In tetrahedron $ABCD$, $AB = 4$, $CD = 7$, and $AC = AD = BC = BD = 5$. Let $I_A, I_B, I_C,$ and $I_D$ denote the incenters of the faces opposite vertices $A, B, C,$ and $D,$ respecitvely. It is provable that $AI_A$ intersects $BI_B$ at a point $X$, and $CI_C$ intersects $DI_D$ at a point $Y$. Compute $XY$.
Can someone provide a diagram for this question because it is hard to visualize it.
Update (by achille)
A diagram based on my understanding of the geometry.
- the red/green/blue arrows are $x$, $y$, $z$ axis.
- the red/green/blue/magenta line segments are $AI_A$, $BI_B$, $CI_C$ and $DI_D$.
- the black/white balls in the middle of diagram are $X$ and $Y$.
Let $u = 4$, $v = 7$ and $w = 5$.
To attack this problem, the key is $AC = AD = BC = BD$ which tell us tetrahedron $ABCD$ possesses certain symmetries. Let $\ell$ be the line passing through the mid-point of $AB$ and the mid-point of $CD$. $ABCD$ will be invariant under
Let $h$ be the distance between line $AB$ and $CD$. These symmetries allow us to choose a coordinate system so that $AB$ is part of the $x$-axis and $\ell$ is the $z$-axis. The vertices will be located at:
$$\begin{cases} \vec{A} &= ( -\frac{u}{2}, 0, 0 )\\ \vec{B} &= ( +\frac{u}{2}, 0, 0 )\\ \vec{C} &= ( 0, +\frac{v}{2}, h )\\ \vec{D} &= ( 0,-\frac{v}{2}, h ) \end{cases} \quad\text{ and }\quad \begin{array}{c} AC = AD = BC = BD = w\\ \Downarrow\\ h = \frac12\sqrt{4w^2 - u^2 - v^2} \end{array} $$ For a picture, please look at the one on question. To compute the coordinates of $I_A, I_B, I_C, I_D$, we use the fact the Barycentric coordinates of the in-center of a triangle is proportional to the length of the sides. From this fact, we can deduce
$$ \begin{cases} \vec{I}_A &= \frac{1}{v + 2w}\left(v \vec{B} + w(\vec{C}+\vec{D})\right)\\ \vec{I}_B &= \frac{1}{v + 2w}\left(v \vec{A} + w(\vec{C}+\vec{D})\right)\\ \vec{I}_C &= \frac{1}{u + 2w}\left(u \vec{D} + w(\vec{A}+\vec{B})\right)\\ \vec{I}_D &= \frac{1}{u + 2w}\left(u \vec{C} + w(\vec{A}+\vec{B})\right) \end{cases} $$ The line $AI_A$ consists of points of the form $$ s \vec{A} + (1-s)\vec{I}_A = s\vec{A} + \frac{1-s}{v+2w}\left(v \vec{B} + w(\vec{C}+\vec{D})\right)\quad\text{ for some } s \in \mathbb{R} $$ Similarly, the line $BI_B$ consists of points of the form $$ t \vec{B} + (1-t)\vec{I}_B = t\vec{B} + \frac{1-t}{v+2w}\left(v \vec{A} + w(\vec{C}+\vec{D})\right)\quad\text{ for some } t \in \mathbb{R} $$ If you compare these two expressions, it is easy to see $AI_A$ and $BI_B$ intersect at $$\vec{X} = \frac{1}{2(v+w)}\left(v (\vec{A}+\vec{B}) + w (\vec{C}+\vec{D})\right) = \left( 0, 0, \frac{wh}{v+w} \right)$$ By a similar argument, $CI_C$ and $DI_D$ intersect at $$\vec{Y} = \frac{1}{2(u+w)}\left(u (\vec{C}+\vec{D}) + w (\vec{A} + \vec{B})\right) = \left( 0, 0, \frac{uh}{u+w} \right)$$ So the distance $XY$ we seek is given by
$$XY = \left|\frac{u}{u+w} - \frac{w}{v+w}\right|h = \frac12\left|\frac{4}{4+5} - \frac{5}{7+5}\right|\sqrt{4\cdot 5^2 - 4^2 - 7^2} = \frac{\sqrt{35}}{72}$$