Problem 1994

Throughout the solution, triangles $AA_{1}D_{1}, BB_{1}A_{1}, CC_{1}B_{1}$, and $DD_{1}C_{1}$ will be referred to as border triangles. We will denote by $[\mathcal{R}]$ the area of a region $\mathcal{R}$.

First, we show that $k \geq 1$. 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 $[BB_{1}A_{1}] = \frac{1}{4}$. Moreover, as $C_{1}, D_{1}, D \rightarrow K$, we get $[A_{1}B_{1}C_{1}D_{1}] \rightarrow [A_{1}B_{1}K] = \frac{1}{4}$, $[AA_{1}D_{1}] \rightarrow [AA_{1}K] = \frac{1}{4}$, $[CC_{1}B_{1}] \rightarrow [CKB_{1}] = \frac{1}{4}$ and $[DD_{1}C_{1}] \rightarrow 0$. 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 $k \geq 1$.

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 $[A_{1}B_{1}C_{1}] \geq \min \{[AC_{1}B_{1}], [BA_{1}C_{1}], [CB_{1}A_{1}]\}$.

Proof. Let $A', B', C'$ 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 $AC'B'$, $BA'C'$, and $CB'A'$. (For convenience, let points $B_{1}$ and $C_{1}$ lie in triangle $AC'B'$; see Figure 2.) Let segments $B_{1}C_{1}$ and $AA_{1}$ intersect at point $X$. Then $X$ also lies in triangle $AC'B'$. Hence $A_{1}X \geq AX$, and we have
$$\frac{[A_{1}B_{1}C_{1}]}{[AC_{1}B_{1}]} = \frac{\frac{1}{2} A_{1}X \cdot B_{1}C_{1} \cdot \sin \angle A_{1}XC_{1}}{\frac{1}{2} AX \cdot B_{1}C_{1} \cdot \sin \angle AXB_{1}} = \frac{A_{1}X}{AX} \geq 1,$$
as required.

Otherwise, each one of triangles $AC'B'$, $BA'C'$, $CB'A'$ contains exactly one of points $A_{1}$, $B_{1}$, $C_{1}$, and we can assume that $BA_{1} < BA'$, $CB_{1} < CB'$, $AC_{1} < AC'$ (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 $\frac{[A_{1}B_{1}C_{1}]}{[A_{1}B_{1}C']} = \frac{C_{1}Y}{C'Y} > 1$; also, lines $A_{1}C'$ and $CA$ intersect at a point $Z$ on the extension of $CA$ beyond point $A$, hence $\frac{[A_{1}B_{1}C']}{[A_{1}B'C']} = \frac{B_{1}Z}{B'Z} > 1$. Finally, since $A_{1}A' \parallel B'C'$, we have $[A_{1}B_{1}C_{1}] > [A_{1}B_{1}C'] > [A_{1}B'C'] = [A'B'C'] = \frac{1}{4}[ABC]$.

Now, from $[A_{1}B_{1}C_{1}] + [AC_{1}B_{1}] + [BA_{1}C_{1}] + [CB_{1}A_{1}] = [ABC]$ we obtain that one of the remaining triangles $AC_{1}B_{1}, BA_{1}C_{1}, CB_{1}A_{1}$ 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 $[BB_{1}A_{1}]$ and $[CC_{1}B_{1}]$; 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, $S_{1} = [A_{1}B_{1}C_{1}] + [C_{1}D_{1}A_{1}] \geq S$. 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 $[A_{1}B_{1}C_{1}] \leq [D_{1}A_{1}B_{1}]$. 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 $[A_{1}B_{1}C_{1}] < [CC_{1}B_{1}] < [LC_{1}B_{1}]$, $[A_{1}B_{1}C_{1}] < [BB_{1}A_{1}]$ (both - since $[A_{1}B_{1}C_{1}]$ is small), and $[A_{1}B_{1}C_{1}] \leq [D_{1}A_{1}B_{1}] < [AA_{1}D_{1}] < [KA_{1}D_{1}] < [KA_{1}C_{1}]$ (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 $[KA_{1}C_{1}] > [A_{1}B_{1}C_{1}]$, and that ray $KC_{1}$ 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 $[A_{1}B_{1}C_{1}] < [CC_{1}B_{1}] < [LC_{1}B_{1}]$ and $[A_{1}B_{1}C_{1}] < [BB_{1}A_{1}]$ since triangle $A_{1}B_{1}C_{1}$ is small. This (together with $[A_{1}B_{1}C_{1}] < [KA_{1}C_{1}]$) contradicts the Lemma again.