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 with a 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 $\boxed{B}$ .

～username2333 ~hashbrown2009

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 $\boxed{B}$ .

~tsun26

Solution 3

~Kathan

Solution 4 (Easy Geometry)

Firstly, scale each of the sides up by $2\sqrt{2}$ , and split the octagon into half. Next, divide the octagon into a rectangle and the isosceles trapezoid, and let $A$ and $D$ be the points where the shapes coincide. Also, let $I$ be the center of the rectangle (which is also the center of the octagon.) By drawing $45-45-90$ triangles in the trapezoid, we see that $AD = 4+2\sqrt{2}$ . Now, consider the length of $AI$ . It is half of the diagonal of the rectangle, and since the diagonal is just $\sqrt{(2\sqrt{2})^{2}+(4+2\sqrt{2})^{2}} = \sqrt{8+24+16\sqrt{2}} = \sqrt{32+16\sqrt{2}}$ , the length of AI is just $\sqrt{8+4\sqrt{2}}$ . Now, consider triangle $ADV$ . The apex of a regular pyramid is defined to be equidistant from every vertex, and since segments $AV$ and $DV$ are perpendicular, then triangle $ADV$ is a $45-45-90$ right triangle with the right angle at $V$ . This means that both $AV$ and $DV$ can be written as $\frac{AD}{\sqrt{2}} = \frac{4+2\sqrt{2}}{\sqrt{2}} = 2\sqrt{2} + 2$ . Finally, since $IV$ is perpendicular to base $ABCDEFGH$ , then $AIV$ is a right triangle. Applying Pythagorean Theorem yields $IV^{2} = (2\sqrt{2} + 2)^{2} - (8+4\sqrt{2}) = 4+4\sqrt{2}$ . However, initially we had scaled each of the sides by $2\sqrt{2}$ , so we have to divide the answer by the square of that, so the real answer is then $\frac{1+\sqrt{2}}{2}$ . Therefore the answer is then $\boxed{\textbf{(B) }\frac{1+\sqrt2}{2}}$

~mathwizard123123

Solution 5 (Vectors)

Consider the vectors $\vec{AV}$ and $\vec{DV}$ . If we use a coordinate plane where one of the axes is parallel to one of the sides of the octagon, we can calculate each of the vectors to be $\vec{AV} = \left\langle \frac{1}{2}, \frac{1+\sqrt{2}}{2}, h \right\rangle$ $\vec{DV} = \left\langle \frac{1}{2}, \frac{-1-\sqrt{2}}{2}, h \right\rangle$ Now, we must have $\vec{AV} \cdot \vec{DV} = 0$ if the vectors are perpendicular to each other, so $\left\langle \frac{1}{2}, \frac{-1-\sqrt{2}}{2}, h \right\rangle \cdot \left\langle \frac{1}{2}, \frac{1+\sqrt{2}}{2}, h \right\rangle = \frac{1}{4} - \frac{3 + 2\sqrt{2}}{4} + h^2 = 0$ $h^2=\frac{2+2\sqrt{2}}{4}=\frac{1+\sqrt{2}}{2}$

Yielding answer choice $\boxed{B}$ .

~tkl

==Only B and D looks normal , guess one using掐头去尾

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=G-PzTyKqqV4

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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