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

Revision as of 14:33, 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:

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

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

~eevee9406

Solution 2

Let the side lengths of one face of the disphenoid be $a, b, c$ . By the definition of a disphenoid with congruent 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 $a \leq b \leq c$ . $a = 2$ 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

@@ Line 14: / Line 14: @@
 ~eevee9406
+==Solution 2==
+Let the side lengths of one face of the disphenoid be <math>a, b, c</math>. By the definition of a disphenoid with congruent 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
+<cmath>p^2 + q^2 = a^2</cmath>
+<cmath>p^2 + r^2 = b^2</cmath>
+<cmath>q^2 + r^2 = c^2</cmath>
+for positive integers <math>a, b, c</math> and positive <math>p, q, r</math>.
+Solving for <math>p, q, r</math>, we have
+<cmath>p^2 = \frac{a^2 + b^2 - c^2}{2}</cmath>
+<cmath>q^2 = \frac{a^2 - b^2 + c^2}{2}</cmath>
+<cmath>r^2 = \frac{-a^2 + b^2 + c^2}{2}</cmath>
+so <math>a^2, b^2, c^2</math> are the side lengths of a triangle.
+WLOG, let <math>a \leq b \leq c</math>. <math>a = 2</math> and <math>a = 3</math> yield no valid solutions, and <math>4^2 + 5^2 > 6^2</math> so <math>a, b, c = 4, 5, 6</math>. Using Heron's Formula, the minimum total surface area of the disphenoid is <math>\boxed{\textbf{(D) }15\sqrt{7}}</math>.
+~babyhamster
 ==See also==
 {{AMC12 box|year=2024|ab=A|num-b=23|num-a=25}}
 {{MAA Notice}}

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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Difference between revisions of "2024 AMC 12A Problems/Problem 24"

Revision as of 14:33, 9 November 2024

Contents

Problem

Solution 1 (Definition of disphenoid)

Solution 2

See also