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Difference between revisions of "2024 AMC 12A Problems/Problem 24"
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==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})}=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>
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=23|num-a=25}}
 
{{AMC12 box|year=2024|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:23, 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$= 4\sqrt{\frac{15}{2}(\frac{7}{2})(\frac{3}{2})(\frac{1}{2})}=\boxed{\textbf{(D) }15\sqrt{7}}$

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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