Given an equilateral triangle $ABC$ and a point $P$ so that the distances $P$ to $A$ and to $C$ are not farther than the distances $P$ to $B$. Prove that $PB = PA + PC$ if and only if $P$ lies on the circumcircle of $\vartriangle ABC$.
Problem
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Tags: geometry, circumcircle, Equilateral
SPHS1234
11.12.2020 17:03
I see , the converse is easy to prove ..
SlimTune
11.12.2020 18:35
First, let $a, b,$ and $c$ be the distances $PA, PB,$ and $PC$ respectively.
Since $PB > PA, PC$, we have that $P$ must be a point on minor arc $AC$.
Now, since quadrilateral $PABC$ is cyclic, we can use Ptolemy's theorem. Let $s$ be the side of the equilateral triangle. We have that following:
$$a \cdot s + c \cdot s = b \cdot s \Rightarrow b = a + c$$so we are done proving the converse of the statement.
@below, that's true. Then, you get that equality occurs when $P$ is on the circumcircle of $ABC$.
Again, use the same notation in the above proof. By Ptolemy's inequality, we have that
$$a \cdot s + c \cdot s \geq b \cdot s$$We can simplify to get the following inequality:
$$a + c \geq b$$However, we are given that $a + c = b$, so we know that the equality condition on the inequality is true, so $P$ must be on the circumcircle.
natmath
11.12.2020 18:38
Can't you just use ptolemy's inequality?
franzliszt
11.12.2020 18:43
Van Schooten’s theorem
SlimTune
11.12.2020 18:45
Are you allowed to cite van Schooten's theorem as "well-known" and finish the proof in one line? @below, it's a pretty fun theorem to say.
franzliszt
11.12.2020 18:47
Idk but this is AoPS and I always wanted to use it I'd be more careful on a real contest
natmath
11.12.2020 19:04
franzliszt wrote: Van Schooten’s theorem I think this theorem only proves the converse of the problem statement.
JustinLee2017
11.12.2020 19:10
SlimTune wrote: Are you allowed to cite van Schooten's theorem as "well-known" and finish the proof in one line? @below, it's a pretty fun theorem to say. Yes. it's a well known, obscure theorem (similar to radiation symbol theorem ). Can be proved by ptolemy's, or by extending stuff
wehaventawake
25.05.2023 17:37
Suppose $P$ lies on the $(ABC)$ then by Ptolemy's theorem we know that $$PA\cdot BC +AB\cdot PC=AC\cdot PC$$Because $AB=BC=AC$ then $$AP+PC=PB$$This equality hold if and only if $P$ lies on $(ABC)$