2025 AMC 12A Problems/Problem 22

Problem 22

Three real numbers are chosen independently and uniformly at random between $0$ and $1$ . What is the probability that the greatest of these three numbers is greater than $2$ times each of the other two numbers? (In other words, if the chosen numbers are $a \geq b \geq c$ , then $a > 2b$ .)

$\textbf{(A)}~\frac{1}{12}\qquad\textbf{(B)}~\frac19\qquad\textbf{(C)}~\frac18\qquad\textbf{(D)}~\frac16\qquad\textbf{(E)}~\frac14$

Problem 22 (Chinese exams)

Three real numbers are chosen independently and uniformly at random between $0$ and $1$ . What is the probability that the greatest of these three numbers is greater than the sum of the other two numbers?

Solution 1

We can solve the problem by approaching it geometrically by mapping each possible triple to a coordinate $(a, b, c)$ in a unit cube. Now, we just have to find the volume of the solution set over $1$ .

WLOG, assume that $a$ is the greatest number. The intersection between $a>2b, a>2c$ and the unit cube is an oblique square pyramid with apex $(0,0,0)$ and base vertices $(1,0,0), (1,0,\frac12), (1,\frac12,0), \text{ and } (1,\frac12, \frac12)$ . It helps to visualize the two planes cutting into the cube and leaving triangular traces in the $xy$ and $xz$ planes. The square base $b=s^2=(\frac12) ^2=\frac14$ , and the height $h=1$ , so the volume $v=\frac13 bh=\frac13(\frac14)(1)=\frac{1}{12}$ .

From here, we can multiply this volume by $3$ to account for $b,c$ being the greatest (all mutually exclusive). Alternatively, we can find $P(a>2b, 2c|a>b,c)$ by taking $\frac{1}{12}$ over the volume of the interesction between $a>b, a>c$ and the unit cube (a pyramid wth base $1$ and height $1$ ). Either way, we get the final probability of $\boxed{\text{(E) } \frac{1}{4}}$ .

~Zak2155

~Revised by Zhixing

Here is a desmos version created for a clear overview: https://www.desmos.com/3d/eidrbnqnvz ~DRA777

Solution 1.1 (same logic but faster)

Call the numbers $x$ , $y$ , and $z$ for parity with the coordinate plane.

For simplicity, assume that $x>y>z$ . This has a $\frac{1}{3!} = \frac{1}{6}$ chance of happening, so in the end we must remember to multiply our answer by $6$ .

From here, we can use geometric probability. All possible triples $(x,y,z)$ lie in a unit cube centered at $(0,0,0)$ , which has a volume of $1$ . However, our desired region is where $x>2y>2z$ . We can graph this and find that the desired triples lie in a right tetrahedron of sides $1$ , $\frac{1}{2}$ , and $\frac{1}{2}$ , Using $v = \frac{1}{6}lwh$ , the volume of our desired region is $\frac{1}{6}(1)(\frac{1}{2})(\frac{1}{2}) = \frac{1}{24}$ . From this, we can see that the probability that we land in this region is $\frac{\frac{1}{24}}{1} = \frac{1}{24}$ .

But remember that we still have to multiply our answer by 6, resulting in $6 \times \frac{1}{24} = \boxed{\text{(E) } \frac{1}{4}}$ . ~mikeypoo5608

Solution 2 (Integration)

Consider the bounds of $a$ , $b$ , and $c$ . We know that $a\geq b\geq c$ and $a\geq 2b$ . For $a$ to be in $[0, 1]$ , $b$ must be in $[0, 0.5]$ . As such, each of $a$ , $b$ , and $c$ must be in the following intervals: \begin{align*} 1\geq &a\geq 2b, \\ 0.5\geq &b\geq 0, \\ b\geq &c\geq 0. \end{align*} Now, we can simply integrate over these bounds: $\int_0^{1/2}\int_{2b}^1\int_0^b dc da db,$ $\int_0^{1/2}\int_{2b}^1 b da db,$ $\int_0^{1/2} b(1-2b) db,$ $\left[\frac{b^2}{2}-\frac{2b^3}{3}\right]_0^{1/2},$ $\frac{1}{24}.$ Since there are $6$ equally likely permutations of $a$ , $b$ , and $c$ , we multiply $6\cdot\frac{1}{24}$ to obtain our answer of $\boxed{\text{(E) } \frac{1}{4}}.$

Note that this is almost the same solution as Solution 1, where we find the volume of a similar polyhedron using a different method (except we do not need to visualize it). This method would remain effective even if the bounds were non-linear or if there were more than three variables.

Solution 3 (Logic)

Say that, WLOG, $a$ is the maximum. Then notice that we need $b \le \frac{a}{2}$ , given that $b \le a$ , and similar for $c$ . This probability is just $\frac{1}{2}$ for $b$ to be less, and $\frac{1}{2}$ for $c$ to be less (note that this is independent of the value of $a$ ). Therefore, the answer is $\boxed{\text{(E) } \frac{1}{4}}$ (technically, the answer is $\int_{0}^{1} \frac{1}{4} da$ but this is constant).

~ScoutViolet

Solution 4 (Chinese exams)

Let the three numbers be $a$ , $b$ , and $c$ . By symmetry, the probability that we want to find is exactly

$3 \Pr(a>b+c)$

Taking the cube $[0,1]^3$ in a Cartesian plane, we see that the region that corresponds to the event $a>b+c$ is the set of points

$\{(x, \, y, \, z) \mid 0 \leq x, \, y, \, z \leq 1, x > y+z\}$

Let $u=1-x$ , $v=y$ , and $w=z$ , then the set above is equivalent to the set

$\{(u, \, v, \, w) \mid 0 \leq u, \, v, \, w \leq 1, u + v + w \leq 1 \}$

It is then obvious that the solid given by this set is a tetrahedron formed by cutting the first octant with the plane $u + v + w = 1$ . The volume of it can be found by integration:

$\int\!\!\int\!\!\int_{u, \, v, \, w \geq 0, \, u + v + w \leq 1} du\, dv\, dw =\frac{1}{6}$

Hence, the answer is

$3 \Pr(a>b+c) = 3 \cdot \frac{1}{6} = \frac{1}{2}$

~Bloggish

Solution 5

Consider three independent uniform random variables $X, Y, Z$ on $[0, 1]$ . The goal is to find the probability that the maximum value is greater than twice each of the other two values. Equivalently, if we order them as $a \geq b \geq c$ , this is the probability that $a > 2b$ (which implies $a > 2c$ since $b \geq c$ ).

Due to symmetry, compute the probability that $X$ is the maximum and satisfies the condition, then multiply by 3. The condition simplifies to $P(X > 2Y \cap X > 2Z)$ , as this ensures $X$ is the maximum (almost surely).

The joint density is 1 over the unit cube. Thus, \begin{align*} P(X > 2Y \cap X > 2Z) &= \int_0^1 \int_0^{x/2} \int_0^{x/2} \, dy \, dz \, dx \\ &= \int_0^1 \left(\frac{x}{2}\right)^2 \, dx \\ &= \int_0^1 \frac{x^2}{4} \, dx \\ &= \frac{1}{4} \cdot \frac{1}{3} \\ &= \frac{1}{12}. \end{align*}The total probability is $3 \times \frac{1}{12} = \frac{1}{4}$ .

Alternatively, using order statistics $U < V < W$ with joint density 6 on $0 < u < v < w < 1$ , \begin{align*} P(W > 2V) &= \int_0^{0.5} \int_{2v}^1 \int_0^v 6 \, du \, dw \, dv \\ &= 6 \int_0^{0.5} v (1 - 2v) \, dv \\ &= 6 \left[ \frac{v^2}{2} - \frac{2v^3}{3} \right]_0^{0.5} \\ &= 6 \times \dfrac{1}{24} \\ &= \dfrac{1}{4}. \end{align*} No matter what entry points on this problem we choose, we can always get the correct answer: $\color{red}\boxed{\color{black}\text{(E) } \frac14}\color{black}$.

~funkCCP

Solution 6

We have $a,b,c \sim U(0,1)$. Thus, if $a$ is the greatest, we must have $b,c < a/2$, which occurs with probability $a^2/4$. We integrate this over all $a$ (note the pdf of $U(0,1)$ is just $1$):

$\int_0^1 \frac{a^2}{4} \, da=\left[\frac{a^3}{12}\right]_0^1=\frac{1}{12}$

Since there are $3$ possibilities for the greatest among $a,b$, and $c$, we multiply this by $3$ to get $3\cdot 1/12=\boxed{\text{(E) } \frac14}$

~Jackson La Vallee

Video Solution 1 by OmegaLearn

https://youtu.be/XxVC6Hx2tIY

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=8wNoOAqd5_o

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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