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Source: Mediterranean MO 2000

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#1 h Aug 26, 2005, 8:20 PM • 2 Y

Y by Adventure10, Mango247

Let $P,Q,R,S$ be the midpoints of the sides $BC,CD,DA,AB$ of a convex quadrilateral, respectively. Prove that
$4(AP^2+BQ^2+CR^2+DS^2)\le 5(AB^2+BC^2+CD^2+DA^2)$

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MysticTerminator

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MysticTerminator

#2 h Aug 26, 2005, 9:46 PM • 2 Y

Y by Adventure10, Mango247

consider the midpoints $M, N$ of $\overline{AC}, \overline{BD}$ . As the distance $\overline{MN} \ge 0$ , we have (using vectors), $4(\vec{A} + \vec{C} - \vec{B} - \vec{D})^2 \ge 0 \Rightarrow$ $4\vec{A}^2 + 4\vec{B}^2 + 4\vec{C}^2 + 4\vec{D}^2 + 8\vec{A} \cdot \vec{C} + 8\vec{B} \cdot \vec{D} \ge 8\vec{A} \cdot \vec{B} + 8\vec{B} \cdot \vec{C} + 8\vec{C} \cdot \vec{D} + 8\vec{D} \cdot \vec{A}$ $\Rightarrow 10\vec{A}^2 + 10\vec{B}^2 + 10\vec{C}^2 + 10\vec{D}^2 - 10\vec{A} \cdot \vec{B} - 10\vec{B} \cdot \vec{C} - 10\vec{C} \cdot \vec{D} - 10\vec{D} \cdot \vec{A} \ge 6\vec{A}^2 + 6\vec{B}^2 + 6\vec{C}^2 + 6\vec{D}^2 - 2\vec{A} \cdot \vec{B} - 2\vec{B} \cdot \vec{C} - 2\vec{C} \cdot \vec{D} - 2\vec{D} \cdot \vec{A} - 8\vec{A} \cdot \vec{C} - 8\vec{B} \cdot \vec{D}$ $\Rightarrow 5(\overline{AB}^2 + \overline{BC}^2 + \overline{CD}^2 + \overline{DA}^2) \ge 4(\overline{AP}^2 + \overline{BQ}^2 + \overline{CR}^2 + \overline{DS}^2)$

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#3 h Sep 11, 2005, 12:12 PM • 2 Y

Y by Adventure10, Mango247

Problem. Let A, B, C, D be four arbitrary points in the plane, and let P, Q, R, S be the midpoints of the segments BC, CD, DA, AB. Prove that

$4\left(AP^2+BQ^2+CR^2+DS^2\right)\leq 5\left(AB^2+BC^2+CD^2+DA^2\right)$ .

Solution. Here is a different way to write MysticTerminator's solution, without explicitely using vectors but with a formula for the length of a median of a triangle:

Lemma 1. If $m_a$ is the length of the median from the vertex A of a triangle ABC, then $4m_a^2=2b^2+2c^2-a^2$ .

Applying Lemma 1 to the triangle ABC, whose median from the vertex A is the segment AP (in fact, P is the midpoint of the side BC of this triangle), we get

$4\cdot AP^2=2\cdot AC^2+2\cdot AB^2-BC^2$ .

Similarly,

$4\cdot BQ^2=2\cdot BD^2+2\cdot BC^2-CD^2$ ;
$4\cdot CR^2=2\cdot AC^2+2\cdot CD^2-DA^2$ ;
$4\cdot DS^2=2\cdot BD^2+2\cdot DA^2-AB^2$ .

Adding these four equations yields

$4\cdot AP^2+4\cdot BQ^2+4\cdot CR^2+4\cdot DS^2=\left(2\cdot AC^2+2\cdot AB^2-BC^2\right)$
$+\left(2\cdot BD^2+2\cdot BC^2-CD^2\right)+\left(2\cdot AC^2+2\cdot CD^2-DA^2\right)+\left(2\cdot BD^2+2\cdot DA^2-AB^2\right)$ .

This simplifies to

$4\left(AP^2+BQ^2+CR^2+DS^2\right)=4\cdot\left(AC^2+BD^2\right)+\left(AB^2+BC^2+CD^2+DA^2\right)$ .

Now, let M and N be the midpoints of the segments AC and BD, respectively. Then, the segment MN is the median from the vertex N of triangle ANC (since M is the midpoint of the segment AC); thus, by Lemma 1, we have

$4\cdot MN^2=2\cdot AN^2+2\cdot CN^2-AC^2$ .

On the other hand, the segment AN is the median from the vertex A of triangle ABD (since N is the midpoint of the segment BD); hence, Lemma 1 yields $4\cdot AN^2=2\cdot AB^2+2\cdot DA^2-BD^2$ . Similarly, $4\cdot CN^2=2\cdot BC^2+2\cdot CD^2-BD^2$ . Adding these two equations, we get

$4\cdot AN^2+4\cdot CN^2=\left(2\cdot AB^2+2\cdot DA^2-BD^2\right)+\left(2\cdot BC^2+2\cdot CD^2-BD^2\right)$
$=2\cdot\left(AB^2+BC^2+CD^2+DA^2-BD^2\right)$ .

Division by 2 yields $2\cdot AN^2+2\cdot CN^2=AB^2+BC^2+CD^2+DA^2-BD^2$ , and thus

$4\cdot MN^2=2\cdot AN^2+2\cdot CN^2-AC^2=AB^2+BC^2+CD^2+DA^2-BD^2-AC^2$
$=\left(AB^2+BC^2+CD^2+DA^2\right)-\left(AC^2+BD^2\right)$ .

Since $MN^2\geq 0$ , we thus have $\left(AB^2+BC^2+CD^2+DA^2\right)-\left(AC^2+BD^2\right)\geq 0$ , so that $AC^2+BD^2\leq AB^2+BC^2+CD^2+DA^2$ . Thus,

$4\left(AP^2+BQ^2+CR^2+DS^2\right)=4\cdot\left(AC^2+BD^2\right)+\left(AB^2+BC^2+CD^2+DA^2\right)$
$\leq 4\cdot\left(AB^2+BC^2+CD^2+DA^2\right)+\left(AB^2+BC^2+CD^2+DA^2\right)=5\left(AB^2+BC^2+CD^2+DA^2\right)$ ,

and the problem is solved.

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