Sum of length of chords equal another

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Sum of length of chords equal another

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Source: CGMO 2016 Q2

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#1 h Aug 13, 2016, 3:10 AM • 2 Y

Y by Adventure10, Mango247

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 $\angle BAC<60^{\circ}$ .

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#2 h Aug 13, 2016, 9:53 AM • 2 Y

Y by Adventure10, Mango247

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 $\angle BAC < 60$ . $\blacksquare$

This post has been edited 1 time. Last edited by rkm0959, Aug 13, 2016, 9:53 AM

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bobthesmartypants

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bobthesmartypants

#3 h Feb 14, 2017, 6:13 PM • 2 Y

Y by Adventure10, Mango247

Too stupid to see rkm's easy solution for part b...

solution

This post has been edited 1 time. Last edited by bobthesmartypants, Feb 14, 2017, 6:14 PM

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