AP BP CP < (1+x)^2 (1-x)

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AP BP CP < (1+x)^2 (1-x)

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Source: South African MO 1995 Q7

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djb86

#1 h May 27, 2012, 3:02 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

The circumcircle of $\triangle ABC$ has radius $1$ and centre $O$ and $P$ is a point inside the triangle such that $OP=x$ . Prove that
$AP\cdot BP\cdot CP\le(1+x)^2(1-x),$
with equality if and only if $P=O$ .

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306 posts

ArefS

#2 h May 27, 2012, 7:04 PM • 1 Y

Y by Adventure10

in case when $P\equiv O$ we have equality. so assume that $P,O$ are distinct.
let $PO$ intersect the circle at $X,Y$ such that $X,P,O,Y$ are in this order.
$P$ is inside $\triangle ABC$ so $P$ lies in one of the triangles: $\triangle AYC,\triangle BYC,\triangle AYB$
WLOG assume that $P$ lies inside $AYB$ .
assume that $AP$ intersects the circle in $D$ and $BP$ intersects the circle in $E$
obviously, the maximum value of $PZ$ where $Z$ in the circle is $PY$ . so we have: $PC\le PY$ so $AP\cdot BP\cdot CP\le AP\cdot BP \cdot PY$
we need to prove that $AP\cdot BP \cdot PY\le (1+x)^2(1-x)$
or equivalently: $AP\cdot BP \le 1-x^2\iff AP\cdot BP\le AP\cdot PD\iff PB\le PD$ .
to prove that $PB\le PD$ we prove that $\angle PBD \ge \angle PDB\iff \overarc{DE}\ge \overarc{AB}$ .
we have that:
$\angle DOY=\angle DPY +\angle ODP$ so $\angle DOY > \angle DPY\iff \overarc{AX} <\overarc{DY}$ .
because of the fact that $B$ lies on arc $\overarc{XD}$ by a similiar argument we get that $\overarc{BX}<\overarc{YE}$
so $\overarc{AX}+\overarc{XB}<\overarc{DY}+\overarc{YE}\iff \angle PDB<\angle PBD$
hence: $\boxed{AP\cdot BP\cdot CP\le (x+1)^2(x-1)}$
by a similar argument we can also prove that: $\boxed{AP\cdot BP\cdot CP\ge (x-1)^2(x+1)}$ .

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