AMC 10 · 2015 · #21

Grade 8 geometry-3d
Archetype Solid Volume by Decomposition └ Sub-archetype Sum of Primitive Solids
pythagorean-theoremspatial-visualizationinteger-pythagorean-triples physical-representation ↑ Prerequisites: pythagorean-theorem
📏 Medium solution 💡 3 insights
Problem
A tetrahedron ABCD has six known edge lengths: AB=5, AC=3, BC=4, BD=4, AD=3, and CD=125\frac{12}{5}√2. Find the volume of the solid.

Pick an answer.

(A)
$3\sqrt2$
(B)
$2\sqrt5$
(C)
$\dfrac{24}5$
(D)
$3\sqrt3$
(E)
$\dfrac{24}5\sqrt2$

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Visualize Spatial Relationships

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.

1STEP 1

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.

3² + 4² = 9 + 16 = 25 = 5²
2STEP 2

Drop both altitudes to AB

Perpendiculars from C and D to AB land at the same foot H, each measuring 125\frac{12}{5}.

CH = DH = (AC · BC)/AB = 3⋅45\frac{3 \cdot 4}{5} = 125\frac{12}{5}
3STEP 3

Show CH and DH are perpendicular

Triangle CHD satisfies the Pythagorean converse, so CH ⊥ DH.

125\frac{12}{5}² + 125\frac{12}{5}² = 2125\frac{12}{5}² = (125\frac{12}{5}√2)² = CD²
4STEP 4

Read 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.

[△ ABC] = 12\frac{1}{2} · 3 · 4 = 6, h = DH = 125\frac{12}{5}
5STEP 5

Compute the volume

One-third base times height gives a volume of 245\frac{24}{5}, answer (C).

V = 13\frac{1}{3} · [△ ABC] · DH = 13\frac{1}{3} · 6 · 125\frac{12}{5} = 7215\frac{72}{15} = 245\frac{24}{5}
Answer
245\frac{24}{5}
The height DH = 125\frac{12}{5} = 2.4 is comfortably shorter than the slant edges (3 and 4), which it must be, and the base area is 6, so the volume should be near one-third of 6×2.4 = 14.4, i.e. about 4.8. The exact value 245\frac{24}{5} = 4.8 matches, and it is the only answer choice that comes out to a clean rational number, which fits the clean 3-4-5 structure.
💡Key takeaway

When 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

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