Angle Between Diagonals?

Geometry Level 4

Find the acute angle θ \theta between the diagonals of a rectangle with perimeter 2 p 2p and area 3 16 p 2 \dfrac{3}{16}p^{2} .

If θ \theta = = a a tan 1 ( b c ) \tan^{-1} \left(\dfrac{b}{c} \right) , where a , b , c a , b , c are coprime positive integers, then enter the value of a + b + c a + b + c .


The answer is 6.

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1 solution

Ahmad Saad
Jun 5, 2016

A typo first line N O T 3 16 p b u t 3 16 p 2 NOT\ \frac 3 {16}*p\ \ but\ \ \frac 3 {16}*p^2 .
A different approach is as under with the same diagram above but not for the given condition.
m + n = p , m n = 3 16 p 2 . D i a g o n a l 2 = m 2 + n 2 = ( m + n ) 2 2 m n = 5 8 p 2 . H a l f D i a g o n a l , ( H D ) = 1 2 5 8 p 2 . A r e a , A 4 o f q u a r t e r r e c t a n g l e = 1 2 ( H D ) 2 S i n θ A = 4 1 8 5 8 p 2 S i n θ = 3 16 p 2 . . . g i v e n . S i n θ = 3 5 , i m p l i e s T a n θ = 3 4 . 1 T a n 1 3 4 . a T a n 1 b c , s i n c e a i s n o t c o p r i m e + t i v e i n t e g e r . T h o u g h t 1 T a n 1 3 4 = 2 T a n 1 1 3 . m+n=p, \ \ \ m*n=\frac 3 {16}p^2.\\ \therefore\ Diagonal^2=m^2+n^2=(m+n)^2-2*m*n=\frac 5 8 *p^2.\\ \implies\ Half\ Diagonal,( HD)=\frac 1 2 *\sqrt{\frac 5 8*p^2.}\\ \therefore\ Area,\ \ \frac A 4\ of\ quarter\ rectangle=\frac 1 2 *( HD)^2*Sin\theta\\ \therefore\ A=4*\frac 1 8 *\frac 5 8*p^2*Sin\theta=\frac 3 {16}p^2...given.\\ \therefore\ Sin\theta=\frac 3 5 , \ implies\ Tan\theta=\frac 3 4.\\ \therefore\ 1*Tan^{-1}\frac 3 4.{\Large\ \color{#D61F06}{\neq}} a*Tan^{-1}\frac b c,\ \ since \ a\ is\ not\ co-prime +tive\ integer.\\ Thought\ \ 1*Tan^{-1}\frac 3 4=2*Tan^{-1}\frac 1 3. .

Niranjan Khanderia - 4 years, 11 months ago

Nice Solution \text{Nice Solution} ( + 1 ) ! +1)!

Rishabh Tiwari - 5 years ago

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