In $\triangle ABC, BC=a, CA=b, AB=c,$ and $\Gamma$ is its circumcircle. $(1)$ Determine a necessary and sufficient condition on $a,b$ and $c$ if there exists a unique point $P(P\neq B, P\neq C)$ on the arc $BC$ of $\Gamma$ not passing through point $A$ such that $PA=PB+PC$. $(2)$ Let $P$ be the unique point stated in $(1)$. If $AP$ bisects $BC$, prove that $\angle BAC<60^{\circ}$.
Problem
Source: CGMO 2016 Q2
Tags: geometry, circumcircle
rkm0959
13.08.2016 12:53
Hmm... Let $x=PA, y=PB, z=PC$. By Ptolemy we have $ax=by+cz$ and now we want $x=y+z$. These two are equivalent to $(a-b)y=(c-a)z$. Now as $\frac{y}{z} = \frac{\sin \angle BCP}{\sin \angle CBP}$, as $P$ moves from $B$ to $C$, this value strictly increases from $0$ to infinity. Therefore, what we want is $\frac{c-a}{a-b} > 0$, with $c \not= a$, $a \not= b$. For (2), since $AP$ bisects $BC$, we have $\frac{c-a}{a-b} = \frac{y}{z} = \frac{ \sin \angle BAP}{\sin \angle CAP} = \frac{b}{c}$. This gives us $ab-b^2=c^2-ca$. Since $(a-b)(c-a) > 0$, we can get that $\cos \angle BAC = \frac{b^2+c^2-a^2}{2bc} = \frac{ab+ac-a^2}{2bc} > \frac{1}{2}$. This is enough to claim that $\angle BAC < 60$. $\blacksquare$
bobthesmartypants
14.02.2017 21:13
Too stupid to see rkm's easy solution for part b...
By Ptolemy $PA\cdot a = PB\cdot b+PC\cdot c = (PB+PC)\cdot a \iff \dfrac{PB}{PC} = \dfrac{a-c}{b-a}$. Thus, $P$ exists and is unique when $\dfrac{a-c}{b-a} > 0\iff b<a<c\text{ or }b>a>c$.
When $AP$ bisects $BC$, by ratio lemma $\dfrac{BP\cdot \sin BPA}{CP\cdot \sin CPA} = 1\implies \dfrac{b}{c}=\dfrac{\sin B}{\sin C} = \dfrac{BP}{CP}=\dfrac{a-b}{b-a}\iff a=\dfrac{b^2+c^2}{b+c}$.
Then $$\cos\angle A = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{b^2+c^2-\left(\frac{b^2+c^2}{b+c}\right)^2}{2bc} > \dfrac{1}{2}$$$$\iff b^2+c^2 > bc+\left(\frac{b^2+c^2}{b+c}\right)^2 \iff (b+c)^2(b^2+c^2) > bc(b+c)^2 + (b^2+c^2)^2$$$$\iff b^3c+c^3b > 2b^2c^2$$which is true by AM-GM. Equality case is not obtained as $b>c$ or $b<c$. $\Box$