What If I take a 2 + b 2 + c 2 = 11 a^2+b^2+c^2=11 out

Geometry Level pending

Consider triangle A B C ABC , such that:

  1. ∠ A = 4 5 ∘ \angle A = 45^\circ

  2. tan ⁡ A , tan ⁡ B , tan ⁡ C \tan A, \tan B, \tan C are in an arithmetic progression.

Find C H 2 △ \dfrac{{CH}^2}{\triangle} to 4 decimal places, where H H is the orthocenter and △ \triangle is the area of triangle A B C ABC .

Inspiration


The answer is 0.3333.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

T a n A = T a n 45 = 1. L e t T a n B = 1 + d , a n d T a n C = 1 + 2 d . B u t i n a △ T a n A + T a n B + T a n C = T a n A ∗ T a n B ∗ T a n C . ∴ 1 + ( 1 + d ) + ( 1 + 2 d ) = 1 ∗ ( 1 + d ) ∗ ( 1 + 2 d ) ⟹ 3 ( 1 + d ) = ( 1 + d ) ( 1 + 2 d ) . I f 1 + d = 0. T a n B = 0 , a n d B = 0. B u t a n a n g l e o f n o n − d e g e n e r a t e d Δ c a n n o t b e 0. S o 1 + d ≠ 0. ∴ 3 = 1 + 2 d , ⟹ d = 1. S o A = T a n − 1 1 , B = T a n − 1 2 C = T a n − 1 3. . . . . . . . . . . ( 1 ) F r o m t h e F i g . : − S i n A = 1 2 , S i n B = 2 5 , C o s C = 1 10 . . . . . . . . . . ( 2 ) S o b y S i n L a w s u b s t i t u t i o n f r o m ( 2 ) a n d s i m p l i f y i n g b c = S i n B S i n C = 2 2 3 . . . . . . . . . . . . ( 3 ) C D = b C o s C , C H = C D C o s D C H = b C o s C S i n B . . . . . . . . . . . ( 4 ) Δ = 1 2 ∗ b c S i n A . . . . . . . . . . . ( 5 ) ∴ B y ( 4 , 5 ) C H 2 Δ = ( b C o s C ) 2 S i n 2 B ∗ 1 2 ∗ b ∗ c ∗ S i n A = b c ∗ 2 ∗ C o s 2 C S i n 2 B ∗ S i n A = B y ( 2 ) a n d ( 3 ) 2 2 3 ∗ 2 10 4 5 ∗ 1 2 ∴ C H 2 Δ = 0.333333. TanA=Tan45=1.\ \ \ Let\ \ TanB=1+d,\ \ \ and\ \ \ TanC=1+2d.\\ But\ in\ a\ \triangle\ TanA+TanB+TanC=TanA*TanB*TanC.\\ \therefore\ \ 1+(1+d)+(1+2d) = 1*(1+d)*(1+2d)\\ \implies\ 3(1+d)=(1+d)(1+2d).\\ If\ 1+d=0.\ TanB=0,\ and \ B=0.\ But\ an\ angle\ of\ non-degenerated\ \Delta\ can\ not\ be\ 0. \\ So\ 1+d \neq 0.\ \ \therefore\ \ 3=1+2d,\ \ \implies\ d=1.\\ So\ A=Tan^{-1}1,\ \ \ B=Tan^{-1}2\ \ \ C=Tan^{-1}3.\color{#3D99F6}{..........(1)}\\ From\ the\ Fig.:-\ \ SinA=\dfrac 1 {\sqrt2},\ \ \ SinB=\dfrac 2{\sqrt5},\ \ \ CosC=\dfrac 1 {\sqrt{10}}\color{#3D99F6}{..........(2)}\\ So\ by\ Sin\ Law\ \ substitution\ from \ (2)\ and\ simplifying\ \ \ \ \ \dfrac b c=\dfrac {SinB}{SinC}=\dfrac{2\sqrt2} 3..\color{#3D99F6}{..........(3)}\\ CD=bCosC,\ \ \ \ CH=\dfrac{CD}{CosDCH}=\dfrac{bCosC}{SinB}.\color{#3D99F6}{..........(4)}\\ \Delta=\frac 1 2 *bcSinA.\color{#3D99F6}{..........(5)}\\ \therefore\ \ By\ (4,5)\ \dfrac{CH^2}{\Delta}= \dfrac{ (bCosC)^2}{ Sin^2 B*\frac 1 2 *b*c*SinA}\\ = \dfrac b c* \dfrac{ 2*Cos^2C}{ Sin^2 B*SinA}\\ =By\ (2)\ and(3)\ \dfrac{ 2\sqrt2} 3 *\dfrac{\dfrac 2 {10} } {\dfrac 4 5 *\dfrac 1 {\sqrt2} } \ \ \ \ \ \ \ \therefore\ \dfrac{CH^2}{\Delta}=\Large\ \ \color{#D61F06}{0.333333}.\\\ \ \ \\

Above is the copy of my solution of the same problem posted before.

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too complex there is a simpler method

anyways good solution! u are the inspiration

A Former Brilliant Member - 4 years, 6 months ago

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Thanks for your comment.

After finding the angles, their trig ratios, I found ratio of b:c.
Next found the required CH from two triangles. Found the area and the required ratio.
Values are then submitted to get the result.
Where have I gone complex ?
I am asking this to improve my solution.
Please post the simpler method.
Thanks.


Niranjan Khanderia - 4 years, 6 months ago

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