Circles and $\infty$ tangents

Geometry Level 4

A circle with center point $O$ and radius $R \space cm$ is drawn. Point $P_1$ is lying out of the circle such that $\overline{OP_1}=3cm$ . $\overline{P_1B}$ is a tangent to the circle.

Point $P_2,P_3...(till \space\infty)$ are taken on $\overline{P_1B}$ such that $\forall n\in\mathbb{N}\space P_{n+1}$ is mid point of $\overline{P_nB}$ .

$\forall n\in\mathbb{N}\space\overline{P_nA_n}$ is a tangent to the circle different from $\overline{P_nB}$ .

$If\space\sum_{n=1}^{\infty}ar(BP_nA_nO)=2\sqrt{14}cm^2$

Find the sum of all possible values of $R^2$

Note:

$ar(X)$ is the area of plane figure $X$

The answer is 9.

1 solution

Zakir Husain
Feb 7, 2021

$\forall n\in\mathbb{N}\space BP_nA_nO$ will be a Right Kite

$\therefore ar(BP_nA_nO)=\overline{OB}\times\overline{BP_n}=R\times\overline{BP_n}..........[1]$

$\because\forall n\in\mathbb{N}\space\overline{BP_{n+1}}=\frac{BP_n}{2}\therefore \overline{BP_n}=\frac{BP_1}{2^{n-1}}..........[2]$

From $[1]$ and $[2]$

$ar(BP_nA_nO)=R\times\frac{BP_1}{2^{n-1}}$ $\Rightarrow \sum_{n=1}^{\infty}ar(BP_nA_nO)=\sum_{n=1}^{\infty}R\times\frac{BP_1}{2^{n-1}}$ $=R\times\overline{BP_1}\times\red{\sum_{n=1}^{\infty}\frac{1}{2^{n-1}}}$ $=R\times\overline{BP_1}\times\red{2}\Rightarrow\boxed{\cancel{2}R\times\overline{BP_1}=\cancel{2}\sqrt{14}}$ $\Rightarrow R\times\overline{BP_1}=\sqrt{14}$ $\Rightarrow R^2\times\overline{BP_1}^2=14..........[3]$ Applying Pythagorus Theorem on $\triangle OBP_1$ $\overline{BP_1}^2=\overline{OP_1}^2-R^2=9-R^2$ Putting this in $[3]$ $R^2(9-R^2)=14$ $\Rightarrow R^4-9R^2+14=0$ $\Rightarrow (R^2)^2-9(R^2)+14=0$ $\Rightarrow (R^2-7)(R^2-2)=0$ $R^2\in\{7,2\}$ $Answer=7+2=\boxed{9}$

Circles and ∞ \infty ∞ tangents

The answer is 9.

1 solution

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Circles and $\infty$ tangents