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

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==Solution==
 
==Solution==
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Notice that any scalene triangle can be the faces of a <math>\textit{disphenoid}</math>. As a result, we simply have to find the smallest area a scalene triangle with integer side lengths can take on. This occurs with a <math>2,3,4</math> triangle (notice that if you decrease the value of any of the sides the resulting triangle will either be isosceles or degenerate). For this triangle, the semiperimeter is <math>\frac{9}{2}</math>, so by Heron’s Formula:
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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:
  
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<cmath>A=\sqrt{\frac{9}{2}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}}</cmath>
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<cmath>A=\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}}</cmath>
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<cmath>=\sqrt{\frac{135}{16}}</cmath>
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<cmath>=\sqrt{\frac{15^2\cdot7}{16}}</cmath>
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<cmath>=\frac{3}{4}\sqrt{15}</cmath>
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<cmath>=\frac{15}{4}\sqrt{7}</cmath>
  
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The surface area is simply four times the area of one of the triangles, or <math>\boxed{\textbf{(B) }3\sqrt{15}}</math>.
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The surface area is simply four times the area of one of the triangles, or <math>\boxed{\textbf{(D) }15\sqrt{7}}</math>.
  
 
~eevee9406
 
~eevee9406

Revision as of 19:59, 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

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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