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

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<math>\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}</math>
 
<math>\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}</math>
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==Solution 1 (Definition of disphenoid)==
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Notice that any scalene <math>\textit{acute}</math> triangle can be the faces of a <math>\textit{disphenoid}</math>. (See proof in Solution 2.)
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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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\begin{align*}
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A&=\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}}\\
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&=\sqrt{\frac{15^2\cdot7}{16}}\\
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&=\frac{15}{4}\sqrt{7}
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\end{align*}
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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>.
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~eevee9406
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==Solution 2 (Disphenoid in Box)==
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Let the side lengths of one face of the disphenoid be <math>a, b, c</math>. 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 <math>p, q, r</math> such that <math>a, b, c</math> are the <math>3</math> 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
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<cmath>p^2 + q^2 = a^2</cmath>
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<cmath>p^2 + r^2 = b^2</cmath>
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<cmath>q^2 + r^2 = c^2</cmath>
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for positive integers <math>a, b, c</math> and positive <math>p, q, r</math>.
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Solving for <math>p, q, r</math>, we have
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<cmath>p^2 = \frac{a^2 + b^2 - c^2}{2}</cmath>
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<cmath>q^2 = \frac{a^2 - b^2 + c^2}{2}</cmath>
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<cmath>r^2 = \frac{-a^2 + b^2 + c^2}{2}</cmath>
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(Notice that, by law of cosines (or by Pythagorean Inequality), these (parallelepiped box side lengths squared) <math>p^2, q^2, r^2</math> each have the same sign as the cosine of the angle at a vertex of a triangular face of the disphenoid. Since they are squares and so non-negative, every angle is non-obtuse. Further, since they are squares of positive side lengths, every angle is acute. So <math>a, b, c</math> are the side lengths of an <math>\textit{acute}</math> triangle.)
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WLOG, let <math>a < b < c</math>.
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For <math>a<4</math>, <math>a^2</math> is less than or equal to the gap between squares greater than <math>a^2</math>, so all such triangles are non-acute and fail.
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The next smallest case works:  <math>4^2 + 5^2 > 6^2</math> so <math>a, b, c = 4, 5, 6</math>.
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Using Heron's Formula, the minimum total surface area of the disphenoid is <math>\boxed{\textbf{(D) }15\sqrt{7}}</math>.
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Bonus: by Heron’s Formula, the area of an <math>x-1, x, x+1</math> triangle is <math>x \sqrt{3 (x-2)(x+2)}/4 = x \sqrt{3 (x^2-2)}/4</math>.
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~babyhamster
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(Acute triangle observations and bonus formula by oinava)
  
 
==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}}

Latest revision as of 10:38, 24 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 $\textit{acute}$ triangle can be the faces of a $\textit{disphenoid}$. (See proof in Solution 2.)

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

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


(Notice that, by law of cosines (or by Pythagorean Inequality), these (parallelepiped box side lengths squared) $p^2, q^2, r^2$ each have the same sign as the cosine of the angle at a vertex of a triangular face of the disphenoid. Since they are squares and so non-negative, every angle is non-obtuse. Further, since they are squares of positive side lengths, every angle is acute. So $a, b, c$ are the side lengths of an $\textit{acute}$ triangle.)

WLOG, let $a < b < c$. For $a<4$, $a^2$ is less than or equal to the gap between squares greater than $a^2$, so all such triangles are non-acute and fail.

The next smallest case works: $4^2 + 5^2 > 6^2$ so $a, b, c = 4, 5, 6$.

Using Heron's Formula, the minimum total surface area of the disphenoid is $\boxed{\textbf{(D) }15\sqrt{7}}$.

Bonus: by Heron’s Formula, the area of an $x-1, x, x+1$ triangle is $x \sqrt{3 (x-2)(x+2)}/4 = x \sqrt{3 (x^2-2)}/4$.


~babyhamster

(Acute triangle observations and bonus formula by oinava)

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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