# If $P_1,P_2,P_3$ lie on the circle $x^2+y^2=1$,then prove that $P_4$ lies on the circle.

Given $4$ points $P_1,P_2,P_3,P_4$ on the coordinate plane with origin $O$ which satisfy the condition $\vec{OP_1}+\vec{OP_3}=\frac{3}{2}\vec{OP_2}$ and $\vec{OP_2}+\vec{OP_4}=\frac{3}{2}\vec{OP_3}$
If $P_1,P_2,P_3$ lie on the circle $x^2+y^2=1$,then prove that $P_4$ lies on the circle.

My Attempt:
Let the coordinates of $P_1,P_2,P_3,P_4$ be $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$.According to the given condition,
$x_1+x_3=\frac{3}{2}x_2$
$y_1+y_3=\frac{3}{2}y_2$
$x_2+x_4=\frac{3}{2}x_3$
$y_2+y_4=\frac{3}{2}y_3$
Since $P_1,P_2,P_3$ lie on the circle $x^2+y^2=1$.
So,$x_1^2+y_1^2=1.......................(1)$
$x_2^2+y_2^2=1...........................(2)$
$x_3^2+y_3^2=1...........................(3)$
Now we need to prove that $x_4^2+y_4^2=1$
Subtracting $(3)$ from $(1)$ we get
$x_1^2-x_3^2+y_1^2-y_3^2=0$
$(x_1+x_3)(x_1-x_3)+(y_1+y_3)(y_1-y_3)=0$
$\frac{3}{2}x_2(x_1-x_3)+\frac{3}{2}y_2(y_1-y_3)=0$
$x_2(x_1-x_3)+y_2(y_1-y_3)=0$

But i am stuck here and cannot move ahead.Please help me.Thanks.

## 1 Answer

Let $P_2'$ be the endpoint of $\frac{3}{2} \overrightarrow{OP_2}$ and $P_3'$ be the endpoint of $\frac{3}{2} \overrightarrow{OP_3}$. The equation $\overrightarrow{OP_1} + \overrightarrow{OP_3} = \frac{3}{2}\overrightarrow{OP_2}$ implies the triangle $\Delta OP_3P_2'$ has side lengths $1$, $1$, and $3/2$, since $OP_i$ are unit vectors for $i = 1,2,3$. This triangle has an angle $\angle P_3OP_2'$ between the sides of length $1$ and $3/2$.

The second equation $\overrightarrow{OP_2} + \overrightarrow{OP_4} = \frac{3}{2}\overrightarrow{OP_3}$ implies the triangle $\Delta OP_2P_3'$ has side lengths of $1$, $||\overrightarrow{OP_4}||$ and $3/2$. This triangle has an angle of $\angle P_3'OP_2$ between the sides of length $1$ and $3/2$.

However since $P_3$ and $P_3'$ lie on the same ray from the origin, and so do $P_2$ and $P_2'$, we know $\angle P_3'OP_2 \cong \angle P_3OP_2'$. So the triangles $\Delta OP_2P_3'$ and $\Delta OP_3P_2'$ are congruent since they have two sides of the same length with an equal angle inbetween. Thus, $||\overrightarrow{OP_4}||$ must be the length of the remaining side of $\Delta OP_2P_3'$, hence $||\overrightarrow{OP_4}|| = 1$, which implies $P_4$ is on the unit circle.

• Beautiful solution,innovative approach without using coordinates! Nov 12, 2015 at 4:15