Y$ are concyclic $\Rightarrow O_1A_2*O_1Y=O_1A_3*O_1X \Rightarrow O_1 \in$ radical axis of the two circles.