$A B C D$ is a tetrahedron and $O$ is any point. If the line joining $O$ to the vertices meets the opposite faces at $P, Q, R$ and $S$ , then the value of
$\frac{O P}{A P} + \frac{O Q}{B Q} + \frac{O R}{C R} + \frac{O S}{D S}$ is

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Solution

Here, $A B C D$ is the tetrahedron.
Let $O$ be the origin and position vectors of $A, B, C$ and $D$ be $\to a, \to b, \to c$ and $\to d$ respectively.
$4$ linearly dependent variable can be expressed as
$x \to a + y \to b + z \to c + t \to d = 0$
where $x, y, z, t$ are scalars.
$\Rightarrow y \to b + z \to c + t \to d = - x \to a \Rightarrow \frac{y \to b + z \to c + t \to d}{y + z + t} = - \frac{x \to a}{y + z + t}$
Here, L.H.S. represents the position vector of a point in $B C D$ plane and R.H.S. is the point on $- - \to A O$
$∴$ There must be a point common to both the plane and the straight line, that is
$- - \to O P = - \frac{x \to a}{y + z + t} - - \to A P = - - \to O P - - - \to O A$
$= - \frac{x \to a}{y + z + t} - \to a = - \frac{x + y + z + t}{y + z + t} (\to a) \Rightarrow \frac{O P}{A P} = \frac{x}{x + y + z + t}$

Simillarly, $\frac{O Q}{B Q} = \frac{y}{x + y + z + t}$ ,
$\frac{O R}{C R} = \frac{z}{x + y + z + t}$ ,
and $\frac{O S}{D S} = \frac{t}{x + y + z + t}$

$∴$ Required value is $1$

Alternate:
Let $O$ be the origin
And the position vectors of $A, B, C, D$ be $a, b, c, d$ respectively,
Any one of these vectors may be expressed in terms of the other three, so
$x \to a + y \to b + z \to c + t \to d = 0 \dots (1)$

Let the equation of the line $A P$ be,
$\to r = - u \to a$
$u$ is positive when it lies on the oposite side of the origin w.r.t. $A$
Using equation $(1)$ ,
$x \to r - u (y \to b + z \to c + t \to d) = 0 \dots (2)$
The point $P$ , which lies on the plane $B C D$ ,
So $\to r, \to b, \to c, \to d$ are coplanar.
Hence, the sum of the coefficients in equation $(2)$ is zero,
$u = \frac{x}{(y + z + t)}$ ,
$\frac{O P}{A P} = \frac{u}{1 + u} = \frac{x}{x + y + z + t}$
Simillarly,
$\frac{O Q}{B Q} = \frac{y}{x + y + z + t}$ ,
$\frac{O R}{C R} = \frac{z}{x + y + z + t}$ ,
$\frac{O S}{D S} = \frac{t}{x + y + z + t}$
Therefore,
$\frac{O P}{A P} + \frac{O Q}{B Q} + \frac{O R}{C R} + \frac{O S}{D S} = 1$

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