2024 AMC 12A Problems/Problem 24
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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:
\begin{align*} A&=\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}}\\ &=\sqrt{\frac{15^2\cdot7}{16}}\\ &=\frac{15}{4}\sqrt{7} \end{align*}
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
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 |
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