48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions

Let $A_0,B_0,C_0$ be the points of intersection of the lines $A{A}^{\prime },B{B}^{\prime }$ and $C{C}^{\prime }$ (see Figure). We claim that area $\left(A_0{B}_0{C}_0\right)=\frac{1}{2}\mathrm{area}(ABC)$, hence it is constant.

Consider the inscribed hexagon $ABC{C}^{\prime }P{A}^{\prime }$. By Pascal's theorem, the points of intersection of its opposite sides (or of their extensions) are collinear. These points are $AB\cap C^{\prime }P=C_1$, $BC\cap P{A}^{\prime }=A_1$, $C{C}^{\prime }\cap A^{\prime }A=B_0$. So point $B_0$ lies on the midline $A_1{C}_1$ of triangle $ABC$. Analogously, points $A_0$ and $C_0$ lie on lines $B_1{C}_1$ and $A_1{B}_1$, respectively.

Lines $AC$ and $A_1{C}_1$ are parallel, so triangles $B_0{C}_0{A}_1$ and $A{C}_0{B}_1$ are similar; hence we have $$\frac{B_0{C}_0}{A{C}_0}=\frac{A_1{C}_0}{B_1{C}_0}.$$ Analogously, from $BC\parallel B_1{C}_1$ we obtain $$\frac{A_1{C}_0}{B_1{C}_0}=\frac{B{C}_0}{A_0{C}_0}.$$ Combining these equalities, we get $$\frac{B_0{C}_0}{A{C}_0}=\frac{B{C}_0}{A_0{C}_0},$$ or $$A_0{C}_0\cdot B_0{C}_0=A{C}_0\cdot B{C}_0.$$ Hence we have $$\mathrm{area}\left(A_0{B}_0{C}_0\right)=\frac{1}{2}{A}_0{C}_0\cdot B_0{C}_0\sin \angle A_0{C}_0{B}_0=\frac{1}{2}A{C}_0\cdot B{C}_0\sin \angle A{C}_{0}B=\mathrm{area}\left(AB{C}_0\right).$$ Since $C_0$ lies on the midline, we have $d\left(C_0,AB\right)=\frac{1}{2}d(C,AB)$ (we denote by $d(X,YZ)$ the distance between point $X$ and line $YZ$). Then we obtain $$\mathrm{area}\left(A_0{B}_0{C}_0\right)=\mathrm{area}\left(AB{C}_0\right)=\frac{1}{2}AB\cdot d\left(C_0,AB\right)=\frac{1}{4}AB\cdot d(C,AB)=\frac{1}{2}\mathrm{area}(ABC).$$

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Alternative solution.

Again, we prove that area $\left(A_0{B}_0{C}_0\right)=\frac{1}{2}\mathrm{area}(ABC)$.

We can assume that $P$ lies on arc $AC$. Mark a point $L$ on side $AC$ such that $\angle CBL=\angle PBA$; then $\angle LBA=\angle CBA-\angle CBL=\angle CBA-\angle PBA=\angle CBP$. Note also that $\angle BAL=\angle BAC=\angle BPC$ and $\angle LCB=\angle APB$. Hence, triangles $BAL$ and $BPC$ are similar, and so are triangles $LCB$ and $APB$.

Analogously, mark points $K$ and $M$ respectively on the extensions of sides $CB$ and $AB$ beyond point $B$, such that $\angle KAB=\angle CAP$ and $\angle BCM=\angle PCA$. For analogous reasons, $\angle KAC=\angle BAP$ and $\angle ACM=\angle PCB$. Hence $△ABK\sim △APC\sim △MBC$, $△ACK\sim △APB$, and $△MAC\sim △BPC$. From these similarities, we have $\angle CMB=\angle KAB=\angle CAP$, while we have seen that $\angle CAP=\angle CBP=\angle LBA$. Hence, $AK\parallel BL\parallel CM$.



Let line $C{C}^{\prime }$ intersect $BL$ at point $X$. Note that $\angle LCX=\angle AC{C}^{\prime }=\angle AP{C}^{\prime }=\angle AP{C}_1$, and $P{C}_1$ is a median in triangle $APB$. Since triangles $APB$ and $LCB$ are similar, $CX$ is a median in triangle $LCB$, and $X$ is a midpoint of $BL$. For the same reason, $A{A}^{\prime }$ passes through this midpoint, so $X=B_0$. Analogously, $A_0$ and $C_0$ are the midpoints of $AK$ and $CM$.

Now, from $A{A}_0\parallel C{C}_0$, we have $$\mathrm{area}\left(A_0{B}_0{C}_0\right)=\mathrm{area}\left(A{C}_0{A}_0\right)-\mathrm{area}\left(A{B}_0{A}_0\right)=\mathrm{area}\left(AC{A}_0\right)-\mathrm{area}\left(A{B}_0{A}_0\right)=\mathrm{area}\left(AC{B}_0\right).$$ Finally, $$\mathrm{area}\left(A_0{B}_0{C}_0\right)=\mathrm{area}\left(AC{B}_0\right)=\frac{1}{2}{B}_{0}L\cdot AC\sin AL{B}_0=\frac{1}{4}BL\cdot AC\sin ALB=\frac{1}{2}\mathrm{area}(ABC).$$