Auswahlwettbewerb Zur IMO, 2002

Solution:

In addition to the given figure we consider the square erected outward on the side $A B$ with center $C_{2}$. Since there is a homothety which maps the inscribed square with center $C_{1}$ onto the outward erected square, $C, C_{1}$ and $C_{2}$ lie on a line.
The altitude $h_{c}$ meets $A B$ at $H_{c}$ and let $|A H_{c}|=p, |H_{c} B|=q$. Further let $M_{c}$ be the midpoint of $A B$ and $C^{\prime}$ the intersection point of $C C_{2}$ with $A B$. Finally we set $|A C^{\prime}|=c_{1}$ and $|C^{\prime} B|=c_{2}$. All partial segments of $A B$ exist because the triangle $A B C$ is acute-angled. For the other sides of the triangle let the corresponding

points and segments be defined analogously.

By the intercept theorems (Strahlensätze) we have $\frac{|H_{c} C^{\prime}|}{|C H_{c}|}=\frac{|C^{\prime} M_{c}|}{|M_{c} C_{2}|}$, i.e. $\frac{c_{1}-p}{h_{c}}=\frac{\frac{1}{2} c-c_{1}}{\frac{1}{2} c} \Rightarrow c_{1}=\frac{c(p+h_{c})}{c+2 h_{c}}$.
Analogously $c_{2}=\frac{c(q+h_{c})}{c+2 h_{c}}$ and we obtain $\frac{c_{1}}{c_{2}}=\frac{\frac{p}{h_{c}}+1}{\frac{q}{h_{c}}+1}$. (For $\alpha=\beta$ we have $\frac{c_{1}}{c_{2}}=1$.)
The angles between the altitudes and the two respective adjacent sides are denoted cyclically by $\gamma_{1}, \gamma_{2}$ resp. $\alpha_{1}, \alpha_{2}$ resp. $\beta_{1}, \beta_{2}$. Then $\alpha_{1}=\gamma_{2}(=90^{\circ}-\beta), \quad \beta_{1}=\alpha_{2}(=90^{\circ}-\gamma)$ and $\gamma_{1}=\beta_{2}(=90^{\circ}-\alpha)$. Further $\frac{p}{h_{c}}=\tan \gamma_{1}$ and $\frac{q}{h_{c}}=\tan \gamma_{2}$, hence $\frac{c_{1}}{c_{2}}=\frac{\tan \gamma_{1}+1}{\tan \gamma_{2}+1}$. Analogously $\frac{a_{1}}{a_{2}}=\frac{\tan \alpha_{1}+1}{\tan \alpha_{2}+1}$ and $\frac{b_{1}}{b_{2}}=\frac{\tan \beta_{1}+1}{\tan \beta_{2}+1}$. Thus we have

$\frac{a_{1}}{a_{2}} \cdot \frac{b_{1}}{b_{2}} \cdot \frac{c_{1}}{c_{2}}=\frac{(\tan \alpha_{1}+1)(\tan \beta_{1}+1)(\tan \gamma_{1}+1)}{(\tan \alpha_{2}+1)(\tan \beta_{2}+1)(\tan \gamma_{2}+1)}=\frac{\tan \alpha_{1}+1}{\tan \gamma_{2}+1} \cdot \frac{\tan \beta_{1}+1}{\tan \alpha_{2}+1} \cdot \frac{\tan \gamma_{1}+1}{\tan \beta_{2}+1}=1 \cdot 1 \cdot 1=1$.
By the converse of Ceva's theorem, the lines $A A^{\prime}=A A_{1}$, $B B^{\prime}=B B_{1}$ and $C C^{\prime}=C C_{1}$ therefore intersect in one point.

Remark: Some participants conjectured that the lines $A A_{1}$ etc. are the angle bisectors of the given triangle. This conjecture is false, as can be seen in a very acute, almost right-angled triangle. Ceva's theorem is not necessary for the proof; there are other possible solutions, e.g. using spiral similarities (rotation-dilations).