Difficult Geometry - Circle #1

Geometry Level 3

Circles $O_1$ and $O_2$ intersect with each other at points $\text{A}$ and $\text{B}$ .

$O_1$ and $O_2$ has a radius of $3$ and $5$ , and has a center of ${\text{C}}_1$ and ${\text{C}}_2$ , respectively.

Point $\text{D}$ internally divides $\overline{\text{AB}}$ with a ratio of $1:3$ .

$\overline{\text{EF}}$ and $\overline{\text{KP}}$ intersect at point $\text{D}$ .

$\overline{\text{EF}}$ is a chord of $O_1$ , with point $\text{F}$ being inside circle $O_1$ and $\text{E}$ being outside it, satisfying $\overline{\text{AE}}<\overline{\text{BE}}$ and $\overline{\text{AF}}>\overline{\text{BF}}$ .

Point $\text{K}$ is on circle $O_2$ and inside circle $O_1$ . Also, it satisfies $\angle\text{EFK}=20^{\circ}$ , $\angle\text{EKF}=130^{\circ}$ and $\angle{\text{C}}_1\text{FK}<\angle{\text{C}}_1\text{FE}$ .

Point $\text{P}$ satisfies $\angle\text{EPF}=50^{\circ}$ .

Find the value of $\overline{{\text{C}}_2 \text{P}}$ .

If you think the given information is too less to figure out the answer, submit $-1$ as your answer. And if you think the given information doesn't make sense, submit $0$ as your answer.

The answer is 5.

1 solution

Boi (보이)
Jun 14, 2017

Carefully drawing the picture, you'll realize that the above picture is correct.

Now for the proof.

Since $\overline{\text{EF}}$ and $\overline{\text{AB}}$ are chords of circle ${\text{C}}_1$ ,

$\overline{\text{ED}}\times\overline{\text{DF}}=\overline{\text{AD}}\times\overline{\text{DB}}$

And as said in the question above,

$\angle\text{EKF}=180^{\circ}-\angle\text{EPF}$

Therefore points $\text{E}$ , $\text{P}$ , $\text{F}$ , $\text{K}$ are on a circle.

Then, according to the power theorem, we can say that:

$\overline{\text{PD}}\times\overline{\text{DK}}=\overline{\text{ED}}\times\overline{\text{DF}}$

Substitute what we've found earlier before.

$\boxed{\overline{\text{PD}}\times\overline{\text{DK}}=\overline{\text{AD}}\times\overline{\text{DB}}}$

Therefore, we may safely say that $\text{P}$ is on circle ${\text{C}}_2$ .

$\overline{{\text{C}}_2 \text{P}}=\boxed{5}$ .

Difficult Geometry - Circle #1

The answer is 5.

1 solution

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