The point $A_1$ on the perimeter of a convex quadrilateral $ABCD$ is such that the line $AA_1$ divides the quadrilateral into two parts of equal area. The points $B_1$, $C_1$, $D_1$ are defined similarly. Prove that the area of the quadrilateral $A_1B_1C_1D_1$ is greater than a quarter of the area of $ABCD$. L. Emelyanov
Problem
Source: Tuymaada 2013, Day 2, Problem 8 Juniors
Tags: inequalities, geometry, perimeter, analytic geometry, trigonometry, geometry proposed
pepat
27.07.2013 02:35
Let $AC$ be horizontal and denote the midpoint of $BD$ by $M$. The area is linear in $y$-coordinate hence $[ABCM]=[ADCM]=\frac12[ABCD]$ and $A_1$, $C_1$ lie on the horizontal line through $M$.
Moreover, $A_1C_1$ lies closer to $AC$ then both the $B$-midline in triangle $ABC$ and $D$-midline in triangle $ADC$ (it lies half the way from $B$ to $D$), therefore $A_1C_1>\frac12 AC$.
Likewise we find $B_1D_1\parallel BD$ and $B_1D_1>\frac12 BD$, hence (denoting the angle between diagonals by $\varphi$) we may finish by \[[A_1B_1C_1D_1] = \frac12 A_1C_1\cdot B_1D_1\cdot \sin\varphi > \frac12 \left(\frac12 AC\right)\cdot \left(\frac12 BD\right)\cdot \sin\varphi = \frac14 [ABCD].\]
IndoMathXdZ
29.09.2020 18:23
Really beautiful and weird problem. $\textbf{Lemma 01.}$ Let $C$ and $D$ be two points such that $[ABC] = [ABD]$, then $CD \parallel AB$. $\textit{Proof.}$ Obvious, well-known. Degenerate case will be left to the interesting readers. Now, we'll apply the above lemma to this problem by proving the crucial claim. $\textbf{Claim 01.}$ $A_1 C_1$ is a line passing through the midpoint of $BD$ and parallel to $AC$. $\textit{Proof.}$ Denote the midpoint of $BD$ as $M_1$. WLOG $M_1$ lies inside $[ACD]$. First, notice that $[ABCM_1] = [AM_1CD] = \frac{1}{2}[ABCD]$. Therefore, by the definition of $A_1, C_1$, we should have $A_1 \in CD$, since otherwise $[ABA_1] < [ABC] < [ABCM_1] = \frac{1}{2} [ABC]$, a contradiction. Therefore, we need to have $A_1 \in CD$. Similarly, $C_1 \in AD$. Now, notice that by the condition, $[ACA_1] = [ACM_1] = [ACC_1]$. Therefore, by our lemma, we have $A_1, M_1, C_1$ collinear and parallel to $AC$. Similarly, $B_1, M_2, D_1$ collinear and parallel to $BD$ where $M_2$ is the midpoint of $AC$. $\textbf{Claim 02.}$ $\frac{B_1D_1}{BD} \ge \frac{1}{2}$. $\textit{Proof.}$ Suppose otherwise, that $\frac{B_1D_1}{BD} < \frac{1}{2}$. Notice that $\triangle B_1 CD_1 \sim \triangle BCD$. Therefore, suppose line $BD$ intersects $AC$ at $O$. Then, by similar triangles, $\frac{CM_2}{CO} = \frac{CB_1}{CD} = \frac{B_1D_1}{BD} < \frac{1}{2}$. By assumption, which means that $\frac{CM_2}{CO} < \frac{CM_2}{CA}$. This gives us $CA < CO$. But, we assume that $ABCD$ is $\textbf{convex}$, which means that $O \in \overline{AC}$, a contradiction. To finish this, suppose $\measuredangle(AC, BD) = \theta$. Then, \[ [A_1 B_1 C_1 D_1] = \frac{1}{2} \cdot A_1 C_1 \cdot B_1 D_1 \cdot \sin \measuredangle (A_1 C_1 , B_1 D_1) \ge \frac{1}{4} \cdot \frac{1}{2} \cdot AC \cdot BD \cdot \sin \measuredangle (AC, BD) = \frac{1}{4} [ABCD] \]as desired.