2024 AMC 12A Problems/Problem 24
Contents
[hide]Problem
A is a tetrahedron whose triangular faces are congruent to one another. What is the least total surface area of a disphenoid whose faces are scalene triangles with integer side lengths?
Solution 1 (Definition of disphenoid)
Notice that any scalene triangle can be the faces of a . (See proof in Solution 2.)
As a result, we simply have to find the smallest area a scalene acute triangle with integer side lengths can take on. This occurs with a triangle (notice that if you decrease the value of any of the sides the resulting triangle will either be isosceles, degenerate, or non-acute). For this triangle, the semiperimeter is , so by Heron’s Formula:
The surface area is simply four times the area of one of the triangles, or .
~eevee9406
Solution 2 (Disphenoid in Box)
Let the side lengths of one face of the disphenoid be . By the definition of a disphenoid with scalene faces, opposite sides must be the same length. Then the disphenoid can be constructed in a rectangular box with dimensions such that are the different diagonal lengths of the faces of the box and no two sides are parallel (someone feel free to insert a diagram). Then we have the system
for positive integers and positive .
Solving for , we have
(Notice that, by law of cosines (or by Pythagorean Inequality), these (parallelepiped box side lengths squared) each have the same sign as the cosine of the angle at a vertex of a triangular face of the disphenoid. Since they are squares and so non-negative, every angle is non-obtuse. Further, since they are squares of positive side lengths, every angle is acute. So are the side lengths of an triangle.)
WLOG, let . For , is less than or equal to the gap between squares greater than , so all such triangles are non-acute and fail.
The next smallest case works: so .
Using Heron's Formula, the minimum total surface area of the disphenoid is .
Bonus: by Heron’s Formula, the area of an triangle is .
~babyhamster
(Acute triangle observations and bonus formula by oinava)
See also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.