Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.7 AMC 10/12 Find the answer South Africa

$ABCD$ is a rectangle and $P$ is a point on $BC$ . If the area of triangle $ABP$ is one third of the area of the rectangle, then the ratio $BP : PC$ is

Pick one

Solutions — 2

Solution 1

$\frac{\text{area } \triangle ABP}{\text{area } ABCD} = \frac{1}{3}$ so $\frac{\frac{1}{2} \text{BP} \cdot \text{AB}}{\text{BC} \cdot \text{AB}} = \frac{1}{3}$ so $\frac{\frac{1}{2} \text{BP}}{\text{BC}} = \frac{1}{3}$ and therefore $\frac{\text{BP}}{\text{BC}} = \frac{2}{3}$
and then $BP : PC$ is $\frac{2}{3} : \frac{1}{3} = 2 : 1$

Solution 2

$\frac{\text{area } \triangle ABP}{\text{area } ABCD} = \frac{1}{3}$ so $\frac{\frac{1}{2} \cdot BP \cdot AB}{BC \cdot AB} = \frac{1}{3}$ so $\frac{\frac{1}{2} BP}{BC} = \frac{1}{3}$ and therefore $\frac{BP}{BC} = \frac{2}{3}$
and then $BP : PC$ is $\frac{2}{3} : \frac{1}{3} = 2 : 1$

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