48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions

Throughout the solution, triangles $A{A}_1{D}_1,B{B}_1{A}_1,C{C}_1{B}_1$, and $D{D}_1{C}_1$ will be referred to as border triangles. We will denote by $[\mathcal{R}]$ the area of a region $\mathcal{R}$.

First, we show that $k\ge 1$. Consider a triangle $ABC$ with unit area; let $A_1,B_1,K$ be the midpoints of its sides $AB,BC,AC$, respectively. Choose a point $D$ on the extension of $BK$, close to $K$. Take points $C_1$ and $D_1$ on sides $CD$ and $DA$ close to $D$ (see Figure 1). We have $[B{B}_1{A}_1]=\frac{1}{4}$. Moreover, as $C_1,D_1,D\to K$, we get $[A_1{B}_1{C}_1{D}_1]\to [A_1{B}_{1}K]=\frac{1}{4}$, $[A{A}_1{D}_1]\to [A{A}_{1}K]=\frac{1}{4}$, $[C{C}_1{B}_1]\to [CK{B}_1]=\frac{1}{4}$ and $[D{D}_1{C}_1]\to 0$. Hence, the sum of the two smallest areas of border triangles tends to $\frac{1}{4}$, as well as $[A_1{B}_1{C}_1{D}_1]$; therefore, their ratio tends to $1$, and $k\ge 1$.

We are left to prove that $k=1$ satisfies the desired property.

Figure 1 Figure 2 Figure 3

Lemma. Let points $A_1,B_1,C_1$ lie respectively on sides $BC,CA,AB$ of a triangle $ABC$. Then $[A_1{B}_1{C}_1]\ge \min \{[A{C}_1{B}_1],[B{A}_1{C}_1],[C{B}_1{A}_1]\}$.

Proof. Let $A^{\prime },B^{\prime },C^{\prime }$ be the midpoints of sides $BC,CA$ and $AB$, respectively.

Suppose that two of points $A_1,B_1,C_1$ lie in one of triangles $A{C}^{\prime }{B}^{\prime }$, $B{A}^{\prime }{C}^{\prime }$, and $C{B}^{\prime }{A}^{\prime }$. (For convenience, let points $B_1$ and $C_1$ lie in triangle $A{C}^{\prime }{B}^{\prime }$; see Figure 2.) Let segments $B_1{C}_1$ and $A{A}_1$ intersect at point $X$. Then $X$ also lies in triangle $A{C}^{\prime }{B}^{\prime }$. Hence $A_{1}X\ge AX$, and we have $$\frac{[A_1{B}_1{C}_1]}{[A{C}_1{B}_1]}=\frac{\frac{1}{2}{A}_{1}X\cdot B_1{C}_1\cdot \sin \angle A_{1}X{C}_1}{\frac{1}{2}AX\cdot B_1{C}_1\cdot \sin \angle AX{B}_1}=\frac{A_{1}X}{AX}\ge 1,$$ as required.

Otherwise, each one of triangles $A{C}^{\prime }{B}^{\prime }$, $B{A}^{\prime }{C}^{\prime }$, $C{B}^{\prime }{A}^{\prime }$ contains exactly one of points $A_1$, $B_1$, $C_1$, and we can assume that $B{A}_1<B{A}^{\prime }$, $C{B}_1<C{B}^{\prime }$, $A{C}_1<A{C}^{\prime }$ (see Figure 3). Then lines $B_1{A}_1$ and $AB$ intersect at a point $Y$ on the extension of $AB$ beyond point $B$, hence $\frac{[A_1{B}_1{C}_1]}{[A_1{B}_1{C}^{\prime }]}=\frac{C_{1}Y}{C^{\prime }Y}>1$; also, lines $A_1{C}^{\prime }$ and $CA$ intersect at a point $Z$ on the extension of $CA$ beyond point $A$, hence $\frac{[A_1{B}_1{C}^{\prime }]}{[A_1{B}^{\prime }{C}^{\prime }]}=\frac{B_{1}Z}{B^{\prime }Z}>1$. Finally, since $A_1{A}^{\prime }\parallel B^{\prime }{C}^{\prime }$, we have $[A_1{B}_1{C}_1]>[A_1{B}_1{C}^{\prime }]>[A_1{B}^{\prime }{C}^{\prime }]=[A^{\prime }{B}^{\prime }{C}^{\prime }]=\frac{1}{4}[ABC]$.

Now, from $[A_1{B}_1{C}_1]+[A{C}_1{B}_1]+[B{A}_1{C}_1]+[C{B}_1{A}_1]=[ABC]$ we obtain that one of the remaining triangles $A{C}_1{B}_1,B{A}_1{C}_1,C{B}_1{A}_1$ has an area less than $\frac{1}{4}[ABC]$, so it is less than $[A_1{B}_1{C}_1]$. $\square$

Now we return to the problem. We say that triangle $A_1{B}_1{C}_1$ is small if $[A_1{B}_1{C}_1]$ is less than each of $[B{B}_1{A}_1]$ and $[C{C}_1{B}_1]$; otherwise this triangle is big (the similar notion is introduced for triangles $B_1{C}_1{D}_1,C_1{D}_1{A}_1,D_1{A}_1{B}_1$). If both triangles $A_1{B}_1{C}_1$ and $C_1{D}_1{A}_1$ are big, then $[A_1{B}_1{C}_1]$ is not less than the area of some border triangle, and $[C_1{D}_1{A}_1]$ is not less than the area of another one; hence, $S_1=[A_1{B}_1{C}_1]+[C_1{D}_1{A}_1]\ge S$. The same is valid for the pair of $B_1{C}_1{D}_1$ and $D_1{A}_1{B}_1$. So it is sufficient to prove that in one of these pairs both triangles are big.

Suppose the contrary. Then there is a small triangle in each pair. Without loss of generality, assume that triangles $A_1{B}_1{C}_1$ and $D_1{A}_1{B}_1$ are small. We can assume also that $[A_1{B}_1{C}_1]\le [D_1{A}_1{B}_1]$. Note that in this case ray $D_1{C}_1$ intersects line $BC$.

Consider two cases.

Figure 4 Figure 5

Case 1. Ray $C_1{D}_1$ intersects line $AB$ at some point $K$. Let ray $D_1{C}_1$ intersect line $BC$ at point $L$ (see Figure 4). Then we have $[A_1{B}_1{C}_1]<[C{C}_1{B}_1]<[L{C}_1{B}_1]$, $[A_1{B}_1{C}_1]<[B{B}_1{A}_1]$ (both - since $[A_1{B}_1{C}_1]$ is small), and $[A_1{B}_1{C}_1]\le [D_1{A}_1{B}_1]<[A{A}_1{D}_1]<[K{A}_1{D}_1]<[K{A}_1{C}_1]$ (since triangle $D_1{A}_1{B}_1$ is small). This contradicts the Lemma, applied for triangle $A_1{B}_1{C}_1$ inside $LKB$.

Case 2. Ray $C_1{D}_1$ does not intersect $AB$. Then choose a "sufficiently far" point $K$ on ray $BA$ such that $[K{A}_1{C}_1]>[A_1{B}_1{C}_1]$, and that ray $K{C}_1$ intersects line $BC$ at some point $L$ (see Figure 5). Since ray $C_1{D}_1$ does not intersect line $AB$, the points $A$ and $D_1$ are on different sides of $KL$; then $A$ and $D$ are also on different sides, and $C$ is on the same side as $A$ and $B$. Then analogously we have $[A_1{B}_1{C}_1]<[C{C}_1{B}_1]<[L{C}_1{B}_1]$ and $[A_1{B}_1{C}_1]<[B{B}_1{A}_1]$ since triangle $A_1{B}_1{C}_1$ is small. This (together with $[A_1{B}_1{C}_1]<[K{A}_1{C}_1]$) contradicts the Lemma again.