For the cube shown above, let
v = cos 2 w + cos 2 x + cos 2 y + cos 2 z .
If v is in the form d c , where c and d are coprime positive integers, find c + d .
Note: w , x , y and z are all angles between 2 diagonals of the cube.
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Let s = side of the cube and p be the intersection of two diagonals. → s 3 will be the length of diagonals. Now, since the diagonals bisect each other, therefore, the lengths of Ap = Bp = Cp = Dp = 2 s 3 .
By Law Of Cosines,
→ ( 2 s 3 ) 2 + ( 2 s 3 ) 2 − 2 ( 2 s 3 ) ( 2 s 3 ) ∗ cos x = s 2 2 ∗ ( 4 3 ) s 2 − 2 ( 4 3 ) s 2 ∗ cos x = s 2 ( 2 1 ) s 2 = ( 2 3 ) s 2 ∗ cos x cos x = cos z = 3 1
and...
→ ( 2 s 3 ) 2 + ( 2 s 3 ) 2 − 2 ( 2 s 3 ) ( 2 s 3 ) ∗ cos a = [ A D ] 2 = [ s 2 ] 2 ( 2 ∗ 4 3 ) s 2 − ( 2 ∗ 4 3 ) s 2 ∗ cos a = 2 s 2 ( − 2 1 ) s 2 = ( 2 3 ) s 2 ∗ cos a cos w = cos y = − 3 1
∴ v = cos 2 x + cos 2 y + cos 2 z + cos 2 w = ( 3 1 ) 2 + ( − 3 1 ) 2 + ( 3 1 ) 2 + ( − 3 1 ) 2 = 9 4 ⟹ c + d = 1 3
Yes I also Did the same bcuz topper forever
The lines A C and B D are diagonals of the rectangle A B C D . By geometry of two intersecting lines in a plane, x = z and w = y . Since the scaling is arbitrary, keep it simple with A B = D C = 1 , then A D = B C = 2 , applying Pythagoras' Theorem to the diagonal of the unit square (corresponding to a face of the unit cube).
Now we have two sides of △ B D C , where it is clear to see that ∠ D C B = 9 0 ∘ . Define the point E as the intersection point where the four angles of interest are made, (the midpoint of line segment B D ), and F as the midpoint of the line segment B C . Then as B D B E = B C B F = 2 1 , and E F ∥ D C so m ∠ E F B = m ∠ D C B = 9 0 ∘ , we show that △ B E F ∼ △ B D C with scale factor 2 1 .
Using Pythagoras' Theorem in △ B D C , B D = 3 ⇒ B E = 2 3 . Since △ B E F and △ C E F share E F , m ∠ E F B = m ∠ E F C = 9 0 ∘ (angles on line B C at point F sum to 1 8 0 ∘ ) and B E = E C = 2 3 by the symmetry of the rectangle A B C D , we conclude that △ B E F ≅ △ C E F .
Therefore m ∠ B E F = 2 y . Taking B F = 2 2 , E F = 2 1 and B E = 2 3 and applying trigonometry in a right triangle:
cos 2 y = 2 3 2 1 = 3 1
By the double angle formula:
cos y = 2 cos 2 2 y − 1 = 2 ( 3 1 ) 2 − 1 = − 3 1 ⇒ cos 2 y = 9 1
The same method can be applied to find cos 2 x , by letting the midpoint of A B be the point G , proving that △ A E G ≅ △ B E G and using the same trigonometric technique. This yields that cos 2 x = 9 1 also.
Now we simply solve for v . Since we established that x = z and w = y , it is implicit that cos 2 x = cos 2 z and cos 2 w = cos 2 y . Therefore:
v = cos 2 w + cos 2 x + cos 2 y + cos 2 z = 2 ( cos 2 x + cos 2 y ) = 2 ( 9 1 + 9 1 ) = 9 4
Finally, c + d = 4 + 9 = 1 3
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Let the intersection of the two diagonals A C and B D be O ( 0 , 0 ) the origin, x - and y -axis be parallel and perpendicular to A B respectively and z -axis vertically up, and the side length of the cube be 2. Then A ( − 1 , − 1 , 1 ) , B ( 1 , − 1 , 1 ) and C ( 1 , 1 , − 1 ) and let a , b and c be the respective vectors. By dot product we have:
a ⋅ b [ − 1 , − 1 , 1 ] ⋅ [ 1 , − 1 , 1 ] − 1 + 1 + 1 ⟹ cos x = ∣ a ∣ ∣ b ∣ cos x = 3 × 3 cos x = 3 cos x = 3 1 = cos z
Similarly,
b ⋅ c [ 1 , − 1 , 1 ] ⋅ [ 1 , 1 , − 1 ] 1 − 1 − 1 ⟹ cos y = ∣ b ∣ ∣ c ∣ cos y = 3 × 3 cos y = 3 cos y = − 3 1 = cos w
Therefore,
v = cos 2 x + cos 2 y + cos 2 z + cos 2 w = ( 3 1 ) 2 + ( − 3 1 ) 2 + ( 3 1 ) 2 + ( − 3 1 ) 2 = 9 4
⟹ c + d = 4 + 9 = 1 3