AIME 2026 I · 第 12 题

Solution 1 (5 min Coordbash)

In plane $P$ let $A$ be the origin and point $B$ be $(6,0)$ and $C$ be $(0,4)$. We see that the centroid is point $\left(2, \frac{4}{3}\right)$. Consequently, the line that is perpendicular to $BC$ passing through the centroid is $y= \frac{3}{2} (x - 2) + \frac{4}{3}$. This intersects $BC$ at $\left(\frac{34}{13}, \frac{88}{39}\right)$. Thus, $D= \left(\frac{42}{13}, \frac{124}{39}\right)$. We see that the centers of the spheres are $(0,0,1)$, $(6,0, 2)$, $(0,4,3)$, $\left(\frac{42}{13}, \frac{124}{39}, r\right)$. The condition about tangent planes show that the four centers are coplanar. By some easy linear algebra and cross product calculations, we obtain that the plane is $-x - 3y + 6z = 6$. Plugging in the coordinates of $D$, we find that $r=z= \frac{122}{39}$ so the sum is $122 + 39 = \boxed{161}$.

~Ddk001

made the parentheses bigger ~math_cool123

Solution 2 (Coordbash using Tangency Condition)

Begin by imposing a coordinate system where point $A$ is at $(0,0)$, point $B$ is at $(6,0)$, and point $C$ is at $(0,4)$. Then the centroid is $G=\left(\frac{6}{3},\frac{4}{3}\right)=\left(2,\frac{4}{3}\right)$.

Consider the sphere tangent to point $A$. Its center, denoted $A'$, will be at $(0,0,1)$ as it has a radius of $1$ and will therefore be a distance of $1$ from point $A$. Similarly, our other spheres will have centers $(6,0,2)$ and $(0,4,3)$ at points $B$ and $C$, respectively. We will also denote the centers of the spheres tangent to points $B$ and $C$ as $B'$ and $C'$, respectively.

Now, consider the additional plane which is tangent to all the spheres. The center of the sphere at $A$ will be a distance of $1$ (its radius) from this plane. Similarly, the center of the sphere at $B$ will be a distance of $2$, and the center of the sphere at $C$ will be a distance of $3$ from this plane. Say our additional plane has equation $ax+by+cz+d=0$, and WLOG, $a^2+b^2+c^2=1$. Then, by the point-to-plane distance formula, we get

$c+d=1$ from $A'$, $6a+2c+d=2$ from $B'$, and $4b+3c+d=3$ from $C'$. Solving this system, we get $d=\frac{36}{23}$, $c=-\frac{13}{23}$, $b=\frac{18}{23}$, and $a=\frac{6}{23}$. Therefore, the equation of our plane is $6x+18y-13z+36=0$ as we scaled up by a factor of $23$.

Now, we must find $D$, the result of reflecting $G$ over $BC$. Line $BC$ has equation $2x+3y=12$, so reflecting gives $D=\left(\frac{42}{13},\frac{124}{39}\right)$. Now, $D'=\left(\frac{126}{39},\frac{124}{39},r\right)$, and it also has a distance of $r$ from the plane $6x+18y-13z+36=0$. By the point-to-plane distance formula, we get $\frac{6(\frac{126}{39})+18(\frac{124}{39})-13r+36}{23}=r$.

Thus, $\frac{2988}{39}-13r+36=23r$, so $\frac{2988}{39}+36=36r$, and $r=\frac{122}{39}$, giving a requested answer of $\boxed{161}$.

~pl246631

Solution 3 (Slightly more Efficient)

Notice that spheres can be replaced with heights from $A,B,C,D$ with lengths $1,2,3,r$. This is because planes are flat and increase linearly based on moving up or right. Now we determine the coordinates of $D$. To start off, assign coordinates as follows:

$$A=(0,0)$$ $$B=(0,6)$$ $$C=(4,0)$$ Note that the centroid is the average of the $3$ coordinates of $A,B,C$:

$$G=\left(\frac{4}{3},2\right)$$ The equation of $BC$ is:

$$y=-\frac{3}{2}x+6$$ The projection of $G$ to $BC$ is the intersection of the following lines:

$$y=-\frac{3}{2}x+6$$ $$y=\frac{2}{3}x+\frac{10}{9}$$ This intersection increases the $x$ coordinate by $\tfrac{11}{12}$. Now we find how much our radius increases when increasing the $x$ and $y$ by $1$. Notice that $AC=4$ and the incerase is $2$, therefore every $1$ added to the $x$ coordinate increases the radius by $\tfrac{1}{2}$. Similarly, every increase in the $y$ coordinate increases the radius by $\tfrac{1}{6}$. The radius of the centroid is $\tfrac{1+2+3}{3}=2$. Therefore our radius is:

$$r=2+2\cdot\frac{12}{13}\cdot\frac{1}{2}+2\cdot\frac{12}{13}\cdot\frac{2}{3}\cdot\frac{1}{6}=\frac{122}{39}$$ This gives us $\boxed{161}$.

~ zhenghua

Solution 4

Try to find that line of intersection of the 2 planes. If we look from a "sideways" perspective of the 2 planes, we would see this:

Set a coordinate system with points $A$,$B$,$C$

$A=(0,0)$ $B=(6,0)$ $C=(0,4)$

By similar triangles, we discovered that the distances from the line of intersection to the 3 bottom points of the spheres have same ratios as their radius. So, in simple words, we are trying to find a line that has distances with ratios of $1:2:3$, corresponding to points $A$,$B$,$C$.

We discovered that this line passes through $(-2,0)$ and is actually parallel to the midline of $\triangle ABC$ from point $B$. The line is

$$x+3y+6=0$$ We get that the centroid of the triangle should be $G(2,\frac{4}{3})$, flipping it across $BC$ makes $D(\frac{42}{13},\frac{124}{39})$. Its distance compared to the distance of $A$ to the intersection line gives us our desired radius. So $\frac{\left|\frac{42}{13}+3\cdot\frac{124}{39}+6\right|}{6} = \frac{122}{39}$, giving us $\boxed{161}$.

~cassphe