AMC 10 · 2024 · #23

Grade 10 geometry-3d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

regular-octagonisosceles-right-trianglepythagorean-theoremcoordinate-geometry spatial-visualizationidentify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremcoordinate-geometryangle-sum-polygon

📏 Long solution 💡 4 insights

Problem

A regular octagon ABCDEFGH with every side of length 1 lies flat, and a single apex V sits directly above the octagon's center, forming a right pyramid. The two edges AV and DV meet at V at a right angle. Find the square of the pyramid's height, that is, the square of the distance from V down to the plane of the octagon.

Pick an answer.

(A)

(B)

$\frac{1+\sqrt2}{2}$

(C)

$\sqrt2$

(D)

$\frac32$

(E)

$\frac{2+\sqrt2}{3}$

How to solve

Strategy Visualize Spatial Relationships

The right angle at V lives in three dimensions, which is hard to hold in the head. But A, V, and D are only three points, and three points always lie in one plane, so the whole condition can be carried down into a single flat triangle. Once the picture is flat the triangle turns out to be a 45-45-90 triangle, which ties the slant edge to the distance AD measured inside the base. That leaves two ordinary octagon measurements to find, and the fastest way to get both at once is to lay the octagon on coordinate axes as a square with its corners sliced off. The last step is then a one-line equation in the square of the height.

1STEP 1

Flatten the right angle into one triangle

Right isosceles means AD squared is twice the edge squared.

VA = VD, ∠ AVD = 90° → AD² = VA² + VD² = 2 VA²

2STEP 2

Build the slant edge from height and radius

The edge squared is radius squared plus height squared.

VA² = R² + h² → AD² = 2(R² + h²)

3STEP 3

Count the gap from A to D

Three steps make a central angle of 135 degrees.

central angle ∠ AOD = 3 × 360°/8 = 135°

4STEP 4

Put the octagon on axes

Coordinates surface half of one plus root two.

w = (1+√2)/2; A(1/2, w), B(w, 1/2), C(w, -1/2), D(1/2, -w), …

5STEP 5

Read off the two lengths

We get AD and the radius squared.

R² = 1/4 + (3+2√2)/4 = (2+√2)/2, AD = 2w = 1+√2, AD² = 3 + 2√2

6STEP 6

Solve for the square of the height

It reduces to one plus root two, over two.

3 + 2√2 = (2+√2) + 2h² → 2h² = 1 + √2 → h² = (1+√2)/2

Answer

(1+√2)/2

Numerically h² = (1+√2)/2 ≈ 1.207, so h ≈ 1.099, and the slant edge satisfies VA² = R² + h² ≈ 1.707 + 1.207 = 2.914, giving VA ≈ 1.707. Then VA² + VD² ≈ 2.914 + 2.914 = 5.828 and AD² = (1+√2)² ≈ 5.828, so the right angle at V checks out exactly. A height of about 1.1 against a base roughly 2.41 wide is a believable squat pyramid, and h being a bit under R ≈ 1.31 matches the fact that the right angle at V demands a fairly open apex. The three-step gap is also confirmed by testing the wrong gaps: if A and D were adjacent then AD = 1 and the equation would force h² = (1 - (2+√2))/2 ＜ 0, which is impossible; two apart would give AD² = 2 + √2 and h² = 0, a flat pyramid; and directly opposite would give AD = 2R and h² = R² = (2+√2)/2, which is not among the choices. Only the true three-step gap lands on a listed answer.

💡Key takeaway

Any three points sit on one flat sheet, so a right angle hanging in space can be dragged down into a single triangle — and then the only thing left to be careful about is counting exactly how many sides apart the two vertices really are.

Flatten the right angle into one triangle
Build the slant edge from height and radius
Count the gap from A to D
Put the octagon on axes
Read off the two lengths
Solve for the square of the height