A junior high school plane geometry problem

A junior high school plane geometry problem

Zhang Weihai (School of Electrical and Automation Engineering, Shandong University of Science and Technology, Qingdao 266590)

I graduated in early 1981 and graduated from high school for about three months. In the following days, while working in the production team, he continued to study high school mathematics. After studying the quadratic curve part of high school, especially the elliptic curve, I found the following junior high school plane geometry theorem and gave a very concise proof. During the Ph.D. study at Zhejiang University from 1995 to 1998, I once told a senior man, he thought it was a very clever proof. I always think that plane geometry in junior high school was my most interesting learning content at first, and it led me to like mathematics. Back then, due to the few exercises and reference books after class, I used to develop the habit of compiling and solving problems by myself. The discovery of the following conclusion may have benefited from this habit. After so many years, the proof process is still fresh in my memory. As a pastime, write it down and share it with everyone, and welcome everyone to provide valuable comments on the proof process. I believe that if the proof is correct (you are welcome to find the wrong one), it is beneficial to middle school students and their parents, and it can be regarded as a good intellectual development question. Of course, if the conclusion or the proof process is similar to that of the predecessors, it is purely coincidental (after all, the subject of plane geometry is too old, and I am not engaged in the teaching of middle school mathematics, worrying that something I don’t know already exists).

Theorem : in $△$ There is no pointPPinside$A$$B$$C$$P$ to the three vertices and the longest distance. That is, there is no point$P$使$PA+PB+PC$ maximum.

Proof: as shown in the picture above, with$B$ and$C$ is the focus, over$P$ make an ellipse,PPon the ellipse$P$ both sides were taken$P_1$And $P_2$Two points. According to the nature of the ellipse, $P_{1}B+P_{1}C=PB+PC=P_{2}B+P_{2}C$ . Therefore, we only need to prove that$P_{1}A$ and$P_{2}A$ has at least one greater than$PA$ can. Connect$P_1$And $P$，$P_2$And $P$ . Consider$△A{P}_{1}P$ and$△AP{P}_2$, Because $\angle P_{1}P{P}_2<180^{.}$, So$\angle P_{1}PA$ sum$\angle P_2$At least one of$P$$A$ is an obtuse angle, and the edge corresponding to the obtuse angle must be the largest edge. The conclusion is thus proved.

Note 1: The above proof combines the nature of the quadratic curve. How to use the knowledge of junior high school plane geometry to prove is an interesting question. In the above proof, ∠ P 1 PP 2 <18 0 is intuitively used $\angle P_{1}P{P}_2<180^{.}$, And this actually uses the convexity of the ellipse.
Note 2: Does the above theorem also hold for convex polygons? How to prove if it is established? Can this derive something from modern mathematics?