AIME 2026 II · 第 12 题

Solution 1

We place the tetrahedron in a coordinate system by symmetry:

$$A(-5, 0, 0),\ B(5, 0, 0),\ C(0, 9, h),\ D(0, -9, h).$$ Given $AC = 5\sqrt{10}$, we have \begin{align*} AC^2 &= 5^2 + 9^2 + h^2 = 250, \\ h^2 &= 144 \implies h = 12. \end{align*} So the vertices are

$$A(-5,0,0),\ B(5,0,0),\ C(0,9,12),\ D(0,-9,12).$$ By symmetry, the circumcenter $S$ lies on the $z$-axis. Let $S=(0,0,a)$. Using $SA=SC$: \begin{align*} SA^2 &= 25 + a^2, \\ SC^2 &= 81 + (12-a)^2, \\ 25 + a^2 &= 225 - 24a + a^2, \\ 24a &= 200 \implies a = \frac{25}{3}. \end{align*} Thus

$$S=\Big(0,0,\frac{25}{3}\Big).$$ By symmetry, the incenter $R$ lies on the $z$-axis. Let $R=(0,0,b)$.

Vectors for plane $ABC$:

$$\overrightarrow{AB}=\begin{pmatrix}10\\0\\0\end{pmatrix},\quad \overrightarrow{AC}=\begin{pmatrix}5\\9\\12\end{pmatrix}.$$ Normal vector by cross product: \begin{align*} \mathbf{n}_1 &= \overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\10&0&0\\5&9&12\end{vmatrix} \\ &= \mathbf{i}(0\cdot12-0\cdot9) -\mathbf{j}(10\cdot12-0\cdot5) +\mathbf{k}(10\cdot9-0\cdot5) \\ &= 0\mathbf{i}-120\mathbf{j}+90\mathbf{k} = 30\begin{pmatrix}0\\-4\\3\end{pmatrix}. \end{align*} We get the equation of the plane:

$$-4y + 3z = 0.$$ Distance from $R$ to plane $ABC$:

$$d_1 = \frac{|3b|}{\sqrt{(-4)^2+3^2}} = \frac{3|b|}{5}.$$ Vectors for plane $ACD$:

$$\overrightarrow{AC}=\begin{pmatrix}5\\9\\12\end{pmatrix},\quad \overrightarrow{AD}=\begin{pmatrix}5\\-9\\12\end{pmatrix}.$$ Normal vector by cross product: \begin{align*} \mathbf{n}_2 &= \overrightarrow{AC}\times\overrightarrow{AD} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&9&12\\5&-9&12\end{vmatrix} \\ &= \mathbf{i}(9\cdot12-12\cdot(-9)) -\mathbf{j}(5\cdot12-12\cdot5) +\mathbf{k}(5\cdot(-9)-9\cdot5) \\ &= 216\mathbf{i}-0\mathbf{j}-90\mathbf{k} = 18\begin{pmatrix}12\\0\\-5\end{pmatrix}. \end{align*} We get the equation of the plane:

$$12x - 5z + 60 = 0.$$ Distance from $R$ to plane $ACD$:

$$d_2 = \frac{|60-5b|}{\sqrt{12^2+(-5)^2}} = \frac{|60-5b|}{13}.$$ We set $d_1=d_2$: \begin{align*} \frac{3b}{5} &= \frac{60-5b}{13}, \\ 39b &= 300-25b, \\ 64b &= 300 \implies b = \frac{75}{16}. \end{align*} Thus

$$R=\Big(0,0,\frac{75}{16}\Big).$$ Hence, the distance $RS$ is: \begin{align*} RS &= \left|\frac{25}{3}-\frac{75}{16}\right| = \left|\frac{400-225}{48}\right| = \frac{175}{48}. \end{align*} Since $\gcd(175,48)=1$, we have $m=175$, $n=48$.

$$m+n = 175+48=\boxed{223}.$$ ~Steven Zheng

Solution 2

~Diagram by Confringo

Choose face $ABC$ as the base. Let $O$ be the midpoint of $AB$ as the origin, with $OA$ along the positive $x$-axis, $OC$ along the positive $y$-axis, and the line through $O$ perpendicular to plane $ABC$ as the $z$-axis.

First, find the coordinates of point $D$. Due to the symmetry of the figure, plane $OCD$ is the $yOz$-plane. Let $D'$ be the projection of $D$ onto $OC$. Let $OD' = y$ and $DD' = z$. Then

$$\begin{cases} y^2+z^2=15^2 \\ (15-y)^2+z^2=18^2 \end{cases}$$ Solving this system yields $y=\dfrac{21}{5}$ and $z=\dfrac{72}{5}$, so $D\left(0,\dfrac{21}{5},\dfrac{72}{5}\right)$.

Next, find the circumcenter. The center of the circumscribed sphere lies on the line perpendicular to the base through the circumcenter of the base triangle. Here $CO=\sqrt{AC^2-AO^2}=15$. We need a point $S'$ on $OC$ such that $CS' = SA$. Let $CS' = AS' = x$, then $S'O = 15-x$. In right triangle $AOS'$, by the Pythagorean theorem,

$$x^2=(15-x)^2+5^2,$$ which gives $x=\dfrac{25}{3}$, so $OS' = 15-x = \dfrac{20}{3}$ (the $y$-coordinate of $S'$). Let $S = \left(0,\dfrac{20}{3},h\right)$. Then $AS = CS$, so

$$\sqrt{\left(\dfrac{25}{3}\right)^2+h^2}=\sqrt{(0-0)^2+\left(\dfrac{21}{5}-\dfrac{20}{3}\right)^2+\left(\dfrac{72}{5}-h\right)^2}.$$ Solving gives $h=5$, so $S = \left(0,\dfrac{20}{3},5\right)$.

Now find the incenter. By volume considerations: connecting each vertex to the incenter divides the tetrahedron into four smaller tetrahedra, each with height equal to the insphere radius $r$ and base area equal to the area of a face of the tetrahedron. Their volumes sum to the total volume of the tetrahedron.

The areas of the two types of faces are $75$ and $117$. The height from $C$ to base $ABD$ is $z_C = \dfrac{72}{5}$. Volume equation:

$$\dfrac{1}{3}\cdot75\cdot\dfrac{72}{5}=\dfrac{1}{3}\cdot(2\cdot75+2\cdot117)\cdot r.$$ Solving gives $r = \dfrac{45}{16}$. Thus $z_R = \dfrac{45}{16}$.

By symmetry, the incenter also lies in the $yOz$-plane. Let $R'$ be the projection of $R$ onto $OC$. Then $\tan\angle ROR' = \dfrac{RR'}{OR'} = \dfrac{MC}{OM} = \dfrac34$. So

$$OR' = \dfrac43 RR' = \dfrac43 \cdot \dfrac{45}{16} = \dfrac{15}{4}.$$ Thus $y_R = \dfrac{15}{4}$.

Finally, $S = \left(0,\dfrac{20}{3},5\right)$ and $R = \left(0,\dfrac{15}{4},\dfrac{45}{16}\right)$. By the Pythagorean theorem, the distance $RS$ is

$$\sqrt{(0-0)^2+\left(\dfrac{20}{3}-\dfrac{15}{4}\right)^2+\left(5-\dfrac{45}{16}\right)^2}=\dfrac{175}{48}.$$ Since $175$ and $48$ are coprime, $m=175$ and $n=48$. Hence $m+n=\boxed{223}$.

~Confringo

Solution 3 (no coordinates)

We let $AB$ be the side of length $10$ and $CD$ of length $18$. Call midpoints of side $AB$ and $CD$ as $M$ and $N$. The faces of our tetrahedron are isosceles triangle so $AM=5$ and $CN=9$, along with $MN\perp AB$ and $MN\perp CD$. By symmetry we know both $R$ and $S$ locate on the segment $MN$.

For the position of $R$, we set $MR=x$. By isosceles we also have $CM\perp AB$ so the plane $DMC$ is perpendicular to the place $ABC$. The distance from $R$ to the plance $ABC$ is the same as the distance between $R$ and the line $CM$. The same happens for the distance between $R$ and the line $AN$. In the triangle $ABC$, we know $CM=\sqrt{25\cdot 10-25}=15$. Then $EF=\sqrt{15^2-9^2}=12$ in the triangle $CMD$. Draw altitude from $R$ to $CM$ and call the foot as $E$. We get $\triangle MRE\sim \triangle MCN$ so $RE=\dfrac{3}{5}x$. In $\triangle ABN$, we also use $RN=12-x$ and similar triangles to get the distance between $R$ and $AN$ is $\dfrac{5}{13}(12-x)$. The two distances are equal so $\dfrac{3}{5}x=\dfrac{5}{13}(12-x)$ and $MR=x=\dfrac{75}{16}$.

For the position of $S$, we set $MS=y$. In $\triangle AMS$ we know $AS=\sqrt{25+y^2}$. In $\triangle CNS$ we know $CS=\sqrt{(12-y)^2+9^2}$. Setting $AS=CS$ we solve $MS=y=\dfrac{25}{3}$.

In summary, $RS=MS-MR=\dfrac{25}{3}-\dfrac{75}{16}=\dfrac{175}{48}$ so $m+n=\boxed{223}$.

~figures are based on the ones from Confringo.

~lanouzhihun