Library / /1265 of 1394

, 2015

Geometry Difficulty 5.9 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a regular tetrahedron with side length $1$ . Let $X$ be the point in triangle $BCD$ such that $[XBC] = 2[XBD] = 4[XCD]$ , where $[\varpi]$ denotes the area of figure $\varpi$ . Let $Y$ lie on segment $AX$ such that $2AY = YX$ . Let $M$ be the midpoint of $BD$ . Let $Z$ be a point on segment $AM$ such that the lines $YZ$ and $BC$ intersect at some point. Find $\frac{AZ}{ZM}$ .

Solution

Solution:

We apply three-dimensional barycentric coordinates with reference tetrahedron $ABCD$ . The given conditions imply that

$\begin{aligned} X & =(0: 1: 2: 4) \\ Y & =(14: 1: 2: 4) \\ M & =(0: 1: 0: 1) \\ Z & =(t: 1: 0: 1) \end{aligned}$

for some real number $t$ . Normalizing, we obtain $Y=\left(\frac{14}{21}, \frac{1}{21}, \frac{2}{21}, \frac{4}{21}\right)$ and $Z=\left(\frac{t}{t+2}, \frac{1}{t+2}, 0, \frac{1}{t+2}\right)$ . If $YZ$ intersects line $BC$ then there exist parameters $\alpha+\beta=1$ such that $\alpha Y+\beta Z$ has zero $A$ and $D$ coordinates, meaning

$\begin{aligned} \frac{14}{21} \alpha+\frac{t}{t+2} \beta & =0 \\ \frac{4}{21} \alpha+\frac{1}{t+2} \beta & =0 \\ \alpha+\beta & =1 \end{aligned}$

Adding twice the second equation to the first gives $\frac{22}{21} \alpha+\beta=0$ , so $\alpha=-22, \beta=21$ , and thus $t=\frac{7}{2}$ . It follows that $Z=(7: 2: 0: 2)$ , and $\frac{AZ}{ZM}=\frac{2+2}{7}=\frac{4}{7}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.