Difference between revisions of "2024 AMC 12A Problems/Problem 24"

(Solution 1 (Definition of disphenoid))
(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>
 +
\newline
 
~ERiccc
 
~ERiccc
  

Revision as of 20:24, 8 November 2024

Problem

A $\textit{disphenoid}$ 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?

$\textbf{(A) }\sqrt{3}\qquad\textbf{(B) }3\sqrt{15}\qquad\textbf{(C) }15\qquad\textbf{(D) }15\sqrt{7}\qquad\textbf{(E) }24\sqrt{6}$

Solution 1 (Definition of disphenoid)

By definition, if a $\textit{disphenoid}$ has sides $x,y,z$ such that $x<y<z$ (since it is scalene), then we must have $x^2+y^2>z^2$. Clearly the smallest triple of $(x,y,z)$ is $(4,5,6)$. Then using Heron's formula gives us Surface area$= 4\sqrt{\frac{15}{2}(\frac{7}{2})(\frac{3}{2})(\frac{1}{2})}=\boxed{\textbf{(D) }15\sqrt{7}}$ \newline ~ERiccc

See also

2024 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png