MHB Cos^2(∠POA)+cos^2(∠POB)+cos^2(∠POC)+cos^2(∠POD) is independent of P.

[lfdahl]

A cube is inscribed in the unit sphere $x^2 +y^2 +z^2 = 1$. Let $A,B,C$ and $D$ denote the
vertices of one face of the cube. Let $O$ denote the center of the sphere, and $P$
denote a point on the sphere. Show that

\[\cos ^2(\angle POA)+\cos ^2(\angle POB)+\cos ^2(\angle POC)+\cos ^2(\angle POD)\]

is independent of $P$.

 

[castor28]

A cube is inscribed in the unit sphere $x^2 +y^2 +z^2 = 1$. Let $A,B,C$ and $D$ denote the
vertices of one face of the cube. Let $O$ denote the center of the sphere, and $P$
denote a point on the sphere. Show that

\[\cos ^2(\angle POA)+\cos ^2(\angle POB)+\cos ^2(\angle POC)+\cos ^2(\angle POD)\]

is independent of $P$.

[sp]
Let us write $S$ for the expression, and $R$ for the radius of the sphere. As $S$ is dimensionless, it is independent of $R$, and we may choose $R$ as we want.

We take the vertices as $A=(1,1,1)$, $B=(-1,1,1)$, $C=(1,-1,1)$ and $D=(-1,-1,1)$; this makes $R=\sqrt3$.

Let the coordinates of $P$ be $(x,y,z)$. We have:

$\cos POA = \dfrac{\langle OA\cdot OP\rangle}{R^2} = \dfrac{x + y + z}{3}$
​

and a similar expression for the other terms ($\langle\cdot\rangle$ is the dot product).

This gives:

$$\begin{align*}
S &=\frac19\left((x+y+z)^2 + (-x+y+z)^2 + (x-y+z)^2 +(-x-y+z)^2\right)\\
&= \frac49\left(x^2+y^2+z^2\right)
\end{align*}
$$
​

and, as $x^2+y^2+z^2=R^2 = 3$, we get $S=\dfrac43$, which is indeed independent of the position of $P$.
[/sp]

 

[lfdahl]

[sp]
Let us write $S$ for the expression, and $R$ for the radius of the sphere. As $S$ is dimensionless, it is independent of $R$, and we may choose $R$ as we want.

We take the vertices as $A=(1,1,1)$, $B=(-1,1,1)$, $C=(1,-1,1)$ and $D=(-1,-1,1)$; this makes $R=\sqrt3$.

Let the coordinates of $P$ be $(x,y,z)$. We have:

$\cos POA = \dfrac{\langle OA\cdot OP\rangle}{R^2} = \dfrac{x + y + z}{3}$
​

and a similar expression for the other terms ($\langle\cdot\rangle$ is the dot procuct).

This gives:

$$\begin{align*}
S &=\frac19\left((x+y+z)^2 + (-x+y+z)^2 + (x-y+z)^2 +(-x-y+z)^2\right)\\
&= \frac49\left(x^2+y^2+z^2\right)
\end{align*}
$$
​

and, as $x^2+y^2+z^2=R^2 = 3$, we get $S=\dfrac43$, which is indeed independent of the position of $P$.
[/sp]

Thankyou, castor28 for such a nice solution and for your participation!(Handshake)