Question

Diameter $\overline{AB}$ of a circle with center $O$ is $10$ units. $C$ is a point $4$ units from $A,$ and on $\overline{AB}.$ $D$ is a point $4$ units from $B,$ and on $\overline{AB}.$ $P$ is any point on the circle. Then the broken-line path from $C$ to $P$ to $D:$

has the same length for all positions of $P$

exceeds $10$ units for all positions of $P$

cannot exceed $10$ units

is shortest when $CPD$ is a right triangle

is longest when $P$ is equidistant from $C$ and $D$

Answer: E

Concepts:circle distance formula optimization symmetry

Difficulty rating: 1990

Small Hint:

Place the center at the origin and the diameter on the $x$ -axis, so $C=(-1,0)$ and $D=(1,0)$

Big Hint:

For $P=(u,v)$ on the circle, compare $CP^2=26+2u$ and $DP^2=26-2u$

Solution:

Place $O=(0,0),$ $C=(-1,0),$ $D=(1,0),$ and $P=(u,v)$ with $u^2+v^2=25.$ Then $\begin{aligned} CP^2&=26+2u,\\ DP^2&=26-2u. \end{aligned}$ Therefore $\begin{aligned} (CP+DP)^2 &=52\\ &\quad+2\sqrt{676-4u^2}. \end{aligned}$ This is largest when $u=0,$ exactly when $CP=DP.$

Therefore, the correct answer is E.

1958 AMC 12 Problem 48

Answer: E

Difficulty rating: 1990

Small Hint:

Big Hint:

Solution:

Problem 48 in Other Years