2024 AMC 12B Problems/Problem 23

Problem

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?

$\textbf{(A) }1 \qquad \textbf{(B) }\frac{1+\sqrt2}{2} \qquad \textbf{(C) }\sqrt2 \qquad \textbf{(D) }\frac32 \qquad \textbf{(E) }\frac{2+\sqrt2}{3} \qquad$

Solution 1

To find the height of the pyramid, we need the length from the center of the octagon (denote as $I$ ) to its vertices and the length of AV.

From symmetry, we know that $\overline{AV} = \overline{DV}$ , therefore $\triangle{AVD}$ is a 45-45-90 triangle. Denote $\overline{AV}$ as $x$ so that $\overline{AD} = x\sqrt{2}$ . Doing some geometry on the isosceles trapezoid $ABCD$ (we know this from the fact that it is a regular octagon) reveals that $\overline{AD}=1+2(\sqrt{2}/2)=1+\sqrt{2}$ and $\overline{AV}=(\overline{AD})/\sqrt{2}=(\sqrt{2}+2)/2$ .

To find the length $\overline{IA}$ , we cut the octagon into 8 triangles, each witha smallest angle of 45 degrees. Using the law of cosines on $\triangle{AIB}$ we find that ${\overline{IA}}^2=(2+\sqrt{2})/2$ .

Finally, using the pythagorean theorem, we can find that ${\overline{IV}}^2={\overline{AV}}^2-{\overline{IA}}^2= {((\sqrt{2}+2)/2)}^2 - (2+\sqrt{2})/2 = \boxed{(1+\sqrt{2})/2}$ which is answer choice $\boxed{B}$ .

Solution 2 (Less computation)

Let $O$ be the center of the regular octagon. Connect $AD$ , and let $I$ be the midpoint of line segment $AD$ . It is easy to see that $VI=\frac{1}{2} AD=\frac{1+\sqrt{2}}{2}$ and $OI=\frac{1}{2}AH=\frac{1}{4}$ . Hence, $VO^2=VI^2-OI^2$ $=\left(\frac{1+\sqrt{2}}{2}\right)^2-\frac{1}{4}$ $=\frac{1+\sqrt{2}}{2}$ Hence, the answer is $\boxed{B}$ .

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2024 AMC 12B Problems/Problem 23

Problem

Solution 1

Solution 2 (Less computation)