Selection tests for the International Mathematical Olympiad 2013

Because $LM$ is parallel to $BC$, the problem is equivalent to proving that $\angle A{A}_{1}L=\angle AML=\angle ACB$. We present two solutions:



First solution. Let $L_1$ be the point on $AB$ such that $\angle A{A}_1{L}_1=\angle ACB$ and let us prove that $L_1=L$.

Quadrilateral $BC{B}_1{C}_1$ is cyclic since $\angle B{C}_{1}C=\angle B{B}_{1}C=90^{\circ }$. Therefore, $\angle B_1{C}_{1}A=\angle ACB=\angle A{A}_1{L}_1$. We deduce that the quadrilateral $A_{1}K{C}_1{L}_1$ is cyclic. Hence, $\angle L_{1}K{A}_1=\angle L_1{C}_1{A}_1$.

Quadrilateral $A{C}_1{A}_{1}C$ is cyclic since $\angle C{A}_{1}A=\angle C{C}_{1}A=90^{\circ }$. We deduce that $\angle L_1{C}_1{A}_1=\angle ACB$. Thus, $\angle L_{1}K{A}_1=\angle A{A}_1{L}_1$. This proves that $L_1{A}_1=L_{1}K$, that is $L_1=L$, the intersection point of the perpendicular bisector of $K{A}_1$ with $AB$.

Second solution. Since $LK=L{A}_1$, the problem is equivalent to proving that $\angle LK{A}_1=\angle ACB$. But $\angle BH{A}_1=90^{\circ }-\angle A_{1}BH=\angle ACB$, where $H$ is the orthocenter of triangle $ABC$. Therefore, the problem is equivalent to proving that $LK$ and $BH$ are parallel.

We know that $\frac{AL}{LB}=\frac{AN}{N{A}_1}$, where $N$ is the midpoint of $K{A}_1$, since $LM$ and $BC$ are parallel. It remains to prove that $\frac{AK}{KH}=\frac{AN}{N{A}_1}$.

To simplify the notations let us write $\alpha =\angle BAC$, $\beta =\angle CBA$ and $\gamma =\angle ACB$.

We know that $\angle K{C}_{1}A=\gamma$, $\angle H{C}_{1}K=90^{\circ }-\gamma$, $\angle C_{1}AK=90^{\circ }-\beta$ and $\angle KH{C}_1=\beta$. We deduce by applying sine laws on triangles $A{C}_{1}K$ and $C_{1}HK$ that $$\frac{AK}{KH}=\frac{\sin \angle K{C}_{1}A\cdot \sin \angle KH{C}_1}{\sin \angle H{C}_{1}K\cdot \sin \angle C_{1}AK}=\tan \beta \cdot \tan \gamma$$ On the other hand, we have $\frac{AN}{N{A}_1}=\frac{AK}{N{A}_1}+1=2\frac{AK}{K{A}_1}+1$. We also know that $\angle A_1{C}_{1}K=180^{\circ }-2\gamma$ and $\angle K{A}_1{C}_1=90^{\circ }-\alpha$. We deduce by applying sine laws on triangles $A{C}_{1}K$ and $C_1{A}_{1}K$ that $$\frac{AK}{K{A}_1}=\frac{\sin \gamma \cdot \cos \alpha }{\sin 2\gamma \cdot \cos \beta }=\frac{\cos \alpha }{2\cos \beta \cdot \cos \gamma }$$ We deduce that $$\frac{AN}{N{A}_1}=\frac{\cos \alpha }{\cos \beta \cdot \cos \gamma }+1=\frac{\cos \beta \cdot \cos \gamma -\cos (\beta +\gamma )}{\cos \beta \cdot \cos \gamma }=\tan \beta \cdot \tan \gamma =\frac{AK}{KH}.$$