# 2 triangles

• Nov 7th 2009, 09:41 PM
thereddevils
2 triangles
THe diagram shows triangle ABC where points A , B , C and P are coplanar . BC=BP=AP and angle BAC =angle ABP = angle PBC = pi/5 rad

Prove that triangle ABC and BPC are similar triangles . Hence deduce that

(1) BC^2=CP.CA

(2) $\displaystyle \cos (\frac{\pi}{5})=\frac{1}{4}(\sqrt{5}+1)$

I am only stucked with (2) , i am ok with the rest . THanks .
• Nov 7th 2009, 11:20 PM
ukorov
if your diagram is right, then BC = BP is wrong.
• Nov 9th 2009, 05:42 AM
thereddevils
Quote:

Originally Posted by ukorov
if your diagram is right, then BC = BP is wrong.

why is it so
• Nov 9th 2009, 10:15 AM
ukorov
Quote:

Originally Posted by thereddevils
why is it so

oh alright not wrong actually, but could have been more detailed and more accurate. (Giggle)
• Nov 10th 2009, 01:38 AM
Hello thereddevils
Quote:

Originally Posted by thereddevils
THe diagram shows triangle ABC where points A , B , C and P are coplanar . BC=BP=AP and angle BAC =angle ABP = angle PBC = pi/5 rad

Prove that triangle ABC and BPC are similar triangles . Hence deduce that

(1) BC^2=CP.CA

(2) $\displaystyle \cos (\frac{\pi}{5})=\frac{1}{4}(\sqrt{5}+1)$

I am only stucked with (2) , i am ok with the rest . THanks .

(2) Suppose $\displaystyle BP=BC=AP = d$. Then in $\displaystyle \triangle BPC$
$\displaystyle \angle BCP = \tfrac{2\pi}{5}$

$\displaystyle \Rightarrow PC = 2d\cos(\angle BCP) = 2d\cos(\tfrac{2\pi}{5})$
So, using the result in (1):
$\displaystyle d^2 = 2d\cos(\tfrac{2\pi}{5})\Big(d+2d\cos(\tfrac{2\pi}{ 5})\Big)$

$\displaystyle \Rightarrow 1 = 2\cos(\tfrac{2\pi}{5})+4\cos^2(\tfrac{2\pi}{5})$
Solving the quadratic, taking the positive root gives
$\displaystyle \cos(\tfrac{2\pi}{5})=\frac{\sqrt5-1}{4}$

$\displaystyle \Rightarrow 2\cos^2(\tfrac{\pi}{5})-1 = \frac{\sqrt5-1}{4}$

$\displaystyle \Rightarrow \cos^2(\tfrac{\pi}{5})= \frac{\sqrt5+3}{8}$
$\displaystyle =\frac{5+2\sqrt5+1}{16}$

$\displaystyle =\frac{(\sqrt5+1)^2}{16}$
$\displaystyle \Rightarrow \cos(\tfrac{\pi}{5}) = \frac{\sqrt5+1}{4}$
• Nov 10th 2009, 06:37 PM
ukorov
Quote:

Solving the quadratic, taking the positive root gives
$\displaystyle \cos(\tfrac{2\pi}{5})=\frac{\sqrt5-1}{4}$

$\displaystyle \Rightarrow 2\cos^2(\tfrac{\pi}{5}) -1 = \frac{\sqrt5-1}{4}$

why $\displaystyle \cos(\tfrac{2\pi}{5}) = 2\cos^2(\tfrac{\pi}{5})-1$ ??
• Nov 10th 2009, 09:58 PM
why $\displaystyle \cos(\tfrac{2\pi}{5}) = 2\cos^2(\tfrac{\pi}{5})-1$ ??
Because $\displaystyle \cos2\theta = 2\cos^2\theta -1$.