Inspired by Omkar kulkarni

Geometry Level 5

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 .
  3. a 2 + b 2 + c 2 = 11 a^2+b^2+c^2=11 , where a a , b b and c c are side lengths opposite vertices A A , B B and C C .

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.

View wiki
A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

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)}\\ \color{#E81990}{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}.\\\ \ \ \\ T h e r e i s n o n e e d f o r a 2 + b 2 + c 2 = 11. \color{#E81990}{There\ is\ no\ need \ for\ \ a^2+b^2+c^2=11.}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Note: You should explain why 1 + d ≠ 0 1 + d \neq 0 . It is not (yet) immediately obvious.

Calvin Lin Staff - 4 years, 6 months ago

Log in to reply

Thanks. I have added the explanation.

Niranjan Khanderia - 4 years, 6 months ago

Log in to reply

I believe what you want is

Since B ≠ 18 0 ∘ , B \neq 180^\circ, hence tan ⁡ B ≠ − 1 \tan B \neq -1 and so 1 + d ≠ 0 1 + d \neq 0 .

Calvin Lin Staff - 4 years, 6 months ago

Log in to reply

@Calvin Lin – Sorry. I did not follow your argument. So I am retaining my explanation. Please let me know what is the correct position.

Niranjan Khanderia - 4 years, 6 months ago

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...