A triangular geometry problem: 2

Geometry Level 3

As shown in the image, a triangle BFG, is inscribed in the square ABCD. Given that the triangle GFD has a perimeter of 2cm, and the square has sides 1cm, find angle GBF to its exact value in degrees. Note that the diagram is not to scale!


The answer is 45.

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4 solutions

A I
Apr 12, 2018

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.

Hosam Hajjir
Apr 12, 2018

let A B G = t \angle ABG = t and C B F = s \angle CBF = s , then the perimeter condition on G F D \triangle GFD implies that

2 = ( 1 tan t ) + ( 1 tan s ) + sec 2 t + sec 2 s 2 sec t sec s cos ( π 2 ( t + s ) ) 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 + sec 2 t + sec 2 s 2 sec t sec s sin ( t + s ) 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 \cos t \cos s results in,

0 = sin t cos s cos t sin s + cos 2 s + cos 2 t 2 cos s cos t sin ( t + s ) 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 ) (\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 ) \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,

c o s 2 s ( s i n 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 ) 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 ) - 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 -2 \cos s \cos t ,

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

Therefore,

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

Which implies

tan ( t + s ) = 1 \tan(t + s) = 1 , so that ( t + s ) = 4 5 (t + s) = 45^{\circ}

Finally, G B F = 9 0 ( t + s ) = 9 0 4 5 = 4 5 \angle GBF = 90^{\circ} - (t + s) = 90^{\circ} - 45^{\circ} = 45^{\circ}

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

x 2 + y 2 = 2 ( x + y ) x 2 + y 2 = 4 + ( x + y ) 2 4 ( x + y ) = 4 + x 2 + y 2 + 2 x y 4 x 4 y 4 y 2 x y = 4 4 x y = 2 ( 1 x ) 2 x \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 G B F = π 2 F B C A B G = π 2 arctan ( 1 x ) arctan ( 1 y ) = \angle GBF = \dfrac{\pi}{2} - \angle FBC - \angle ABG = \dfrac{\pi}{2} - \arctan(1 - x) - \arctan(1 - y) =

π 2 ( arctan ( 1 x ) + arctan ( 1 2 ( 1 x ) 2 x ) ) = π 2 ( arctan ( 1 x ) + arctan ( x 2 x ) ) \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 ( u + v 1 u v ) \arctan(u) + \arctan(v) = \arctan\left(\dfrac{u + v}{1 - uv}\right) we find that

arctan ( 1 x ) + arctan ( x 2 x ) = arctan ( 1 x + x 2 x 1 x ( 1 x ) 2 x ) = arctan ( ( 1 x ) ( 2 x ) + x 2 x x ( 1 x ) ) = arctan ( 2 2 x + x 2 2 2 x + x 2 ) = arctan ( 1 ) = π 4 \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 G B F = π 2 π 4 = π 4 = 4 5 \angle GBF = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4} = \boxed{45^{\circ}} .

Rab Gani
Feb 8, 2019

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

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