Question

$ABCDE$ is a regular pentagon. $AP,$ $AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD,$ $CB$ extended and $DE$ extended, respectively. Let $O$ be the center of the pentagon. If $OP=1,$ then $AO+AQ+AR$ equals

$3$

$1+\sqrt5$

$4$

$2+\sqrt5$

$5$

Answer: C

Concepts:regular pentagon area decomposition altitude

Difficulty rating: 2320

Small Hint:

Let $s$ be the side length and compute the pentagon’s area from its five central triangles

Big Hint:

Also split the pentagon into triangles $ABC,$ $ACD$ and $ADE$ with altitudes $AQ,AP,AR$

Solution:

Let the side length be $s.$ Since the apothem $OP=1,$ the five central triangles give pentagon area $\frac{5s}{2}.$ The same pentagon is the union of triangles $ABC,$ $ACD$ and $ADE,$ whose respective altitudes to side-length bases are $AQ,AP,AR.$ Hence $\frac{s}{2}(AQ+AP+AR)=\frac{5s}{2},$ so $AQ+AP+AR=5.$ Also $AP=AO+OP=AO+1.$ Therefore $AO+AQ+AR=4.$

Thus the correct answer is C.

1986 AMC 12 Problem 28

Answer: C

Difficulty rating: 2320

Small Hint:

Big Hint:

Solution:

Problem 28 in Other Years