2025 AMC 12A Problems/Problem 22: Difference between revisions
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<imath>\textbf{(A)}~\frac{1}{12}\qquad\textbf{(B)}~\frac19\qquad\textbf{(C)}~\frac18\qquad\textbf{(D)}~\frac16\qquad\textbf{(E)}~\frac14</imath> | <imath>\textbf{(A)}~\frac{1}{12}\qquad\textbf{(B)}~\frac19\qquad\textbf{(C)}~\frac18\qquad\textbf{(D)}~\frac16\qquad\textbf{(E)}~\frac14</imath> | ||
==Solution | ==Solution 1== | ||
We can solve the problem by approaching it geometrically, where each possible outcome is a coordinate in a 1 by 1 by cube (a, b, c). Now we just have to find the area of the solution set over 1. Let's assume that a is the greatest number, and then multiply by 3 afterwards to account for b or c also possible being the largest. It can be seen that the possible values of a change linearly with b and c changing values, so to find the figure, we can just find the vertices and then connect them. By doing this, we can determine the solution set with a being the biggest value is the volume of the figure with coordinates (1,0,0), (0,0,0), (1,1/2,0), (1,0,1/2), which forms a tetrahedron which the volume can easily be calculated to be 1/12 with the formula V=1/3bh and the base is an isosceles right triangle with side length 1/2, and the height is just the height of the cube which is 1. Now just multiplying this value by 3 to account for b or c also being the maximum gives us the answer of <imath>\boxed{\frac{1}{4},\textbf{E}}.</imath> | We can solve the problem by approaching it geometrically, where each possible outcome is a coordinate in a 1 by 1 by cube (a, b, c). Now we just have to find the area of the solution set over 1. Let's assume that a is the greatest number, and then multiply by 3 afterwards to account for b or c also possible being the largest. It can be seen that the possible values of a change linearly with b and c changing values, so to find the figure, we can just find the vertices and then connect them. By doing this, we can determine the solution set with a being the biggest value is the volume of the figure with coordinates (1,0,0), (0,0,0), (1,1/2,0), (1,0,1/2), which forms a tetrahedron which the volume can easily be calculated to be 1/12 with the formula V=1/3bh and the base is an isosceles right triangle with side length 1/2, and the height is just the height of the cube which is 1. Now just multiplying this value by 3 to account for b or c also being the maximum gives us the answer of <imath>\boxed{\frac{1}{4},\textbf{E}}.</imath> | ||
~Kevin Wang | |||
==See Also== | ==See Also== | ||
Revision as of 18:14, 6 November 2025
Problem 22
Three real numbers are chosen independently and uniformly at random between 
and 
. What is the probability that the greatest of these three numbers is greater than 
times each of the other two numbers? (In other words, if the chosen numbers are 
, then 
.)
Solution 1
We can solve the problem by approaching it geometrically, where each possible outcome is a coordinate in a 1 by 1 by cube (a, b, c). Now we just have to find the area of the solution set over 1. Let's assume that a is the greatest number, and then multiply by 3 afterwards to account for b or c also possible being the largest. It can be seen that the possible values of a change linearly with b and c changing values, so to find the figure, we can just find the vertices and then connect them. By doing this, we can determine the solution set with a being the biggest value is the volume of the figure with coordinates (1,0,0), (0,0,0), (1,1/2,0), (1,0,1/2), which forms a tetrahedron which the volume can easily be calculated to be 1/12 with the formula V=1/3bh and the base is an isosceles right triangle with side length 1/2, and the height is just the height of the cube which is 1. Now just multiplying this value by 3 to account for b or c also being the maximum gives us the answer of 
~Kevin Wang
See Also
| 2025 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by 21 |
Followed by Problem 23 |
| 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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