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Source: Tuymaada 2013, Day 2, Problem 8 Juniors

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#1 h Jul 26, 2013, 8:33 AM • 2 Y

Y by Adventure10, Mango247

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

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#2 h Jul 26, 2013, 11:35 PM • 2 Y

Y by Adventure10, Mango247

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#3 h Sep 29, 2020, 3:23 PM

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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.

This post has been edited 1 time. Last edited by IndoMathXdZ, Sep 29, 2020, 3:24 PM

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