Revival Of This Problem. Part 1

Geometry Level 4

For the cube shown above, let

v = cos 2 w + cos 2 x + cos 2 y + cos 2 z . v = \cos^2 w + \cos^2 x + \cos^2 y + \cos^2 z.

If v v is in the form c d \frac{c}{d} , where c c and d d are coprime positive integers, find c + d c+d .

Note: w w , x x , y y and z z are all angles between 2 diagonals of the cube.


The answer is 13.

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3 solutions

Chew-Seong Cheong
Jan 25, 2017

Let the intersection of the two diagonals A C AC and B D BD be O ( 0 , 0 ) O(0,0) the origin, x x - and y y -axis be parallel and perpendicular to A B AB respectively and z z -axis vertically up, and the side length of the cube be 2. Then A ( 1 , 1 , 1 ) A (-1,-1,1) , B ( 1 , 1 , 1 ) B(1,-1,1) and C ( 1 , 1 , 1 ) C(1,1,-1) and let a \textbf{a} , b \textbf{b} and c \textbf{c} be the respective vectors. By dot product we have:

a b = a b cos x [ 1 , 1 , 1 ] [ 1 , 1 , 1 ] = 3 × 3 cos x 1 + 1 + 1 = 3 cos x cos x = 1 3 = cos z \begin{aligned} \textbf{a} \cdot \textbf{b} & = |\textbf{a}||\textbf{b}| \cos x \\ [-1,-1,1] \cdot [1,-1,1] & = \sqrt 3 \times \sqrt 3 \cos x \\ -1+1+1 & = 3 \cos x \\ \implies \cos x & = \frac 13 = \cos z \end{aligned}

Similarly,

b c = b c cos y [ 1 , 1 , 1 ] [ 1 , 1 , 1 ] = 3 × 3 cos y 1 1 1 = 3 cos y cos y = 1 3 = cos w \begin{aligned} \textbf{b} \cdot \textbf{c} & = |\textbf{b}||\textbf{c}| \cos y \\ [1,-1,1] \cdot [1,1,-1] & = \sqrt 3 \times \sqrt 3 \cos y \\ 1-1-1 & = 3 \cos y \\ \implies \cos y & = - \frac 13 = \cos w \end{aligned}

Therefore,

v = cos 2 x + cos 2 y + cos 2 z + cos 2 w = ( 1 3 ) 2 + ( 1 3 ) 2 + ( 1 3 ) 2 + ( 1 3 ) 2 = 4 9 \begin{aligned} v & = \cos^2 x + \cos^2 y + \cos^2 z + \cos^2 w \\ & = \left(\frac 13 \right)^2 + \left(-\frac 13 \right)^2 + \left(\frac 13 \right)^2 + \left(-\frac 13 \right)^2 \\ & = \frac 49 \end{aligned}

c + d = 4 + 9 = 13 \implies c+d = 4+9 = \boxed{13}

Christian Daang
Jan 25, 2017

Let s = side of the cube and p be the intersection of two diagonals. s 3 \rightarrow s \sqrt{3} will be the length of diagonals. Now, since the diagonals bisect each other, therefore, the lengths of Ap = Bp = Cp = Dp = s 3 2 \cfrac{s \sqrt{3}}{2} .

By Law Of Cosines,

( s 3 2 ) 2 + ( s 3 2 ) 2 2 ( s 3 2 ) ( s 3 2 ) cos x = s 2 2 ( 3 4 ) s 2 2 ( 3 4 ) s 2 cos x = s 2 ( 1 2 ) s 2 = ( 3 2 ) s 2 cos x cos x = cos z = 1 3 \rightarrow \left( \cfrac{s \sqrt{3}}{2}\right )^2 + \left( \cfrac{s \sqrt{3}}{2}\right )^2 - 2\left( \cfrac{s \sqrt{3}}{2}\right )\left( \cfrac{s \sqrt{3}}{2}\right ) * \cos x = s^2 \\ 2*\left( \cfrac{3}{4}\right )s^2 - 2\left( \cfrac{3}{4}\right )s^2 *\cos x = s^2 \\ \left( \cfrac{1}{2}\right )s^2 = \left( \cfrac{3}{2}\right )s^2 * \cos x \\ \cos x = \cos z = \ \cfrac{1}{3}

and...

( s 3 2 ) 2 + ( s 3 2 ) 2 2 ( s 3 2 ) ( s 3 2 ) cos a = [ A D ] 2 = [ s 2 ] 2 ( 2 3 4 ) s 2 ( 2 3 4 ) s 2 cos a = 2 s 2 ( 1 2 ) s 2 = ( 3 2 ) s 2 cos a cos w = cos y = 1 3 \rightarrow \left( \cfrac{s \sqrt{3}}{2}\right )^2 + \left( \cfrac{s \sqrt{3}}{2}\right )^2 - 2\left( \cfrac{s \sqrt{3}}{2}\right )\left( \cfrac{s \sqrt{3}}{2}\right ) * \cos a = [\overline{AD}]^2 = [s \sqrt{2}]^2 \\ \left( 2*\cfrac{3}{4}\right )s^2 - \left( 2*\cfrac{3}{4}\right )s^2 *\cos a = 2s^2 \\ \left( -\cfrac{1}{2}\right )s^2 = \left( \cfrac{3}{2}\right )s^2 *\cos a \\ \cos w = \cos y = \ -\cfrac{1}{3}

v = cos 2 x + cos 2 y + cos 2 z + cos 2 w = ( 1 3 ) 2 + ( 1 3 ) 2 + ( 1 3 ) 2 + ( 1 3 ) 2 = 4 9 c + d = 13 \therefore v = \cos^2 x + \cos^2 y + \cos^2 z + \cos^2 w \\ = \left( \cfrac{1}{3}\right )^2 + \left( -\cfrac{1}{3}\right )^2 + \left( \cfrac{1}{3}\right )^2 + \left( -\cfrac{1}{3}\right )^2 \\ = \cfrac{4}{9} \\ \implies c + d = \ \boxed{13}

Yes I also Did the same bcuz topper forever

Topper Forever - 4 years, 3 months ago
David Orrell
Jan 27, 2017

The lines A C \overline{AC} and B D \overline{BD} are diagonals of the rectangle A B C D ABCD . By geometry of two intersecting lines in a plane, x = z x=z and w = y w=y . Since the scaling is arbitrary, keep it simple with A B = D C = 1 \overline{AB}=\overline{DC}=1 , then A D = B C = 2 \overline{AD}=\overline{BC}=\sqrt{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 \triangle BDC , where it is clear to see that D C B = 9 0 \angle DCB = 90^{\circ} . Define the point E as the intersection point where the four angles of interest are made, (the midpoint of line segment B D \overline{BD} ), and F as the midpoint of the line segment B C \overline{BC} . Then as B E B D = B F B C = 1 2 \frac{BE}{BD} = \frac{BF}{BC} = \frac{1}{2} , and E F D C \overline{EF} \parallel \overline{DC} so m E F B = m D C B = 9 0 m\angle EFB = m\angle DCB = 90^{\circ} , we show that B E F B D C \triangle BEF \sim \triangle BDC with scale factor 1 2 \frac{1}{2} .

Using Pythagoras' Theorem in B D C \triangle BDC , B D = 3 B E = 3 2 BD = \sqrt{3} \Rightarrow BE = \frac{\sqrt{3}}{2} . Since B E F \triangle BEF and C E F \triangle CEF share E F \overline{EF} , m E F B = m E F C = 9 0 m\angle EFB = m\angle EFC = 90^{\circ} (angles on line B C BC at point F F sum to 18 0 180^{\circ} ) and B E = E C = 3 2 BE = EC = \frac{\sqrt{3}}{2} by the symmetry of the rectangle A B C D ABCD , we conclude that B E F C E F \triangle BEF \cong \triangle CEF .

Therefore m B E F = y 2 m\angle BEF = \frac{y}{2} . Taking B F = 2 2 BF = \frac{\sqrt{2}}{2} , E F = 1 2 EF = \frac{1}{2} and B E = 3 2 BE = \frac{\sqrt{3}}{2} and applying trigonometry in a right triangle:

cos y 2 = 1 2 3 2 = 1 3 \cos \frac{y}{2} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}

By the double angle formula:

cos y = 2 cos 2 y 2 1 = 2 ( 1 3 ) 2 1 = 1 3 cos 2 y = 1 9 \cos y = 2\cos^{2} \frac{y}{2} - 1 = 2(\frac{1}{\sqrt{3}})^{2} - 1 = -\frac{1}{3} \Rightarrow \cos^{2}y = \frac{1}{9}

The same method can be applied to find cos 2 x \cos^{2}x , by letting the midpoint of A B \overline{AB} be the point G G , proving that A E G B E G \triangle AEG \cong \triangle BEG and using the same trigonometric technique. This yields that cos 2 x = 1 9 \cos^{2}x = \frac{1}{9} also.

Now we simply solve for v v . Since we established that x = z x=z and w = y w=y , it is implicit that cos 2 x = cos 2 z \cos^{2}x=\cos^{2}z and cos 2 w = cos 2 y \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 ( 1 9 + 1 9 ) = 4 9 v = \cos^{2}w + \cos^{2}x + \cos^{2}y + \cos^{2}z = 2(\cos^{2}x + \cos^{2}y) = 2(\frac{1}{9} + \frac{1}{9}) = \frac{4}{9}

Finally, c + d = 4 + 9 = 13 c + d = 4 + 9 = \boxed{13}

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