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

(→‎Solution 1 (Definition of disphenoid))
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==Solution 1 (Definition of disphenoid)==
 
==Solution 1 (Definition of disphenoid)==
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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 such <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>
+
Notice that any scalene acute triangle can be the faces of a <math>\textit{disphenoid}</math>. 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 <math>4,5,6</math> 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 <math>\frac{15}{2}</math>, so by Heron’s Formula:
  
−
~ERiccc
+
<cmath>A=\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}}</cmath>
 +
<cmath>=\sqrt{\frac{15^2\cdot7}{16}}</cmath>
 +
<cmath>=\frac{15}{4}\sqrt{7}</cmath>
  
 +
The surface area is simply four times the area of one of the triangles, or <math>\boxed{\textbf{(D) }15\sqrt{7}}</math>.
 +
 +
~eevee9406
 
==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 20:36, 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)

Notice that any scalene acute triangle can be the faces of a $\textit{disphenoid}$. 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 $4,5,6$ 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 $\frac{15}{2}$, so by Heron’s Formula:

\[A=\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}}\] \[=\sqrt{\frac{15^2\cdot7}{16}}\] \[=\frac{15}{4}\sqrt{7}\]

The surface area is simply four times the area of one of the triangles, or $\boxed{\textbf{(D) }15\sqrt{7}}$.

~eevee9406

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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