AMC 10 · 2024 · #24
Grade 10 geometry-2dcountingPick an answer.
Altitudes are an awkward thing to search over directly, so the first job is to trade them for something searchable. Naming the area converts each altitude into a side length, and the sides turn out to be the reciprocals 1/a, 1/b, 1/c up to one common scale factor. That single move does two things at once: it says exactly when a triangle with those altitudes exists, and it collapses the inradius into one clean identity. After that the problem is pure inequality-squeezing on integers, and the squeeze is where the Extreme Principle earns its place — the smallest altitude a and the largest altitude c each get pinned between two multiples of r, which cuts the search down to three values of r and a handful of cases. In each surviving case the sum of reciprocals is already at its maximum or its minimum, so equality forces every entry at once.
Trade altitudes for side lengths
Each side is area over altitude.
A fixed area spread over a longer base needs less height, so the sides are the altitudes turned upside down.
A fixed area spread over a longer base needs less height, so the sides are the altitudes turned upside down.
▸ Why?
An area is half the base times the height, so with the area fixed the two are locked against each other.
▸ Why?
With their product fixed, one is the reciprocal of the other up to a constant factor.
Cut the triangle at the incenter
The three reciprocals add to one over the inradius.
The incircle gives all three sub-triangles the same height r, so the area splits into one tidy sum in which the area itself cancels out.
10.G-C.A.3Identify SubproblemsWhen does such a triangle exist
Any three positive altitudes work after rescaling.
Shape is fixed by the ratios of the sides and only the size is free, so one dial — the scale factor — can be turned until the altitudes land exactly on target.
10.G-SRT.A.2Solve An Easier Related ProblemNon-degeneracy is one inequality
It is the triangle inequality on reciprocals.
Only the longest side can be too long to close up, and reciprocating a sorted list turns the smallest altitude into the longest side.
7.G.A.2Convert To AlgebraPin the smallest and largest altitude
Squeezing leaves the inradius as 1, 2, or 3.
Swapping all three reciprocals for the biggest one, then for the smallest one, traps the sum between two simple fractions and squeezes both a and c against multiples of r.
9.A-CED.A.3Extreme PrincipleSweep the three values
Each case forces all three altitudes equal.
In every case the required reciprocal sum already sits at the extreme the window allows, and a sum can only hit its own bound when all its terms are equal.
5.NF.A.1Make A Systematic ListCount them and read the pattern
That leaves 3 triples.
In an equilateral triangle the inradius is always one third of the altitude, so the answers are just the multiples of 3 that stay under the cap of 9.
7.RP.A.2Look For A PatternAltitudes are sides turned upside down, and once you write 1/r = 1/a + 1/b + 1/c the geometry is gone: three unit fractions have to add to another unit fraction while staying close enough in size to still close into a triangle, and only the equilateral ones manage it.
- Trade altitudes for side lengths
- Cut the triangle at the incenter
- When does such a triangle exist?
- Non-degeneracy is one inequality
- Pin the smallest and largest altitude
- Sweep the three values of r
- Count them and read the pattern