# On Proof (1 Viewer)

#### CM_Tutor

##### Moderator
Moderator
I encourage all MX2 students to have a look at the following thread in the MX1 forum:

The third question sounds simple but I believe it has been set in error as there is no simple solution. The posts in the thread do explore some aspects of the problem and could be adapted to a problem in proof.

Perhaps as something like:

If $\bg_white O$ is the origin, $\bg_white B$ is a variable point satisfying $\bg_white \left|\overrightarrow{OB}\right| = 4\sqrt{5}$, and $\bg_white POB$ is a triangle of area 10 u2, show that $\bg_white \overrightarrow{OP}$ can be located anywhere on the number plane so long as $\bg_white \left|\overrightarrow{OP}\right| \geqslant k, k\in\mathbb{R}^+$, and find the value of $\bg_white k$.

#### notme123

##### Member
If you have triangle POB:
$\bg_white \frac{1}{2} |OP||OB|\sin(\theta)=10$ theta is angle between OB and OP
$\bg_white |OP||OB|\sin(\theta)=20$
Since $\bg_white |OB|=4\sqrt5$
$\bg_white |OP|=\frac{20}{4(\sqrt5)\sin(\theta)}$
$\bg_white |OP|=\frac{\sqrt5}{\sin(\theta)}$
and since $\bg_white \csc(\theta)\geqslant1$
therfore, $\bg_white |OP|\geqslant\sqrt5$
$\bg_white k=\sqrt5$
Works as long as B and P aren't collinear i.e. $\bg_white \theta \neq 0,180...$
But since B is variable, for every point P there will be a non-collinear B.
Therefore OP can be anywhere on the number plane as long as $\bg_white |OP|\geqslant\sqrt5$

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#### CM_Tutor

##### Moderator
Moderator
Works as long as B and P are not collinear i.e. $\bg_white \theta = 0,180...$
But since B is variable, for every point P there will be a non-collinear B.
B, O, and P can't be collinear and be the vertices of a triangle of non-zero area.

Otherwise, good job!