AMC 10 · 2011 · #22
Grade 8 geometry-3dA pyramid has a square base with sides of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A pyramid has a square base of side $1$, and every triangular side face is equilateral. A cube sits inside with its bottom face flat on the base, and the four edges of its top face each lie on a slanted face of the pyramid. Find the cube's volume.
Givens: The base is a square with side length $1$.; Each lateral face is an equilateral triangle, so every slanted edge of the pyramid also has length $1$.; The cube's bottom face rests on the base; the cube shares the base's center.; Each of the four top edges of the cube lies on a lateral face of the pyramid.; Answer choices: (A) $5\sqrt{2}-7$, (B) $7-4\sqrt{3}$, (C) $\frac{2\sqrt{2}}{27}$, (D) $\frac{\sqrt{2}}{9}$, (E) $\frac{\sqrt{3}}{9}$.
Unknowns: The edge length of the cube, and from it the cube's volume.
Understand
Restated: A pyramid has a square base of side $1$, and every triangular side face is equilateral. A cube sits inside with its bottom face flat on the base, and the four edges of its top face each lie on a slanted face of the pyramid. Find the cube's volume.
Givens: The base is a square with side length $1$.; Each lateral face is an equilateral triangle, so every slanted edge of the pyramid also has length $1$.; The cube's bottom face rests on the base; the cube shares the base's center.; Each of the four top edges of the cube lies on a lateral face of the pyramid.; Answer choices: (A) $5\sqrt{2}-7$, (B) $7-4\sqrt{3}$, (C) $\frac{2\sqrt{2}}{27}$, (D) $\frac{\sqrt{2}}{9}$, (E) $\frac{\sqrt{3}}{9}$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #17 Visualize Spatial Relationships, #4 Introduce a Variable, #9 Solve an Easier Related Problem, #7 Identify Subproblems
A 3D pyramid-and-cube picture is hard to reason about directly, so the winning move (Tool #1, backed by Tool #9) is to slice the solid with a single flat cut and turn the whole problem into a 2D triangle diagram. The right cut passes through the apex and two opposite corners of the base; in that plane the pyramid becomes a clean triangle and the cube becomes a rectangle. Spatial visualization (Tool #17) tells us which cut to take and why the cube's tall corners land on the pyramid's slanted edges. Then naming the cube's edge $s$ (Tool #4) turns 'corner sits on the slant' into one equation we can solve.
Execute — Answer: A
8.G.B.7 Step 1 Find the pyramid's height
- Every lateral face is equilateral, so all four slanted edges have length $1$.
- The base is a $1\times 1$ square, so its diagonal is $\sqrt{2}$ and the center of the base sits $\frac{\sqrt{2}}{2}$ from each corner.
- The apex is straight above the center.
- The apex, the center, and a base corner form a right triangle with slanted edge $1$ as the hypotenuse and horizontal leg $\frac{\sqrt{2}}{2}$.
- So the height is $H=\sqrt{1^2-\left(\frac{\sqrt{2}}{2}\right)^2}=\sqrt{1-\frac12}=\frac{\sqrt{2}}{2}$.
💡 The slanted edge is the hypotenuse of a right triangle, so its height comes straight from the Pythagorean theorem.
8.G.B.7 Step 2 Slice through the apex and two opposite corners
- Cut the pyramid with a vertical plane through the apex and two opposite base corners.
- In this plane the base becomes the diagonal, a segment of length $\sqrt{2}$, and the two cut edges are slanted edges of length $1$.
- A triangle with sides $1$, $1$, and $\sqrt{2}$ satisfies $1^2+1^2=(\sqrt{2})^2$, so it is an isosceles right triangle: the apex angle is $90^\circ$ and both base angles are $45^\circ$.
💡 One flat slice trades a hard 3D solid for an easy 2D triangle you can measure.
8.G.A.4 Step 3 Place the cube in the 2D picture
- Put the diagonal on the $x$-axis from $(0,0)$ to $(\sqrt{2},0)$, with the apex at $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$.
- The left slanted side runs along the line $y=x$.
- Let $s$ be the cube's edge.
- Since the cube is centered and this cut runs along the cube's base diagonal, the cube shows up as a rectangle that is $s\sqrt{2}$ wide (a face diagonal) and $s$ tall.
- Its top-left corner sits at $\left(\frac{\sqrt{2}}{2}-\frac{s\sqrt{2}}{2},\; s\right)$, and that corner must lie on the slanted side $y=x$.
💡 The corner riding up the slanted edge is the one fact that pins down the cube's size.
8.EE.C.7 Step 4 Solve for the cube's edge
- Setting the corner's $y$ equal to its $x$ gives $s=\frac{\sqrt{2}}{2}-\frac{s\sqrt{2}}{2}$.
- Collect the $s$ terms: $s+\frac{s\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$, so $s\cdot\frac{2+\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$, giving $s=\frac{\sqrt{2}}{2+\sqrt{2}}$.
- Multiply top and bottom by $2-\sqrt{2}$: $s=\frac{\sqrt{2}(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}=\frac{2\sqrt{2}-2}{2}=\sqrt{2}-1$.
💡 One corner on one line gives one equation, and one equation pins one unknown.
6.EE.A.1 Step 5 Cube the edge to get the volume
- The volume is $s^3=(\sqrt{2}-1)^3$.
- Expand step by step: $(\sqrt{2}-1)^2=2-2\sqrt{2}+1=3-2\sqrt{2}$, then $(3-2\sqrt{2})(\sqrt{2}-1)=3\sqrt{2}-3-4+2\sqrt{2}=5\sqrt{2}-7$.
- So the cube's volume is $5\sqrt{2}-7$, which is choice (A).
💡 Volume of a cube is just its edge multiplied by itself three times.
8.G.B.7 Every lateral face is equilateral, so all four slanted edges have length $1$. Th 8.G.B.7 Cut the pyramid with a vertical plane through the apex and two opposite base cor 8.G.A.4 Put the diagonal on the $x$-axis from $(0,0)$ to $(\sqrt{2},0)$, with the apex a 8.EE.C.7 Setting the corner's $y$ equal to its $x$ gives $s=\frac{\sqrt{2}}{2}-\frac{s\sq 6.EE.A.1 The volume is $s^3=(\sqrt{2}-1)^3$. Expand step by step: $(\sqrt{2}-1)^2=2-2\sqr Review
Reasonableness: Estimate the size: $\sqrt{2}\approx1.414$, so the edge $s=\sqrt{2}-1\approx0.414$ and the volume $\approx0.414^3\approx0.071$. That is comfortably less than $1$, which it must be since the cube fits inside a pyramid whose base is only $1\times1$. Checking the answer choices, $5\sqrt{2}-7\approx5(1.414)-7\approx0.071$ matches, while (C), (D), (E) are near $0.10$–$0.19$ (too big for a cube of edge $0.414$) and (B) $7-4\sqrt{3}\approx0.072$ is close but comes from a $30^\circ$-flavored setup, not the $45^\circ$ triangle we actually get.
Alternative: Skip the diagonal slice and use similar pyramids instead. The cube of edge $s$ leaves a smaller pyramid of height $H-s$ sitting on top, similar to the whole pyramid of height $H=\frac{\sqrt{2}}{2}$. The top of the cube must match that smaller pyramid's square cross-section, whose side is $\frac{H-s}{H}$. Setting the cube's top width equal to that shrunk square, $s=1\cdot\frac{H-s}{H}$, gives $sH=H-s$, so $s=\frac{H}{H+1}=\frac{\sqrt{2}/2}{\sqrt{2}/2+1}=\sqrt{2}-1$, the same edge and the same volume $5\sqrt{2}-7$.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the pyramid's height $H=\frac{\sqrt{2}}{2}$ and recognizing the $1,1,\sqrt{2}$ cross-section as a right isosceles triangle.)8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Reading the cube's rectangle inside the similar-shrinking triangle and placing its top corner on the slanted side.)8.EE.C.7Solve linear equations in one variable (Solving $s=\frac{\sqrt{2}}{2}-\frac{s\sqrt{2}}{2}$ for the edge $s=\sqrt{2}-1$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Cubing the edge, $(\sqrt{2}-1)^3=5\sqrt{2}-7$, to get the volume.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Approximating $\sqrt{2}-1\approx0.414$ and $5\sqrt{2}-7\approx0.071$ to sanity-check the answer against the choices.)
⭐ Slice a hard 3D shape with one flat cut, and it often turns into a simple 2D triangle you can actually solve.
⭐ Slice a hard 3D shape with one flat cut, and it often turns into a simple 2D triangle you can actually solve.
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