Difference between revisions of "2024 AMC 12A Problems/Problem 24"
Ericzzzqwq (talk | contribs) (→Solution 1 (Definition of disphenoid)) |
Ericzzzqwq (talk | contribs) (→Solution 1 (Definition of disphenoid)) |
||
| Line 6: | Line 6: | ||
==Solution 1 (Definition of disphenoid)== | ==Solution 1 (Definition of disphenoid)== | ||
By definition, if a <math>\textit{disphenoid}</math> has sides <math>x,y,z</math> such that <math>x<y<z</math> (since it is scalene), then we must have <math>x^2+y^2>z^2</math>. Clearly the smallest triple of <math>(x,y,z)</math> is <math>(4,5,6)</math>. Then using Heron's formula gives us Surface area<math>= 4\sqrt{\frac{15}{2}(\frac{7}{2})(\frac{3}{2})(\frac{1}{2})}=\boxed{\textbf{(D) }15\sqrt{7}}</math> | By definition, if a <math>\textit{disphenoid}</math> has sides <math>x,y,z</math> such that <math>x<y<z</math> (since it is scalene), then we must have <math>x^2+y^2>z^2</math>. Clearly the smallest triple of <math>(x,y,z)</math> is <math>(4,5,6)</math>. Then using Heron's formula gives us Surface area<math>= 4\sqrt{\frac{15}{2}(\frac{7}{2})(\frac{3}{2})(\frac{1}{2})}=\boxed{\textbf{(D) }15\sqrt{7}}</math> | ||
| − | + | ||
~ERiccc | ~ERiccc | ||
Revision as of 20:24, 8 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)
By definition, if a has sides 
such that 
(since it is scalene), then we must have 
. Clearly the smallest triple of 
is 
. Then using Heron's formula gives us Surface area
~ERiccc
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. 
