Difference between revisions of "2024 AMC 12A Problems/Problem 24"
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==Solution== | ==Solution== | ||
| − | Notice that any scalene triangle works for this. 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 | + | Notice that any scalene triangle works for this. 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: |
<cmath>A=\sqrt{\frac{9}{2}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}}</cmath> | <cmath>A=\sqrt{\frac{9}{2}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}}</cmath> | ||
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<cmath>=\frac{3}{4}\sqrt{15}</cmath> | <cmath>=\frac{3}{4}\sqrt{15}</cmath> | ||
| − | The surface area is simply four times the area of one of the triangles, or <math>\boxed\textbf{(B) }3\sqrt{15}}</math>. | + | The surface area is simply four times the area of one of the triangles, or <math>\boxed{\textbf{(B) }3\sqrt{15}}</math>. |
~eevee9406 | ~eevee9406 | ||
Revision as of 19:57, 8 November 2024
Problem
A 
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?
Solution
Notice that any scalene triangle works for this. 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 
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 
, so by Heron’s Formula:
![\[A=\sqrt{\frac{9}{2}\cdot\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}}\]](http://latex.artofproblemsolving.com/6/4/a/64a60f34ce97bea4bc3b1a1a347ec4c463277160.png)
![\[=\sqrt{\frac{135}{16}}\]](http://latex.artofproblemsolving.com/9/4/4/944c2aa7cfd8a27ee5c9e2233c8df405062dfa42.png)
![\[=\frac{3}{4}\sqrt{15}\]](http://latex.artofproblemsolving.com/8/4/6/846283605e7b07d786216ebdc6ea3913d4346eb6.png)
The surface area is simply four times the area of one of the triangles, or 
.
~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
