AMC 10 · 2024 · #24
Grade 11 geometry-3dnumber-theoryPick an answer.
There are two separate demands hiding in this problem, and the whole thing turns on noticing the second one. The first is easy: four congruent faces means the total surface area is just four times one triangle's area, so the search is over triangles. The second is the trap. Not every triangle can be the face of a disphenoid. Taking four copies of any old scalene triangle and trying to fold them into a tetrahedron usually fails, so the phrase "a disphenoid exists" must first be translated into a condition on the triangle's three sides. The cleanest way to get that translation is to picture the disphenoid sitting inside a rectangular box, using four of the box's eight corners. That picture turns the three face sides into three box edges, and the box edges must have positive length. Working that requirement backwards produces exactly one extra inequality on a, b, c. After that the problem becomes a bounded integer search: find the smallest side lengths that pass both the triangle test and the new test, compute the area with Heron's formula, and then prove with a lower bound on area that nothing else can do better.
Four faces, one triangle
The surface is four times one face.
When every face is a copy of the same shape, the whole surface is controlled by one flat piece.
6.G.A.4Identify SubproblemsPut the tetrahedron inside a box
The sides become face diagonals of a box.
A disphenoid is just a box with four alternating corners kept and the rest sliced away, so its shape is really controlled by three box edges.
10.G-MG.A.1Visualize Spatial RelationshipsRecover the box from the triangle
The three sides recover the three edges.
If the box is completely determined by the triangle, then the triangle works only when that forced box turns out to be real.
8.G.B.7Work BackwardsPositive edges means an acute triangle
For real edges the face must be acute.
The box edge that survives the fold is measuring how far the largest angle sits from a right angle, so a right or obtuse triangle leaves nothing to fold up with.
The surviving box edge measures how far the largest angle sits from a right angle, so a non-acute triangle cannot fold up.
▸ Why?
The squares on the two shorter sides beat the square on the longest exactly when the largest angle is acute.
▸ Why?
A length has to be strictly positive, so the moment that comparison flips the box stops existing.
Put a floor under the area
Acuteness makes the area grow with the product of sides.
An acute triangle can never be squashed thin, so its area cannot fall far below the product of its two shorter sides.
11.G-SRT.D.9Extreme PrincipleThe shortest side is at least 4
Scalene plus acute forces the short side to 4.
Integers cannot squeeze between b and √(a²+b²) unless that gap is wider than 1, and that only happens once the short side is big enough.
7.EE.B.4Extreme PrincipleMeasure the 4-5-6 triangle
Heron gives a surface of fifteen root seven.
Once the shape is pinned to one triple, the area is a single Heron computation that simplifies because 1575 = 3² · 5² · 7.
11.N-RN.A.2Identify SubproblemsRule out everything else
No smaller candidate survives.
A lower bound that grows with ab turns an infinite search into a list of one pair.
6.EE.B.5Eliminate PossibilitiesCheck the box is real
All three squared edges are positive, so it is 15√(7).
A minimum is only real if some object actually attains it, so the box has to be exhibited, not just argued for.
8.G.B.7Work BackwardsFour copies of a triangle only fold up into a real solid when that triangle is acute, so the whole problem is a hunt for the smallest acute triangle with three different whole-number sides — and no triangle smaller than 4-5-6 survives both tests.
- Four faces, one triangle
- Put the tetrahedron inside a box
- Recover the box from the triangle
- Positive edges means an acute triangle
- A floor under the area
- The shortest side is at least 4
- Measure the 4-5-6 triangle
- Rule out everything else
- Check the box is real