Difference between revisions of "2024 AMC 12A Problems/Problem 24"
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so <math>a^2, b^2, c^2</math> are the side lengths of a triangle. | so <math>a^2, b^2, c^2</math> are the side lengths of a triangle. | ||
− | WLOG, let <math>a | + | WLOG, let <math>a < b < c</math>. <math>a = 2</math> and <math>a = 3</math> yield no valid solutions, and <math>4^2 + 5^2 > 6^2</math> so <math>a, b, c = 4, 5, 6</math>. Using Heron's Formula, the minimum total surface area of the disphenoid is <math>\boxed{\textbf{(D) }15\sqrt{7}}</math>. |
~babyhamster | ~babyhamster |
Revision as of 15:33, 9 November 2024
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 acute triangle can be the faces of a . 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
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
so are the side lengths of a triangle.
WLOG, let . and yield no valid solutions, and so . Using Heron's Formula, the minimum total surface area of the disphenoid is .
~babyhamster
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 |
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