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!
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let ∠ A B G = t and ∠ C B F = s , then the perimeter condition on △ G F D implies that
2 = ( 1 − tan t ) + ( 1 − tan s ) + sec 2 t + sec 2 s − 2 sec t sec s cos ( 2 π − ( t + s ) )
Therefore,
0 = − tan t − tan s + 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 + 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,
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 )
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 ) = 4 5 ∘
Finally, ∠ G B F = 9 0 ∘ − ( t + s ) = 9 0 ∘ − 4 5 ∘ = 4 5 ∘
Let ∣ D F ∣ = x and ∣ D G ∣ = y . Then we require that x + y + 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 − x 2 ( 1 − x ) .
Now ∠ G B F = 2 π − ∠ F B C − ∠ A B G = 2 π − arctan ( 1 − x ) − arctan ( 1 − y ) =
2 π − ( arctan ( 1 − x ) + arctan ( 1 − 2 − x 2 ( 1 − x ) ) ) = 2 π − ( arctan ( 1 − x ) + arctan ( 2 − x x ) ) .
Next, using the arctan addition formula arctan ( u ) + arctan ( v ) = arctan ( 1 − u v u + v ) we find that
arctan ( 1 − x ) + arctan ( 2 − x x ) = arctan ⎝ ⎜ ⎜ ⎛ 1 − 2 − x x ( 1 − x ) 1 − x + 2 − x x ⎠ ⎟ ⎟ ⎞ = arctan ( 2 − x − x ( 1 − x ) ( 1 − x ) ( 2 − x ) + x ) = arctan ( 2 − 2 x + x 2 2 − 2 x + x 2 ) = arctan ( 1 ) = 4 π .
So finally ∠ G B F = 2 π − 4 π = 4 π = 4 5 ∘ .
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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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.