Theorem: Let $ABC$ be an equilateral triangle, let $P$ be a point on the plain, let $P_a,P_b,P_c$ be projection foot of $P$ on three sidelines $BC,CA,AB$ respectively. Then: $BP_b+CP_c > AP_a$.
I call result above is theorem because this theorem similarly: https://en.wikipedia.org/wiki/Pompeiu%27s_theorem
Leo Giugiuc's proof:
Let $b=e^\frac{2\pi i}{3}$ and $c=e^\frac{4\pi i}{3}$ then $b^3=c^3=1$ and $b^2=c$ and $c^2=b$ and $b+c=-1$ and $bc=1$.
We choose $A=1, B=b, C=c$ and $P=z$ where $b,c$ define above. Let $R_a$ be the reflection of $P$ in $BC$, then $\frac{R_a-b}{c-b}=\frac{\overline{z-b}}{c-b}$ $ \Rightarrow $ $\frac{R_a-b}{c-b}=\frac{\frac{1}{b}-\overline{z}}{\frac{1}{b}-\frac{1}{c}} $ $ \Rightarrow $ $R_a=b+c-bc.\overline{z}=-1-\overline{z}$. Since $P_a$ is the midpoint of $PR_a$ show that: $P_a=\frac{-1+z-\overline{z}}{2}$, similarly we have: $P_b=\frac{1+c+z-c\overline{z}}{2}$ and $P_c=\frac{1+b+z-b\overline{z}}{2}$ $ \Rightarrow $ $2AP_a=|3-z+\overline{z}|=|3-ki|=\sqrt{9+k^2}$ (1), where $z-\overline{z}=ki, k \in R$
$2BP_b=|2b-c-1-z+c.\overline{z}|$ and $2CP_c=|2c-b-1-z+b.\overline{z}|$
We have:
$2BP_b+2CP_c=|2b-c-1-z+c.\overline{z}|+|2c-b-1-z+b.\overline{z}|\\ =|c|.|2b-c-1-z+c.\overline{z}|+|b|.|2c-b-1-z+b.\overline{z}|\\= |2bc-c^2-c-zc+c^2.\overline{z}|+|2bc-b^2-b-bz+b^2.\overline{z}|\\=|2-b-c-zc+b.\overline{z}|+|2-c-b-bz+c.\overline{z}| \\= |3-zc+b.\overline{z}|+|3-bz+c.\overline{z}| \ge |6+z-\overline{z}|=\sqrt{36+k^2} $, (2)
Since (1) and (2), the proof of theorem are complete.
I call result above is theorem because this theorem similarly: https://en.wikipedia.org/wiki/Pompeiu%27s_theorem
Leo Giugiuc's proof:
Let $b=e^\frac{2\pi i}{3}$ and $c=e^\frac{4\pi i}{3}$ then $b^3=c^3=1$ and $b^2=c$ and $c^2=b$ and $b+c=-1$ and $bc=1$.
We choose $A=1, B=b, C=c$ and $P=z$ where $b,c$ define above. Let $R_a$ be the reflection of $P$ in $BC$, then $\frac{R_a-b}{c-b}=\frac{\overline{z-b}}{c-b}$ $ \Rightarrow $ $\frac{R_a-b}{c-b}=\frac{\frac{1}{b}-\overline{z}}{\frac{1}{b}-\frac{1}{c}} $ $ \Rightarrow $ $R_a=b+c-bc.\overline{z}=-1-\overline{z}$. Since $P_a$ is the midpoint of $PR_a$ show that: $P_a=\frac{-1+z-\overline{z}}{2}$, similarly we have: $P_b=\frac{1+c+z-c\overline{z}}{2}$ and $P_c=\frac{1+b+z-b\overline{z}}{2}$ $ \Rightarrow $ $2AP_a=|3-z+\overline{z}|=|3-ki|=\sqrt{9+k^2}$ (1), where $z-\overline{z}=ki, k \in R$
$2BP_b=|2b-c-1-z+c.\overline{z}|$ and $2CP_c=|2c-b-1-z+b.\overline{z}|$
We have:
$2BP_b+2CP_c=|2b-c-1-z+c.\overline{z}|+|2c-b-1-z+b.\overline{z}|\\ =|c|.|2b-c-1-z+c.\overline{z}|+|b|.|2c-b-1-z+b.\overline{z}|\\= |2bc-c^2-c-zc+c^2.\overline{z}|+|2bc-b^2-b-bz+b^2.\overline{z}|\\=|2-b-c-zc+b.\overline{z}|+|2-c-b-bz+c.\overline{z}| \\= |3-zc+b.\overline{z}|+|3-bz+c.\overline{z}| \ge |6+z-\overline{z}|=\sqrt{36+k^2} $, (2)
Since (1) and (2), the proof of theorem are complete.
