Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.7 AMC 10/12 Find the answer China

In convex quadrilateral $ABCD$ , $\vec{BC} = 2\vec{AD}$ . Point $P$ is on the plane of quadrilateral $ABCD$ , satisfying $\vec{PA} + 2020\vec{PB} + 2020\vec{PC} = 2020\vec{PD} = \vec{0}$ . Let $s$ and $t$ be the areas of quadrilateral $ABCD$ and $\triangle PAB$ , respectively. Then the value of $\frac{t}{s}$ is ______.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

$\overrightarrow{PA} + \overrightarrow{PC} = 2\overrightarrow{PY}, \quad \overrightarrow{PB} + \overrightarrow{PD} = 2\overrightarrow{PX},$
combining the given conditions, we know that $\overrightarrow{PY} + 2020\overrightarrow{PX} = \overrightarrow{0}$ . Hence, point $P$ lies on segment $XY$ , and $PX = \frac{1}{2021}$ . Let the distance from $A$ to $MN$ be $h$ . By the area formula, we can get

$\begin{aligned} \frac{t}{s} &= \frac{S_{\triangle PAB}}{S_{ABCD}} = \frac{PM \cdot h}{MN \cdot 2h} = \frac{PM}{2MN} \\ &= \frac{1 + \frac{1}{2021}}{2 \times 3} = \frac{337}{2021}. \end{aligned}$ $\quad \square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.