A triangular geometry problem: 2

Thanks to all for writing a solution! I am actually surprised to see a such detailed and complex solution, as I did this problem in a very different way: 1. Take the lengths GD and DF as length a and b respectively. 2. We deduce that GF is 2-a-b 3.imagine rotating triangle BAG by 90 degrees anti clockwise, forming a new triangle BFG'. 4. FC is 1-b, and CG' is 1-a, hence FG' is equal to FG, deducing that BFG and BFG' is actually congruent. 5.angle GBG' is 90 degrees and bisected by angle GBF, so GBF=90/2=45.

let $\angle ABG = t$ and $\angle CBF = s$ , then the perimeter condition on $\triangle GFD$ implies that

$2 = (1 - \tan t) + (1 - \tan s) + \sqrt{ \sec^2 t + \sec^2 s - 2 \sec t \sec s \cos( \dfrac{\pi}{2} - (t+s) ) }$

Therefore,

$0 = - \tan t - \tan s + \sqrt{ \sec^2 t + \sec^2 s - 2 \sec t \sec s \sin(t+s) }$

Multiplying through by $\cos t \cos s$ results in,

$0 = - \sin t \cos s - \cos t \sin s + \sqrt{ \cos^2 s + \cos^2 t - 2 \cos s \cos t \sin(t+s) }$

Hence,

$(\sin t \cos s + \cos t \sin s)^2 = \cos^2 s + \cos^2 t - 2 \cos s \cos t \sin(t+s)$

Expanding the left hand side,

$\sin^2 t \cos^2 s + \cos^2 t \sin^2 s + 2 \sin t \sin s \cos t \cos s = \cos^2 s + \cos^2 t - 2 \cos s \cos t \sin(t+s)$

Collecting terms,

$cos^2 s (sin^2 s - 1) + \cos^2 t (\sin^2 s - 1) + 2 \sin t \sin s \cos t \cos s = -2 \cos s \cos t \sin(t+s)$

Hence,

$- 2 \cos^2 s \cos^2 t + 2 \sin t \sin s \cos t \cos s = -2 \cos s \cos t \sin(t + s)$

Dividing through by $-2 \cos s \cos t$ ,

$\cos s \cos t - \sin t \sin s = \sin(t + s)$

Therefore,

$\cos(t + s) = \sin(t + s)$

Which implies

$\tan(t + s) = 1$ , so that $(t + s) = 45^{\circ}$

Finally, $\angle GBF = 90^{\circ} - (t + s) = 90^{\circ} - 45^{\circ} = 45^{\circ}$

Let $|DF| = x$ and $|DG| = y$ . Then we require that $x + y + \sqrt{x^{2} + y^{2}} = 2$ , and so

$\sqrt{x^{2} + y^{2}} = 2 - (x + y) \Longrightarrow x^{2} + y^{2} = 4 + (x + y)^{2} - 4(x + y) = 4 + x^{2} + y^{2} + 2xy - 4x - 4y \Longrightarrow 4y - 2xy = 4 - 4x \Longrightarrow y = \dfrac{2(1 - x)}{2 - x}$ .

Now $\angle GBF = \dfrac{\pi}{2} - \angle FBC - \angle ABG = \dfrac{\pi}{2} - \arctan(1 - x) - \arctan(1 - y) =$

$\dfrac{\pi}{2} - \left(\arctan(1 - x) + \arctan\left(1 - \dfrac{2(1 - x)}{2 - x}\right)\right) = \dfrac{\pi}{2} - \left(\arctan(1 - x) + \arctan\left(\dfrac{x}{2 - x}\right)\right)$ .

Next, using the arctan addition formula $\arctan(u) + \arctan(v) = \arctan\left(\dfrac{u + v}{1 - uv}\right)$ we find that

$\arctan(1 - x) + \arctan\left(\dfrac{x}{2 - x}\right) = \arctan\left(\dfrac{1 - x + \dfrac{x}{2 - x}}{1 - \dfrac{x(1 - x)}{2 - x}}\right) = \arctan\left(\dfrac{(1 - x)(2 - x) + x}{2 - x - x(1 - x)}\right) = \arctan\left(\dfrac{2 - 2x + x^{2}}{2 - 2x + x^{2}}\right) = \arctan(1) = \dfrac{\pi}{4}$ .

So finally $\angle GBF = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4} = \boxed{45^{\circ}}$ .

Let AG=a, and FC=b.By using Pythagoras GF=((1-a)^2 +(1-b)^2)^0.5. The perimeter : (1-a)+(1-b) +((1-a)^2 +(1-b)^2)^0.5 =2, We get (a+b)^2 = (1-a)^2 +(1-b)^2, or b=(1-a)/(1+a). Tan (ABG + FBC) = [a + (1-a)/(1+a)]/[1- a * (1-a)/(1+a)] =[a^2+1]/[a^2+1]=1. So the angle of (ABG + FBC) = 45.Finally the angle GBF = 90- 45 = 45