AMC 10 · 2012 · #23
Grade 8 geometry-3dA solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?
Pick an answer.
AMC 10 2012 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 wooden unit cube has a corner sliced off by one flat cut through three of its vertices: two diagonally opposite corners of one face and the far corner of the opposite face that touches neither of those two. The sliced-off tetrahedron is thrown away, and the leftover solid is set on a table resting on the flat triangular cut. How tall does the leftover solid stand?
Givens: The cube has edge length 1.; The cut passes through two nonadjacent (diagonal) vertices of one face and one vertex of the opposite face that is adjacent to neither of them.; The sliced-off tetrahedron is discarded; the rest rests on the table with the triangular cut face down.
Unknowns: The height of the resting leftover solid (its vertical extent above the table).
Understand
Restated: A wooden unit cube has a corner sliced off by one flat cut through three of its vertices: two diagonally opposite corners of one face and the far corner of the opposite face that touches neither of those two. The sliced-off tetrahedron is thrown away, and the leftover solid is set on a table resting on the flat triangular cut. How tall does the leftover solid stand?
Givens: The cube has edge length 1.; The cut passes through two nonadjacent (diagonal) vertices of one face and one vertex of the opposite face that is adjacent to neither of them.; The sliced-off tetrahedron is discarded; the rest rests on the table with the triangular cut face down.
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #1 Draw a Diagram, #4 Introduce a Variable, #7 Identify Subproblems
The hard part is picturing which corner gets shaved and how the flipped solid sits. Once you see that the slice removes a single cube corner and that the cube's long diagonal runs straight through that corner perpendicular to the cut face, the height is just that diagonal minus the sliver the cut removes. Setting coordinates and breaking the sliver's height out of a two-way volume turns the spatial picture into a short calculation.
Execute — Answer: D
7.G.B.6 Step 1 Locate the sliced-off corner
- Place the unit cube as [0,1] in each direction.
- The cut goes through P=(1,0,0) and Q=(0,1,0), two diagonally opposite corners of the bottom face, and R=(1,1,1), the top corner that touches neither P nor Q.
- These three points are exactly the three neighbors of the single corner O=(1,1,0).
- So the discarded piece is the small tetrahedron OPQR sitting in that corner, and the flat triangular face PQR is what the leftover solid will rest on.
💡 The three cut points are the three corners next to one cube corner, so the slice just shaves off that corner.
8.G.B.7 Step 2 Measure the cut triangle
- The three edges OP, OQ, OR each have length 1 and are mutually perpendicular, since they run along the three cube directions.
- Each side of the cut face PQR joins two of these neighbors, so by the Pythagorean theorem every side is the square root of 1 squared plus 1 squared, which is the square root of 2.
- The cut face PQR is therefore an equilateral triangle with side the square root of 2.
💡 Two perpendicular unit legs give a hypotenuse of the square root of 2, and all three sides are built the same way.
8.G.B.7 Step 3 Find the tall diagonal through the corner
- Straight across the cube from the sliced corner O=(1,1,0) is the opposite corner S=(0,0,1).
- The segment OS is a space diagonal of the cube, with length the square root of 1 squared plus 1 squared plus 1 squared, which is the square root of 3.
- Because the corner tetrahedron has three-fold symmetry about OS, this diagonal is perpendicular to the equilateral face PQR, so OS is the exact line along which heights are measured once the solid is flipped onto that face.
💡 The long diagonal from the cut corner pierces the slanted cut face at a right angle, so it becomes the natural up-direction after flipping.
6.G.A.1 Step 4 Area of the equilateral cut face
- For an equilateral triangle with side s the area is the square root of 3 over 4, times s squared.
- With s equal to the square root of 2, the side squared is 2, so the area is the square root of 3 over 4 times 2, which is the square root of 3 over 2.
💡 Square the side, take the square root of 3 over 4 of it, and the equilateral area appears.
8.G.C.9 Step 5 Height of the discarded tetrahedron
- Compute the corner tetrahedron OPQR's volume two ways.
- Using the three perpendicular unit legs, the volume is one third of the base area (half of 1 times 1) times the height 1, giving one sixth.
- Using the equilateral face PQR as the base and calling its height h (the distance from O to that face, measured along OS), the same volume is one third times the square root of 3 over 2 times h.
- Setting the two equal, one sixth equals the square root of 3 over 6 times h, so h equals 1 over the square root of 3, which is the square root of 3 over 3.
💡 Computing one volume two different ways lets the unknown height pop out on its own.
7.G.B.6 Step 6 Subtract along the diagonal
- The whole space diagonal OS has length the square root of 3.
- The cut face crosses it a distance h equal to the square root of 3 over 3 from the sliced corner O.
- When the tetrahedron is removed and the solid rests on face PQR, the leftover height is everything on the far side of the cut: the square root of 3 minus the square root of 3 over 3, which is three roots of 3 minus one root of 3, over 3, equal to two roots of 3 over 3.
- So the object stands two roots of 3 over 3 tall, which is answer (D).
💡 The cut chops off only the top square-root-of-3-over-3 slice of the long diagonal, leaving the rest as the height.
7.G.B.6 Place the unit cube as [0,1] in each direction. The cut goes through P=(1,0,0) a 8.G.B.7 The three edges OP, OQ, OR each have length 1 and are mutually perpendicular, si 8.G.B.7 Straight across the cube from the sliced corner O=(1,1,0) is the opposite corner 6.G.A.1 For an equilateral triangle with side s the area is the square root of 3 over 4, 8.G.C.9 Compute the corner tetrahedron OPQR's volume two ways. Using the three perpendic 7.G.B.6 The whole space diagonal OS has length the square root of 3. The cut face crosse Review
Reasonableness: The height 2√3/3 is about 1.155 — a little taller than the cube's own edge of 1, yet well below the full space diagonal √3 ≈ 1.732. That fits: only a small corner sliver was removed, so the flipped object should stand almost as tall as the whole diagonal, but not quite. It also equals 2/√3, matching a direct check that the farthest leftover vertex S=(0,0,1) sits a perpendicular distance |0+0-1-1|/√3 = 2/√3 from the cutting plane x+y-z=1.
Alternative: Skip the volume step and stay in coordinates. The cutting plane through P, Q, R is x+y-z=1 with normal vector (1,1,-1). The standing height is the largest perpendicular distance from this plane to any leftover cube vertex. Checking all vertices, the farthest is S=(0,0,1), giving distance |0+0-1-1|/√(1²+1²+1²) = 2/√3 = 2√3/3 — the same answer, (D).
CCSS standards used (min grade 8)
7.G.B.6Solve real-world problems involving area, surface area, and volume (Identifying the sliced-off corner tetrahedron and subtracting along the diagonal to get the standing height)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the √2 edges of the equilateral cut face and the √3 space diagonal of the cube)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area √3/2 of the equilateral cut face)8.G.C.9Know the formulas for volumes of cones, cylinders, and spheres (Computing the corner tetrahedron's volume two ways to solve for its height √3/3)
⭐ Slice a corner off a cube, and the leftover block stands as tall as the cube's long diagonal minus the little bit the slice removed.
⭐ Slice a corner off a cube, and the leftover block stands as tall as the cube's long diagonal minus the little bit the slice removed.
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