Locus and Planes in 3D
DAILY_CHALLENGE
Grade None
Question:
Let $\mathbb{R}^3$ denote the three-dimensional space. Take two points $P=(1,2,3)$ and $Q=(4,2,7)$. Let
$$S=\{X\in\mathbb{R}^3:(\text{dist}(X,P))^2-(\text{dist}(X,Q))^2=50\}$$
and
$$T=\{Y\in\mathbb{R}^3:(\text{dist}(Y,Q))^2-(\text{dist}(Y,P))^2=50\}.$$
Then which of the following statements is (are) TRUE?
There is a triangle whose area is 1 and all of whose vertices are from $S$.
There are two distinct points $L$ and $M$ in $T$ such that each point on the line segment $LM$ is also in $T$.
There are infinitely many rectangles of perimeter 48, two of whose vertices are from $S$ and the other two vertices are from $T$.
There is a square of perimeter 48, two of whose vertices are from $S$ and the other two vertices are from $T$.
Step-by-Step Solution
Key Concept: The locus equation simplifies to a plane; S and T are parallel planes at distance 10 apart
For $S$: let $X=(x,y,z)$.
$(x-1)^2+(y-2)^2+(z-3)^2-[(x-4)^2+(y-2)^2+(z-7)^2]=50$
$\Rightarrow 6x+8z-55=50\Rightarrow 6x+8z=105$. $S$ is a plane.
For $T$: similarly $-6x-8z+55=50\Rightarrow 6x+8z=5$. $T$ is a parallel plane.
Distance between planes: $|105-5|/\sqrt{36+64}=100/10=10$.
(A) TRUE: $S$ is an infinite plane; triangles of any area exist within it.
(B) TRUE: $T$ is a plane (convex set); any two points in $T$ have their entire connecting segment in $T$.
(C) Rectangles with 2 vertices on $S$ and 2 on $T$, perimeter 48 (sides $a$ and $b$, $2a+2b=48$, $a+b=24$): the sides crossing between planes have length $\geq10$. For any valid rectangle width $a\geq10$, $b=24-a\leq14$. Infinitely many such rectangles exist (varying position and orientation). TRUE.
(D) Square with side 12, perimeter 48: two sides cross planes at angle $\theta$ where $12\sin\theta=10$, $\theta=\arcsin(5/6)$. Such a square exists. TRUE.
Correct Answer: A, B, C, D