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

(→‎Solution 2)
m (→‎Solution 1 (Definition of disphenoid): formatting)
 
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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:
 
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:
  
−
<cmath>A=\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}}</cmath>
+
\begin{align*}
−
<cmath>=\sqrt{\frac{15^2\cdot7}{16}}</cmath>
+
A&=\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}}\\
−
<cmath>=\frac{15}{4}\sqrt{7}</cmath>
+
&=\sqrt{\frac{15^2\cdot7}{16}}\\
 +
&=\frac{15}{4}\sqrt{7}
 +
\end{align*}
  
 
The surface area is simply four times the area of one of the triangles, or <math>\boxed{\textbf{(D) }15\sqrt{7}}</math>.
 
The surface area is simply four times the area of one of the triangles, or <math>\boxed{\textbf{(D) }15\sqrt{7}}</math>.

Latest revision as of 15:42, 9 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:

\begin{align*} 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} \end{align*}

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

~eevee9406

Solution 2 (Disphenoid in Box)

Let the side lengths of one face of the disphenoid be $a, b, c$. By the definition of a disphenoid with scalene faces, opposite sides must be the same length. Then the disphenoid can be constructed in a rectangular box with dimensions $p, q, r$ such that $a, b, c$ are the $3$ different diagonal lengths of the faces of the box and no two sides are parallel (someone feel free to insert a diagram). Then we have the system

\[p^2 + q^2 = a^2\] \[p^2 + r^2 = b^2\] \[q^2 + r^2 = c^2\]

for positive integers $a, b, c$ and positive $p, q, r$.

Solving for $p, q, r$, we have

\[p^2 = \frac{a^2 + b^2 - c^2}{2}\] \[q^2 = \frac{a^2 - b^2 + c^2}{2}\] \[r^2 = \frac{-a^2 + b^2 + c^2}{2}\]

so $a^2, b^2, c^2$ are the side lengths of a triangle.

WLOG, let $a < b < c$. $a = 2$ and $a = 3$ yield no valid solutions, and $4^2 + 5^2 > 6^2$ so $a, b, c = 4, 5, 6$. Using Heron's Formula, the minimum total surface area of the disphenoid is $\boxed{\textbf{(D) }15\sqrt{7}}$.

~babyhamster

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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