Toolkit 119

Circle Distance Inequalities

Point $P$ and the circle are fixed. $A$ lies on the circle and can move.

\min\{PA\}=PB,\qquad \max\{PA\}=PC

The two circles are fixed. Points $P$ and $Q$ lie on the circles $\omega_1$ and $\omega_2$ , respectively, and can move.

\min\{PQ\}=BC,\qquad \max\{PQ\}=AD

Proof

PA\geq PB

\iff PA+R\geq PB+R

\iff PA+OA\geq PO

By Toolkit 88 — Triangle Inequality in $\triangle APO$ , this is correct.

Equality holds when $P$ , $A$ , and $O$ are collinear, so $A=B$ .

PA\leq PC

\iff PA-R\leq PC-R

\iff PA-R\leq PO

\iff PA\leq PO+R=PO+OA

By Toolkit 88 — Triangle Inequality in $\triangle APO$ , this is correct.

Equality holds when $P$ , $A$ , and $O$ are collinear, so $A=C$ .

By the previous part, $PQ\geq QL$ .

Now we want to prove $QL\geq BC$ .

QL\geq BC\Longleftrightarrow QL+R_1+R_2\geq BC+R_1+R_2

\Longleftrightarrow QO_1+QO_2\geq O_1O_2

By Toolkit 88 — Triangle Inequality in $\triangle QO_1O_2$ , this is correct.

Equality holds when $Q$ , $O_1$ , and $O_2$ are collinear, so $Q=C$ and $P=B$

By the previous theorem, $PQ\leq QN$ .

Now we want to prove $QN\leq AD$ .

QN\leq AD\Longleftrightarrow QO_1+R_1\leq O_1O_2+R_1+R_2

\Longleftrightarrow QO_1\leq O_1O_2+R_2 \Longleftrightarrow QO_1\leq O_1O_2+QO_2

By Toolkit 88 — Triangle Inequality in $\triangle QO_1O_2$ , this is correct.

Equality holds when $Q=D$ and $P=A$ .