An Equal-Ratio Fermat Point

★ 8/10plane geometry Law of Cosines Triangle Geometry vectors

Problem

In $\triangle ABC$ , $AC = 2AB = 2$ and $BC = \sqrt{3}$ . A point $P$ inside the triangle satisfies

\frac{S_{\triangle PAB}}{\overrightarrow{PA}\cdot\overrightarrow{PB}} = \frac{S_{\triangle PBC}}{\overrightarrow{PB}\cdot\overrightarrow{PC}} = \frac{S_{\triangle PCA}}{\overrightarrow{PC}\cdot\overrightarrow{PA}},

where $S_{\triangle}$ denotes area. Find $PA + PB + PC$ .

Answer

Solution

Difficulty	8/10
Topics	plane geometry, Law of Cosines, Triangle Geometry, vectors

Whiteboard

Your sketch is saved only in this browser. To share it, export your drawing as an image (whiteboard menu → Export as → PNG), then upload that image in the comments below.

Discussion

Ask questions, share alternate solutions, and use LaTeX freely.

0 comments

No comments yet.