In a triangle ABC if $B-C = \frac\pi 4$ and .......... What will be the value of $\frac {a+b+c}{a}$

Question

In a triangle ABC if $B-C = \frac\pi 4$ . Consider the following determinant.

|-2      cosC    cosB  |
|                      |
|cosC    -1      cosA  |   = P
|                      |
|cosB    cosA    -1    |

If $ P + \cos ^2 A = 0$ then what is the value of $\frac {a+b+c}{a}$?
(Where $A$,$B$,$C$ represent the angles opposite to the sides $a$,$b$,$c$)

I tried to expand the determinant and got $\cos ^2 B +2\cos ^2 A +\cos ^2 C + 2\cos A\cos B\cos C - 2$. I don't think that helps?

Just to solve it I assumed the triangle ABC with $B = 90^0$, $A = 45^0$, $C = 45^0$. And lucky me, the triangle satisfied the given equation. But I wouldn't call that a real method.

All help will be appreciated.

Answer 1

Hint: with $$B=\frac{\pi}{4}+C$$,$$A=\frac{3}{4}\pi-2C$$ you can express all angles by $$C$$!

Answer 2

►The function $f(x)$ where $x=C$ solution of your problem can be find out without using your determinant. The variable $x$ should be such that $0\lt x\lt\dfrac{3\pi}{8}(=135^{\circ})$, as you can verify by drawing triangles appropriate to the problem, and the function is increasing from $2$ to $\infty$. You have

$$f(x)=\frac{\sin(x+\frac{\pi}{4})+\sin( x)}{\sin(\frac{3\pi}{4}-2x)}+1 \left(\text {this is }f(x)=\frac {a+b+c}{a}\right)$$ and its (blue) graphic where figure the red point $\left(\dfrac{\pi}{4},2+\sqrt2\right)$ corresponding to your verification with $C=45^{\circ}$.

► Another thing (more difficult) is to use the $p$ value of your determinant.

We have to calculate $\dfrac {b+c}{a}+1$.

One has $$\frac {b+c}{a}=\frac{\sin B+\sin C}{\sin A}\space\space \qquad (1)$$ and the determinant equal to $p$ above is equal to $$p=2\cos A\cos B\cos C-2+2\cos^2A+\cos^2B+\cos^2C$$ or $$\\2\cos A\cos B\cos C-\sin^2B-\sin^2C=3p\qquad (2)$$

Noting $C=x$ we have $B=x+\dfrac{\pi}{4}$ and $A=\dfrac{3\pi}{4}-2x$ it follows
$\sin A=\dfrac{\sqrt2}{2}(\sin(2x)+\cos(2x))\Rightarrow \sin^2 A=\dfrac 12(1+\sin 4x)=P$ where $0\lt P=-p$.

Therefore $\sin 4x=2P-1\Rightarrow \boxed{x=\dfrac{\arcsin(2P-1)}{4}}$. This solves already the problem because this is to find a relationship between $ x $ and $ P $ but if you want to use $(2)$ for the purpose of finding a simpler relationship between $x$ and $P$, you have $$2\cos A\cos B\cos C=(\sin 2x-\cos 2x)(\sin x-\cos x)\cos x=(\cos x-\sin 3x)\cos x\\\sin^2B=\frac 12(1+\sin 2x)$$ from which you get in $(2)$ $$(\cos x-\sin 3x)\cos x-\frac 12(1+\sin 2x)-\sin^2x=-3P\\$$

I stop here. Whatever you can deduce, you will find at the end an expression for the angle x equal to the boxed above. I add an easy exercise to find algebraic expressions for trigonometric functions of the variable $x$. In the attached figure is drawn a right triangle in which a construction of the angle $ 4x $ is presented. By the geometric theorem of the bisector you can find expressions for trigonometric functions of $ x $ and $2x$ as a function of $ P $. The result, if you get it, will be a quite complicated algebraic expression for $ f (x) $.