AMC 10 · 2015 · #21
Grade 8 geometry-3dPick an answer.
AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The six edges hide two identical 3-4-5 right triangles glued along the edge AB. The whole problem turns into picturing how the two triangles fold across AB. If we can find the height of D above the plane of triangle ABC, the volume is just one-third base times height, so the spatial picture is the key, supported by breaking the solid into familiar right triangles.
Spot the two 3-4-5 triangles
Faces ABC and ABD both have sides 3, 4, 5, so both are right triangles sharing hypotenuse AB.
Any triangle whose sides are 3, 4, 5 is automatically a right triangle.
8.G.B.6Identify SubproblemsDrop both altitudes to AB
Perpendiculars from C and D to AB land at the same foot H, each measuring 12/5.
From the right angle, the shortest path to the hypotenuse is the product of the legs divided by the hypotenuse.
8.G.B.7Draw A DiagramShow CH and DH are perpendicular
Triangle CHD satisfies the Pythagorean converse, so CH ⊥ DH.
If two segments satisfy a² + b² = c², the corner between them is a perfect right angle.
8.G.B.6Visualize Spatial RelationshipsRead off the height of the solid
DH is perpendicular to two crossing lines of plane ABC, so it stands as the tetrahedron's height above base ABC.
A segment perpendicular to two crossing lines of a flat surface stands straight up out of that surface.
6.G.A.1Visualize Spatial RelationshipsCompute the volume
One-third base times height gives a volume of 24/5, answer (C).
A pyramid holds exactly one-third of the box with the same base and height.
8.G.C.9Visualize Spatial RelationshipsWhen edges hide 3-4-5 right triangles, find where the two altitudes meet, then the volume is one-third base times height.
- Spot the two 3-4-5 triangles
- Drop both altitudes to AB
- Show CH and DH are perpendicular
- Read off the height of the solid
- Compute the volume