2024 AMC 12B Problem 23

Question

A right pyramid has regular octagon

ABCDEFGH

with side length

1

as its base and apex

V.

Segments

\overline{AV}

and

\overline{DV}

are perpendicular. What is the square of the height of the pyramid?

1

\dfrac{1 + \sqrt2}{2}

\sqrt2

\dfrac{3}{2}

\dfrac{2 + \sqrt2}{3}

Each lateral edge has length

L

with

L^2 = h^2 + R^2,

where

R

is the octagon’s circumradius; the right angle at

V

gives

AD^2 = 2L^2

A

and

D

are three vertices apart, so the central angle is

135^\circ

and

AD^2 = R^2(2 + \sqrt2);

also

R^2 = \dfrac{2 + \sqrt2}{2}.

Solve for

h^2

Let

R

be the circumradius of the octagon and

L

the length of each lateral edge, so

L^2 = h^2 + R^2.

Since

\angle AVD = 90^\circ,

AD^2 = 2L^2.

Vertices

A

and

D

are three steps apart, a central angle of

135^\circ,

so

AD^2

= 2R^2(1 - \cos 135^\circ)

= R^2(2 + \sqrt2).

Setting

R^2(2 + \sqrt2) = 2(h^2 + R^2)

gives

2h^2 = R^2\sqrt2.

For a regular octagon of side

1,

R^2 = \dfrac{1}{2\sin^2(22.5^\circ)} = \dfrac{2 + \sqrt2}{2}.

Therefore

h^2

= \dfrac{R^2\sqrt2}{2}

= \dfrac{(2 + \sqrt2)\sqrt2}{4}

= \dfrac{2\sqrt2 + 2}{4}

= \dfrac{1 + \sqrt2}{2}.

Thus, the correct answer is B.

Problem 23 in other years