Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 65

In triangle ABC, AB = 5, AC = 4, and BC = 3. Let P be the point on the circumcircle of triangle ABC so that \angle PCA = 45^\circ. Find CP.

This is pretty hard. I got to where AB is a diameter (ACB is 90 degrees) and the radius is 2.5, but what next?

thanks!

Good. You can read.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,291

hi thedarktiger,

The following method will work but I'll keep looking for a quicker way.

You're right about AB being a diameter so I've marked the centre as O, and there are lots of radii that must therefore be 2.5 long.

In my diagram I have also marked the midpoints of AC as D and of PC as E.

Triangles OAC and separately POC, are isosceles and the lines from O to those midpoints will cut each isosceles triangle down the middle. That means there will be a right angled triangle in each half.

Now lots of trigonometry.

Use acos(2/2.5) to get angle OCD and hence calculate angle OCE.

Now you can use that to calculate EC and double this to get PC.

Bob

*Last edited by bob bundy (2014-02-08 22:08:57)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Pages: **1**